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Mind-Expanding Exercises 2-168. Let E denote a read error and let S, O, B, P denote skewed, off-center, both, and proper alignments,

respectively.P(E) = P(E|S)P(S) + P(E|O)P(O) + P(E|B)P(B) + P(E|P)P(P)= 0.01(0.10) + 0.02(0.05) + 0.06(0.01) + 0.001(0.84) = 0.00344

2-169. Let n denote the number of washers selected.

a) The probability that non are thicker than, that is, all are less than the target is 0 4 , by independence.. n

n0 4. n

1 0.42 0.16

3 0.064Therefore, n = 3b) The requested probability is the complement of the probability requested in part a. Therefore, n = 3

2-29



2-170. Let x denote the number of kits produced.Revenue at each demand

0 50 100 200

0 5≤ ≤ 0 x -5x 100x 100x 100x

Mean profit = 100x(0.95)-5x(0.05)-20x

50 100≤ ≤ x -5x 100(50)-5(x-50) 100x 100x

Mean profit = [100(50)-5(x-50)](0.4) + 100x(0.55)-5x(0.05)-20x

100 200≤ ≤ x -5x 100(50)-5(x-50) 100(100)-5(x-100) 100x

Mean profit = [100(50)-5(x-50)](0.4) + [100(100)-5(x-100)](0.3) + 100x(0.25) - 5x(0.05) - 20x

Mean Profit Maximum Profit

0 5≤ ≤ 0 x 74.75 x $ 3737.50 at x=50

50 100≤ ≤ x 32.75 x + 2100 $ 5375 at x=100

100 200≤ ≤ x 1.25 x + 5250 $ 5500 at x=200

Therefore, profit is maximized at 200 kits. However, the difference in profit over 100 kits is small.

2-171. Let E denote the probability that none of the bolts are identified as incorrectly torqued. The requestedprobability is P(E'). Let X denote the number of bolts in the sample that are incorrect. Then,P(E) = P(E|X=0)P(X=0) + P(E|X=1) P(X=1) + P(E|X=2) P(X=2) + P(E|X=3) P(X=3) + P(E|X=4)P(X=4)and P(X=0) = (15/20)(14/19)(13/18)(12/17) = 0.2817. The remaining probability for x can be determinedfrom the counting methods in Appendix B-1. Then,

( )( )( )

( )( )( )

( )( )( )

P X

P X

P X

( )

!

! !

!

! !

!

! !

! ! ! !

! ! ! !.

( )

!

! !

!

! !

!

! !

.

( )

!

! !

!

! !

!

! !

.

= = =

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

= =

= = =

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

=

= = =

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

=

1

5

4 1

15

3 12

20

4 16

5 15 4 16

4 3 12 200 4696

2

5

3 2

15

2 13

20

4 16

0 2167

3

5

3 2

15

1 14

20

4 16

0 0309

15

315

420

25

215

420

35

115

420

P(X=4) = (5/20)(4/19)(3/18)(2/17) = 0.0010 and P(E|X=0) = 1, P(E|X=1) = 0.05, P(E|X=2) =

, P(E|X=3) = , P(E|X=4) = . Then,0 05 0 00252. .= 0 05 125 103. .= × −4 0 05 6 25 104 6. .= × −

P E( ) ( . ) . ( . ) . ( . ) . ( . )

. ( . )

.

= + + + ×

+ ×

=

−

−

10 2817 005 04696 00025 02167 125 10 00309

6 25 10 0 0010

0 306

4

6

and P(E') = 0.6942-172.

P A B P A B P A B

P A P B P A B

P A P B P A P B

P A P BP A P B

( ' ' ) ([ ' ' ]' ) ( )

[ ( ) ( ) ( )]

( ) ( ) ( ) ( )

[ ( )][ ( )]( ' ) ( ' )

∩ = − ∩ = − ∪

= − + − ∩

= − − +

= − −

=

1 1

1

1

1 1

2-30



2-173. The total sample size is ka + a + kb + b = (k + 1)a + (k +1)b.

P Ak a b

k a k bP B

ka a

k a k b

and

P A Bka

k a k b

ka

k a b

Then

P A P Bk a b ka a

k a k b

k a b k a

k a b

ka

k a bP A B

( )( )

( ) ( ), ( )

( ) ( )

( )( ) ( ) ( )( )

,

( ) ( )( )( )

[( ) ( ) ]

( )( )

( ) ( ) ( )( )( )

=+

+ + +=

+

+ + +

∩ =+ + +

=+ +

=+ +

+ + +=

+ +

+ +=

+ += ∩

1 1 1 1

1 1 1

1 1

1

1 12 2 2

2-31



CHAPTER 2

Section 2-1

2-1. Let "a", "b" denote a part above, below the specification

{ }S aaa aab aba abb baa bab bba bbb= , , , , , , ,

2-2. Let "e" denote a bit in error

Let "o" denote a bit not in error ("o" denotes okay)

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

=

oooooeooeoooeeoo

oooeoeoeeooeeeoe

ooeooeeoeoeoeeeo

ooeeoeeeeoeeeeee

S

,,,

,,,,

,,,,

,,,,

2-3. Let "a" denote an acceptable power supply

Let "f" ,"m","c" denote a supply with a functional, minor, or cosmetic error, respectively.

{ }S a f m c= , , ,

2-4. = set of nonnegative integers{S = 0 12, , , .. .

}

2-5. If only the number of tracks with errors is of interest, then { }S = 0 12 24, , , .. .,

2-6. A vector with three components can describe the three digits of the ammeter. Each digit can be

0,1,2,...,9. Then S is a sample space of 1000 possible three digit integers, { }S = 000 001 999, ,...,

2-7. S is the sample space of 100 possible two digit integers.

2-8. Let an ordered pair of numbers, such as 43 denote the response on the first and second question. Then, S

consists of the 25 ordered pairs { }1112 55, , .. .,

2-9. in ppb.{ }091,...,2,1,0 E S =

2-10. in milliseconds{ ,...,2,1,0=S }

}

2-11. { }0.14,2.1,1.1,0.1 …=S

2-12. s = small, m = medium, l = large; S = {s, m, l, ss, sm, sl, ….}

2-13 in milliseconds.{ ,...,2,1,0=S

2-14.

automatictransmission transmission

standard

withoutwithair

without

airairair

with

whitered blue black whitered blue black whitered blue black whitered blue black

2-1



2-15.

PRESS

CAVITY

1 2

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8

2-16.

memory

disk storage

4 8 12

200 300 400 200 300 400 200 300 400 2-17. c = connect, b = busy, S = {c, bc, bbc, bbbc, bbbbc, …}

2-18. { }S s fs ffs fffS fffFS fffFFS fffFFFA= , , , , , ,

2-19 a)

b)

c)

2-2



d)

e)

2.20 a)

2-3



b)

c)

d)

e)

2-4



2-21. a) S = nonnegative integers from 0 to the largest integer that can be displayed by the scale.

Let X represent weight.

A is the event that X > 11 B is the event that X ≤ 15 C is the event that 8 ≤ X <12

S = {0, 1, 2, 3, …}

b) S

c) 11 < X ≤ 15 or {12, 13, 14, 15}

d) X ≤ 11 or {0, 1, 2, …, 11}

e) Sf) A ∪ C would contain the values of X such that: X ≥ 8

Thus (A ∪ C)′ would contain the values of X such that: X < 8 or {0, 1, 2, …, 7}

g) ∅

h) B′ would contain the values of X such that X > 15. Therefore, B′ ∩ C would be the empty set. They

have no outcomes in common or ∅

i) B ∩ C is the event 8 ≤ X <12. Therefore, A ∪ (B ∩ C) is the event X ≥ 8 or {8, 9, 10, …}

2-22. a)

A B

C

b)

A B

C

c)

d)

A B

C

2-5



e) If the events are mutually exclusive, then A∩B is equal to zero. Therefore, the process does not

produce product parts with X =50 cm and Y =10 cm. The process would not be successful.

2-23. Let "d" denoted a distorted bit and let "o" denote a bit that is not distorted.

a) S

dddd dodd oddd oodd

dddo dodo oddo oodo

ddod dood odod ooodddoo dooo odoo oooo

=

⎧

⎨

⎪⎪

⎩⎪⎪

⎫

⎬

⎪⎪

⎭⎪⎪

, , , ,

, , , ,

, , , ,

, , ,

b) No, for example { }A A dddd dddo ddod ddoo1 2∩ = , , ,

c)

⎪⎪

⎭

⎪⎬

⎫

⎪⎪

⎩

⎪⎨

⎧

=

doooddoo

dood ddod

dododddo

dodd dddd

A

,

,

,

,,

1

d)

⎪⎪

⎭

⎪⎬

⎫

⎪⎪

⎩

⎪⎨

⎧=

ooooodoo

oood odod

oodooddooodd oddd

A

,

,,

,,,,

1

e) }{4321 dddd A A A A =∩∩∩

f) { }ddoooodd ddod oddd dddododd dddd A A A A ,,,,,,)()( 4321 =∩∪∩

2-24

Let w denote the wavelength. The sample space is {w | w = 0, 1, 2, …}

(a) A={w | w = 675, 676, …, 700 nm}(b) B={ w | w = 450, 451, …, 500 nm}

(c) Φ=∩ B A

(d) =∪ B A {w | w = 450, 451, …, 500, 675, 676, …, 700 nm}

2-25

Let P denote being positive and let N denote being negative.

The sample space is {PPP, PPN, PNP, NPP, PNN, NPN, NNP, NNN}.

(a) A={ PPP }

(b) B={ NNN }

(c) Φ=∩ B A

(d) =∪ B A { PPP , NNN }

2-26.A ∩ B = 70, A′ = 14, A ∪ B = 95

2-27. a) B A ∩′ = 10, =10, B′ B A ∪ = 92

b)

2-6



Surface 1

E

G

G

E

GEdge 1

E

E

E E

G

GG

E

EE

G

GG

E

EE

G

GG

E

Surface 2Edge 2

EE

G

GG

2-28. B A ∩′ = 55, B′ =23, B A ∪ = 85

2-29. a) A′ = {x | x ≥ 72.5}

b) B′ = {x | x ≤ 52.5}

c) A ∩ B = {x | 52.5 < x < 72.5}

d) A ∪ B = {x | x > 0}

2.30 a) {ab, ac, ad, bc, bd, cd, ba, ca, da, cb, db, dc}

b) {ab, ac, ad, ae, af, ag, bc, bd, be, bf, bg, cd, ce, cf, cg, ef, eg, fg, ba, ca, da, ea, fa, ga, cb, db,eb, fb, gb, dc, ec, fc, gc, fe, ge, gf }c) Let d = defective, g = good; S = {gg, gd, dg, dd }

d) Let d = defective, g = good; S = {gd, dg, gg}

2.31 Let g denote a good board, m a board with minor defects, and j a board with major defects.

a.) S = {gg, gm, gj, mg, mm, mj, jg, jm, jj}

b) S={gg,gm,gj,mg,mm,mj,jg,jm}

2-32.a.) The sample space contains all points in the positive X-Y plane.

b)

10

c)

B

20

d)

2-7



10

20

A

B

e)

10

20

A

B

2-33 a)

b)

c)

2-8



d)

2-34.212

= 4096

2-35. From the multiplication rule, the answer is 5 3 4 2 120× × × =

2-36. From the multiplication rule, 3 4 3 36× × =

2-37. From the multiplication rule, 3 4 3 4 144× × × =

2-38. From equation 2-1, the answer is 10! = 3,628,800

2-39. From the multiplication rule and equation 2-1, the answer is 5!5! = 14,400

2-40. From equation 2-3,7

3 435

!

! != sequences are possible

2-41. a) From equation 2-4, the number of samples of size five is ( ) 528,965,416!135!5

!140140

5 ==

b) There are 10 ways of selecting one nonconforming chip and there are ( ) 880,358,11!126!4

!130130

4 ==

ways of selecting four conforming chips. Therefore, the number of samples that contain exactly one

nonconforming chip is 10 ( ) 800,588,113130

4 =×

c) The number of samples that contain at least one nonconforming chip is the total number of samples

minus the number of samples that contain no nonconforming chips( )5140 ( )5

130 .

2-9



That is ( - =)5140 ( )5

130 752,721,130!125!5

!130

!135!5

!140=−

2-42. a) If the chips are of different types, then every arrangement of 5 locations selected from the 12 results in a

different layout. Therefore, 040,95!7

!1212

5==P layouts are possible.

b) If the chips are of the same type, then every subset of 5 locations chosen from the 12 results in a different

layout. Therefore, ( )512 12

5 7792= =

!

! !layouts are possible.

2-43. a) 21!5!2

!7= sequences are possible.

b) 2520!2!1!1!1!1!1

!7= sequences are possible.

c) 6! = 720 sequences are possible.

2-44. a) Every arrangement of 7 locations selected from the 12 comprises a different design.

P712 12

53991680= =

!

!designs are possible.

b) Every subset of 7 locations selected from the 12 comprises a new design. 792!7!5

!12= designs are

possible.

c) First the three locations for the first component are selected in ( ) 220!9!3

!1212

3 == ways. Then, the four

locations for the second component are selected from the nine remaining locations in ( ) 126!5!4

!994 ==

ways. From the multiplication rule, the number of designs is 720,27126220 =×

2-45. a) From the multiplication rule, 103 1000= prefixes are possible

b) From the multiplication rule, are possible8 2 10 160× × =c) Every arrangement of three digits selected from the 10 digits results in a possible prefix.

P310 10

7720= =

!

!prefixes are possible.

2-46. a) From the multiplication rule, bytes are possible2 2568 =

b) From the multiplication rule, bytes are possible2 1287 =

2-47. a) The total number of samples possible is ( ) .626,10!20!4

!24244 == The number of samples in which

exactly one tank has high viscosity is ( )( ) 4896!15!3

!18

!5!1

!618

3

6

1 =×= . Therefore, the probability is

461.010626

4896=

2-10



b) The number of samples that contain no tank with high viscosity is ( ) .3060!14!4

!1818

4 == Therefore, the

requested probability is 1 712.010626

3060=− .

c) The number of samples that meet the requirements is ( )( )( ) 2184!12!2

!14

!3!1

!4

!5!1

!614

2

4

1

6

1=××= .

Therefore, the probability is 206.010626

2184=

2-48. a) The total number of samples is ( )312 12

3 9220= =

!

! !. The number of samples that result in one

nonconforming part is ( )( ) .90!8!2

!10

!1!1

!210

2

2

1 =×= Therefore, the requested probability is

90/220 = 0.409.

b) The number of samples with no nonconforming part is ( ) .120!7!3

!1010

3 == The probability of at least one

nonconforming part is 1 455.0220

120=− .

2-49. a) The probability that both parts are defective is 0082.049

4

50

5=×

b) The total number of samples is ( )2

4950

!48!2

!5050

2

×== . The number of samples with two defective

parts is ( )2

45

!3!2

!55

2

×== . Therefore, the probability is 0082.0

4950

45

24950

245

=×

×=

×

×

.

2-11



Section 2-3

2-66. a) P(A') = 1- P(A) = 0.7

b) P ( ) = P(A) + P(B) - P(A B∪ A B∩ ) = 0.3+0.2 - 0.1 = 0.4

c) P( ) + P( ) = P(B). Therefore, P(′ ∩A B A B∩ ′ ∩A B ) = 0.2 - 0.1 = 0.1

d) P(A) = P( ) + P( ). Therefore, P(A B∩ A B∩ ′ A B∩ ′ ) = 0.3 - 0.1 = 0.2

e) P(( )') = 1 - P( ) = 1 - 0.4 = 0.6A B∪ A B∪

f) P( ) = P(A') + P(B) - P(′ ∪A B ′ ∩A B ) = 0.7 + 0.2 - 0.1 = 0.8

2-67. a) P( ) = P(A) + P(B) + P(C), because the events are mutually exclusive. Therefore,

P( ) = 0.2+0.3+0.4 = 0.9

C B A ∪∪

C B A ∪∪

b) P ( ) = 0, becauseC B A ∩∩ A B C∩ ∩ = ∅

2-13



c) P( B A ∩ ) = 0 , because =A B∩ ∅

d) P( ) = 0, becauseC B A ∩∪ )( C B A ∩∪ )( = ∅=∩∪∩ )()( C BC A

e) P( A B C ′ ′∩ ∩ ′ ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1

2-68. (a) P(Caused by sports)= P(Caused by contact sports or by noncontact sports)

= P(Caused by contact sports) + P(Caused by noncontact sports)

=0.46+0.44

=0.9

(b) 1- P(Caused by sports)=0.1.

2-69.a) 70/100 = 0.70

b) (79+86-70)/100 = 0.95

c) No, P( ) ≠ 0A B∩

2-70. (a) P(High temperature and high conductivity)= 74/100 =0.74

(b)P(Low temperature or low conductivity)

= P(Low temperature) + P(Low conductivity) – P(Low temperature and low conductivity)

=(8+3)/100 + (15+3)/100 – 3/100

=0.26

(c)

No, they are not mutually exclusive. Because

P(Low temperature) + P(Low conductivity)

=(8+3)/100 + (15+3)/100

=0.29, which is not equal to P(Low temperature or low conductivity).

2-71. a) 350/370

b) 345 5 12370

362370

+ + =

c)345 5 8

370

358

370

+ +=

d) 345/370

2-72.a) 170/190 = 17/19

b) 7/190

2-73.a) P(unsatisfactory) = (5+10-2)/130 = 13/130

b) P(both criteria satisfactory) = 117/130 = 0.90, No

2-74. (a) 5/36

(b) 5/36

(c) 01929.0)()()( ==∩ BP AP B AP

(d) 2585.0)()()( =+=∪ BP AP B AP



Section 2-4

2-75. a) P(A) = 86/100 b) P(B) = 79/100

2-14



c) P( A B ) =79

70

100 / 79

100 / 70

)(

)( =∩ BP

B AP

d) P( B A ) =86

70

100 / 86

100 / 70

)(

)( =∩ AP

B AP

2-76. (a) 39.0

100

327)( =

+= AP

(b) 2.0100

713)( =

+= BP

(c) 35.0100 / 20

100 / 7

)(

)()|( ==

∩=

BP

B AP B AP

(d) 1795.0100 / 39

100 / 7

)(

)()|( ==

∩=

AP

B AP A BP

2-77. Let A denote the event that a leaf completes the color transformation and let B denote the event

that a leaf completes the textural transformation. The total number of experiments is 300.

(a) 903.0300 / )26243(

300 / 243

)(

)()|( =+=

∩

= AP

B AP

A BP

(b) 591.0300 / )2618(

300 / 26

)'(

)'()'|( =

+=

∩=

BP

B AP B AP

2-78.a) 0.82

b) 0.90

c) 8/9 = 0.889

d) 80/82 = 0.9756

e) 80/82 = 0.9756

f) 2/10 = 0.20

2-79. a) 12/100 b) 12/28 c) 34/122

2-80.

a) P(A) = 0.05 + 0.10 = 0.15

b) P(A|B) = 153.072.0

07.004.0

)(

)( =+=∩ BP

B AP

c) P(B) = 0.72

d) P(B|A) = 733.015.0

07.004.0

)(

)(=

+=

∩

AP

B AP

e) P(A ∩ B) = 0.04 +0.07 = 0.11

f) P(A ∪ B) = 0.15 + 0.72 – 0.11 = 0.76

2-81. Let A denote the event that autolysis is high and let B denote the event that putrefaction is high. The

total number of experiments is 100.

(a) 5625.0100 / )1814(

100 / 18

)(

)'()|'( =

+=

∩=

AP

B AP A BP

(b) 1918.0100 / )5914(

100 / 14

)(

)()|( =

+=

∩=

BP

B AP B AP

2-15



(c) 333.0100 / )918(

100 / 9

)'(

)''()'|'( =

+=

∩=

BP

B AP B AP

2-82.a) P(gas leak) = (55 + 32)/107 = 0.813

b) P(electric failure|gas leak) = (55/107)/(87/102) = 0.632

c) P(gas leak| electric failure) = (55/107)/(72/107) = 0.764

2-83. a) 20/100

b) 19/99

c) (20/100)(19/99) = 0.038

d) If the chips are replaced, the probability would be (20/100) = 0.2

2-84. a) 4/499 = 0.0080

b) (5/500)(4/499) = 0.000080

c) (495/500)(494/499) = 0.98

d) 3/498 = 0.0060

e) 4/498 = 0.0080

f) 5

500

4

499

3

4984 82 10 7⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟ = −. x

2-85. a)P=(8-1)/(350-1)=0.020b)P=(8/350)× [(8-1)/(350-1)]=0.000458

c)P=(342/350) × [(342-1)/(350-1)]=0.9547

2-86. (a)736

1

(b))36(5

16

(c)5)36(5

15

2-87. No, if B , then P(A/B) =A⊂P A B

P B

P B

P B

( )

( )

( )

( )

∩= = 1

A

B

2-88.

AB

C



Section 2-5

2-16



2-89. a) P A B P AB P B( ) ( ) ( ) ( . )( . ) .∩ = = =0 4 0 5 0 20

b) P A B P A B P B( ) ( ) ( ) ( . )( . ) .′ ∩ = ′ = =0 6 0 5 0 30

2-90.

P A P A B P A B

P AB P B P AB P B

( ) ( ) ( )

( ) ( ) ( ) (( . )( . ) ( . )( . )

. . .

= ∩ + ∩ ′

= + ′ ′

= +

= + =

0 2 0 8 0 3 0 2

016 0 06 0 22

)

2-91. Let F denote the event that a connector fails.

Let W denote the event that a connector is wet.

P F P F W P W P F W P W( ) ( ) ( ) ( ) ( )

( . )( . ) ( . )( . ) .

= + ′ ′

= + =0 05 010 0 01 0 90 0 014

2-92. Let F denote the event that a roll contains a flaw.

Let C denote the event that a roll is cotton.

P F P F C P C P F C P C( ) ( ) ( ) ( ) ( )

( . )( . ) ( . )( . ) .

= + ′ ′

= + =0 02 0 70 0 03 0 30 0 023

2-93. Let R denote the event that a product exhibits surface roughness. Let N,A, and W denote the events that the

blades are new, average, and worn, respectively. Then,

P(R)= P(R|N)P(N) + P(R|A)P(A) + P(R|W)P(W)

= (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15)

= 0.028

2-94. Let A denote the event that a respondent is a college graduate and let B denote the event that a

voter votes for Bush.

P(B)=P(A)P(B|A)+ P(A’)P(B|A’)=0.38× 0.53+0.62× 0.5=0.5114

2-95.a) (0.88)(0.27) = 0.2376

b) (0.12)(0.13+0.52) = 0.0.078

2-96. a)P=0.13×0.73=0.0949

b)P=0.87× (0.27+0.17)=0.3828

2-97. Let A denote a event that the first part selected has excessive shrinkage.

Let B denote the event that the second part selected has excessive shrinkage.

a) P(B)= P(B A )P(A) + P(B A ')P(A')

= (4/24)(5/25) + (5/24)(20/25) = 0.20

b) Let C denote the event that the third part selected has excessive shrinkage.

20.0

25

20

24

19

23

5

25

20

24

5

23

4

25

5

24

20

23

4

25

5

24

4

23

3

)''()''()'()'(

)'()'()()()(

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =

∩∩+∩∩+

∩∩+∩∩=

B AP B AC P B AP B AC P

B AP B AC P B AP B AC PC P

2-98. Let A and B denote the events that the first and second chips selected are defective, respectively.

2-17



a) P(B) = P(B|A)P(A) + P(B|A')P(A') = (19/99)(20/100) + (20/99)(80/100) = 0.2

b) Let C denote the event that the third chip selected is defective.

00705.0

100

20

99

19

98

18

)()()()()()(

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ =

∩=∩∩=∩∩ AP A BP B AC P B AP B AC PC B AP

2-99.

Open surgery

success failuresample

sizesample

percentageconditional success

rate

large stone 192 71 263 0.751428571 0.73%

small stone 81 6 87 0.248571429 0.93%

overall summary 273 77 350 1 0.78

PN

success failuresample

sizesample

percentageconditional success

rate

large stone 55 25 80 0.228571429 69%

small stone 234 36 270 0.771428571 83%

overall summary 289 61 350 1 0.83

The reason is that the overall success rate is dependent on both the success rates conditioned on the two

groups and the probability of the groups. It is the weighted average of the group success rate weighted by

the group size; instead of the direct average.

P(overall success)=P(large stone)P(success| large stone)+ P(small stone)P(success| small stone).

For open surgery, the dominant group (large stone) has a smaller success rate while for PN, the dominant

group (small stone) has a larger success rate.



Section 2-6

2-100. Because P( AB ) ≠ P(A), the events are not independent.

2-101. P(A') = 1 - P(A) = 0.7 and P( A B' ) = 1 - P( AB ) = 0.7

Therefore, A' and B are independent events.

2-102. If A and B are mutually exclusive, then P( A B∩ ) = 0 and P(A)P(B) = 0.04.

Therefore, A and B are not independent.

2-103. a) P(BA ) = 4/499 and

500 / 5)500 / 495)(499 / 5()500 / 5)(499 / 4()'()'()()()(=+=+= AP A BP AP A BP BP

Therefore, A and B are not independent.

b) A and B are independent.

2-104. P( ) = 70/100, P(A) = 86/100, P(B) = 77/100.A B∩Then, P( A ) ≠ P(A)P(B), so A and B areB∩ not independent.

2-105. a) P( A )= 22/100, P(A) = 30/100, P(B) = 77/100, Then P(B∩ A B∩ )≠ P(A)P(B), therefore, A and B are

not independent.

b) P(B|A) = P(A ∩ B)/P(A) = (22/100)/(30/100) = 0.733

2-18



2-106. (a)62 10)001.0( −==P

(b) 002.0)999.0(1 2 =−=P

2-107. It is useful to work one of these exercises with care to illustrate the laws of probability. Let Hi denote the

event that the ith sample contains high levels of contamination.

a) P H H H H H P H P H P H P H P H( ) ( ) ( ) ( ) ( ) ( )' ' ' ' ' ' ' ' ' '1 2 3 4 5 1 2 3 4 5∩ ∩ ∩ ∩ =

by independence. Also, P H . Therefore, the answer isi( ) .' = 0 9 0 9 0 595. .=

b) A H H H H H1 1 2 3 4 5= ∩ ∩ ∩ ∩( )' ' ' '

A H H H H H2 1 2 3 4 5= ∩ ∩ ∩ ∩( )' ' ' '

A H H H H H3 1 2 3 4 5= ∩ ∩ ∩ ∩( )' ' ' '

A H H H H H4 1 2 3 4 5= ∩ ∩ ∩ ∩( )' ' ' '

A H H H H H5 1 2 3 4 5= ∩ ∩ ∩ ∩( )' ' ' '

The requested probability is the probability of the union and these eventsA A A A A1 2 3 4

∪ ∪ ∪ ∪5

are mutually exclusive. Also, by independence P A . Therefore, the answer isi( ) . ( . ) .= =0 9 01 0 06564

5(0.0656) = 0.328.

c) Let B denote the event that no sample contains high levels of contamination. The requested

probability is P(B') = 1 - P(B). From part (a), P(B') = 1 - 0.59 = 0.41.

2-108. Let A denote the event that the ith bit is a one.i

a) By independence P A A A P A P A P A( ... ) ( ) ( )... ( ) ( ) .1 2 10 1 2 10101

20 000976∩ ∩ ∩ = = =

b) By independence, P A A A P A P A P Ac( . . . ) ( ) ( ). . . ( ) ( ) .' ' ' ' '1 2 10 1 2 10

101

20 000976∩ ∩ ∩ = = =

c) The probability of the following sequence is

P A A A A A A A A A A( )' ' ' ' '1 2 3 4 5 6 7 8 9

10

101

2

∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ = ( ) , by independence. The number of

sequences consisting of five "1"'s, and five "0"'s is ( )510 10

5 5252= =

!

! !. The answer is252

1

20 246

10⎛

⎝ ⎜

⎞

⎠⎟ = .

2-109. (a) =0.0048)2.0(3 4

(b) =0.0768)8.0*2.0*4(3 3

2-110. (a) 4096.0)8.0( 4 ==P

(b) 64.02.08.02.01 =×−−=P

(c) Probability defeats all four in a game = 0.84

= 0.4096. Probability defeats all four at least

once = 1 – (1 – 0.4096)3

= 0.7942

2-111. (a) The probability that one technician obtains equivalence at 100 mL is 0.1.So the probability that both technicians obtain equivalence at 100 mL is .01.01.0 2 =(b) The probability that one technician obtains equivalence between 98 and 104 mL is 0.7.

So the probability that both technicians obtain equivalence between 98 and 104 mL is

.49.07.0 2 =(c) The probability that the average volume at equivalence from the technician is 100 mL is

.09.0)1.0(9 2 =

2-19



2-112. (a)10

16

6

1010

10 −==P

(b) 020833.0)12

1(25.0 =×=P

2-113. Let A denote the event that a sample is produced in cavity one of the mold.

a) By independence, P A A A A A( ) ( )1 2 3 4 5

51

80 00003∩ ∩ ∩ ∩ = = .

b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually

exclusive, P B B B P B P B P B( ... ) ( ) ( ) ... ( )1 2 8 1 2 8∪ ∪ ∪ = + + +

From part a., P Bi( ) ( )=1

8

5 . Therefore, the answer is 81

80 000245( ) .=

c) By independence, P A A A A A( ) ( ) ( )'1 2 3 4 5

41

8

7

8∩ ∩ ∩ ∩ = . The number of sequences in

which four out of five samples are from cavity one is 5. Therefore, the answer is 51

8

7

80 001074( ) ( ) .= .

2-114. Let A denote the upper devices function. Let B denote the lower devices function.P(A) = (0.9)(0.8)(0.7) = 0.504

P(B) = (0.95)(0.95)(0.95) = 0.8574

P(A∩B) = (0.504)(0.8574) = 0.4321

Therefore, the probability that the circuit operates = P(A∪B) = P(A) +P(B) − P(A∩B) = 0.9293

2-115. [1-(0.1)(0.05)][1-(0.1)(0.05)][1-(0.2)(0.1)] = 0.9702



Section 2-7

2-116. Because, P( A B ) P(B) = P( ) = P(A B∩ B A ) P(A),

28.05.0

)2.0(7.0

)(

)()()( ===

AP

BP B AP A BP

2-117.' '

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )

0.4 0.80.89

0.4 0.8 0.2 0.2

P A B P B P A B P BP B A

P A P A B P B P A B P B= =

+

×= =

× + ×

2-118. Let F denote a fraudulent user and let T denote a user that originates calls from two or more

metropolitan areas in a day. Then,

003.0)9999(.01.0)0001.0(30.0

)0001.0(30.0

)'()'()()(

)()()( =

+

=

+

=

F PF T PF PF T P

F PF T PT F P

2-119. (a) P=(0.31)(0.978)+(0.27)(0.981)+(0.21)(0.965)+(0.13)(0.992)+(0.08)(0.959)

= 0.97638

(b) 207552.097638.0

)965.0)(21.0(==P

2-120. Let A denote the event that a respondent is a college graduate and let B denote the event that a

voter votes for Bush.

2-20



%3821.39)5.0)(62.0()53.0)(38.0(

)53.0)(38.0(

)'|()'()|()(

)|()(

)(

)()|( =

+

=

+

=∩

=

A BP AP A BP AP

A BP AP

BP

B AP B AP

2-121. Let G denote a product that received a good review. Let H, M, and P denote products that were high,

moderate, and poor performers, respectively.

a)

P G P G H P H P G M P M P G P P P( ) ( ) ( ) ( ) ( ) ( ) ( )

. ( . ) . ( . ) . ( . )

.

= + +

= + +

=

0 95 0 40 0 60 0 35 0 10 0 25

0 615

b) Using the result from part a.,

P H GP G H P H

P G( )

( ) ( )

( )

. ( . )

..= = =

0 95 0 40

0 6150 618

c) P H GP G H P H

P G( ' )

( ' ) ( )

( ' )

. ( . )

..= =

−

=0 05 0 40

1 0 6150 052

2-122. a) P(D)=P(D|G)P(G)+P(D|G’)P(G’)=(.005)(.991)+(.99)(.009)=0.013865

b) P(G|D’)=P(G∩D’)/P(D’)=P(D’|G)P(G)/P(D’)=(.995)(.991)/(1-.013865)=0.9999

2-123.a) P(S) = 0.997(0.60) + 0.9995(0.27) + 0.897(0.13) = 0.9847

b) P(Ch|S) =(0.13)( 0.897)/0.9847 = 0.1184



Section 2-8

2-124. Continuous: a, c, d, f, h, i; Discrete: b, e, and g



Supplemental Exercises

2-125. Let B denote the event that a glass breaks.

Let L denote the event that large packaging is used.

P(B)= P(B|L)P(L) + P(B|L')P(L')

= 0.01(0.60) + 0.02(0.40) = 0.014

2-126. Let "d" denote a defective calculator and let "a" denote an acceptable calculator

a

a) { }aaadaaaad dad adaddaadd ddd S ,,,,,,,=

b) { } daadad ddaddd A ,,,=

c) { } adaadd ddaddd B ,,,=d) { }ddaddd B A ,=∩

e) { }aad dad adaadd ddaddd C B ,,,,,=∪

2-21



2-127. Let A = excellent surface finish; B = excellent length

a) P(A) = 82/100 = 0.82

b) P(B) = 90/100 = 0.90

c) P(A') = 1 – 0.82 = 0.18

d) P(A∩B) = 80/100 = 0.80

e) P(A∪B) = 0.92f) P(A’∪B) = 0.98

2-128. a) (207+350+357-201-204-345+200)/370 = 0.9838

b) 366/370 = 0.989

c) (200+163)/370 = 363/370 = 0.981

d) (201+163)/370 = 364/370 = 0.984

2-129. If A,B,C are mutually exclusive, then P( ) = P(A) + P(B) + P(C) = 0.3 + 0.4 + 0.5 =A B C∪ ∪1.2, which greater than 1. Therefore, P(A), P(B),and P(C) cannot equal the given values.

2-130. a) 345/357 b) 5/13

2-131 (a) P(the first one selected is not ionized)=20/100=0.2

(b)P(the second is not ionized given the first one was ionized)

=20/99=0.202

(c)

P(both are ionized)

=P(the first one selected is ionized) × P(the second is ionized given the first one was ionized)

=(80/100)× (79/99)=0.638

(d)

If samples selected were replaced prior to the next selection,

P(the second is not ionized given the first one was ionized)

=20/100=0.2.

The event of the first selection and the event of the second selection are independent.

2-132. a) P(A) = 15/40

b) P(B A ) = 14/39

c) P( A ) = P(A) P(B/A) = (15/40) (14/39) = 0.135B∩

d) P( ) = 1 – P(A’ and B’) =A B∪ 25 241 0.615

40 39

⎛ ⎞⎛ ⎞− =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

A = first is local, B = second is local, C = third is local

e) P(A ∩ B ∩ C) = (15/40)(14/39)(13/38) = 0.046

f) P(A ∩ B ∩ C’) = (15/40)(14/39)(25/39) = 0.089

2-133. a) P(A) = 0.03

b) P(A') = 0.97

c) P(B|A) = 0.40

d) P(B|A') = 0.05

e) P( ) = P(A B∩ B A )P(A) = (0.40)(0.03) = 0.012

f) P( ') = P(A B∩ B A' )P(A) = (0.60)(0.03) = 0.018

g) P(B) = P(B A )P(A) + P(B A ')P(A') = (0.40)(0.03) + (0.05)(0.97) = 0.0605

2-134. Let U denote the event that the user has improperly followed installation instructions.

2-22



Let C denote the event that the incoming call is a complaint.

Let P denote the event that the incoming call is a request to purchase more products.

Let R denote the event that the incoming call is a request for information.

a) P(U|C)P(C) = (0.75)(0.03) = 0.0225

b) P(P|R)P(R) = (0.50)(0.25) = 0.125

2-135. (a) 18143.0)002.01(1100

=−−=P

(b) 005976.0002.0)998.0( 21

3 == C P

(c) 86494.0])002.01[(1 10100 =−−=P

2-136. P( ) = 80/100, P(A) = 82/100, P(B) = 90/100.A B∩

Then, P( A ) ≠ P(A)P(B), so A and B areB∩ not independent.

2-137. Let Ai denote the event that the ith readback is successful. By independence,

P A A A P A P A P A( ) ( ) ( ) ( ) ( . ) .' ' ' ' ' '1 2 3 1 2 3

30 02 0 000008∩ ∩ = = = .

2-138.

backup main-storage

0.25 0.75

life > 5 yrslife > 5 yrs

life < 5 yrs life < 5 yrs

0.95(0.25)=0.2375 0.05(0.25)=0.0125 0.995(0.75)=0.74625 0.005(0.75)=0.00375

a) P(B) = 0.25

b) P( AB ) = 0.95

c) P( AB ') = 0.995

d) P( ) = P(A B∩ AB )P(B) = 0.95(0.25) = 0.2375

e) P( ') = P(A B∩ AB ')P(B') = 0.995(0.75) = 0.74625

f) P(A) = P( ) + P( ') = 0.95(0.25) + 0.995(0.75) = 0.98375A B∩ A B∩

g) 0.95(0.25) + 0.995(0.75) = 0.98375.

h)

769.0)75.0(005.0)25.0(05.0

)25.0(05.0

)'()''()()'(

)()'()'( =

+=

+=

BP B AP BP B AP

BP B AP A BP

2-139. (a) =∩ B A' 50

(b) B’=37(c) =∪ B A 93

2-140. a) 0.25

b) 0.75

2-141. Let Di denote the event that the primary failure mode is type i and let A denote the event that a board passes

the test.

2-23



The sample space is S = { } .A A D A D A D A D A D, ' , ' , ' , ' , '1 2 3 4 5

2-142. a) 20/200 b) 135/200 c) 65/200

2-24



2-143.a) P(A) = 19/100 = 0.19

b) P(A ∩ B) = 15/100 = 0.15

c) P(A ∪ B) = (19 + 95 – 15)/100 = 0.99

d) P(A′∩ B) = 80/100 = 0.80

e) P(A|B) = P(A ∩ B)/P(B) = 0.158

2-144. Let A denote the event that the ith order is shipped on time.i

a) By independence, 857.0)95.0()()()()( 3

321321 ===∩∩ AP AP AP A A AP

b) Let

B A A A

B A A A

B A A A

1 1 2 3

2 1 2 3

3 1 2 3

= ∩ ∩

= ∩ ∩

= ∩ ∩

'

'

'

Then, because the B's are mutually exclusive,

P B B B P B P B P B( ) ( ) ( )

( . ) ( . )

.

1 2 3 1 2 3

23 0 95 0 05

0135

∪ ∪ = + +

=

=

( )

3= ∩ ∩

= ∩ ∩

= ∩ ∩

= ∩ ∩

' '

' '

' '

' ' '

( )

c) Let

B A A A

B A A A

B A A A

B A A A

1 1 2

2 1 2 3

3 1 2 3

4 1 2 3

Because the B's are mutually exclusive,

P B B B B P B P B P B P B( ) ( ) ( ) ( )

( . ) ( . ) ( . )

.

1 2 3 4 1 2 3 4

2 33 0 05 0 95 0 05

0 00725

∪ ∪ ∪ = + + +

= +

=

2-145. a) No, P(E1 ∩ E2 ∩ E3) ≠ 0

b) No, E1′ ∩ E2′ is not ∅

c) P(E1′ ∪ E2′ ∪ E3′) = P(E1′) + P(E2′) + P(E3′) – P(E1′∩ E2′) - P(E1′∩ E3′) - P(E2′∩ E3′)+ P(E1′ ∩ E2′ ∩ E3′)

= 40/240

d) P(E1 ∩ E2 ∩ E3) = 200/240

e) P(E1 ∪ E3) = P(E1) + P(E3) – P(E1 ∩ E3) = 234/240

f) P(E1 ∪ E2 ∪ E3) = 1 – P(E1′ ∩ E2′ ∩ E3′) = 1 - 0 = 1

2-146.a) (0.20)(0.30) +(0.7)(0.9) = 0.69

2-25



2-147. Let Ai denote the event that the ith bolt selected is not torqued to the proper limit.

a) Then,

282.020

15

19

14

18

13

17

12

)()()()(

)()()(

1122133214

32132144321

=⎟ ⎠

⎞

⎜⎝

⎛

⎟ ⎠

⎞

⎜⎝

⎛

⎟ ⎠

⎞

⎜⎝

⎛

⎟ ⎠

⎞

⎜⎝

⎛

=

∩∩∩=

∩∩∩∩=∩∩∩

AP A AP A A AP A A A AP

A A AP A A A AP A A A AP

b) Let B denote the event that at least one of the selected bolts are not properly torqued. Thus, B' is the

event that all bolts are properly torqued. Then,

P(B) = 1 - P(B') = 115

20

14

19

13

18

12

170718−

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟

⎛

⎝ ⎜

⎞

⎠⎟ = .

2-148. Let A,B denote the event that the first, second portion of the circuit operates. Then, P(A) =

(0.99)(0.99)+0.9-(0.99)(0.99)(0.9) = 0.998

P(B) = 0.9+0.9-(0.9)(0.9) = 0.99 and

P( ) = P(A) P(B) = (0.998) (0.99) = 0.988A B∩

2-149.A1 = by telephone, A2 = website; P(A1) = 0.92, P(A2) = 0.95;

By independence P(A1 ∪ A2) = P(A1) + P(A2) - P(A1 ∩ A2) = 0.92 + 0.95 - 0.92(0.95) = 0.996

2-150.P(Possess) = 0.95(0.99) +(0.05)(0.90) = 0.9855

2-151. Let D denote the event that a container is incorrectly filled and let H denote the event that a container is

filled under high-speed operation. Then,

a) P(D) = P(DH )P(H) + P(D H ')P(H') = 0.01(0.30) + 0.001(0.70) = 0.0037

b) 8108.00037.0

)30.0(01.0

)(

)()()( ===

DP

H P H DP D H P

2-152.a) P(E’ ∩ T’ ∩ D’) = (0.995)(0.99)(0.999) = 0.984

b) P(E ∪ D) = P(E) + P(D) – P(E ∩ D) = 0.005995

2-153.D = defective copy

a) P(D = 1) = 0778.0

73

2

74

72

75

73

73

72

74

2

75

73

73

72

74

73

75

2=

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛+

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛+

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛

b) P(D = 2) = 00108.073

1

74

2

75

73

73

1

74

73

75

2

73

73

74

1

75

2=⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛

c) Let A represent the event that the two items NOT inspected are not defective. Then,

P(A)=(73/75)(72/74)=0.947.

2-154. The tool fails if any component fails. Let F denote the event that the tool fails. Then, P(F') = 0 9 by

independence and P(F) = 1 - 0 9 = 0.0956

910.

910.

2-155. a) (0.3)(0.99)(0.985) + (0.7)(0.98)(0.997) = 0.9764

b) 3159.09764.01

)30.0(02485.0

)(

)1()1()1( =

−==

E P

routeProute E P E routeP

2-26



2-156. a) By independence, 0 15 7 59 105 5. .= × −

b) Let A denote the events that the machine is idle at the time of your ith request. Using independence,i

the requested probability is

0022.0

)85.0)(15.0(5

)85.0(15.0)85.0(15.0)85.0(15.0)85.0(15.0)85.0(15.0

)(

4

44444

5432

'

1543

'

2154

'

3215

'

4321

'

54321

=

=++++=

A A A A Aor A A A A Aor A A A A Aor A A A A Aor A A A A AP

c) As in part b, the probability of 3 of the events is

0244.0

)85.0)(15.0(10

)

(

23

543

'

2

'

154

'

32

'

15

'

432

'

1

'

5432

'

154

'

3

'

21

5

'

43

'

21

'

543

'

215

'

4

'

321

'

54

'

321

'

5

'

4321

=

=

A A A A Aor A A A A Aor A A A A Aor A A A A Aor A A A A A

or A A A A Aor A A A A Aor A A A A Aor A A A A Aor A A A A AP

For the probability of at least 3, add answer parts a) and b) to the above to obtain the requested probability.

Therefore, the answer is 0.0000759 + 0.0022 + 0.0244 = 0.0267

2-157. Let A denote the event that the ith washer selected is thicker than target.i

a) 207.048

28

49

29

50

30=⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛

b) 30/48 = 0.625

c) The requested probability can be written in terms of whether or not the first and second washer selected

are thicker than the target. That is,

60.0

49

19

50

20

48

30

49

30

50

20

48

29

49

30

50

20

48

29

49

29

50

30

48

28

)()()()()()(

)()()()()()(

)()()()'(

)()()()(

)()(

'

1

'

1

'

2

'

2

'

13

'

1

'

122

'

13

11

'

2

'

213112213

'

2

'

1

'

2

'

132

'

1213

'

21

'

21321213

3

'

2

'

132

'

13

'

213213

=

⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ =

++

+=

++

+=

=

AP A AP A A AP AP A AP A A AP

AP A AP A A AP AP A AP A A AP

A AP A A AP A AP A A AP

A AP A A AP A AP A A AP

A AorA A AorA A AorA A A AP AP

2-158. a) If n washers are selected, then the probability they are all less than the target is20

50

19

49

20 1

50 1⋅ ⋅

− +

− +...

n

n.

n probability all selected washers are less than target

1 20/50 = 0.42 (20/50)(19/49) = 0.155

3 (20/50)(19/49)(18/48) = 0.058

Therefore, the answer is n = 3

b) Then event E that one or more washers is thicker than target is the complement of the event that all are

less than target. Therefore, P(E) equals one minus the probability in part a. Therefore, n = 3.

2-27



2-159.

547.0940

514)''()

881.0940

24668514)'()

262.0940

246)()

453.0940

24668112)()

==∩

=++=∪

==∩

=++

=∪

B APd

B APc

B APb

B APa

. e) P( AB ) =P A B

P B

( )

( )

/

/ .

∩= =

246 940

314 9400783

f) P(B A ) = P B A

P A

( )

( )

/

/ .

∩= =

246 940

358 9400 687

2-160. Let E denote a read error and let S,O,P denote skewed, off-center, and proper alignments, respectively.

Then,

a) P(E) = P(E|S) P(S) + P(E|O) P (O) + P(E|P) P(P)

= 0.01(0.10) + 0.02(0.05) + 0.001(0.85)= 0.00285

b) P(S|E) =P E S P S

P E

( ) ( )

( )

. ( . )

..= =

0 01 0 10

0 002850 351

2-161. Let A denote the event that the ith row operates. Then,

.i

P A P A P A P A( ) . , ( ) ( . )( . ) . , ( ) . , ( ) .1 2 3 40 98 0 99 0 99 0 9801 0 9801 0 98= = = = =

The probability the circuit does not operate is7'

4'3

'2

'1 1058.1)02.0)(0199.0)(0199.0)(02.0()()()()( −×== AP AP AP AP

2-162. a) (0.4)(0.1) + (0.3)(0.1) +(0.2)(0.2) + (0.4)(0.1) = 0.15

b) P(4 or more | provided info) = (0.4)(0.1)/0.15 = 0.267

2-163. (a) P=(0.93)(0.91)(0.97)(0.90)(0.98)(0.93)=0.67336

(b) P=(1-0.93)(1-0.91)(1-0.97)(1-0.90)(1-0.98)(1-0.93)=2.646×10-8

(c) P=1-(1-0.91)(1-0.97)(1-0.90)=0.99973

2-164. (a) P=(24/36)(23/35)(22/34)(21/33)(20/32)(19/31)=0.069

(b) P=1-0.069=0.931

2-165. (a) 367

(b) Number of permutations of six letters is 266. Number of ways to select one number = 10.

Number of positions among the six letters to place the one number = 7. Number of passwords =

266 × 10 × 7

(c) 265102

2-166. (a) 087.01000

20730255)( =

++++= AP

(b) 032.01000

725)( =

+=∩ B AP

(c) 20.01000

8001)( =−=∪ B AP

2-28



(d) 113.01000

153563)'( =

++=∩ B AP

(e) 2207.01000 / )357156325(

032.0

)(

)()|( =

++++=

∩=

BP

B AP B AP

(f) 005.01000

5

==P

2-167. (a) Let A denote that a part conforms to specifications and let B denote for a part a simple

component.

For supplier 1,

P(A)=P(A|B)P(B)+ P(A|B’)P(B’)

= [(1000-2)/1000](1000/2000)+ [(1000-10)/1000](1000/2000)

= 0.994

For supplier 2,

P(A)= P(A|B)P(B)+ P(A|B’)P(B’)

= [(1600-4)/1600](1600/2000)+ [(400-6)/400](400/2000)

= 0.995

(b)

For supplier 1, P(A|B’)=0.99

For supplier 2, P(A|B’)=0.985

(c)

For supplier 1, P(A|B)=0.998

For supplier 2, P(A|B)=0.9975

(d) The unusual result is that for both a simple component and for a complex assembly, supplier 1

has a greater probability that a part conforms to specifications. However, for overall parts,

supplier 1 has a lower probability. The overall conforming probability is dependent on both the

conforming probability conditioned on the part type and the probability of each part type. Supplier

1 produces more of the compelx parts so that overall conformance from supplier 1 is lower.



Mind Expanding Exercises

3-153. The binomial distribution

P(X=x) =( )!xn!x

!n

− px(1-p)n-x

If n is large, then the probability of the event could be expressed as λ/n, that is λ=np. Wecould re-write the probability mass function as:

P(X=x) =( )!xn!x

!n

−[λ/n]x[1 – (λ/n)]n-x

Where p = λ/n.

P(X=x)n

1)x(n3)......(n2)(n1)(nnx

+−×−×−×−×x!

λ x(1 – (λ/n))n-x

Now we can re-express:

3-29



[1 – (λ/n)]n-x = [1 – (λ/n)]n[1 – (λ/n)]-x And the first part of this can be re-expressed further as

[1 – (λ/n)]n = ( )( )nλ -1n/λ

λ −

−

So:

P(X=x)= n

1)x(n3)......(n2)(n1)(nnx

+−×−×−×−×

x!

λ x

( )( )nλ -1n/λ

λ −

−

[1 – (λ/n)]-x

Now:

In the limit as n → ∞

n

1)x(n3)......(n2)(n1)(nnx

+−×−×−×−× ≅ 1

In the limit as n → ∞

[1 – (λ/n)]-x ≅ 1Thus:

P(X=x) =x!

λ x

( )( )nλ -1n/λ

λ −

−

We know from exponential series that:

Limit z → ∞ ⎟ ⎠

⎞⎜⎝

⎛ +

z

11

z

= e ≅ 2.7183

In our case above –n/λ = z, so ( )nλ -1n/λ −

= e. Thus,

P(X=x) =x!

λ x

eλ −

It is the case, therefore, that

Limit n → ∞ ( )!xn!x

!n

− px(1 – p)n-x =

!x

e xλ

λ−

The distribution of the probability associated with this process is known as the Poisson distribution.

The pmf can be expressed as:

f(x) =!x

e xλ

λ−

3-154. Sow that using an infinite sum.∑∞

=

− =−1

1 1)1(i

i p p

To begin, , by definition of an infinite sum this can be rewritten

as

∑∑∞

=

−∞

=

− −=−1

1

1

1 )1()1(i

i

i

i p p p p

1

)1(1

)1(

1

1 ==−−

=−∑∞

=

−

p

p

p

p p p

i

i

3-30



3-155.

[ ]

1

1 1

2 2

2 2

2

( ) [( ( 1) ... ]( 1)

( 1) ( 1)

2 2( 1) ( 1)

( ) ( )( 1)

2 2( 1) ( 1)

( )

2

( 1)(( )

( )1

b a

i i

b bb

b a

i a i ai a

E X a a b b a

b b a ai i

b a b a

b a b a b a b a

b a b a

b a

b a b ai b a ii

V X b a

−

= =

+

= ==

= + + + + − +

⎡ ⎤ + −⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= =− + − +

⎡ ⎤− + + + − +⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= =

− + − +

+=

− + +− + +−

= =+ −

∑ ∑

∑ ∑∑2

2

2

)

4

1

( 1)(2 1) ( 1) (2 1) ( 1) ( 1) ( 1)( )( )6 6 2 4

1

( 1) 1

12

b a

b b b a a a b b a a b a b ab a

b a

b a

⎡ ⎤⎢ ⎥⎣ ⎦

+ −

+ + − − + − − − + +⎡ ⎤− − + +⎢ ⎥⎣ ⎦=

− +

− + −=

3-156. Let X denote the number of nonconforming products in the sample. Then, X is approximately binomial with

p = 0.01 and n is to be determined.

If , then90.0)1( ≥≥ X P 10.0)0( ≤= X P .

Now, P(X = 0) = ( ) nnn p p p )1()1(0

0 −=− . Consequently, , and( ) .1 0− ≤p n 10

11.229)1ln(

10.0ln =−

≤ p

n . Therefore, n = 230 is required

3-157. If the lot size is small, 10% of the lot might be insufficient to detect nonconforming product. For example, if

the lot size is 10, then a sample of size one has a probability of only 0.2 of detecting a nonconforming

product in a lot that is 20% nonconforming.

If the lot size is large, 10% of the lot might be a larger sample size than is practical or necessary. For

example, if the lot size is 5000, then a sample of 500 is required. Furthermore, the binomialapproximation to the hypergeometric distribution can be used to show the following. If 5% of the lot of size

5000 is nonconforming, then the probability of zero nonconforming product in the sample is approximately

. Using a sample of 100, the same probability is still only 0.0059. The sample of size 500 might7 10 12× −

be much larger than is needed.

3-31



3-158. Let X denote the number of panels with flaws. Then, X is a binomial random variable with n =100 and p is

the probability of one or more flaws in a panel. That is, p = = 0.095.1.01 −− e

( ) ( ) ( )

( ) ( )

034.0

)1()1(

)1()1()1(

)4()3()2()1()0()4()5(

9641004

9731003

9821002

9911001

10001000

=

−+−+

−+−+−=

=+=+=+=+==≤=<

p p p p

p p p p p p

X P X P X P X P X P X P X P

3-159. Let X denote the number of rolls produced.

Revenue at each demand

0 1000 2000 3000

0 1000≤ ≤x 0.05x 0.3x 0.3x 0.3x

mean profit = 0.05x(0.3) + 0.3x(0.7) - 0.1x

1000 2000≤ ≤x 0.05x 0.3(1000) +0.05(x-1000)

0.3x 0.3x

mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x-1000)](0.2) + 0.3x(0.5) - 0.1x

2000 3000≤ ≤x 0.05x 0.3(1000) +

0.05(x-1000)

0.3(2000) +

0.05(x-2000)

0.3x

mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000) + 0.05(x-2000)](0.3) + 0.3x(0.2) - 0.1x

3000 ≤ x 0.05x 0.3(1000) +

0.05(x-1000)

0.3(2000) +

0.05(x-2000)

0.3(3000)+

0.05(x-3000)

mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000)+0.05(x-2000)]0.3 + [0.3(3000)+0.05(x-

3000)]0.2 - 0.1x

Profit Max. profit

0 1000≤ ≤x 0.125 x $ 125 at x = 1000

1000 2000≤ ≤x 0.075 x + 50 $ 200 at x = 2000

2000 3000≤ ≤x 200 $200 at x = 3000

3000 ≤ x -0.05 x + 350 $200 at x = 3000

The bakery can make anywhere from 2000 to 3000 and earn the same profit.

3-160.Let X denote the number of acceptable components. Then, X has a binomial distribution with p = 0.98 and

n is to be determined such that P X .( ) .≥ ≥100 0 95

n P X( )≥ 100

102 0.666

103 0.848

104 0.942

105 0.981

Therefore, 105 components are needed.

3-32



CHAPTER 3

Section 3-1

3-1. The range of X is { } 1000,...,2,1,0

3-2. The range of X is { } 0 12 50, , ,...,

3-3. The range of X is { } 0 1 2 99999, , ,...,

3-4. The range of X is { } 0 1 2 3 4 5, , , , ,

3-5. The range of X is { . Because 490 parts are conforming, a nonconforming part must be

selected in 491 selections.

}

}

491,...,2,1

3-6. The range of X is { . Although the range actually obtained from lots typically might not

exceed 10%.

0 12 100, , ,...,

3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is{ }0 12, , ,...

3-8. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is

{ }0 12, , ,...

3-9. The range of X is { } 15,...,2,1,0

3-10. The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2,

1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8.

Therefore the range of X is1

4

3

8

1

2

5

8

6

8, , , ,

⎧⎨⎩

⎫⎬⎭

3-11. The range of X is }10000,,2,1,0{ …

3-12.The range of X is {100, 101, …, 150}

3-13.The range of X is {0,1,2,…, 40000)



Section 3-2

3-14.

6/1)3(

6/1)2(

3/1)5.1()5.1(

3/16/16/1)0()0(

=

=

===

=+===

X

X

X

X

f

f

X P f

X P f

a) P( X = 1.5) = 1/3

b) P(0.5< X < 2.7) = P( X = 1.5) +P( X = 2) = 1/3 + 1/6 = 1/2

c) P( X > 3) = 0

d) (0 2) ( 0) ( 1.5) 1/ 3 1/ 3 2 / 3P X P X P X ≤ < = = + = = + =e) P( X = 0 or X = 2) = 1/3 + 1/6 = 1/2

3-15. All probabilities are greater than or equal to zero and sum to one.

3-1



a) P(X ≤ 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1

b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8

c) P(-1 ≤ X ≤ 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4

d) P(X ≤ -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2

3-16. All probabilities are greater than or equal to zero and sum to one.

a) P(X≤ 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.5714=0.4286

c) P(2<X<6)=P(X=3)=0.1429

d) P(X≤1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1

3-17. Probabilities are nonnegative and sum to one.

a) P(X = 4) = 9/25

b) P(X ≤ 1) = 1/25 + 3/25 = 4/25

c) P(2 ≤ X < 4) = 5/25 + 7/25 = 12/25

d) P(X > −10) = 1

3-18. Probabilities are nonnegative and sum to one.

a) P(X = 2) = 3/4(1/4)2 = 3/64

b) P(X ≤ 2) = 3/4[1+1/4+(1/4)2] = 63/64

c) P(X > 2) = 1 − P(X ≤ 2) = 1/64d) P(X ≥ 1) = 1 − P(X ≤ 0) = 1 − (3/4) = 1/4

3-19. X = number of successful surgeries.

P(X=0)=0.1(0.33)=0.033

P(X=1)=0.9(0.33)+0.1(0.67)=0.364

P(X=2)=0.9(0.67)=0.603

3-20. P(X = 0) = 0.023

= 8 x 10-6

P(X = 1) = 3[0.98(0.02)(0.02)]=0.0012

P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576

P(X = 3) = 0.983 = 0.9412

3-21. X = number of wafers that pass

P(X=0) = (0.2)3 = 0.008P(X=1) = 3(0.2)2(0.8) = 0.096

P(X=2) = 3(0.2)(0.8)2 = 0.384

P(X=3) = (0.8)3 = 0.512

3-22. X: the number of computers that vote for a left roll when a right roll is appropriate.

p=0.0001.

P(X=0)=(1-p)4=0.99994=0.9996

P(X=1)=4*(1-p)3 p=4*0.999930.0001=0.0003999

P(X=2)=C42(1-p)2 p2=5.999*10-8

P(X=3)=C43(1-p)1 p3=3.9996*10-12

P(X=4)=C40(1-p)0 p4=1*10-16

3-23. P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2

3-24. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1

3-25. P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1

3-26. X = number of components that meet specifications

P(X=0) = (0.05)(0.02) = 0.001

P(X=1) = (0.05)(0.98) + (0.95)(0.02) = 0.068

P(X=2) = (0.95)(0.98) = 0.931

3-2



3-27. X = number of components that meet specifications

P(X=0) = (0.05)(0.02)(0.01) = 0.00001

P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167

P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663

P(X=3) = (0.95)(0.98)(0.99) = 0.92169



Section 3-3

3-28. where

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≤

<≤

<≤

<≤

<

=

x

x

x

x

x

xF

31

326/5

25.13/2

5.103/1

0,0

)(

6/1)3(

6/1)2(

3/1)5.1()5.1(

3/16/16/1)0()0(

=

=

===

=+===

X

X

X

X

f

f

X P f

X P f

3-29.

where

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

≤

<≤

<≤

<≤−

−<≤−−<

=

x

x

x

x

x

x

xF

21

218/7

108/5

018/3

128/1

2,0

)(

8/1)2(

8/2)1(

8/2)0(

8/2)1(

8/1)2(

=

=

=

=−

=−

X

X

X

X

X

f

f

f

f

f

a) P(X ≤ 1.25) = 7/8

b) P(X ≤ 2.2) = 1

c) P(-1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4

d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8

3-30. where

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

≤

<≤

<≤

<≤

<≤

<

=

x

x

x

x

x

x

xF

41

4325/16

3225/9

2125/4

1025/1

0,0

)(

25/9)4(

25/7)3(

25/5)2(

25/3)1(

25/1)0(

=

=

=

=

=

X

X

X

X

X

f

f

f

f

f

a) P(X < 1.5) = 4/25

b) P(X ≤ 3) = 16/25

c) P(X > 2) = 1 − P(X ≤ 2) = 1 − 9/25 = 16/25

d) P(1 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 1) = 9/25 − 4/25 = 5/25 = 1/5

3-31.

3-3



⎪

⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪

⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≤

<≤

<≤

<≤

<

=

x

x

x

x

x

xF

3,1

32,488.0

21,104.0

10,008.0

0,0

)(

where

,512.0)8.0()3(

,384.0)8.0)(8.0)(2.0(3)2(

,096.0)8.0)(2.0)(2.0(3)1(

,008.02.0)0(

.

3

3

==

==

==

==

f

f

f

f

3-32.

0, 00.9996, 0 1

( ) 0.9999, 1 3

0.99999, 3 4

1, 4

x x

F x x

x

x

<⎧ ⎫⎪ ⎪≤ <⎪ ⎪⎪ ⎪

= ≤ <⎨ ⎬⎪ ⎪≤ <⎪ ⎪

≤⎪ ⎪⎩ ⎭

where

4

3

8

12

16

.

(0) 0.9999 0.9996,

(1) 4(0.9999 )(0.0001) 0.0003999,

(2) 5.999 *10 ,

(3) 3.9996*10 ,

(4) 1*10

f

f

f

f

f

−

−

−

= =

= =

=

=

=

3-33.

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

≤

<≤

<≤

<

=

x

x

x

x

xF

50,1

5025,5.0

2510,2.0

10,0

)(

where P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2

3-4



3-34.

⎪

⎪

⎭

⎪⎪

⎬

⎫

⎪

⎪

⎩

⎪⎪

⎨

⎧

≤

<≤

<≤

<

=

x

x

x

x

xF

10,1

105,7.0

51,1.0

1,0

)(

where P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1

3-35. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero;

pmf: f (1) = 0.5, f (3) = 0.5

a) P(X ≤ 3) = 1

b) P(X ≤ 2) = 0.5

c) P(1 ≤ X ≤ 2) = P(X=1) = 0.5

d) P(X>2) = 1 − P(X≤2) = 0.5


pmf: f(1) = 0.7, f(4) = 0.2, f(7) = 0.1

a) P(X ≤ 4) = 0.9

b) P(X > 7) = 0

c) P(X ≤ 5) = 0.9

d) P(X>4) = 0.1

e) P(X≤2) = 0.7


pmf: f(-10) = 0.25, f(30) = 0.5, f(50) = 0.25

a) P(X≤50) = 1

b) P(X≤40) = 0.75

c) P(40 ≤ X ≤ 60) = P(X=50)=0.25

d) P(X<0) = 0.25

e) P(0≤X<10) = 0

f) P(−10<X<10) = 0

3-5




pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1

a) P(X≤1/18) = 0

b) P(X≤1/4) = 0.9

c) P(X≤5/16) = 0.9

d) P(X>1/4) = 0.1

e) P(X≤1/2) = 1



Section 3-4

3-39. Mean and Variance

2)2.0(4)2.0(3)2.0(2)2.0(1)2.0(0

)4(4)3(3)2(2)1(1)0(0)(

=++++=

++++== f f f f f X E μ

22)2.0(16)2.0(9)2.0(4)2.0(1)2.0(0

)4(4)3(3)2(2)1(1)0(0)(

2

222222

=−++++=

−++++= μ f f f f f X V

3- 40. Mean and Variance for random variable in exercise 3-14

333.1)6/1(3)6/1(2)3/1(5.1)3/1(0

)3(3)2(2)5.1(5.1)0(0)(

=+++=

+++== f f f f X E μ

139.1333.1)6/1(9)6/1(4)3/1(25.2)3/1(0

)3(3)2(2)1(5.1)0(0)(

2

22222

=−+++=

−+++= μ f f f f X V

3-41. Determine E(X) and V(X) for random variable in exercise 3-15

.0)8/1(2)8/2(1)8/2(0)8/2(1)8/1(2

)2(2)1(1)0(0)1(1)2(2)(

=+++−−=

+++−−−−== f f f f f X E μ

5.10)8/1(4)8/2(1)8/2(0)8/2(1)8/1(4

)2(2)1(1)0(0)1(1)2(2)(

2

222222

=−++++=

−+++−−−−= μ f f f f f X V

3-42. Determine E(X) and V(X) for random variable in exercise 3-16

( ) 1 (1) 2 (2) 3 (3)

1(0.5714286) 2(0.2857143) 3(0.1428571)

1.571429

E X f f f μ = = + +

= + +

=

2 2 2( ) 1 (1) 2 (2) 3 (3)

1.428571

V X f f f 2

μ = + + + −

=

3-43. Mean and variance for exercise 3-17

8.2)36.0(4)28.0(3)2.0(2)12.0(1)04.0(0

)4(4)3(3)2(2)1(1)0(0)(

=++++=

++++== f f f f f X E μ

36.18.2)36.0(16)28.0(9)2.0(4)12.0(1)04.0(0

)4(4)3(3)2(2)1(1)0(0)(

2

222222

=−++++=

−++++= μ f f f f f X V

3-6




3

1

4

1

4

3

4

1

4

3)(

10

=⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛ = ∑∑

∞

=

∞

= x

x

x

x

x x X E

The result uses a formula for the sum of an infinite series. The formula can be derived from the fact that the

series to sum is the derivative of a

aaah x

x

−== ∑∞

= 1)(

1

with respect to a.

For the variance, another formula can be derived from the second derivative of h(a) with respect to a.

Calculate from this formula

9

5

4

1

4

3

4

1

4

3)(

1

2

0

22 =⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛ = ∑∑

∞

=

∞

= x

x

x

x

x x X E

Then [ ]9

4

9

1

9

5)()()(

22 =−=−= X E X E X V

3-45. Mean and variance for random variable in exercise 3-19

( ) 0 (0) 1 (1) 2 (2)0(0.033) 1(0.364) 2(0.603)

1.57

E X f f f μ = = + += + +

=

2 2 2

2

( ) 0 (0) 1 (1) 2 (2)

0(0.033) 1(0.364) 4(0.603) 1.57

0.3111

V X f f f 2

μ = + + −

= + + −

=


6

( ) 0 (0) 1 (1) 2 (2) 3 (3)

0(8 10 ) 1(0.0012) 2(0.0576) 3(0.9412)

2.940008

E X f f f f μ

−

= = + + +

= × + + +

=

2 2 2 2( ) 0 (0) 1 (1) 2 (2) 3 (3)

0.05876096

V X f f f f 2μ = + + + −

=

3-47. Determine x where range is [0,1,2,3,x] and mean is 6.

24

2.08.4

2.02.16

)2.0()2.0(3)2.0(2)2.0(1)2.0(06

)()3(3)2(2)1(1)0(06)(

=

=

+=

++++=

++++===

x

x

x

x

x xf f f f f X E μ

3-7



3-48. (a) F(0)=0.17

Nickel Charge: X CDF

0 0.17

2 0.17+0.35=0.52

3 0.17+0.35+0.33=0.854 0.17+0.35+0.33+0.15=1

(b)E(X) = 0*0.17+2*0.35+3*0.33+4*0.15=2.29

V(X) =

42

1

( )( )i i

i

f x x μ

=

−∑ = 1.5259

3-49. X = number of computers that vote for a left roll when a right roll is appropriate.

µ = E(X)=0*f(0)+1*f(1)+2*f(2)+3*f(3)+4*f(4)

= 0+0.0003999+2*5.999*10-8+3*3.9996*10-12+4*1*10-16= 0.0004

V(X)=

5

2

1

( )( )i i

i

f x x μ

=

−∑ = 0.00039996

3-50. µ=E(X)=350*0.06+450*0.1+550*0.47+650*0.37=565

V(X)= =6875∑=

−4

1

2))((i

i x x f μ

σ= )( X V =82.92

3-51. (a)

Transaction Frequency Selects: X f(X)

New order 43 23 0.43

Payment 44 4.2 0.44

Order status 4 11.4 0.04

Delivery 5 130 0.05

Stock level 4 0 0.04

total 100

µ =

E(X) = 23*0.43+4.2*0.44+11.4*0.04+130*0.05+0*0.04 =18.694

V(X) = = 735.964∑=

−5

1

2))((i

i x x f μ 1287.27)( == X V σ

(b)

Transaction Frequency All operation: X f(X) New order 43 23+11+12=46 0.43

Payment 44 4.2+3+1+0.6=8.8 0.44

Order status 4 11.4+0.6=12 0.04

Delivery 5 130+120+10=260 0.05

Stock level 4 0+1=1 0.04

total 100

µ = E(X) = 46*0.43+8.8*0.44+12*0.04+260*0.05+1*0.04=37.172

3-8



V(X) = =2947.996∑=

−5

1

2))((i

i x x f μ 2955.54)( == X V σ



Section 3-5

3-52. E(X) = (0+100)/2 = 50, V(X) = [(100-0+1)2-1]/12 = 850

3-53. E(X) = (3+1)/2 = 2, V(X) = [(3-1+1)2 -1]/12 = 0.667

3-54. X=(1/100)Y, Y = 15, 16, 17, 18, 19.

E(X) = (1/100) E(Y) = 17.02

1915

100

1=⎟

⎠

⎞⎜⎝

⎛ +mm

0002.0

12

1)11519(

100

1)(

22

=⎥

⎦

⎤⎢

⎣

⎡ −+−⎟

⎠

⎞⎜

⎝

⎛ = X V mm2

3-55. 33

14

3

13

3

12)( =⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ = X E

in 100 codes the expected number of letters is 300

( ) ( ) ( ) ( )3

23

3

14

3

13

3

12)(

2222 =−⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ = X V

in 100 codes the variance is 6666.67

3-56. X = 590 + 0.1Y, Y = 0, 1, 2, ..., 9

E(X) = 45.5902

901.0590 =⎟ ⎠ ⎞⎜

⎝ ⎛ ++ mm

0825.012

1)109()1.0()(

22 =⎥

⎦

⎤⎢⎣

⎡ −+−= X V mm2

3-57. a = 675, b = 700

(a) µ = E(X) = (a+b)/2= 687.5

V(X) = [(b – a +1)2

– 1]/12= 56.25

(b) a = 75, b = 100

µ = E(X) = (a+b)/2 = 87.5

V(X) = [(b – a + 1)2 – 1]/12= 56.25

The range of values is the same, so the mean shifts by the difference in the two minimums (or

maximums) whereas the variance does not change.

3-58. X is a discrete random variable. X is discrete because it is the number of fields out of 28 that has an error.

However, X is not uniform because P(X=0) ≠ P(X=1).

3-59. The range of Y is 0, 5, 10, ..., 45, E(X) = (0+9)/2 = 4.5

E(Y) = 0(1/10)+5(1/10)+...+45(1/10)

= 5[0(0.1) +1(0.1)+ ... +9(0.1)]

= 5E(X)

= 5(4.5)

= 22.5

V(X) = 8.25, V(Y) = 52(8.25) = 206.25, σY = 14.36

3-9



3-60. ,∑∑ === x x

X cE x xf c xcxf cX E )()()()(

∑∑ =−=−= x x

X cV x f xc x f ccxcX V )()()()()()( 222μ μ



Section 3-7

3-81 . a) 5.05.0)5.01()1( 0 =−== X P

b) 0625.05.05.0)5.01()4( 43 ==−== X P

c) 0039.05.05.0)5.01()8( 87 ==−== X P

d) 5.0)5.01(5.0)5.01()2()1()2( 10 −+−==+==≤ X P X P X P

75.05.05.02 =+=

e.) 25.075.01)2(1)2( =−=≤−=> X P X P

3-82. E(X) = 2.5 = 1/p giving p = 0.4

a) 4.04.0)4.01()1( 0 =−== X P

b) 0864.04.0)4.01()4( 3 =−== X P

c) 05184.05.0)5.01()5( 4 =−== X P

d) )3()2()1()3( =+=+==≤ X P X P X P X P

7840.04.0)4.01(4.0)4.01(4.0)4.01( 210 =−+−+−=e) 2160.07840.01)3(1)3( =−=≤−=> X P X P

3-83. Let X denote the number of trials to obtain the first success.

a) E(X) = 1/0.2 = 5

b) Because of the lack of memory property, the expected value is still 5.

3-84. a) E(X) = 4/0.2 = 20

b) P(X=20) = 0436.02.0)80.0(3

19416 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

3-15



c) P(X=19) = 0459.02.0)80.0(3

18415 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

d) P(X=21) = 0411.02.0)80.0(3

20417 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

e) The most likely value for X should be near μX. By trying several cases, the most likely value is x = 19.

3-85. Let X denote the number of trials to obtain the first successful alignment. Then X is a geometric random

variable with p = 0.8

a) 0064.08.02.08.0)8.01()4( 33 ==−== X P

b) )4()3()2()1()4( =+=+=+==≤ X P X P X P X P X P

8.0)8.01(8.0)8.01(8.0)8.01(8.0)8.01( 3210 −+−+−+−=

9984.08.02.0)8.0(2.0)8.0(2.08.0 32 =+++=c) )]3()2()1([1)3(1)4( =+=+=−=≤−=≥ X P X P X P X P X P

]8.0)8.01(8.0)8.01(8.0)8.01[(1 210 −+−+−−=

008.0992.01)]8.0(2.0)8.0(2.08.0[1

2

=−=++−=

3-86. Let X denote the number of people who carry the gene. Then X is a negative binomial random variable with

r=2 and p = 0.1

a) )]3()2([1)4(1)4( =+=−=<−=≥ X P X P X P X P

972.0)018.001.0(11.0)1.01(1

21.0)1.01(

1

11 2120 =+−=⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

b) 201.0/2/)( === pr X E

3-87. Let X denote the number of calls needed to obtain a connection. Then, X is a geometric random variable

with p = 0.02.

a) 0167.002.098.002.0)02.01()10( 99 ==−== X Pb) )]5()4()3()2()1([1)4(1)5( =+=+=+=+=−=≤−=> X P X P X P X P X P X P X P

)]02.0(98.0)02.0(98.0)02.0(98.0)02.0(98.0)02.0(98.002.0[1 4332 +++++−=9039.00961.01 =−=

May also use the fact that P(X > 5) is the probability of no connections in 5 trials. That is,

9039.098.002.00

5)5( 50 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ => X P

c) E(X) = 1/0.02 = 50

3-88. X: # of opponents until the player is defeated.

p=0.8, the probability of the opponent defeating the player.

(a) f(x) = (1 – p)x – 1 p = 0.8(x – 1)*0.2

(b) P(X>2) = 1 – P(X=1) – P(X=2) = 0.64

(c) µ = E(X) = 1/p = 5

(d) P(X≥4) = 1-P(X=1)-P(X=2)-P(X=3) = 0.512

(e) The probability that a player contests four or more opponents is obtained in part (d), which is

po = 0.512.

Let Y represent the number of game plays until a player contests four or more opponents.

Then, f(y) = (1-po)y-1 po.

3-16



µY = E(Y) = 1/po = 1.95

3-89. p=0.13

(a) P(X=1) = (1-0.13)1-1*0.13=0.13.

(b) P(X=3)=(1-0.13)3-1*0.13 =0.098

(c) µ=E(X)= 1/p=7.69≈

8

3-90. X: # number of attempts before the hacker selects a user password.

(a) p=9900/366=0.0000045

µ=E(X) = 1/p= 219877

V(X)= (1-p)/p2 = 4.938*1010

σ= )( X V =222222

(b) p=100/363=0.00214

µ=E(X) = 1/p= 467

V(X)= (1-p)/p2 = 217892.39

σ= )( X V =466.78

Based on the answers to (a) and (b) above, it is clearly more secure to use a 6 character password.

3-91 . p = 0.005 , r = 8

a.)198 1091.3005.0)8( −=== x X P

b). 200005.0

1)( === X E μ days

c) Mean number of days until all 8 computers fail. Now we use p=3.91x10-19

18

911056.2

1091.3

1)( x

xY E ===

−μ days or 7.01 x10

15years

3-92. Let Y denote the number of samples needed to exceed 1 in Exercise 3-66. Then Y has a geometric

distribution with p = 0.0169.

a) P(Y = 10) = (1 − 0.0169)9(0.0169) = 0.0145

b) Y is a geometric random variable with p = 0.1897 from Exercise 3-66.

P(Y = 10) = (1 − 0.1897)9(0.1897) = 0.0286

c) E(Y) = 1/0.1897 = 5.27

3-93. Let X denote the number of transactions until all computers have failed. Then, X is negative binomial

random variable with p = 10-8 and r = 3.

a) E(X) = 3 x 108

b) V(X) = [3(1−10-80]/(10-16) = 3.0 x 1016

3-94. (a) p6=0.6, p=0.918

(b) 0.6*p

2

=0.4, p=0.816

3-95 . Negative binomial random variable: f(x; p, r) = .r r x

p pr

x −−⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

−)1(

1

1

When r = 1, this reduces to f(x; p, r) = (1− p)x-1 p, which is the pdf of a geometric random variable.

Also, E(X) = r/p and V(X) = [r(1− p)]/p2

reduce to E(X) = 1/p and V(X) = (1− p)/p2, respectively.

3-17



3-96. Let X denote a geometric random variable with parameter p. Let q = 1-p.

1 1

21 1

1 1( ) (1 )

x x

x x

E X x p p p xq p p p

∞ ∞− −

= =

⎛ ⎞= − = = =⎜ ⎟⎝ ⎠

∑ ∑

( )

2 2

2

2

2

2 2

2 1 21 1

1 1

2 1 1 11

1 1 1

2 1 2 1

1

2 1 1

1

2 3 1

2 1

1

(1 )

( ) ( ) (1 ) 2 (1 )

2

2 3 ...

(1 2 3 ...)

2

1 x x

p p

x x

x x x

p

x x x

x

p p x

x

p x

d dq p

d dq p

qd dq q p

V X x p p px x p

p x q xq q

p x q

p x q

p q q q

p q q q

p

∞ ∞− −

= =

∞ ∞ ∞− − −

= = =

∞−

=

∞−

=

−

= − − = − + −

= − +

= − +

= −

⎡ ⎤= + + + −⎣ ⎦

⎡ ⎤= + + + −⎣ ⎦

⎡ ⎤= − =⎣ ⎦

∑ ∑

∑ ∑ ∑

∑∑

[ ]

2

3 2 1

2 2 2

(1 ) (1 )

2(1 ) 1 (1 )

p pq q p q

p p p q

p p p

− −− + − −

− + − −= = =



Section 3-8

3-97. X has a hypergeometric distribution N=100, n=4, K=20

a.)( ) ( )( )

4191.03921225

)82160(20)1(

100

4

80

3

20

1 ==== X P

b.) , the sample size is only 40)6( == X P

c.)( ) ( )( )

001236.03921225

)1(4845)4(

100

4

80

0

20

4 ==== X P

d.) 8.0

100

204)( =⎟

⎠

⎞⎜

⎝

⎛ ===

N

K nnp X E

6206.099

96)8.0)(2.0(4

1)1()( =⎟

⎠

⎞⎜⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛

−

−−=

N

n N pnp X V

3-18



3-98. a)( ) ( )( )

4623.024/)17181920(

6/)1415164()1(

20

4

16

3

4

1 =×××

×××=== X P

b)( ) ( )( )

00021.024/)17181920(

1)4(

20

4

16

0

4

4 =×××

=== X P

c)

( )( )( )

( )( )( )

( )( )( )

9866.0

)2()1()0()2(

24

17181920

215166

61415164

2413141516

20

4

16

2

4

2

20

4

16

3

4

1

20

4

16

4

4

0

==

++=

=+=+==≤

⎟ ⎠

⎞⎜⎝

⎛ ×××

⎟ ⎠ ⎞⎜

⎝ ⎛ ××+×××+×××

X P X P X P X P

d) E(X) = 4(4/20) = 0.8

V(X) = 4(0.2)(0.8)(16/19) = 0.539

3-99. N=10, n=3 and K=4

0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.5

x

P ( x )

3-100. (a) f(x) = /⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−⎟⎟ ⎠

⎞⎜⎜⎝

⎛

x x 3

1224⎟⎟ ⎠

⎞⎜⎜⎝

⎛

3

36

(b) µ=E(X) = np= 3*24/36=2V(X)= np(1-p)(N-n)/(N-1) =2*(1-24/36)(36-3)/(36-1)=0.629

(c) P(X≤2) =1-P(X=3) =0.717

3-19



3-101. Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high

blood pressure. N=800, K=240 n=10

a) n=10

( )( )( )

( )( )1201.0)1(

!790!10!800

!551!9!560

!239!1!240

800

10

560

9

240

1 ==== X P

b) n=10

)]1()0([1)1(1)1( =+=−=≤−=> X P X P X P X P

( )( )( )

( )( )0276.0)0(

!790!10!800

!550!10!560

!240!0!240

800

10

560

10

240

0 ==== X P

8523.0]1201.00276.0[1)1(1)1( =+−=≤−=> X P X P

3-102. Let X denote the number of cards in the sample that are defective.

a)

( )( )( )

9644.00356.01)1(

0356.0)0(

)0(1)1(

!120!20!140

!100!20!120

140

20

120

20

20

0

=−=≥

=====−=≥

X P

X P

X P X P

b)

( )( )( )

5429.04571.01)1(

4571.0!140!115

!120!135)0(

)0(1)1(

!120!20!140

!115!20!135

140

20

135

20

5

0

=−=≥

=====

=−=≥

X P

X P

X P X P

3-103. N=300

(a) K=243, n=3, P(X=1)=0.087

(b) P(X≥1)=0.9934

(c) K=26+13=39

P(X=1)=0.297

(d) K=300-18=282

P(X≥1)=0.9998

3-104. Let X denote the count of the numbers in the state's sample that match those in the player's

sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6.

a)( )( )( )

7

1

40

6

34

0

6

6 1061.2!34!6

!40)6( −

−

×=⎟ ⎠

⎞⎜⎝

⎛ === X P

b)( )( )( ) ( )

5

40

6

40

6

34

1

6

5 1031.5346

)5( −×=×

=== X P

c)( )( )( )

00219.0)4(40

6

34

2

6

4 === X P

d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p =

3-20



380,838,3

1and E(Y) = 1/p = weeks. This is more than 738 centuries!380,838,3

3-105. Let X denote the number of blades in the sample that are dull.

a)

( ) ( )( )

7069.0)0(1)1(

2931.0!33!48

!43!38)0(

)0(1)1(

!43!5

!48

!33!5

!38

48

5

38

5

10

0

==−=≥

=====

=−=≥

X P X P

X P

X P X P

b) Let Y denote the number of days needed to replace the assembly.

P(Y = 3) = 0607.0)7069.0(2931.0 2 =

c) On the first day,( )

8005.0!41!48

!43!46)0(

!43!5!48

!41!5!46

48

5

46

5

2

0 ===== X P

On the second day,

( )( )( ) 4968.0!37!48

!43!42

)0(!43!5

!48

!37!5!42

485

42

5

6

0

===== X P

On the third day, P(X = 0) = 0.2931 from part a. Therefore,

P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811.

3-106. a) For Exercise 3-97, the finite population correction is 96/99.

For Exercise 3-98, the finite population correction is 16/19.

Because the finite population correction for Exercise 3-97 is closer to one, the binomial approximation to

the distribution of X should be better in Exercise 3-97.

b) Assuming X has a binomial distribution with n = 4 and p = 0.2,

( )

( ) 0016.08.02.0)4(

4096.08.02.0)1(

044

4

314

1

===

===

X P

X P

The results from the binomial approximation are close to the probabilities obtained in Exercise

3-97.

c) Assume X has a binomial distribution with n = 4 and p = 0.2. Consequently, P(X = 1) and

P(X = 4) are the same as computed in part b. of this exercise. This binomial approximation is

not as close to the true answer as the results obtained in part b. of this exercise.

d) From Exercise 3-102, X is approximately binomial with n = 20 and p = 20/140 = 1/7.

( )( ) ( ) 9542.00458.01)0(1)1(20

760

7120

0 =−===−=≥ X P X P

finite population correction is 120/139=0.8633

From Exercise 3-92, X is approximately binomial with n = 20 and p = 5/140 =1/28

( )( ) ( ) 5168.04832.01)0(1)1(20

28270

28120

0 =−===−=≥ X P X P

finite population correction is 120/139=0.8633

3-21



Section 3-9

3-107. a) P Xe

e( )!

.= = = =−

−04

000183

4 04

b) P X P X P X P X( ) ( ) ( ) (≤ = = + = )+ =2 0 1 2

= + +

=

−− −

e e e44 1 4 2

41

42

0 2381

! !

.

c) P Xe

( )!

.= = =−

44

401954

4 4

d) P Xe

( )!

.= = =−

84

80 0298

4 8

3-108 a) P X e( ) ..= = =−0 0 67030 4

b) P X ee e

( )( . )

!

( . )

!..

. .

≤ = + + =−− −

20 4

1

0 4

20 99210 4

0 4 0 4 2

c) P X

e

( )

( . )

! .

.

= = =

−

4

0 4

4 0 000715

0 4 4

d) P Xe

( )( . )

!.

.

= = = ×−

−80 4

8109 10

0 4 88

3-109. . Therefore, λ = −ln(0.05) = 2.996.P X e( ) .= = =−0 0λ 05

Consequently, E(X) = V(X) = 2.996.

3-110 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with λ = 10.

P Xe

( )!

.= = =−

510

50 0378

10 5

.

b) P X ee e e

( )

! ! !

.≤ = + + + =−− − −

310

1

10

2

10

3

0 01031010 10 2 10 3

c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with

λ = 20. P Ye

( )!

.= = =−

1520

150 0516

20 15

d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with

λ = 5. P We

( )!

.= = =−

55

501755

5 5

3-111. λ =1, Poisson distribution. f(x) =e- λ λ x/x!

(a) P(X≥2)= 0.264

(b) In order that P(X≥1) = 1-P(X=0)=1-e- λ exceed 0.95, we need λ =3.

Therefore 3*16=48 cubic light years of space must be studied.

3-112. (a) λ =14.4, P(X=0)=6*10-7

(b) λ =14.4/5=2.88, P(X=0)=0.056

(c) λ =14.4*7*28.35/225=12.7

P(X≥1)=0.999997

(d) P(X≥28.8) =1-P(X ≤ 28) = 0.00046. Unusual.

3-113. (a) λ =0.61. P(X≥1)=0.4566

(b) λ =0.61*5=3.05, P(X=0)= 0.047.

3-22



3-114. a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable

with = 0.1.λ 0045.0!2

)1.0()2(

21.0

===−

e X P

b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable

with = 1.λ 3679.0!11)1( 1

11

==== −

−

eeY P

c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable

with λ = 2. 1353.0)0( 2 === −eW P

d) )1()0(1)1(1)2( =−=−=≤−=≥ Y PY PY PY P

2642.0

1 11

=

−−= −−ee

3-115.a) 2.0)( == λ X E errors per test area

b) 9989.0

!2

)2.0(

!1

2.0)2(

22.02.02.0 =++=≤

−−− ee

e X P

99.89% of test areas

3-116. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with

= 10.λ510 1054.4)0( −− ×=== e X P

b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with

λ = 1. 6321.01)0(1)1( 1 =−==−=≥ −eY PY P

c) The assumptions of a Poisson process require that the probability of a count is constant for all intervals.

If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson

process are not valid. Separate Poisson random variables might be appropriate for the heavy and light

load sections of the highway.

3-117. a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random

variable with = 0.5.λ 6065.0)0( 5.0 === −e X P

b) Let Y denote the number of cars with no flaws,

0067.0)3935.0()6065.0(10

10)10( 010 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ==Y P

c) Let W denote the number of cars with surface flaws. Because the number of flaws has a

Poisson distribution, the occurrences of surface flaws in cars are independent events with

constant probability. From part a., the probability a car contains surface flaws is 1−0.6065 =

0.3935. Consequently, W is binomial with n = 10 and p = 0.3935.

0 10

1 9

10( 0) (0.3935) (0.6065) 0.0067

0

10( 1) (0.3935) (0.6065) 0.0437

1

( 1) 0.0067 0.0437 0.0504

P W

P W

P W

⎛ ⎞= = =⎜ ⎟

⎝ ⎠⎛ ⎞

= = =⎜ ⎟⎝ ⎠

≤ = + =

3-23



3-118. a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with λ = 0.16.

8521.0)0( 16.0 === −e X P

b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with

λ = 0.48. 3812.01)0(1)1( 48 =−==−=≥ −eY PY P




3-119.4

1

3

1

8

3

3

1

4

1

3

1

8

1)( =⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ = X E ,

0104.04

1

3

1

8

3

3

1

4

1

3

1

8

1)(

2222

=⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ = X V

3-120. a) 3681.0)999.0(001.01

1000)1( 9991 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ == X P

( )

( ) ( )

999.0)999.0)(001.0(1000)(

1)001.0(1000)()d

9198.0

999.0001.02

1000999.0001.0

1

1000999.0001.0

0

1000)2()c

6319.0999.0001.00

10001)0(1)1() b

9982999110000

9990

==

==

=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ =≤

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −==−=≥

X V

X E

X P

X P X P

3-121. a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np.

b) ( ) ( ) ( ) 112.09.01.02

509.01.01

509.01.00

50)2(

482

491

500

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =≤ X P

c) ( ) ( ) 48050149 1051.49.01.050

509.01.0

49

50)49(

−×=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ =≥ X P

3-122. (a)Binomial distribution, p=0.01, n=12.

(b) P(X>1)=1-P(X≤1)= 1- - =0.0062120 )1(

0

12 p p −⎟⎟

⎠

⎞⎜⎜⎝

⎛ 141 )1(1

12 p p −⎟⎟

⎠

⎞⎜⎜⎝

⎛

(c) µ =E(X)= np =12*0.01 = 0.12

V(X)=np(1-p) = 0.1188 σ= )( X V = 0.3447

3-123. (a)12(0.5) 0.000244=

(b) = 0.22566 6

12 (0.5) (0.5)C 6

(c) 4189.0)5.0()5.0()5.0()5.0( 6612

6

7512

5 =+C C

3-124. (a) Binomial distribution, n=100, p=0.01.

(b) P(X≥1) = 0.634

(c) P(X≥2)= 0.264

3-24



(d) µ=E(X)= np=100*0.01=1

V(X)=np(1-p) = 0.99

σ= )( X V =0.995

(e) Let pd= P(X≥2)= 0.264,

Y: # of messages requires two or more packets be resent.

Y is binomial distributed with n=10, pm=pd*(1/10)=0.0264

P(Y≥1) = 0.235

3-125 Let X denote the number of mornings needed to obtain a green light. Then X is a geometric random

variable with p = 0.20.

a) P(X = 4) = (1-0.2)30.2= 0.1024

b) By independence, (0.8)10

= 0.1074. (Also, P(X > 10) = 0.1074)

3-126 Let X denote the number of attempts needed to obtain a calibration that conforms to specifications. Then, X

is geometric with p = 0.6.

P(X ≤ 3) = P(X=1) + P(X=2) + P(X=3) = 0.6 + 0.4(0.6) + 0.42(0.6) = 0.936.

3-127. Let X denote the number of fills needed to detect three underweight packages. Then X is a negative

binomial random variable with p = 0.001 and r = 3.

a) E(X) = 3/0.001 = 3000

b) V(X) = [3(0.999)/0.0012] = 2997000. Therefore, σX = 1731.18

3-128. Geometric with p=0.1

(a) f(x)=(1-p)x-1 p=0.9(x-1)0.1

(b) P(X=5) = 0.94*0.1=0.0656

(c) µ=E(X)= 1/p=10

(d) P(X≤10)=0.651

3-129. (a) λ =6*0.5=3.P(X=0) = 0.0498

(b) P(X≥3)=0.5768

(c) P(X≤x) ≥0.9, x=5

(d) σ2= λ =6. Not appropriate.

3-130. Let X denote the number of totes in the sample that do not conform to purity requirements. Then, X has a

hypergeometric distribution with N = 15, n = 3, and K = 2.

3714.0!15!10

!12!131

3

15

3

13

0

2

1)0(1)1( =−=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−==−=≥ X P X P

3-131. Let X denote the number of calls that are answered in 30 seconds or less. Then, X is a binomial random

variable with p = 0.75.

a) P(X = 9) = 1877.0)25.0()75.0(9

1019 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

b) P(X ≥ 16) = P(X=16) +P(X=17) + P(X=18) + P(X=19) + P(X=20)

3-25



4148.0)25.0()75.0(20

20)25.0()75.0(

19

20

)25.0()75.0(18

20)25.0()75.0(

17

20)25.0()75.0(

16

20

020119

218317416

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ =

c) E(X) = 20(0.75) = 15

3-132. Let Y denote the number of calls needed to obtain an answer in less than 30 seconds.

a) 0117.075.025.075.0)75.01()4( 33 ==−==Y P

b) E(Y) = 1/p = 1/0.75 = 4/3

3-133. Let W denote the number of calls needed to obtain two answers in less than 30 seconds. Then, W has a

negative binomial distribution with p = 0.75.

a) P(W=6) =5

10 25 0 75 0 01104 2⎛

⎝ ⎜

⎞

⎠⎟ =( . ) ( . ) .

b) E(W) = r/p = 2/0.75 = 8/3

3-134. a) Let X denote the number of messages sent in one hour. 1755.0

!5

5)5(

55

===−e

X P

b) Let Y denote the number of messages sent in 1.5 hours. Then, Y is a Poisson random variable with

λ =7.5. 0858.0!10

)5.7()10(

105.7

===−e

Y P

c) Let W denote the number of messages sent in one-half hour. Then, W is a Poisson random variable with

= 2.5.λ 2873.0)1()0()2( ==+==< W PW PW P

3-135X is a negative binomial with r=4 and p=0.0001

400000001.0/4/)( === pr X E requests

3-136. X ∼ Poisson(λ = 0.01), X ∼ Poisson(λ = 1)

9810.0!3)1(

!2)1(

!1)1()3(

3121111 =+++=≤−−−− eeeeY P

3-137. Let X denote the number of individuals that recover in one week. Assume the individuals are independent.

Then, X is a binomial random variable with n = 20 and p = 0.1. P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.8670 =

0.1330.

3-138. a.) P(X=1) = 0 , P(X=2) = 0.0025, P(X=3) = 0.01, P(X=4) = 0.03, P(X=5) = 0.065

P(X=6) = 0.13, P(X=7) = 0.18, P(X=8) = 0.2225, P(X=9) = 0.2, P(X=10) = 0.16

b.) P(X=1) = 0.0025, P(X=1.5) = 0.01, P(X=2) = 0.03, P(X=2.5) = 0.065, P(X=3) = 0.13

P(X=3.5) = 0.18, P(X=4) = 0.2225, P(X=4.5) = 0.2, P(X=5) = 0.16

3-139. Let X denote the number of assemblies needed to obtain 5 defectives. Then, X is a negative binomial

random variable with p = 0.01 and r=5.

a) E(X) = r/p = 500.

b) V(X) =(5* 0.99)/0.012 = 49500 and σX = 222.49

3-140. Here n assemblies are checked. Let X denote the number of defective assemblies. If P(X ≥ 1) ≥ 0.95, then

3-26



P(X=0) ≤ 0.05. Now,

P(X=0) = and 0.99nn

n99)99.0()01.0(

0

0 =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ n ≤ 0.05. Therefore,

07.298)95.0ln(

)05.0ln(

)05.0ln())99.0(ln(

=≥

≤

n

n

This would require n = 299.

3-141. Require f(1) + f(2) + f(3) + f(4) = 1. Therefore, c(1+2+3+4) = 1. Therefore, c = 0.1.

3-142. Let X denote the number of products that fail during the warranty period. Assume the units are

independent. Then, X is a binomial random variable with n = 500 and p = 0.02.

a) P(X = 0) = 4.1 x 10=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ 5000 )98.0()02.0(0

500-5

b) E(X) = 500(0.02) = 10

c) P(X >2) = 1 − P(X ≤ 2) = 0.9995

3-143. 16.0)3.0)(3.0()7.0)(1.0()0( =+= X f

19.0)3.0)(4.0()7.0)(1.0()1( =+= X f

20.0)3.0)(2.0()7.0)(2.0()2( =+= X f

31.0)3.0)(1.0()7.0)(4.0()3( =+= X f

14.0)3.0)(0()7.0)(2.0()4( =+= X f

3-144. a) P(X ≤ 3) = 0.2 + 0.4 = 0.6

b) P(X > 2.5) = 0.4 + 0.3 + 0.1 = 0.8

c) P(2.7 < X < 5.1) = 0.4 + 0.3 = 0.7

d) E(X) = 2(0.2) + 3(0.4) + 5(0.3) + 8(0.1) = 3.9

e) V(X) = 22(0.2) + 32(0.4) + 52(0.3) + 82(0.1) − (3.9)2 = 3.09

3-145.

x 2 5.7 6.5 8.5

f(x) 0.2 0.3 0.3 0.2

3-146. Let X denote the number of bolts in the sample from supplier 1 and let Y denote the number of bolts in the

sample from supplier 2. Then, x is a hypergeometric random variable with N = 100, n = 4, and K = 30.

Also, Y is a hypergeometric random variable with N = 100, n = 4, and K = 70.

a) P(X=4 or Y=4) = P(X = 4) + P(Y = 4)

2408.0

4

1004

70

0

30

4

1000

70

4

30

=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ⎟⎟

⎠

⎞⎜⎜

⎝

⎛

+

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ⎟⎟

⎠

⎞⎜⎜

⎝

⎛

=

3-27



b) P[(X=3 and Y=1) or (Y=3 and X = 1)]= 4913.0

4

100

3

70

1

30

1

70

3

30

=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

=

3-147. Let X denote the number of errors in a sector. Then, X is a Poisson random variable with λ = 0.32768.

a) P(X>1) = 1 − P(X≤1) = 1 − e-0.32768

− e-0.32768

(0.32768) = 0.0433

b) Let Y denote the number of sectors until an error is found. Then, Y is a geometric random variable and

P = P(X ≥ 1) = 1 − P(X=0) = 1 − e-0.32768 = 0.2794

E(Y) = 1/p = 3.58

3-148. Let X denote the number of orders placed in a week in a city of 800,000 people. Then X is a Poisson

random variable with λ = 0.25(8) = 2.

a) P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [e-2 + e-2(2) + (e-222)/2!] = 1 − 0.6767 = 0.3233.

b) Let Y denote the number of orders in 2 weeks. Then, Y is a Poisson random variable with λ = 4, and

P(Y>2) =1- P(Y ≤ 2) = e-4

+ (e-4

41)/1!+ (e

-44

2)/2! =1 - [0.01832+0.07326+0.1465] = 0.7619.

3-149.a) hypergeometric random variable with N = 500, n = 5, and K = 125

2357.0115524.2

100164.6

5

500

5

375

0

125

)0( ==

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= E

E f X

3971.0115525.2

)810855.8(125

5

500

4

375

1

125

)1( ==

⎟⎟

⎠

⎞⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= E

E f X

2647.0115524.2

)8718875(7750

5

500

3

375

2

125

)2( ==

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= E

f X

0873.0115524.2

)70125(317750

5

500

2

375

3

125

)3( ==

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= E

f X

01424.0115524.2

)375(9691375

5

500

1

375

4

125

)4( ==

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= E

f X

3-28



00092.0115524.2

83453.2

5

500

0

375

5

125

)5( ==

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= E

E f X

b)

x 0 1 2 3 4 5 6 7 8 9 10

f(x) 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000

3-150. Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial

random variable with n = 30. We are to determine p.

If P(X ≥ 1) = 0.9, then P(X = 0) = 0.1. Then , giving 30ln(1− p)=ln(0.1),

which results in p = 0.0739.

1.0)1()(0

30300 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛ p p

3-151. Let t denote an interval of time in hours and let X denote the number of messages that arrive in time t.Then, X is a Poisson random variable with λ = 10t.

Then, P(X=0) = 0.9 and e-10t

= 0.9, resulting in t = 0.0105 hours = 37.8 seconds

3-152. a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with

λ = 50(0.02) = 1. P(X = 0) = e-1

= 0.3679.

b) Let Y denote the number of flaws in one panel, then

P(Y ≥ 1) = 1 − P(Y=0) = 1 − e-0.02

= 0.0198. Let W denote the number of panels that need to be

inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and

E(W) = 1/0.0198 = 50.51 panels.

c) 0198.01)0(1)1( 02.0 =−==−=≥ −eY PY P

Let V denote the number of panels with 1 or more flaws. Then V is a binomial random

variable with n=50 and p=0.0198

491500 )9802.0(0198.01

50)9802(.0198.00

50)2( ⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ +⎟⎟

⎠ ⎞

⎜⎜⎝ ⎛ =≤V P

9234.0)9802.0(0198.02

50482 =⎟⎟

⎞⎜⎜⎛

+



5-47

Mind-Expanding Exercises

5-94 By the independence,

)()...()(

)(...)()(

...)()...()(...),...,,(

2211

2211

21212211

2

2

1

1

21

21

p p

p p X

A

X

A

X

A

p p X X X

A A A

p p

A X P A X P A X P

dx x f dx x f dx x f

dxdxdx x f x f x f A X A X A X P

p

p

p

p

∈∈∈=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

=∈∈∈

∫∫∫

∫∫∫

5-95 ....)( 2211 p pcccY E μ μ μ +++=

Also,



5-48

[ ]

[ ]p p X X X p p p

p X X X p p p p

dxdxdx x f x f x f xc xc

dxdx x f x f x f ccc xc xc xcY V

p

p

...)()...()()(...)(

...)()...()()...(...)(

2121

2

111

2121

2

22112211

21

21

μ μ

μ μ μ

−++−=

+++−+++=

∫∫

Now, the cross-term

[ ][ ] 0)()()()(

...)()...()())((

2222111121

212122112

1

21

21

=−−=

−−

∫∫∫

dx x f xdx x f xcc

dxdxdx x f x f x f x xcc

X X

p p X X X p

μ μ

μ μ

from the definition of the mean. Therefore, each cross-term in the last integral for V(Y)is zero and

).(...)(

)()(...)()()(

2

1

2

1

22

11

2

11

2

1 1

p p

p p X p p p X

X V c X V c

dx x f xcdx x f xcY V p

++=

−−= ∫∫ μ μ

5-96 cabcdydxdydx y x f

ba

XY

ba

== ∫∫∫∫0000

),( . Therefore, c = 1/ab. Then,

a

b

X cdy x f 1

0

)( == ∫ for 0 < x < a, and f y cdxY

a

b( ) = ∫ =

0

1 for 0 < y < b. Therefore,

(y)(x)f (x,y)=f f Y X XY for all x and y and X and Y are independent.

5-97 The marginal density of X is

.)()()()()()()(000

duuhk where xkgduuh xgduuh xg x f

bbb

X ∫∫∫ ==== Also,

.)()()(0

dvvglwhere ylh y f

a

Y

∫== Because (x,y) f XY is a probability density function,

.1)()()()(0000

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡= ∫∫∫∫ duuhdvvgdydx yh xg

baba

Therefore, kl = 1 and

(y)(x)f (x,y)=f f Y X XY for all x and y.



5-1

CHAPTER 5

Section 5-1

5-1. First, f(x,y) ≥ 0. Let R denote the range of (X,Y).

Then,

∑=++++=

R

y x f 1),(8

1

4

1

4

1

8

1

4

1

a) P(X < 2.5, Y < 3) = f(1.5,2)+f(1,1) = 1/8+1/4=3/8

b) P(X < 2.5) = f (1.5, 2) + f (1.5, 3) +f(1,1)= 1/8 + 1/4+1/4 = 5/8

c) P(Y < 3) = f (1.5, 2)+f(1,1) = 1/8+1/4=3/8

d) P(X > 1.8, Y > 4.7) = f ( 3, 5) = 1/8

e) E(X) = 1(1/4)+ 1.5(3/8) + 2.5(1/4) + 3(1/8) = 1.8125

E(Y) = 1(1/4)+2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875V(X)=E(X2)-[E(X)]2=[12(1/4)+1.52(3/8)+2.52(1/4)+32(1/8)]-1.81252=0.4961

V(Y)= E(Y2)-[E(Y)]2=[12(1/4)+22(1/8)+32(1/4)+42(1/4)+52(1/8)]-2.8752=1.8594

f) marginal distribution of X

x f(x)

1 ¼

1.5 3/8

2.5 ¼

3 1/8

g))5.1(

),5.1()(

5.1

X

XY

Y f

y f y f = and )5.1( X f = 3/8. Then,

y )(5.1y f

Y

2 (1/8)/(3/8)=1/3

3 (1/4)/(3/8)=2/3

h))2(

)2,()(

2

Y

XY

X f

x f x f = and )2(Y f = 1/8. Then,

x )(2y f

X

1.5 (1/8)/(1/8)=1

i) E (Y | X =1.5) = 2(1/3)+3(2/3) =2 1/3

j) Since f Y|1.5(y)≠f Y(y), X and Y are not independent

5-2 Let R denote the range of (X,Y). Because

1)654543432(),( =++++++++=∑ c y x f R

, 36c = 1, and c = 1/36

a) 4/1)432()3,1()2,1()1,1()4,1(36

1 =++=++=<= XY XY XY f f f Y X P

b) P(X = 1) is the same as part a. = 1/4

c) 3/1)543()2,3()2,2()2,1()2(36

1 =++=++== XY XY XY f f f Y P

d) 18/1)2()1,1()2,2(361 ===<< XY f Y X P



5-2

e)

[ ] [ ]

[ ]

( ) ( ) ( ) 167.26/13321

)3,3()2,3()1,3(3

)3,2()2,2()1,2(2)3,1()2,1()1,1(1)(

36

15

36

12

36

9 ==×+×+×=

+++

+++++=

XY XY XY

XY XY XY XY XY XY

f f f

f f f f f f X E

639.0)3()2()1()( 36

152

6

13

36

122

6

13

36

92

6

13

=−+−+−= X V

639.0)(

167.2)(

=

=

Y V

Y E


x )3,()2,()1,()( x f x f x f x f XY XY XY X ++=

1 1/4

2 1/3

3 5/12

g))1(

),1()( X

XY X Y

f y f y f =

y f yY X ( )

1 (2/36)/(1/4)=2/9

2 (3/36)/(1/4)=1/3

3 (4/36)/(1/4)=4/9

h))2(

)2,()(

Y

XY

Y X f

x f x f = and 3/1)2,3()2,2()2,1()2(

3612 ==++= XY XY XY Y f f f f

x f xX Y ( )

1 (3/36)/(1/3)=1/4

2 (4/36)/(1/3)=1/33 (5/36)/(1/3)=5/12

i) E (Y | X =1) = 1(2/9)+2(1/3)+3(4/9) = 20/9

j) Since f XY(x,y) ≠f X(x)f Y(y), X and Y are not independent.

5-3. ∑ =≥ R

y x f and y x f 1),(0),(

a)8

3

4

1

8

1)1,5.0()2,1()5.1,5.0( =+=−−+−−=<< XY XY f f Y X P

b)8

3)1,5.0()2,1()5.0( =−−+−−=< XY XY f f X P

c)8

7)1,5.0()1,5.0()2,1()5.1( =+−−+−−=< XY XY XY f f f Y P

d)8

5)2,1()1,5.0()5.4,25.0( =+=<> XY XY f f Y X P



5-3

e)

41

81

21

41

81

81

81

21

41

81

)(2)(1)(1)(2)(

)(1)(5.0)(5.0)(1)(

=++−−=

=++−−=

Y E

X E

V(X)=(-1-1/8)2(1/8)+(-0.5-1/8)2(1/4)+(0.5-1/8)2(1/2)+(1-1/8)2(1/8)=0.4219

V(Y)=(-2-1/4)2(1/8)+(-1-1/4)2(1/4)+(1-1/4)2(1/2)+(2-1/4)2(1/8)=1.6875


x )( x f X

-1 1/8

-0.5 ¼

0.5 ½

1 1/8

g))1(

),1()(

X

XY

X Y f

y f y f =

y f yY X ( )

2 1/8/(1/8)=1

h))1(

)1,()(

Y

XY

Y X f

x f x f =

x f xX Y ( )

0.5 ½/(1/2)=1

i) E ( X |Y =1) = 0.5

j) no, X and Y are not independent

5-4. Because X and Y denote the number of printers in each category,

40,0 =+≥≥ Y X and Y X

5-5. a) The range of (X,Y) is

3

2

10

y

x1 2 3

The problem needs probabilities to total one. Modify so that the probability of moderate

distortion is 0.04.



5-4

x,y f xy(x,y)

0,0 0.857375

0,1 0.1083

0,2 0.00456

0,3 0.000064

1,0 0.027075

1,1 0.00228

1,2 0.000048

2,0 0.000285

2,1 0.000012

3,0 0.000001

b)

x f x(x)

0 0.970299

1 0.029403

2 0.0002973 0.000001

c) E(X)=0(0.970299)+1(0.029403)+2(0.000297)+3*(0.000001)=0.03

(or np=3*0.01)

d))1(

),1()(

1

X

XY

Y f

y f y f = , f x(1) = 0.029403

y f Y|1(x)

0 0.920824

1 0.077543

2 0.001632

e) E(Y|X=1)=0(.920824)+1(0.077543)+2(0.001632)=0.080807

g) No, X and Y are not independent because, for example, f Y(0)≠f Y|1(0).

5-6. a) The range of (X,Y) is 40,0 ≤+≥≥ Y X and Y X . Here X is the number of pages

with moderate graphic content and Y is the number of pages with high graphic output out

of 4.

x=0 x=1 x=2 x=3 x=4y=4 5.35x10-05 0 0 0 0y=3 0.00184 0.00092 0 0 0

y=2 0.02031 0.02066 0.00499 0 0y=1 0.08727 0.13542 0.06656 0.01035 0y=0 0.12436 0.26181 0.19635 0.06212 0.00699

b)

x=0 x=1 x=2 x=3 x=4f(x) 0.2338 0.4188 0.2679 0.0725 0.0070



5-5

c)

E(X)=

2.1)0070.0(4)7248.0(3)2679.0(2)4188.0(1)2338.0(0)(4

0

==+++=∑ ii x f x

d) )3(

),3(

)(3 X

XY

Y f

y f

y f =

, f x(3) = 0.0725

Y f Y|3(y)

0 0.857

1 0.143

2 0

3 0

4 0

e) E(Y|X=3) = 0(0.857)+1(0.143) = 0.143

f) V(Y|X=3) = 02(0.857)+12(0.143)- 0.1432= 0.123

g) No, X and Y are not independent

5-7 a) The range of (X,Y) is 40,0 ≤+≥≥ Y X and Y X . X is the number of defective

items found with inspection device 1 and Y is the number of defective items found with

inspection device 2.

x=0 x=1 x=2 x=3 x=4y=0 1.94x10-19 1.10x10-16 2.35x10-14 2.22x10-12 7.88x10-11

y=1 2.59x10-16 1.47x10-13 3.12x10-11 2.95x10-9 1.05x10-7

y=2 1.29x10-13 7.31x10-11 1.56x10-8 1.47x10-6 5.22x10-5

y=3 2.86x10-11 1.62x10-8 3.45x10-6 3.26x10-4 0.0116y=4 2.37x10-9 1.35x10-6 2.86x10-4 0.0271 0.961

For x = 1,2,3,4 and y = 1,2,3,4

b)

x=0 x=1 x=2 x=3 x=4

f(x) 2.40 x 10-9 1.36 x 10-6 2.899 x 10-4 0.0274 0.972

c) Because X has binomial distribution E ( X )= n( p) = 4*(0.993)=3.972

d) )()2(

),2()(2| y f

f

y f y f

X

XY

Y == , f x(2) = 2.899 x 10-4

y f Y|1(y)=f(y

⎥⎦

⎤⎢⎣

⎡⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ ⎥

⎦

⎤⎢⎣

⎡⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ = −− y y x x

y x y x f 44 )003.0()997.0(

4)007.0()993.0(

4),(

1,2,3,4for )007.0()993.0(4

),( 4 =⎥⎦

⎤⎢⎣

⎡⎟⎟ ⎠

⎞⎜⎜⎝

⎛ = − x

x y x f x x



5-6

)

0 8.1 x 10-11

1 1.08 x 10-7

2 5.37 x 10-5

3 0.0119

4 0.988

e) E(Y|X=2) = E(Y)= n( p)= 4(0.997)=3.988

f) V(Y|X=2) = V(Y)=n( p)(1- p)=4(0.997)(0.003)=0.0120

g) Yes, X and Y are independent.

5-8. a) 5.0)2,2,2()1,2,2()2,1,2()1,1,2()2( =+++== XYZ XYZ XYZ XYZ f f f f X P

b) 35.0)2,2,1()1,2,1()2,1( =+=== XYZ XYZ f f Y X P

c) c) 5.0)1,2,2()1,1,2()1,2,1()1,1,1()5.1( =+++=< XYZ XYZ XYZ XYZ f f f f Z P

d)

7.03.05.05.0)2,1()2()1()21(=−+===−=+====

Z X P Z P X P Z or X P

e) E(X) = 1(0.5) + 2(0.5) = 1.5

f) 3.005.01.02.015.0

10.005.0

)1(

)1,1()1|1( =

+++

+=

=

=====

Y P

Y X PY X P

g) 2.005.015.02.01.0

1.0

)2(

)2,1,1()2|1,1( =

+++=

=

========

Z P

Z Y X P Z Y X P

h) 4.015.010.0

10.0

)2,1(

)2,1,1()2,1|1( =

+=

==

=======

Z Y P

Z Y X P Z Y X P

i))2,1(

)2,1,()(

YZ

XYZ

YZ X f

x f x f = and 25.0)2,1,2()2,1,1()2,1( =+= XYZ XYZ YZ f f f

x )( x f YZ X

1 0.10/0.25=0.4

2 0.15/0.25=0.6

5-9. (a) f XY(x,y)= (10%)x(30%)y (60%)4-x-y , for X+Y<=4



5-7

f XY(x,y) x y

0.1296 0 0

0.0648 0 1

0.0324 0 2

0.0162 0 30.0081 0 4

0.0216 1 0

0.0108 1 1

0.0054 1 2

0.0027 1 3

0.0036 2 0

0.0018 2 1

0.0009 2 2

0.0006 3 0

0.0003 3 1

0.0001 4 0

(b) f X(x)= P(X=x) = ∑≤+ 4

),(Y X

XY y x f .

f X(x) x

0.2511 0

0.0405 1

0.0063 2

0.0009 3

0.0001 4

(c) E(X)=∑ )( x xf X =0*0.2511+1*0.0405+2*0.0063+3*0.0009+4*0.0001= 0.0562

(d) f(y|X=3) = P(Y=y, X=3)/P(X=3)

P(Y=0, X=3) = C24

1 C43/ C

404 P(Y=1, X=3) = C12

1 C43/ C

404

P(X=3) = C361 C4

3/ C40

4, from the hypergeometric distribution with N=40, n=4, k=4, x=3

Therefore

f(0|X=3) = [C241 C4

3/ C40

4]/[ C36

1 C43/ C

404] = C24

1/ C36

1 = 2/3

f(1|X=3) = [C121 C4

3/ C40

4]/[ C36

1 C43/ C

404] = C12

1/ C36

1 = 1/3

f Y|3(y) y x

2/3 0 3

1/3 1 3

0 2 3

0 3 30 4 3

(e) E(Y|X=3)=0(0.6667)+1(0.3333)=0.3333

(f) V(Y|X=3)=(0-0.3333)2(0.6667)+(1-0.3333)2(0.3333)=0.0741

(g) f X(0)= 0.2511, f Y(0)=0.1555, f X(0) * f Y(0)= 0.039046 ≠ f XY(0,0) = 0.1296

X and Y are not independent.

5-10. (a) P(X<5) = 0.44+0.04=0.48



5-8

(b) E(X)= 0.43*23+0.44*4.2+0.04*11.4+0.05*130+0.04*0=18.694

(c) PX|Y=0(X) = P(X=x,Y=0)/P(Y=0) = 0.04/0.08 = 0.5 for x=0 and 11.4

(d) P(X<6|Y=0) = P(X=0|Y=0) = 0.5

(e) E(X|Y=0)=11.4*0.5+0*0.5 = 5.7

5-11 a) percentage of slabs classified as high = p1 = 0.05

percentage of slabs classified as medium = p2 = 0.85 percentage of slabs classified as low = p3 = 0.10

b) X is the number of voids independently classified as high X ≥ 0

Y is the number of voids independently classified as medium Y ≥ 0

Z is the number of with a low number of voids and Z ≥ 0 and X +Y + Z = 20

c) p1 is the percentage of slabs classified as high.

d) E(X)=np1 = 20(0.05) = 1

V(X)=np1 (1- p1)= 20(0.05)(0.95) = 0.95

e) 0)3,17,1( ==== Z Y X P Because the point 20)3,17,1( ≠ is not in the range of (X,Y,Z).

f)

07195.0010.085.005.0!3!17!0

!20

)3,17,1()3,17,0()3,17,1(

3170 =+=

===+======≤ Z Y X P Z Y X P Z Y X P

Because the point 20)3,17,1( ≠ is not in the range of (X,Y,Z).

g) Because X is binomial, ( ) ( ) 7358.095.005.095.005.0)1( 19120

1

20020

0 =+=≤ X P

h) Because X is binomial E(Y) = np = 20(0.85) = 17

i) The probability is 0 because x+y+z>20

j))17(

)17,2()17|2(

=

=====

Y P

Y X PY X P . Now, because x+y+z = 20,

P(X=2, Y=17) = P(X=2, Y=17, Z=1) = 0540.010.085.005.0!1!17!2

!20 1172 =

2224.02428.0

0540.0

)17(

)17,2()17|2( ==

=

=====

Y P

Y X PY X P

k)

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

=

==+⎟⎟

⎠

⎞⎜⎜⎝

⎛

=

==+

⎟⎟

⎠

⎞⎜⎜

⎝

⎛

=

==+⎟⎟

⎠

⎞⎜⎜

⎝

⎛

=

====

)17(

)17,3(3

)17(

)17,2(2

)17(

)17,1(1

)17(

)17,0(0)17|(

Y P

Y X P

Y P

Y X P

Y P

Y X P

Y P

Y X PY X E

1

2428.0

008994.03

2428.0

05396.02

2428.0

1079.01

2428.0

07195.00)17|(

=

⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ==Y X E



5-9

5-12. a) The range consists of nonnegative integers with x+ y+ z = 4.

b) Because the samples are selected without replacement, the trials are not independent

and the joint distribution is not multinomial.

c))2(

)2,()2|(Y

XY

f x f Y x X P ===

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )0440.0)2and2(

1758.0)2and1(

1098.0)2and0(

3296.00440.01758.01098.0)2(

15

4

6

1

5

2

4

1

15

4

6

1

5

2

4

1

15

4

6

2

5

2

4

0

15

4

6

0

5

2

4

2

15

4

6

1

5

2

4

1

15

4

6

2

5

2

4

0

====

====

====

=++=++==

Y X P

Y X P

Y X P

Y P

x )2,( x f XY

0 0.1098/0.3296=0.3331

1 0.1758/0.3296=0.5334

2 0.0440/0.3296=0.1335

d)

P(X=x, Y=y, Z=z) is the number of subsets of size 4 that contain x printers with graphics

enhancements, y printers with extra memory, and z printers with both features divided by

the number of subsets of size 4.

( )( )( )( )15

4

654

),,(z y x

z Z yY x X P ==== for x+y+z = 4.

( )( )( )( )

1758.0)1,2,1(15

4

6

1

5

2

4

1 ===== Z Y X P

e)( )( )( )

( )2198.0)2,1,1()1,1(

15

4

6

2

5

1

4

1 ======== Z Y X PY X P

f) The marginal distribution of X is hypergeometric with N = 15, n = 4, K = 4. Therefore,

E(X) = nK/N = 16/15 and V(X) = 4(4/15)(11/15)[11/14] = 0.6146.g)

( )( )( )( )[ ] ( )( )

( )[ ] 4762.0

)1(/)1,2,1()1|2,1(

154

93

61

154

61

52

41 ==

======== Z P Z Y X P Z Y X P

h)



5-10

( )( )( )( )[ ] ( ) ( )

( )[ ] 1334.0

)2(/)2,2()2|2(

154

102

52

154

60

52

42 ==

====== Y PY X PY X P

i) Because X+Y+Z = 4, if Y = 0 and Z = 3, then X = 1. Because X must equal 1,

1)1(=

YZ X f .

5-13 a) The probability distribution is multinomial because the result of each trial (a dropped

oven) results in either a major, minor or no defect with probability 0.6, 0.3 and 0.1

respectively. Also, the trials are independent

b) Let X, Y, and Z denote the number of ovens in the sample of four with major, minor,

and no defects, respectively.

1944.01.03.06.0!0!2!2

!4)0,2,2( 022 ===== Z Y X P

c) 0001.01.03.06.0!4!0!0

!4)4,0,0( 400 ===== Z Y X P

d) f x y f x y zXY XYZR

( , ) ( , , )= ∑ where R is the set of values for z such that x+y+z = 4. That is,

R consists of the single value z = 4-x-y and

.41.03.06.0)!4(!!

!4),( 4 ≤+

−−= −−

y x for y x y x

y x f y x y x

XY

e) 4.2)6.0(4)( 1 === np X E

f) 2.1)3.0(4)( 2 === npY E

g) === )2|2( Y X P 7347.02646.0

1944.0

)2(

)2,2(==

=

==

Y P

Y X P

2646.07.03.02

4)2( 42 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ==Y P from the binomial marginal distribution of Y

h) Not possible, x+ y+ z=4, the probability is zero.

i) )2|2(),2|1(),2|0()2|( ======== Y X PY X PY X PY X P

0204.02646.01.03.06.0!2!2!0

!4

)2(

)2,0()2|0( 220 =⎟

⎠

⎞⎜⎝

⎛ =

=

=====

Y P

Y X PY X P

2449.02646.01.03.06.0!1!2!1

!4

)2(

)2,1()2|1( 121 =⎟

⎠

⎞⎜⎝

⎛ =

=

=====

Y P

Y X PY X P

7347.02646.01.03.06.0!0!2!2

!4

)2(

)2,2()2|2( 022 =⎟

⎠

⎞⎜⎝

⎛ =

=

=====

Y P

Y X PY X P

j) E(X|Y=2) = 0(0.0204)+1(0.2449)+2(0.7347) = 1.7143



5-11

5-14 Let X, Y, and Z denote the number of bits with high, moderate, and low distortion. Then,

the joint distribution of X, Y, and Z is multinomial with n =3 and

95.0,04.0,01.0 321 === pand p p .

a)

5012 102.195.004.001.0!0!1!2

!3)0,1,2()1,2(

−×======== Z Y X PY X P

b) 8574.095.004.001.0!3!0!0

!3)3,0,0( 300 ===== Z Y X P

c) X has a binomial distribution with n = 3 and p = 0.01. Then, E(X) = 3(0.01) = 0.03

and V(X) = 3(0.01)(0.99) = 0.0297.

d) First find )2|( =Y X P

0046.095.0)04.0(01.0!1!2!0

!395.0)04.0(01.0

!0!2!1

!3

)1,2,0()0,2,1()2(

12002 =+=

===+===== Z Y X P Z Y X PY P

98958.0004608.095.004.001.0!1!2!0

!3

)2(

)2,0()2|0( 120 =⎟

⎠

⎞⎜⎝

⎛ =

=

=====

Y P

Y X PY X P

01042.0004608.095.004.001.0!1!2!1

!3

)2(

)2,1()2|1( 021 =⎟

⎠

⎞⎜⎝

⎛ =

=

=====

Y P

Y X PY X P

)2|( =Y X E 01042.0)01042.0(1)98958.0(0 =+=

01031.0)01042.0(01042.0))(()()2|( 222 =−=−== X E X E Y X V

5-15. (a) f XYZ(x,y,z)

f XYZ(x,y,z)

Selects(X) Updates(Y)

Inserts(Z)

0.43 23 11 12

0.44 4.2 3 1

0.04 11.4 0 0

0.05 130 120 0

0.04 0 0 0

(b)PXY|Z=0

PXY|Z=0(x,y)

Selects(X) updates(Y)

Inserts(Z)

4/13 = 0.3077 11.4 0 0

5/13 = 0.3846 130 120 0

4/13= 0.3077 0 0 0

(c) P(X<6, Y<6|Z=0)=P(X=Y=0)=0.3077



5-12

(d) E(X|Y=0,Z=0) =0.5*11.4+0.5*0=5.7

5-16 a) Let X, Y, and Z denote the risk of new competitors as no risk, moderate risk,

and very high risk. Then, the joint distribution of X, Y, and Z is multinomial with n =12

and 15.0,72.0,13.0 321 === pand p p . X, Y and Z ≥ 0 and x+y+z=12

b) )1,3,1( === Z Y X P = 0, not possible since x+y+z≠12

c)

7358.02924.03012.01422.0

85.015.02

1285.015.0

1

1285.015.0

0

12)2( 102111120

=++=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ =≤ Z P

d) 0)10,1|2( ==== X Y Z P

e)

8988

201011100210

1089.61004.21097.11072.4

15.072.013.0!2!0!10

!1215.072.013.0

!1!1!10

!1215.072.013.0

!0!2!10

!12

)2,0,10()1,1,10()0,2,10()10(

−−−− =++=

++=

===+===+=====

x x x x

Z Y X P Z Y X P Z Y X P X P

9698.0

1089.615.072.013.010!1!1!

12!1089.615.072.013.0

10!2!0!

12!

)10(

)10,1,1(

)10(

)10,2,0()10|1(

111080210

=

+=

=

===+

=

=====≤

−− x x

X P

X Y Z P

X P

X Y Z P X Z P

f)

2852.0

1089.615.072.013.010!1!1!

12!

)10(

)10,1,1()10|1,1(

81110

=

=

=

=====≤≤

− x

X P

X Y Z P X Z Y P

g)

345.0

1089.6/))1004.2(2)1097.1(1)1072.4(0()10|( 8988

=

++== −−−− x x x x X Z E



5-13

Section 5-2

5-17 Determine c such that .)5.4(4

813

02

3

0

3

0

3

02

3

0

22

ccdy yc xydxdycy x ===∫ ∫∫

Therefore, c = 4/81.

a) 4444.0))(2()2()3,2(2

9

81

4

3

0

3

0

81

4

2

0

81

4 ====<< ∫ ∫∫ ydy xydxdyY X P

b) P(X < 2.5) = P(X < 2.5, Y < 3) because the range of Y is from 0 to 3.

6944.0)125.3()125.3()3,5.2(29

814

3

0

3

0

814

5.2

0

814 ====<< ∫ ∫∫ ydy xydxdyY X P

c) 5833.0)5.4()5.21(5.2

1281

18

5.2

1

5.2

1

814

3

0

814

2

====<< ∫ ∫∫y

ydy xydxdyY P

d)

3733.0)88.2()88.2()5.21,8.1(

5.2

1

5.2

1

2

)15.2(

814814

3

8.1

814

2

∫ ∫∫ ====<<>−

ydy xydxdyY X P

e) 29)(3

029

4

3

0

3

0

81

4

3

0

2

81

42

==== ∫ ∫∫y

ydy ydxdy x X E

f) ∫∫ ∫ ===<<4

0

4

0

0

0

814 00)4,0( ydy xydxdyY X P

g) 30)5.4(),()(9

2

81

4

3

0

81

4

3

0

<<==== ∫∫ x for x ydy xdy y x f x f x XY X

.

h) y

y

f

y f

y f X

XY

Y 9

2

9

2

81

4

5.1)5.1(

)5.1(

)5.1(

),5.1(

)( === for 0 < y < 3.

i) E(Y|X=1.5) = 227

2

9

2

9

23

0

3

0

32

3

0

===⎟ ⎠

⎞⎜⎝

⎛ ∫∫

ydy ydy y y

j)9

40

9

4

9

1

9

2)()5.1|2(

2

0

2

2

0

5.1| =−=====< ∫ y ydy y f X Y P Y

k) x x

f

x f x f

Y

XY

X 9

2

9

2

81

4

2)2(

)2(

)2(

)2,()( === for 0 < x < 3.



5-14

5-18.

[ ]

( ) [ ] c x xdx xc

dx x x x

dx xydydx y xc

x x

x

x

y

x

x

242224

)2(

)(

3

0

2

3

0

3

0

2

2

2

)2(

3

0

3

0

2

2

2

22

2

=+=+=

−−++=

+=+

∫

∫

∫ ∫∫

+

++

Therefore, c = 1/24.

a) P(X < 1, Y < 2) equals the integral of ),( y x f XY over the following region.

0

0 1 2

2

x

y

Then,

10417.02241

2224

1

24

1)(

24

1)2,1(

1

02

2

3

0

2

3

1

0

1

0

2

2

2

3

22

=⎥⎦⎤⎢

⎣⎡ −+

=−+=+=+=<< ∫∫ ∫∫

x

x

x

y

x

x x

dx xdx xydydx y xY X P

b) P(1 < X < 2) equals the integral of ),( y x f XY over the following region.

00 1 2

2

x

y

∫

∫ ∫∫

=⎥⎦

⎤⎢⎣

⎡+=+=

+=+=<<++

3

0

6

12

1

2

2

1

2

1

2

2

2

.2224

1)24(

24

1

24

1)(

24

1)21(

2

x xdx x

dx xydydx y x X P x

x

y x

x



5-15

c) P(Y > 1) is the integral of f x yXY ( , ) over the following region.

9792.002083.01

2

1

2

1

224

11

2

3

2

1

24

11

)2

(24

11)(1)1(1)1(

1

0

321

0

2

1

0

1

0

121

241

=−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+−=−+−=

+−=+−=≤−=>

∫

∫ ∫∫

x x

dx x x

y xydydx y xY PY P

x x

d) P(X < 2, Y < 2) is the integral of f x yXY( , ) over the following region.

0

02

2

x

y

∫

∫ ∫∫

=⎥⎦

⎤⎢⎣

⎡+=+=

+=+=++

3

0

3

0

23

2

3

0

3

0

2

2

2

2

8

15

3

4

24

1)24(

24

1

24

1)(

24

1)(

2

x x

dx x x

dx y xdydx y x x X E x

x

xy

x

x

e)

∫

∫ ∫∫

=⎥⎦

⎤⎢⎣

⎡+=+=

+=+=++

3

0

3

0

23

2

3

0

3

0

2

2

2

2

8

15

3

4

24

1)24(

24

1

24

1)(

24

1)(

2

x x

dx x x

dx y xdydx y x x X E x

x

xy

x

x



5-16

f)

320

31707

8

15

202

3

4

4

3

24

1

8

15)

4

443(

24

1

8

15

24

1

8

15)(

24

1)(

23

0

52

34

23

0

423

23

0

3

0

2

2

3

2 2

222

=⎟ ⎠

⎞⎜⎝

⎛ −⎥

⎦

⎤⎢⎣

⎡−++=

⎟

⎠

⎞⎜

⎝

⎛ −−++=

⎟ ⎠

⎞⎜⎝

⎛ −+=⎟

⎠

⎞⎜⎝

⎛ −+=

∫

∫ ∫∫++

x x

x x

dx x

x x x

dx y xdydx y x x X V x

x

y x

x

x

g) )( x f X is the integral of ),( y x f XY over the interval from x to x+2. That is,

12

1

624

1)(

24

1)(

2

2

22

+=⎥⎦

⎤⎢⎣

⎡+=+=

++

∫x

xydy y x x f x

x

y

x

x

X for 0 < x < 3.

h)

6

1)(

12

1

6

1

)1(24

1

)1(

),1(

1

y y f

y

f

y f

Y X

XY +

===+

+

for 1 < y < 3.

See the following graph,

0

0 1 2

2

x

y

1

f (y) defined over this line segmentY|1

i) E(Y|X=1) = 111.2326

1)(

6

1

6

13

1

3

1

323

1

2

∫ ∫ =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=+=⎟

⎠

⎞⎜⎝

⎛ + y ydy y ydy

y y

j) ==> )1|2( X Y P 5833.026

1)1(

6

1

6

13

2

3

2

23

2

∫ ∫ =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=+=⎟

⎠

⎞⎜⎝

⎛ + y ydy ydy

y

k) )2(

)2,(

2)(

Y

XY

f

x f

X x f = . Here )( y f Y is determined by integrating over x. There are

three regions of integration. For 20 ≤< y the integration is from 0 to y. For 32 ≤< y

the integration is from y-2 to y. For 53 << y the integration is from y to 3. Because

the condition is y=2, only the first integration is needed.

160

2

0

22

24

1)(

24

1)(

y y

x

y

Y xydx y x y f =⎥⎦

⎤⎢⎣

⎡+=+= ∫ for 20 ≤< y .



5-17

0

0 1 2

2

x

y

1

f (x) defined over this line segmentX|2

Therefore, 4/1)2( =Y f and6

2

4/1

)2(24

1

)(2

+=

+=

x x

x f X

for 0 < x < 3

5-19 .8

81

822

3

0

3

0

43

0

3

0

2

0

c x

dx x

cdx y

xc x xydyd c

x x

∫ ∫ ∫∫ === Therefore, c = 8/81

a) P(X<1,Y<2)= .81

1

8

1

81

8

281

8

81

81

0

1

0

3

0 =⎟ ⎠

⎞

⎜⎝

⎛

==∫ ∫∫ dx

x

x xydyd

x

b) P(1<X<2) = .27

5

8

)12(

81

8

881

8

281

8

81

8 42

1

2

1

422

1 0

=−

⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛ == ∫∫ ∫

xdx

x x x xydyd

x

c)

01235.081

1

4

1

8

1

4

3

8

3

81

8

4881

8

2281

8

2

1

81

8

81

8)1(

2424

3

1

3

1

2433

1

23

1 1

==⎥⎦

⎤⎢⎣

⎡⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −==> ∫∫∫ ∫

x xdx

x x xd

x x x xydyd Y P

x

d) P(X<2,Y<2) = .81

16

8

2

81

8

281

8

81

8 42

0

2

0

3

0

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ==∫ ∫∫ dx

x x xydyd

x

e)

5

12

10

3

81

8

281

8

281

8

81

8)(

81

8)(

5

3

0

3

0

3

0

43

0

22

0

2

0

=⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =

==== ∫ ∫ ∫∫∫∫ xd x

xd x x

x ydyd x xdyd xy x X E

x x

f)

5

8

15

3

81

8

381

8

381

8

81

8)(

81

8)(

53

0

4

3

0

33

0

3

0 0

2

0

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ ==

===

∫

∫∫ ∫ ∫∫

dx x

dx x

x xdyd xy xdyd xy yY E

x x



5-18

g) 3081

4

81

8)(

3

0

<<== ∫ x x

xydy x f

x

h) 102

81)1(4

)1(81

8

)1(

),1()(

31| <<==== y y

y

f

y f y f xY

i) ∫ ====1

0

1

0

2 12)1|( y ydy X Y E

j) 0)1|2( ==> X Y P this isn’t possible since the values of y are 0< y < x.

k) 9

4

81

8)(

3

0

y xydx y f == ∫ , therefore

30

9

2

9)2(4

)2(81

8

)2(

)2,()(2| <<==== x

x x

f

x f x f Y X

5-20. Solve for c

( )

10 .10

1

5

1

2

1

3

31

30

52

0

32

0 0

32

==⎟ ⎠

⎞⎜⎝

⎛ −

=−=−= ∫∫∫ ∫∞

−−∞

−−∞

−−

ccc

xd eec

xd eec

xdyd ecx x x x

x

y x

a)

77893.0253

10

3

10

)1(3

10

10)2,1(

1

0

25

5

1

0

23

1

0

2

1

0 0

32

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

−=−==<<

−−

−−−−−−

∫∫∫ ∫ x x

x x x x

x

y x

ee

dyeedyee xdyd eY X P

b)

19057.0253

10

3

1010)21(

2

1

25

2

1

2

1

52

0

32

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

−==<<

−−

−−−−

∫ ∫∫

x x

x x

x

y x

ee

xd ee xdyd e X P

c)

7

3

295

3

39

1

3

2

3

32

10059.3253

10

)(3

1010)3(

−

∞−−−

∞−−

∞−−−

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

−==> ∫ ∫∫

xeee

dyeee xdyd eY P

x x

x x

x

y x



5-19

d)

9695.0

253

10)1(

3

1010)2,2(

2

0

2

0

4103

2

0

2

0

32

=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=−==<< ∫ ∫∫

−−−−−− ee

dxee xdyd eY X Px x

x

y x

e) E(X) =10710

0 0

32

∫ ∫∞

−− = xdyd xe

x

y x

f) E(Y) =5

110

0 0

32

∫ ∫∞

−− = xdyd ye

x

y x

g) )(3

10)1(

3

1010)(

5232

0

32 x x x z x

y x eeee

dye x f −−−−

−− −=−== ∫ for 0 < x

h) y

y

X

Y X

X Y e

ee

e

f

y f y f 3

52

32,

1\ 157.3

)(

3

10

10

)1(

),1()( −

−−

−−

= =

−

== 0 < y < 1

i) 2809.0157311

0

3 =dy ye.)= E(Y|X=y-

∫

j)42

10

62,

2| 25

10

2

2)( +−

−

−−

= === x x

Y

Y X

Y X ee

e

)( f

)(x, f x f for 2 < x,

where f ( y) = 5e-5y for 0 < y

5-21 cdxec

dxeec

dydxeecx x x

x

y x

15

1

3)(

30

5

0

32

0

32

∫∫∫ ∫∞

−∞

−−∞ ∞

−− === c = 15

a)

9879.0)1(2

5155

)(515)2,1(

265

1

0

26

1

0

5

1

0

1

0

632

2

32

=−+−=−=

−==<<

−−−−−−

−−−−−

∫∫

∫ ∫∫

eeedxeedxe

xd eee xdyd eY X P

x x

x x

x

y x

b) P(1 < X < 2) = 0067.0)(515 105

2

1

2

1

532 =−== −−∞

−−

∫ ∫∫ ee xdyd e xdyd ex

x

y x

c)

000308.02

5

2

3

5515)3(

915

3

5

3

0

29

3

0 3

32

3

32

=+−=

+=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=>

−−

∞−−−

∞ ∞−−

∞−−

∫∫∫ ∫ ∫∫

ee

dxedxeedydxedydxeY Px x

x

y x y x



5-20

d)

( ) ( ) 9939.01

2

5155

)(515)2,2(

4610

2

0

2

2

0

65

2

0

2

0

632

2

32

=−+−=−=

=−==<<

−−−−−−

−−−−−

∫∫

∫ ∫∫

eeedxeedxe

xd eee xdyd eY X P

x x

x x

x

y x

e) E(X) = 04.05

1515

2

0 0

532 ===∫ ∫∫∞ ∞

−∞

−−dx xe xdyd xe

x

x

y x

f)

15

8

6

5

10

3

32

55

2

315)(

0 0

3

0

532

=+−=

+−

== ∫ ∫∫∫∞ ∞

−∞

−∞

−−dy yedy ye xdyd yeY E

y y

x

y x

g) x x z

x

y xeedye x f

53232 5)(3

1515)( −−−

∞−− === ∫ for x > 0

h)55)1( −= e f X

y

XY e y f 3215),1( −−=

y y

X Y ee

e y f 33

5

32

1| 35

15)( −

−

−−

= == for 1 < y

i) 3/43)1|( 33

11

3333

1

=+−=== −∞∞−−

∞

∫∫ dyee ydy ye X Y E y y y

j) 9502.013 32

1

33=−= −−

∫ edye y for 0 < y, 6

215)2( −= e f Y

k) For y > 0 x

x

Y X e

e

e y f 2

6

62

2| 2

2

15

15)( −

−

−−

= == for 0 < x < 2

5-22 a) f Y|X=x(y), for x = 2, 4, 6, 8



5-21

0 1 2 3 4

0

1

2

3

4

5

y

f ( y 2 )

b) ∫ ===< −2

0

2 9817.02)2|2( dye X Y Py

c) ∫∞

− ===0

2 2/12)2|( dy ye X Y E y

(using integration by parts)

d) ∫∞

− ===0

/1)|( xdy xye x X Y E xy


e) Use f X(x) =10

11=

− ab,

xy

X Y xe y x f −=),(| , and the relationship

)(

),(),(|

x f

y x f y x f

X

XY X Y =

Therefore,10

),(and10/1

),( xy

XY

XY xy xe y x f

y x f xe

−− ==

f) f Y(y) = ∫−−− −−

=10

0 2

1010

10

101

10 y

e yedx

xey y xy


5-23. The graph of the range of (X, Y) is

0

1

2

3

4

5

1 2 3 4

y

x



5-22

15.76

2)1(

1

23

4

1

1

0

4

1

1

1

1

0

1

0

==+=

++=

=+

∫∫

∫ ∫∫ ∫+

−

+

ccc

dxcdx xc

cdydxcdydx

x

x

x

Therefore, c = 1/7.5=2/15

a)30

1

5.0

0

5.0

0

5.7

1)5.0,5.0( ==<< ∫ ∫ dydxY X P

b)12

1

8

5

15

2

5.0

0

5.7

1

5.0

0

1

0

5.7

1 )()1()5.0( ==+==< ∫∫ ∫+

dx xdydx X P

x

c)

9

19

5.7

2

6

5

15

12

4

1

5.7

2

1

0

2

5.7

1

4

1

1

15.7

1

0

1

05.7

)5.7()()()(

)(

=+=++=

+=

∫∫∫ ∫∫ ∫

+

−

+

dx xdx x x

dydxdydx X E

x

x

x

x

x

d)

( ) ( )4597

151

37

151

4

1

151

1

0

2151

4

1

2

)1()1(

5.71

1

0

2

)1(

5.71

4

1

1

1

5.71

1

0

1

0

5.71

30

4)12(

)(

222

=+=

+++=

+=

+=

∫∫

∫∫

∫ ∫∫ ∫

−−++

+

−

+

xdxdx x x

dxdx

ydydx ydydxY E

x x x

x

x

x

e)

41for 5.7

2

5.7

)1(1

5.7

1)(

,10for 5.7

1

5.7

1)(

1

1

1

0

<<=⎟ ⎠

⎞⎜⎝

⎛ −−+==

<<⎟ ⎠

⎞⎜⎝

⎛ +==

∫

∫+

−

+

x x x

dy x f

x x

dy x f

x

x

x

f)

20for 5.0)(

5.05.7/2

5.7/1

)1(

),1()(

1|

1|

<<=

===

=

=

y y f

f

y f y f

X Y

X

XY

X Y

g) ∫ ====2

0

2

0

2

142

)1|(y

dy y

X Y E



5-23

h) 25.05.05.0)1|5.0(

5.0

0

5.0

0

====< ∫ ydy X Y P

5-24 Let X, Y, and Z denote the time until a problem on line 1, 2, and 3, respectively.

a)3

)]40([)40,40,40(>=>>>

X P Z Y X P because the random variables are independent with the same distribution. Now,

1

40

40/

40

40/

40

1)40( −∞

−∞

− =−==> ∫ eedxe X Px x

and the answer is

( ) 0498.0331 == −−ee .

b)3)]4020([)4020,4020,4020( <<=<<<<<< X P Z Y X P and

2387.0)4020( 15.040

20

40/ =−=−=<< −−−eee X P

x.

The answer is .0136.02387.0 3 =

c) The joint density is not needed because the process is represented by three independentexponential distributions. Therefore, the probabilities may be multiplied.

5-25 μ=3.2 λ =1/3.2

0439.0

2.3)24.10/1()5,5(

2.3

5

2.3

5

2.3

5

5 5

2.3

5

2.32.3

=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ =

⎟⎟

⎠

⎞⎜⎜⎝

⎛ ==>>

−−

−∞ ∞

−∞

−−

∫ ∫∫

ee

dxeedydxeY X P

x y x

0019.0

2.3)24.10/1()10,10(

2.3

10

2.3

10

2.3

10

10 10

2.3

10

2.32.3

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

==>>

−−

−∞ ∞

−∞

−−

∫ ∫∫

ee

dxeedydxeY X P

x y x

b) Let X denote the number of orders in a 5-minute interval. Then X is a Poisson random

variable with λ = 5/3.2 = 1.5625.

256.0!2

)5625.1()2(

25625.1

===−

e X P

For both systems, 0655.0256.0)2()2(2

==== Y P X P

c) The joint probability distribution is not necessary because the two processes are

independent and we can just multiply the probabilities.

5-26. (a) X: the life time of blade. Y: the life time of bearing



5-24

f(x) = 3e-3x f(y) = 4e-4y

P(X≥5, Y≥5)=P(X≥5)P(Y≥5)=e-3(5)e-4(5)≈0

(b) P(X>t, Y>t)=e-3te-4t=e-7t=0.95→t=ln(.95)/-7=0.00733

5-27. a) ∫ ∫ ∫∫ ∫ ∫ =====<5.0

0

1

0

5.0

0

25.0

0

5.0

0

1

0

1

0

25.0)2()4()8()5.0( xdx xdydx xydzdydx xyz X P

b)

∫ ∫ ∫

∫ ∫ ∫

====

=<<

5.0

0

5.0

0

5.0

04

5.0

0

5.0

0

5.0

0

1

0

0625.0)5.0()4(

)8()5.0,5.0(

2 xdx xdydx xy

dzdydx xyzY X P

c) P(Z < 2) = 1, because the range of Z is from 0 to 1.

d) P(X < 0.5 or Z < 2) = P(X < 0.5) + P(Z < 2) - P(X < 0.5, Z < 2). Now, P(Z < 2) =1 and

P(X < 0.5, Z < 2) = P(X < 0.5). Therefore, the answer is 1.

e) 3/2)2()8()(1

032

1

0

1

0

2

1

0

1

0

2 3

==== ∫ ∫∫ ∫ xdx xdzdydx yz x X E

f) )5.05.0( =< Y X P is the integral of the conditional density )( x f Y X

. Now,

)5.0(

)5.0,()(

5.0

Y

XY

X f

x f x f = and x xdz z x x f XY 25.04))05.0(8()5.0,(

1

0

=== ∫ for 0 < x

< 1 and

0 < y < 1. Also, ∫ ∫ ==1

0

1

0

2)8()( ydxdz xyz y f Y for 0 < y < 1.

Therefore, x x

x f X

21

2)(

5.0== for 0 < x < 1.

Then, 25.02)5.05.0(5.0

0

===< ∫ xdxY X P .

g) )8.05.0,5.0( =<< Z Y X P is the integral of the conditional density of X and Y.

Now, z z f Z 2)( = for 0 < z < 1 as in part a) and

xy xy

z f

z y x f y x f

Z

XYZ

Z XY 4

)8.0(2

)8.0(8

)(

),,(),( === for 0 < x < 1 and 0 < y < 1.

Then, ∫ ∫∫ =====<<5.0

0

16

15.0

0

5.0

0

0625.0)2/()4()8.05.0,5.0( dx xdydx xy Z Y X P

h) yzdx xyz z y f YZ 4)8(),(1

0

== ∫ for 0 < y < 1 and 0 < z < 1.

Then, x x

z y f

z y x f x f

YZ

XYZ

YZ X 2

)8.0)(5.0(4

)8.0)(5.0(8

),(

),,()( === for 0 < x < 1.



5-25

i) Therefore, 25.02)8.0,5.05.0(5.0

0

====< ∫ xdx Z Y X P

5-28 ∫∫ ∫≤+ 4

4

022

y xcdzdydx = the volume of a cylinder with a base of radius 2 and a height of 4 =

π π 164)2( 2 = . Therefore,π 16

1=c

a) )2( 22 <+ Y X P equals the volume of a cylinder of radius 2 and a height of 4 (

=8π) times c. Therefore, the answer is .2/116

8=

π

π

b) P(Z < 2) equals half the volume of the region where ),,( z y x f XYZ is positive times

1/c. Therefore, the answer is 0.5.

c dx x xcdx xycdzdydx X E

x

x

x

x

c

x

)48(4)(2

2

2

2

2

4

4

2

2

4

4

4

0

2

2

2

2 −=⎥⎥⎦

⎤

⎢⎢⎣

⎡

== ∫∫∫ ∫ ∫ −−

−

−−−

−

−−. Using

substitution,24 xu −= , du = -2 x dx, and 0)4(4)(

2

2

2

324 2

3

=−==−

−∫ xduuc X E c

.

d) 4),()1(

)1,()( 22

414

4

0

1<+==== ∫ y x for dzc y x f and

f

x f x f

c XY

Y

XY

X π .

Also,2

4

4

4

0

48)(

2

2

ycdzdxc y f

y

y

Y −== ∫ ∫−

−−

for -2 < y < 2.

Then, 248

4

)( yc

c

x f y X

−= evaluated at y = 1. That is, 32

1

1 )(=

x f X for

33 <<− x .

Therefore, 7887.032

31)1|1(

32

1

1

3

=+

==<< ∫−

dxY X P

e) ∫ ∫∫− −

−

−−

−===2

2

2

2

2

4

4

142)(

)1(

)1,,(),(

2

2

dx xccdydx z f and f

y x f y x f

x

x

Z

Z

XYZ

XY

Because )( z f Z is a density over the range 0 < z < 4 that does not depend on Z,

)( z f Z =1/4 for

0 < z < 4. Then,π 41

4/1),(

1== c y x f

XY for 422 <+ y x .

Then, 4/14

1)1|1(

2222 =

<+==<+

π

y xinarea Z Y X P .



5-26

f)),(

),,()(

y x f

z y x f z f

XY

XYZ

xy Z = and from part 5-59 a., 4),( 22

4

1 <+= y x for y x f XY π .

Therefore, 4/1)(4

1

16

1

==π

π z f xy Z

for 0 < z < 4.

5-29 Determine c such that c xyz f =)( is a joint density probability over the region x>0, y>0and z>0 with x+y+z<1

( )

6.cTherefore, .6

1

622

1

2

1

2

)1()1()1(

)2

()1()(

1

0

321

0

21

0

2

1

0

21

0

1

0

1

0

1

0

1

0

1

0

==

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +−=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−−−−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−=−−==

∫∫

∫∫ ∫∫ ∫ ∫−−− −−

c

x x xcdx

xcdx

x x x xc

dx y

xy ycdydx y xcdzdydxc xyz f

x x x y x

a) ⇒=<<< ∫ ∫ ∫− −−1

0

1

0

1

0

6)5.0,5.0,5.0(

x y x

dzdydx Z Y X P The conditions x<0.5, y<0.5,

z<0.5 and x+y+z<1 make a space that is a cube with a volume of 0.125. Therefore the

probability of 75.0)125.0(6)5.0,5.0,5.0( ==<<< Z Y X P

b)

( )

∫

∫ ∫ ∫

=⎟ ⎠

⎞⎜⎝

⎛ −=⎟

⎠

⎞⎜⎝

⎛ −=

−−=−−=<<

5.0

0

5.0

0

2

5.0

0

5.0

0

5.0

0

5.0

0

2

4/32

3

4

93

4

9

366)1(6)5.0,5.0(

x xdx x

dx y xy ydydx y xY X P

c)

( ) 875.033)2

1

2(6

)2

(6)1(66)5.0(

5.0

0

23

5.0

0

2

1

0

25.0

0

51.0

0

1

0

5.0

0

1

0

1

0

=+−=+−=

−−=−−==<

∫

∫∫ ∫∫ ∫ ∫−−− −−

x x xdx x x

y xy ydydx y xdzdydx X P

x x x y x

d)



5-27

25.02

32

4

3)

22(6

)2

(6)1(66)(

1

0

23

42

1

0

3

1

0

25.0

0

1

0

1

0

1

0

1

0

1

0

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +−=+−=

−−=−−==

∫

∫∫ ∫∫ ∫ ∫−−− −−

x x

xdx

x x

x

y xy y xdydx y x x xdzdydx X E

x x x y x

e)

10for )1(3)2

1

2(6

26)1(66)(

22

1

0

21

0

1

0

1

0

<<−=+−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−=−−==

−−− −−

∫∫ ∫

x x x x

y xy ydy y xdzdy x f

x x x y x

f)

100for

)1(66),(

1

0

<+>>

−−== ∫−−

yand x , y x

y xdz y x f

y x

g)

16

6

)5.0,5.0(

)5,0,5.0,()5.0,5.0|( ==

==

=====

z y f

z y x f z y x f for x > 0

h) The marginal )( y f Y is similar to )( x f X and2)1(3)( y y f Y −= for 0 < y < 1.

5.0for )21(4

)25.0(3

)5.0(6

)5.0(

)5.0,()5.0|(| <−=

−== x x

x

f

x f x f

Y

Y X

5-30 Let X denote the production yield on a day. Then,

84134.0)1()()1400(10000

15001400 =−>=>=> − Z P Z P X P .

a) Let Y denote the number of days out of five such that the yield exceeds 1400. Then, by

independence, Y has a binomial distribution with n = 5 and p = 0.8413. Therefore, the

answer is ( ) 4215.0)8413.01(8413.0)5( 055

5 =−==Y P .

b) As in part (a), the answer is

( ) 8190.04215.0)8413.01(8413.0

)5()4()4(

145

4 =+−=

=+==≥ Y PY PY P

5-31a) Let X denote the weight of a brick. Then,

84134.0)1()()75.2(25.0

375.2 =−>=>=> − Z P Z P X P .

Let Y denote the number of bricks in the sample of 20 that exceed 2.75 pounds. Then, byindependence, Y has a binomial distribution with n = 20 and p = 0.84134. Therefore, the

answer is ( ) 032.084134.0)20( 2020

20 ===Y P .



b) Let A denote the event that the heaviest brick in the sample exceeds 3.75 pounds.

Then, P(A) = 1 - P(A') and A' is the event that all bricks weigh less than 3.75 pounds. As

in part a., P(X < 3.75) = P(Z < 3) and

0267.099865.01)]3([1)( 2020 =−=<−= Z P AP .

5-32 a) Let X denote the grams of luminescent ink. Then,

022750.0)2()()14.1(3.0

2.114.1 =−<=<=< − Z P Z P X P .

Let Y denote the number of bulbs in the sample of 25 that have less than 1.14 grams.

Then, by independence, Y has a binomial distribution with n = 25 and p = 0.022750.

Therefore, the answer is

( ) 4375.05625.01)97725.0(02275.0)0(1)1( 25025

0 =−===−=≥ Y PY P .

b)

( ) ( ) ( )

( ) ( ) ( ) 199997.00002043.0002090.001632.009146.03274.05625.0)97725.0(02275.0)97725.0(02275.0)97725.0(02275.0

)97725.0(02275.0)97725.0(02275.0)97725.0(02275.0

)5()4()3(()2()1()0()5(

20525

5

21425

4

22325

3

23225

2

24125

1

25025

0

≅=+++++=

+++

++=

=+=+=+=+=+==≤ Y PY PY PY PY PY PY P

.

c) ( ) 5625.0)97725.0(02275.0)0( 25025

0 ===Y P

d) The lamps are normally and independently distributed, therefore, the probabilities can

be multiplied.



5-28

Section 5-3

5-33 E(X) = 1(3/8)+2(1/2)+4(1/8)=15/8 = 1.875

E(Y) = 3(1/8)+4(1/4)+5(1/2)+6(1/8)=37/8 = 4.625

9.375=75/8=

(1/8)]6[4+(1/2)]5[2+(1/4)]4[1+(1/8)]3[1=E(XY) ××××××××

703125.0)625.4)(875.1(375.9)()()( =−=−= Y E X E XY E XY σ

V(X) = 12(3/8)+ 2

2(1/2) +4

2(1/8)-(15/8)

2= 0.8594

V(Y) = 32(1/8)+ 42(1/4)+ 52(1/2) +62(1/8)-(37/8)2 = 0.7344

8851.0)7344.0)(8594.0(

703125.0 ===Y X

XY XY

σ σ

σ ρ

5-34 125.0)8/1(1)2/1(5.0)4/1)(5.0()8/1(1)( =++−+−= X E

25.0)8/1(2)2/1(1)4/1)(1()8/1(2)( =++−+−=Y E

0.875(1/8)]2[1+(1/2)]1[0.5+(1/4)]-1[-0.5+(1/8)]2[-1=E(XY) =×××××××−×V(X) = 0.4219

V(Y) = 1.6875



5-29

16875.14219.0

8438.0

8438.0)25.0)(125.0(875.0

===

=−=

Y X

XY

XY

XY

σ σ

σ

ρ

σ

5-35

0435.0

36

23

36

23

36

1

36

23)()(

3

16)(

3

16)(

36

1

6

13

3

14

3

14)(

6

13)(

6

13)(

36/1,36)(

22

2

3

1

3

1

−=

−

=

====

−=⎟

⎠

⎞⎜⎝

⎛ −====

==+∑∑= =

ρ

σ

Y V X V Y E X E

XY E Y E X E

cc y xc

xy

x y

5-36 99.0)99.0(1)01.0(0)( =+= X E

98.0)98.0(1)02.0(0)( =+=Y E

0.9702(0.9702)]1[1+(0.0198)]0[1+(0.0098)]1[0+(0.002)]0[0=E(XY) =××××××××V(X) = 0.99-0.992=0.0099V(Y) = 0.98-0.982=0.0196

00196.00099.0

0

0)98.0)(99.0(9702.0

===

=−=

Y X

XY

XY

XY

σ σ

σ

ρ

σ

5-37E(X1) = np1 = 3(1/3)=1

E(X2) = np2= 3(1/3)= 1

V(X1) = 3p1(1-p1)=3(1/3)(2/3)=2/3

V(X2) = 3p2(1-p2)=3(1/3)(2/3)=2/3

E(X1X2) =n(n-1)p1 p2 =3(2)(1/3)(1/3)=2/3

5.0)3/2)(3/2(

3/13/113/2 2 −=

−=−=−= XY XY and ρ σ

The sign is negative.

For another example assume that n = 20E(X1) = np1 = 20(1/3)=6.67

E(X2) = np2=20(1/3)=6.67

V(X1) = np1(1-p1)=20(1/3)(2/3)=4.44

V(X2) = np2(1-p2)=20(1/3)(2/3)=4.44

E(X1X2) =n(n-1)p1 p2 =20(19)(1/3)(1/3)=42.22

51.0)44.4)(44.4(

267.2267.267.622.42 2 −=

−=−=−= XY XY and ρ σ



5-30

5-38.

Transaction FrequencySelects(X)

Updates(Y)

Inserts(Z)

New Order 43 23 11 12

Payment 44 4.2 3 1

Order Status 4 11.4 0 0

Delivery 5 130 120 0Stock Level 4 0 0 0

Mean Value 18.694 12.05 5.6

(a) COV(X,Y) = E(XY)-E(X)E(Y) = 23*11*0.43 + 4.2*3*0.44 + 11.4*0*0.04 + 130*120*0.05

+ 0*0*0.04 - 18.694*12.05=669.0713

(b) V(X)=735.9644, V(Y)=630.7875; Corr(X,Y)=cov(X,Y)/(V(X)*V(Y) )0.5

= 0.9820

(c) COV(X,Z)=23*12*0.43+4.2*1*0.44+0-18.694*5.6 = 15.8416

(d) V(Z)=31; Corr(X,Z)=0.1049

5-39 From Exercise 5-19, c = 8/81, E ( X ) = 12/5, and E (Y ) = 8/5

418

3

81

8

3818

3818

818)(

818)(

6

3

0

3

0

3

0

53

0

2

3

0

22

0

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =

==== ∫ ∫ ∫∫∫∫ xd x xd x x xdyd y x xdyd xy xy XY E

x x

4924.044.024.0

16.0

44.0)(,24.0)(

3)(6)(

16.05

8

5

124

22

==

==

==

=⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ −=

ρ

σ

Y V xV

Y E X E

xy

5-40 Similar to Exercise 5-23, 19/2=c

∫ ∫∫ ∫+

−

+

=+=5

1

1

1

1

0

1

0

614.219

2

19

2)(

x

x

x

xdydx xdydx X E

∫ ∫∫ ∫+

−

+

=+=5

1

1

1

1

0

1

0

649.219

2

19

2)(

x

x

x

ydydx ydydxY E

Now, ∫ ∫∫ ∫+

−

+

=+=5

1

1

1

1

0

1

0

7763.8

19

2

19

2)(

x

x

x

xydydx xydydx XY E

9206.0097.2930.1

852.1

0968.2)(,930.1)(

11403.9)(7632.8)(

85181.1)649.2)(614.2(7763.8

22

==

==

==

=−=

ρ

σ

Y V xV

Y E X E

xy



5-31

5-41 a) E(X) = 1 E(Y) = 1

)()(

)(

00

0 0

Y E X E

dy yedx xe

dxdy xye XY E

y x

y x

=

=

=

∫∫

∫ ∫∞

−∞

−

∞ ∞−−

Therefore, σ ρXY XY= = 0 .

5-42

E(X) = 333.33, E(Y)= 833.33

E(X2)=222,222.2

V(X)=222222.2-(333.33)2=111,113.31

E(Y2)=1,055,556

V(Y)=361,117.11

5547.011.36111731.111113

01.115,111

01.115,111)33.833)(33.333(9.888,388

9.888,388106)(0

002.001.6

==

=−=

=×= ∫ ∫∞ ∞

−−−

ρ

σ xy

x

y xdydx xye XY E

5-43 0)4/1(1)4/1(1)( =+−= X E

0)4/1(1)4/1(1)( =+−=Y E

0(1/4)]1[0+(1/4)]0[1+(1/4)]0[-1+(1/4)]0[-1=E(XY) =××××××××

V(X) = 1/2

V(Y) = 1/2

02/12/1

0

0)0)(0(0

===

=−=

Y X

XY

XY

XY

σ σ

σ

ρ

σ

The correlation is zero, but X and Y are not independent, since, for example, if y = 0, X

must be –1 or 1.

5-44 If X and Y are independent, then )()(),( y f x f y x f Y X XY = and the range of

(X, Y) is rectangular. Therefore,

)()()()()()()( Y E X E dy y yf dx x xf dxdy y f x xyf XY E Y X Y X === ∫∫∫∫

hence σXY=0

5-45 Suppose the correlation between X and Y is ρ. for constants a, b, c, and d, what is the

correlation between the random variables U = aX+b and V = cY+d? Now, E(U) = a E(X) + b and E(V) = c E(Y) + d.

Also, U - E(U) = a[ X - E(X) ] and V - E(V) = c[ Y - E(Y) ]. Then,

XY UV acY E Y X E X acE V E V U E U E σ σ =−−=−−= )]}()][({[)]}()][({[



Also,222222222 )]([)]([ Y V X U cand a X E X E aU E U E σ σ σ σ ==−=−= . Then,

⎪

⎪⎨⎧

==signindiffer candaif -

signsametheof arecandaif

2222 XY

XY

Y X

XY

UV

ca

ac

ρ

ρ

σ σ

σ ρ



5-32

Section 5-4

5-46 a)

10

-2

0.00

x0-1

0.01

0.02

0

0.03

0.04

1 2

z(0)

3-10

4y

b)



5-33

10

-2

0.00

x

0.01

0

-1

0.02

0.03

0.04

0

0.05

0.06

0.07

1 2

z(.8)

3-10

4y

c)

10

-2

0.00

x

0.01

0

-1

0.02

0.03

0.04

0

0.05

0.06

0.07

12

z(-.8)

3

-10

4y

5-47 Because 0= ρ and X and Y are normally distributed, X and Y are independent.

Therefore,



5-34

(a) P(2.95 < X < 3.05) = )(04.0

305.304.0

395.2 −− << Z P = 0.7887

(b) P(7.60 < Y < 7.80) = )(08.0

70.780.708.0

70.760.7 −− << Z P = 0.7887

(c) P(2.95 < X < 3.05, 7.60 < Y < 7.80) = P(2.95 < X < 3.05) P(7.60 < Y < 7.80) =

6220.07887.0)()( 2

08.070.780.7

08.070.760.7

04.0305.3

04.0395.2 ==<<<< −−−− Z P Z P

5-48 (a) ρ = cov(X,Y)/σxσy =0.6

cov(X,Y)= 0.6*2*5=6

(b) the marginal probability distribution of X is normal with mean µx , σx.

(c) P(X<116) =P(X-120<-4)=P((X_120)/5<-0.8)=P(Z<-0.8) = 0.21

(d) The conditional probability distribution of X given Y=102 is bivariate normal distribution with

mean and variance

µX|y=102 = 120 – 100*0.6*(5/2) +(5/2)*0.6(102) = 123

σ2X|y=102 = 25(1-0.36) =16

(e) P(X<116|Y=102)=P(Z<(116-123)/4)=0.040

5-49 Because ρ = 0 and X and Y are normally distributed, X and Y are independent.Therefore, μX = 0.1 mm, σX=0.00031 mm, μY = 0.23 mm, σY=0.00017 mm

Probability X is within specification limits is

0.8664)5.1()5.1()5.15.1(

00031.0

1.0100465.0

00031.0

1.0099535.0)100465.0099535.0(

=−<−<=<<−=

⎟ ⎠

⎞⎜⎝

⎛ −<<

−=<<

Z P Z P Z P

Z P X P

Probability that Y is within specification limits is

0.9545)2()2()22(

00017.0

23.023034.0

00017.0

23.022966.0)23034.022966.0(

=−<−<=<<−=

⎟ ⎠

⎞⎜⎝

⎛ −<<

−=<<

Z P Z P Z P

Z P X P

Probability that a randomly selected lamp is within specification limits is(0.8664)(0.9594) = 0.8270

5-50 a) By completing the square in the numerator of the exponent of the bivariate normal PDF, the joint

PDF can be written as

2

)1(2

)1())(((1

2

| 2

2

2

2

2

2

2

1

12

1

)(

),(

⎥⎦

⎤⎢⎣

⎡ −

−

−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−+⎥

⎦

⎤⎢⎣

⎡−+−

−

=

−==

x

x

x

x x

X

Y Y

Y

x

x

x x y

y x

X

XY x X Y

e

e

x f

y x f f

σ

μ

ρ

σ

μ ρ μ

σ

σ ρ μ

σ

σ π

ρ σ πσ

Also, f x(x) =

2

21

2

x

x

x

x

e

μ

σ

π σ

⎡ ⎤−⎢ ⎥⎣ ⎦−

By definition,



5-35

)1(2

))(((1

2

)1(2

)1(2

)1())(((1

2

|

2

2

2

2

2

2

2

2

2

2

12

1

2

1

12

1

)(

),(

ρ−

⎥⎦

⎤⎢⎣

⎡μ−

σ

σρ+μ−

σ−

ρ−

⎥⎦

⎤⎢⎣

⎡

σ

μ−

−

ρ−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

σ

μ−ρ−+⎥

⎦

⎤⎢⎣

⎡μ−

σ

σρ+μ−

σ−

=

ρ−σπ=

σπ

ρ−σπσ==

x

X

Y Y

Y

x

x

x

x x

X

Y Y

Y

x y

y

x

x

x x y

y x

X

XY x X Y

e

e

e

x f

y x f f

Now f Y|X=x is in the form of a normal distribution.

b) E(Y|X=x) = )( x x

y y x μ−

σ

σρ+μ . This answer can be seen from part a). Since the PDF is in the

form of a normal distribution, then the mean can be obtained from the exponent.

c) V(Y|X=x) = )1( 22 ρ−σ y . This answer can be seen from part a). Since the PDF is in the form of a

normal distribution, then the variance can be obtained from the exponent.

5-51

∫∫

∫ ∫∫ ∫

∞

∞−

⎥⎦

⎤⎢⎣

⎡ −−∞

∞−

⎥⎦

⎤⎢⎣

⎡ −−

∞

∞−

∞

∞−

⎥⎦

⎤⎢⎣

⎡ −+

−−∞

∞−

∞

∞−

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

=⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

dyedxe

dxdyedxdy y x f

Y

Y

Y

X

X

X

Y

Y

X

X

Y X

y x

y x

XY

2

2)(

2

1

2

2)(

2

1

2

2)(

2

2)(

2

1

2

1

2

1

21),(

σ

σ

σ

σ

σ σ σ σ

μ

π

μ

π

μ μ

π

and each of the last two integrals is recognized as the integral of a normal probability

density function from −∞ to ∞. That is, each integral equals one. Since

)()(),( y f x f y x f XY = then X and Y are independent.

5-52 Let )1(2

))((2

2

2

22

12

1),( ρ−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

σ

μ−+

σσ

μ−μ−ρ−⎟⎟

⎠

⎞⎜⎜⎝

⎛

σ

μ−

−

ρ−σπσ=

Y

Y

Y X

X X

X

X Y Y X X

y x

XY e y x f

Completing the square in the numerator of the exponent we get:

2

2

222

)1())((2

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+

−−−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

X

X

X

X

Y

Y

Y

Y

Y X

X X

X

X X X Y Y Y X X

σ

μ ρ

σ

μ ρ

σ

μ

σ

μ

σ σ

μ μ ρ

σ

μ

But,

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−−=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛ −−⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛ −))(((

1)()(

1 x

X

Y Y

Y x

X

Y Y

Y X

X

Y

Y X Y X Y X Y

μ σ

σ ρ μ

σ μ

σ

σ ρ μ

σ σ

μ ρ

σ

μ

Substituting into f XY(x,y), we get



5-36

22

2

2

2

2

2 2 2

1( ( )) (1 )

2(1 )

2

( ( ))

2 (1 )1

2

2

1( , )

2 1

1 1

2 2 1

xY Y x

X xY

y y x

x

x x

x

x y x

XY

x y

y x

x

x y

f x y e dydx

e dx e dy

μ σ μ ρ μ ρ

σ σ σ

ρ

σ μ ρ μ

σ

σ ρ μ

σ

πσ σ ρ

π σ π σ ρ

⎡ ⎤⎛ ⎞⎡ ⎤ −⎢ ⎥− + − + − ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦−∞ ∞ −

−∞ −∞

⎡ ⎤⎛ ⎞⎢ ⎥− + −⎜ ⎟

⎢ ⎥⎝ ⎠−⎢ ⎥⎛ ⎞ −− ⎢ ⎥− ⎜ ⎟ ⎢ ⎥∞ ∞⎝ ⎠ ⎢ ⎥⎣ ⎦

−∞ −∞

=−

= ×−

∫ ∫

∫ ∫

The integrand in the second integral above is in the form of a normally distributed random

variable. By definition of the integral over this function, the second integral is equal to 1:

2

2 2 2

2

( ( ))

2 (1 )1

2

2

1

2

1 1

2 2 1

11

2

y y x

x

x x

x

x

x

y x

x

x y

x

x

e dx e dy

e dx

σ μ ρ μ

σ

σ ρ μ

σ

μ

σ

π σ π σ ρ

π σ

⎡ ⎤⎛ ⎞⎢ ⎥− + −⎜ ⎟⎢ ⎥⎝ ⎠−⎢ ⎥

⎛ ⎞ −− ⎢ ⎥− ⎜ ⎟ ⎢ ⎥∞ ∞⎝ ⎠ ⎢ ⎥⎣ ⎦

−∞ −∞

⎛ ⎞−− ⎜ ⎟∞ ⎝ ⎠

−∞

×−

= ×

∫ ∫

∫

The remaining integral is also the integral of a normally distributed random variable and therefore,

it also integrates to 1, by definition. Therefore,

1),(∫ ∫∞

∞−

∞

∞−= y x f XY

5-53

dyee

dyee

dye x f

X

X

Y

Y

Y

X

X

X

X

X

X

X

Y

Y

Y

X

X

X

Y

Y

Y X

Y X

X

X

Y X

x y x

x x y x

y y x x

X

2

)()(

21

2

2

2)(

2)(

2)()(

21

2

2

2)(

21

2

2)())((

2

2)(

21

2

5.0

12

15.0

2

1

5.0

12

1

5.0

2

1

25.0

12

1)(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−−−

∞

∞− −

−−

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦⎤

⎢⎣⎡ −

−⎥⎦⎤

⎢⎣⎡ −

−−−∞

∞−−

−−

∞

∞−

⎥⎦

⎤⎢⎣

⎡ −+

−−−

−−

−

−

−−

−

∫

∫

∫

=

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

σ σ

σ

σ

σ

σ σ σ

σ

σ

σ

σ σ σ σ

σ πσ

μ ρ μ

ρ

ρ π

μ

π

μ ρ μ ρ μ

ρ

ρ π

μ

ρ

π

μ μ μ ρ μ

ρ

ρ

The last integral is recognized as the integral of a normal probability density with mean

X

X Y x

Y σ σ

μ μ ρ )( −+ and variance )1( 22

ρ σ −Y . Therefore, the last integral equals one and

the requested result is obtained.Section 5-5



5-54 a) E(2X + 3Y) = 2(0) + 3(10) = 30

b) V(2X + 3Y) = 4V(X) + 9V(Y) = 97

c) 2X + 3Y is normally distributed with mean 30 and variance 97. Therefore,

5.0)0()()3032(97

3030 =<=<=<+ − Z P Z PY X P

d) 8461.0)02.1()()4032( 97

3040

=<=<=<+−

Z P Z PY X P 5-55

(a) E(3X+2Y)= 3*2+2*6=18

(b) V(3X+2Y) = 9*5+4*8 =77

(c) 3X+2Y ~ N( 18, 77)

P(3X+2Y<18) = P(Z<(18-18)/770.5)=0.5

(d) P(3X+2Y<28) = P(Z<(28-18)/770.5)=P(Z<1.1396) =0.873



5-37

5-54 a) E(2X + 3Y) = 2(0) + 3(10) = 30 b) V(2X + 3Y) = 4V(X) + 9V(Y) = 97

c) 2X + 3Y is normally distributed with mean 30 and variance 97. Therefore,

5.0)0()()3032(97

3030 =<=<=<+ − Z P Z PY X P

d) 8461.0)02.1()()4032(97

3040 =<=<=<+ − Z P Z PY X P

5-55

(a) E(3X+2Y)= 3*2+2*6=18

(b) V(3X+2Y) = 9*5+4*8 =77

(c) 3X+2Y ~ N( 18, 77)

P(3X+2Y<18) = P(Z<(18-18)/770.5)=0.5

(d) P(3X+2Y<28) = P(Z<(28-18)/770.5)=P(Z<1.1396) =0.873

Section 5-5

5-56 Y = 10X and E(Y) =10E(X) = 50mm.

V(Y)=102

V(X)=25mm2

5-57 a) Let T denote the total thickness. Then, T = X + Y and E(T) = 4 mm,

V(T) = 0 1 0 1 0 022 2 2. . .+ = mm , and σT = 0 1414. mm.

b)

0170.0983.01)12.2(1

)12.2(1414.0

43.4)3.4(

=−=<−=

>=⎟ ⎠

⎞⎜⎝

⎛ −>=>

Z P

Z P Z PT P

5-58 (a) X: time of wheel throwing. X~N(40,4)

Y : time of wheel firing. Y~N(60,9)

X+Y ~ N(100, 13)

P(X+Y≤ 95) = P(Z<(95-100)/130.5) =P(Z<1.387) =0.917

(b) P(X+Y>110) = 1- P(Z<(110-100)/ 130.5) =1-P(Z<2.774) = 1-0.9972 =0.0028

5-59 a) X ∼ N(0.1, 0.00031) and Y ∼ N(0.23, 0.00017) Let T denote the total thickness.

Then, T = X + Y and E(T) = 0.33 mm,

V(T) =2722 1025.100017.000031.0 mm x

−=+ , and 000354.0=T

σ mm.

0)272(000354.0

33.02337.0)2337.0( ≅−<=⎟

⎠

⎞⎜⎝

⎛ −<=< Z P Z P T P

b)

1)253(1)253(000345.0

33.02405.0)2405.0( ≅<−=−>=⎟

⎠

⎞⎜⎝

⎛ −>=> Z P Z P Z P T P

5-60 Let D denote the width of the casing minus the width of the door. Then, D is normally

distributed.

a) E(D) = 1/8 V(D) =25652

1612

81 )()( =+ 5 0.1398

256T σ = =



5-38

b) 187.0)89.0()()(256

5

8

1

4

1

41 =>=>=>

− Z P Z P DP

c) 187.0)89.0()()0(256

5

8

10

=−<=<=<−

Z P Z P DP

5-61 X : time of ACL reconstruction surgery for high-volume hospitals.

X ~ N(129,196).

E(X1+X2+…+X10) = 10* 129 =1290

V(X1+X2+…+X10) = 100*196 =19600

5-62 a) Let X denote the average fill-volume of 100 cans. 05.01005.0

2

== X

σ .

b) E( X ) = 12.1 and 023.0)2(05.0

1.1212)12( =−<=⎟

⎠

⎞⎜⎝

⎛ −<=< Z P Z P X P

c) P( X < 12) = 0.005 implies that .005.005.0

12=⎟

⎠

⎞⎜⎝

⎛ −<

μ Z P

Then05.0

12 μ −= -2.58 and 129.12=μ .

d) P( X < 12) = 0.005 implies that .005.0100/

1.1212=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −<σ

Z P

Then100/

1.1212

σ

− = -2.58 and 388.0=σ .

e) P( X < 12) = 0.01 implies that .01.0/5.0

1.1212=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −<

n Z P

Thenn/5.0

1.1212− = -2.33 and 13672.135 ≅=n .

5-63 Let X denote the average thickness of 10 wafers. Then, E( X ) = 10 and V( X ) = 0.1.

a) 998.0)16.316.3()()119(1.0

1011

1.0

109 =<<−=<<=<< −− Z P Z P X P .

The answer is 1 − 0.998 = 0.002

b)n X

and X P 101.0)11( ==> σ .

Therefore, 01.0)()11(1

1011 =>=> −

n

Z P X P , 11 101−

n

= 2.33 and n = 5.43 which is

rounded up to 6.

c)10

0005.0)11( σ σ ==> X

and X P .

Therefore, 0005.0)()11(10

1011 =>=> −σ

Z P X P ,10

1011

σ

− = 3.29

9612.029.3/10 ==σ

5-64 X~N(160, 900)

a) Let Y = X1 + X2 + ... + X25, E(Y) = 25E(X) = 4000, V(Y) = 252(900) = 22500

P(Y > 4300) =

0227.09773.01)2(1)2(22500

40004300=−=<−=>=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −> Z P Z P Z P



b) c) P(Y > x) = 0.0001 implies that .0001.022500

4000=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −>x

Z P

Then1504000− x = 3.72 and 4558= x

5-65 W: weights of parts E: measurement error.

W~ N(µw, σw2) , E ~ N(0, σe

2) ,W+E ~ N(µw, σw2+σe

2) .

Wsp = weights of the specification P

(a) P(W > µw + 3σw) + P(W < µw – 3σw) = P(Z > 3) + P(Z < -3) = 0.0027

(b) P(W+E > µw + 3σw) + P( W+E < µw – 3σw)

= P (Z > 3σw / (σw2+σe

2)1/2) + P (Z < -3σw / (σw2+σe

2)1/2)

Because σe2 = 0.5σw

2 the probability is

= P (Z > 3σw / (1.5σw2)1/2) + P (Z < -3σw / (1.5σw

2)1/2)

= P (Z > 2.45) + P (Z < -2.45) = 2(0.0072) = 0.014

No.

(c) P( E + µw + 2σw > µw + 3σw) = P(E > σw) = P(Z > σw/(0.5σw2)1/2)

= P(Z > 1.41) = 0.079

Also, P( E + µw + 2σw < µw – 3σw) = P( E < – 5σw)

= P(Z < -5σw/(0.5σw2)1/2) = P(Z < -7.07) ≈ 0

5-66 D = A - B - C

a) E(D) = 10 - 2 - 2 = 6 mm

mm

mm DV

D 1225.0

015.005.005.01.0)( 2222

=

=++=

σ

b) P(D < 5.9) = P(Z <1225.0

69.5 −) = P(Z < -0.82) = 0.206.



5-39

Section 5-6

5-674

1)( = y f Y at y = 3, 5, 7, 9 from equation 5-40.

5-68 Because X ≥ 0 , the transformation is one-to-one; that is2 x y = and y x = . From equation 5-40,

( y y

y X Y p p y f y f −−== 33 )1()()( for y = 0, 1, 4, 9.

If p = 0.25, ( ) y y

yY y f −= 33 )75.0()25.0()( for y = 0, 1, 4, 9.

5-69 a)72

10

2

1

2

10)(

−=⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ −=

y y f y f X Y for 10 ≤ ≤y 22

b) ( ) 1872

1

72

10)(

22

102

10

3

22

10

223

=−=−

= ∫y y

dy y y

Y E

5-70 Because y = -2 ln x, xe y

=−

2 . Then, 222

21

21)()(

y y y

eee f y f X Y

−−−

=−= for ≤≤−

20 y

e 1 or

0≥ y , which is an exponential distribution with λ =1/2 (which equals a chi-square distribution with k = 2

degrees of freedom).



5-71 a) If 2 x y = , then y x = for x ≥ 0 and y ≥ 0 . Thus,

y

e y y f y f

y

X Y 2

)()( 2

1

2

1

−−

== for

y > 0.

b) If 2/1 x y = , then

2 y x = for 0≥ x and 0≥ y . Thus,2

22)()( 2 y

X Y ye y y f y f −== for y >

0.c) If y = ln x, then x ey= for 0≥ x . Thus,

y ye ye y y y

X Y eeeee f y f −− === )()( for

∞<<∞− y .

5-72 a) Now, dveavbv−

∞

∫0

2must equal one. Let u = bv, then 1 .)(

0

2

3

0

2 dueub

a

b

duea uu

b

u −∞

−∞

∫∫ == From

the definition of the gamma function the last expression is33

2)3(

b

a

b

a=Γ . Therefore,

2

3ba = .

b) If 2

2mvw = , then

m

wv

2= for 0,0 ≥≥ wv .

( )

m

w

m

w

b

b

mw

V W

ewmb

mwem

wb

dw

dv f w f

2

2

2/12/33

2/13

2

2

)2(2

2)(

−−

−−

=

==

for 0≥w .

5-73 If xe y = , then x = ln y for 1

21 and 2 e ye x ≤≤≤≤ . Thus, y y

y f y f X Y

11)(ln)( == for

1 2ln ≤≤ y . That is,

y

y f Y 1

)( = for 2e ye ≤≤ .

5-74 If y =2)2( − x , then x = y−2 for 20 ≤≤ x and x = y+2 for 42 ≤≤ x . Thus,

( ) 40for

16

2

16

2

||)2(||)2()(

2/1

4

1

2/1

2

12/1

2

1

≤≤=

++

−=

++−−=

−

−−

y y

y

y

y

y

y y f y y f y f X X Y



5-40


5-75 The sum of ∑∑ = x y

y x f 1),( , ( ) ( ) ( ) ( ) ( ) 14

14

18

18

14

1 =++++

and 0),( ≥ y x f XY

a) 8/34/18/1)0,0()1,0()5.1,5.0( =+=+=<< XY XY f f Y X P .

b) 4/3)1,1()0,1()1,0()0,0()1( =+++=≤ XY XY XY XY f f f f X P



5-41

c) 4/3)1,1()0,1()1,0()0,0()5.1( =+++=< XY XY XY XY f f f f Y P

d) 8/3)1,1()0,1()5.1,5.0( =+=<> XY XY f f Y X P

e) E(X) = 0(3/8) + 1(3/8) + 2(1/4) = 7/8.

V(X) = 02(3/8) + 12(3/8) + 22(1/4) - 7/82 =39/64

E(Y) = 1(3/8) + 0(3/8) + 2(1/4) = 7/8.

V(Y) = 12

(3/8) + 02

(3/8) + 22

(1/4) - 7/82

=39/64

f) 4 / 1)2(,8 / 3)1(,8 / 3)0(),()( ==== ∑ X X X

y

XY X f f f and y x f x f .

g) 3/2)1(,3/1)0()1(

),1()(

8/3

4/1

18/3

8/1

11=====

Y Y

X

XY

Y f f and

f

y f y f .

h) 3/2)3/2(1)3/1(0)()1|(1

1| =+=== ∑=

= x

X Y y yf X Y E

i) As is discussed after Example 5-19, because the range of (X, Y) is not rectangular, X

and Y are not independent.

j) E( XY ) = 1.25, E( X ) = E(Y )= 0.875 V(X) = V(Y) = 0.6094COV( X ,Y )=E( XY )-E( X )E(Y )= 1.25-0.8752=0.4844

7949.06094.06094.0

4844.0== XY ρ

5-76 a) 0.063170.020.010.0!14!4!2

!20)14,4,2( 1442 ===== Z Y X P

b) 0.121690.010.0)0( 200 ===X P

c) 2)10.0(20)( 1 === np X E

8.1)9.0)(10.0(20)1()( 11==−= pnp X V

d))(

),()19|(|

z f

z x f Z X f

Z

XZ

z Z X ==

z z x x

XZ z x z x

xz f 7.02.01.0)!20(!!

!20)( 20 −−

−−=

z z

Z z z

z f 7.03.0)!20(!

!20)( 20−

−=

z x x

z

z x x

Z

XZ

z Z X z x x

z

z x x

z

z f

z x f Z X f

−−

−

−−

= ⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ −−

−=

−−

−==

20

20

20

|3

2

3

1

)!20(!

)!20(

3.0

2.01.0

)!20(!

)!20(

)(

),()19|(

Therefore, X is a binomial random variable with n=20- z and p=1/3. When z=19,

3

2)0(19| = X f and

3

1)1(19| = X f .

e)3

1

3

11

3

20)19|( =⎟

⎠

⎞⎜⎝

⎛ +⎟ ⎠

⎞⎜⎝

⎛ == Z X E

5-77 Let X, Y, and Z denote the number of bolts rated high, moderate, and low. Then, X, Y,

and Z have a multinomial distribution.



5-42

a) 0560.01.03.06.0!2!6!12

!20)2,6,12( 2612 ===== Z Y X P .

b) Because X, Y, and Z are multinomial, the marginal distribution of Z is binomial with

n = 20 and p = 0.1.c) E(Z) = np = 20(0.1) = 2.

d) P(low>2)=1-P(low=0)-P(low=1)-P(low=2)=1-0.1*0.920- 0.1*0.919-0.12*0.918=0.863

e))16(

),16()(

16|

X

XZ Z

f

z f z f = and

z z x x

XZ z x z x

z x f 1.03.06.0)!20(!!

!20),( )20( −−

−−= for

z xand z x ≤≤≤+ 0,020 . Then,

( ) ( ) z z

z z

z z

z z

Z z f

4.01.04

4.03.0

)!4(!!4

416

!4!16!20

)4(16

)!4(!!16!20

164.06.0

1.03.06.0)(

−

−

−− ==

for 40 ≤≤ z . That is the distribution of Z given X = 16 is binomial with n = 4 and

p = 0.25.

f) From part a., E(Z) = 4 (0.25) = 1.

g) Because the conditional distribution of Z given X = 16 does not equal the marginaldistribution of Z, X and Z are not independent.

5-78 Let X, Y, and Z denote the number of calls answered in two rings or less, three or

four rings, and five rings or more, respectively.

a) 0649.005.025.07.0!1!1!8

!10)1,1,8( 118 ===== Z Y X P

b) Let W denote the number of calls answered in four rings or less. Then, W is a binomial

random variable with n = 10 and p = 0.95.

Therefore, P(W = 10) = ( ) 5987.005.095.0 01010

10 = .

c) E(W) = 10(0.95) = 9.5.

d))8(

),8()(

8

X

XZ

Z f

z f z f = and

z z x x

XZ z x z x

z x f 05.025.070.0)!10(!!

!10),( )10( −−

−−= for

z xand z x ≤≤≤+ 0,010 . Then,

( ) ( ) z z

z z

z z

z z

Z z f

30.005.02

30.025.0

)!2(!!2

28

!2!8!10

)2(8

)!2(!!8

!10

830.070.0

05.025.070.0)(

−

−

−− ==

for 0 2≤ ≤z . That is Z is binomial with n =2 and p = 0.05/0.30 = 1/6.

e) E(Z) given X = 8 is 2(1/6) = 1/3.f) Because the conditional distribution of Z given X = 8 does not equal the marginal

distribution of Z, X and Z are not independent.

5-79 ∫ ∫∫ ===3

0

3

03

3

0

2

02

2

2

0

2 18232

ccdxcx ydydxcx x y. Therefore, c = 1/18.

a) ∫ ∫∫ ====<<1

0

108

11

0336

1

1

0

1

02

2

18

1

1

0

2

18

132

)1,1( x ydx x ydydx xY X P



5-43

b) ∫ ∫∫ ====<5.2

0

5.2

039

1

5.2

0

2

02

2

18

1

2

0

2

18

1 5787.0)5.2(32

x ydx x ydydx x X P

c) ∫ ∫∫ ====<<3

0

4

33

0312

1

3

0

2

12

2

18

1

2

1

2

18

132

)5.21( x ydx x ydydx xY P

d)

2199.0

)5.11,2(

432

95

3

2

3

23144

5

3

2

5.1

12

2

18

1

5.1

1

2

18

132

==

===<<> ∫ ∫∫ x ydx x ydydx xY X P

e) ∫ ∫∫ ====3

0

4

93

049

1

3

0

3

18

1

2

0

3

18

14

2)( xdx x ydydx x X E

f) ∫ ∫∫ ====3

0

3

43

0327

4

3

0

3

82

18

1

2

0

22

18

13

)( xdx xdydx y xY E

g) 30)( 2

9

1

2

0

2

18

1 <<== ∫ x for x ydy x x f X

h)2)1(

),1()(

9

1

18

1 y y

f

y f y f

X

XY

X Y === for 0 < y < 2.

i))1()1(

)1,()(

2

18

1

1

Y Y

XY

X f

x

f

x f x f == and 20)(

2

3

0

2

18

1 <<== ∫ y for ydx x y f y

Y .

Therefore,2

9

1

2

18

1

12/1

)( x x

x f X

== for 0 < x < 3.

5-80 The region x y2 2 1+ ≤ and 0 < z < 4 is a cylinder of radius 1 ( and base area π ) and

height 4. Therefore, the volume of the cylinder is 4π and f x y zXYZ( , , ) =1

4πfor x y2 2 1+ ≤

and 0 < z < 4.

a) The region X Y2 2 0 5+ ≤ . is a cylinder of radius 0 5. and height 4. Therefore,

2/1)5.0(4

)5.0(422 ==≤+π

π Y X P .

b) The region X Y2 2 0 5+ ≤ . and 0 < z < 2 is a cylinder of radius 0 5. and height 2.

Therefore,

4/1)2,5.0( 4

)5.0(222

==<≤+ π

π

Z Y X P

c))1(

)1,,(),(

1

Z

XYZ

XY f

y x f y x f = and 4/1)(

1

41

22

== ∫∫≤+

dydx z f

y x

Z π

for 0 < z < 4. Then, 14/1

4/1),( 221

1≤+== y x for y x f

XY π

π .



5-44

d)222

4

0

21

41

1

1

4

0

11)(

2

2

xdz xdydz x f

x

x

X −=−== ∫∫∫−

−−

π π π for -1 < x < 1

e)

)0,0(

),0,0()(

0,0

XY

XYZ

Z

f

z f z f = and π

π

/1),(4

0 4

1 ==

∫dz y x f XY for 122 ≤+ y x . Then,

4/1/1

4/1)(

0,0==

π

π z f

Z for 0 < z < 4 and 2

0,0=

Z μ .

f) 4/1/1

4/1

),(

),,()( ===

π

π

y x f

z y x f z f

XY

XYZ

xy Z for 0 < z < 4. Then, E(Z) given X = x and Y = y is

24

0

4=∫ dz z

.

5-81 c y x f XY =),( for 0 < x < 1 and 0 < y < 1. Then,

∫∫=

1

0

1

0

1cdxdy and c = 1. Because

),( y x f XY is constant, )5.0( <− Y X P is the area of the shaded region below

0

0.5

0.5

1

1

That is, )5.0( <− Y X P = 3/4.

5-82 a) Let X X X1 2 6, ,..., denote the lifetimes of the six components, respectively. Because of

independence,P X X X P X P X P X( , , . .. , ) ( ) ( )... ( )1 2 6 1 2 65000 5000 5000 5000 5000 5000> > > = > > >

If X is exponentially distributed with mean θ , then λθ

= 1 and

θ θ θ

θ

///1)( x

x

t

x

t eedt e x X P

−∞

−∞

− =−==> ∫ . Therefore, the answer is

0978.0325.22.025.025.05.05.08/5

==−−−−−−−

eeeeeee . b) The probability that at least one component lifetime exceeds 25,000 hours is

the same as 1 minus the probability that none of the component lifetimes exceed

25,000 hours. Thus,

1-P(Xa<25,000, X2<25,000, …, X6<25,000)=1-P(X1<25,000)…P(X6<25,000)

=1-(1-e-25/8

)(1-e-2.5

)(1-e-2.5

)(1-e-1.25

)(1-e-1.25

)(1-e-1

)=1-.2592=0.7408



5-45

5-83 Let X, Y, and Z denote the number of problems that result in functional, minor, and no

defects, respectively.

a) 085.03.05.02.0)3,5,2()5,2( 352

!3!5!2

!10 ======== Z Y X PY X P

b) Z is binomial with n = 10 and p = 0.3.c) E(Z) = 10(0.3) = 3.

5-84 a) Let X denote the mean weight of the 25 bricks in the sample. Then, E( X ) = 3 and

05.025

25.0 == X

σ . Then, P( X < 2.95) = P(Z < 2 95 30 05.

.− ) = P (Z < -1) = 0.159.

b) 99.0)05.

3()( =

−>=>

x Z P x X P . So, =

−05.0

3 x-2.33 and x=2.8835.

5-85 a)

Because ,25.025.5

75.4

25.18

75.17

ccdydx =∫∫ c = 4. The area of a panel is XY and P(XY > 90) is the

shaded area times 4 below,

5.25

4.75

17.25 18.25

That is, 499.0)ln9025.5(425.54425.18

75.17

25.18

75.17

90

25.5

/90

25.18

75.17

=−=−= ∫∫∫ x xdxdydx x

x

b) The perimeter of a panel is 2X+2Y and we want P(2X+2Y >46)

5.0)2

75.17(4)75.17(4

)23(25.544

25.18

75.17

225.18

75.17

25.18

75.17

25.5

23

25.18

75.17

=+−=+−=

−−=

∫

∫∫∫−

x xdx x

dx xdydx x

5-86 a) Let X denote the weight of a piece of candy and X∼ N(0.1, 0.01). Each package has 16

candies, then P is the total weight of the package with 16 pieces and E( P ) = 16(0.1)=1.6ounces and V(P) = 162(0.012)=0.0256 ounces2

b) 5.0)0()()6.1(16.0

6.16.1 =<=<=< − Z P Z PPP .

c) Let Y equal the total weight of the package with 17 pieces, E(Y) = 17(0.1)=1.7 ounces

and V(Y) = 172(0.012)=0.0289 ounces2

2776.0)59.0()()6.1(0289.0

7.16.1 =−<=<=< − Z P Z PY P .

5-87 Let X denote the average time to locate 10 parts. Then, E( X ) =45 and10

30= X

σ

a) 057.0)58.1()()60(10/30

4560 =>=>=> − Z P Z P X P



5-46

b) Let Y denote the total time to locate 10 parts. Then, Y > 600 if and only if X > 60.

Therefore, the answer is the same as part a.

5-88 a) Let Y denote the weight of an assembly. Then, E(Y) = 4 + 5.5 + 10 + 8 = 27.5 and

V(Y)= 7.05.02.05.04.0 2222 =+++ .

0084.0)39.2()()5.29( 7.0

5.275.29

=>=>=>−

Z P Z PY P b) Let X denote the mean weight of 8 independent assemblies. Then, E( X ) = 27.5 and

V( X ) = 0.7/8 = 0.0875. Also, 0)07.5()()29(0875.0

5.2729 =>=>=> − Z P Z P X P .

5-89

10

-2

0.00

x

0.01

0

-1

0.02

0.03

0.04

0

0.05

0.06

0.07

1 2

z(-.8)

3-10

4y

5-90

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+−−−−

−

−

⎥⎦

⎤⎢⎣

⎡−+−−−−

−

⎥⎦

⎤⎢⎣

⎡ −+−−−−−

−=

=

=

})2()2)(1)(8(.2)1{()8.01(2

1

2

})2()2)(1(6.1)1{()36.0(2

1

})2()2)(1(6.1)1{(72.0

1

22

2

22

22

8.12

1),(

36.2

1),(

2.1

1),(

y y x x

XY

y y x x

XY

y y x x

XY

e y x f

e y x f

e y x f

π

π

π

1)( = X E , 2)( =Y E 1)( = X V 1)( =Y V and ρ = 0.8

5-91 Let T denote the total thickness. Then, T = X1 + X2 anda) E(T) = 0.5+1=1.5 mm



a. V(T)=V(X1) +V(X2) + 2Cov(X1X2)=0.01+0.04+2(0.014)=0.078mm2

i. where Cov(XY)=ρσXσY=0.7(0.1)(0.2)=0.014

b) 0367.0)79.1(078.0

5.11)1( =−<=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −<=< Z P Z PT P

c) Let P denote the total thickness. Then, P = 2X1 +3 X2 andE(P) =2(0.5)+3(1)=4 mm

V(P)=4V(X1) +9V(X2) +

2(2)(3)Cov(X1X2)=4(0.01)+9(0.04)+2(2)(3)(0.014)=0.568mm2

where Cov(XY)=ρσXσY=0.7(0.1)(0.2)=0.014

5-92 Let T denote the total thickness. Then, T = X1 + X2 + X3 anda) E(T) = 0.5+1+1.5 =3 mm

V(T)=V(X1) +V(X2) +V(X3)+2Cov(X1X2)+ 2Cov(X2X3)+

2Cov(X1X3)=0.01+0.04+0.09+2(0.014)+2(0.03)+ 2(0.009)=0.246mm2

where Cov(XY)=ρσXσY

b) 0)10.6(246.0

35.1)5.1( ≅−<=⎟

⎠ ⎞⎜

⎝ ⎛ −<=< Z P Z PT P

5-93 Let X and Y denote the percentage returns for security one and two respectively.

If ½ of the total dollars is invested in each then ½X+ ½Y is the percentage return.

E(½X+ ½Y) = 0.05 (or 5 if given in terms of percent)V(½X+ ½Y) = 1/4 V(X)+1/4V(Y)+2(1/2)(1/2)Cov(X,Y)

where Cov(XY)=ρσXσY=-0.5(2)(4) = -4

V(½X+ ½Y) = 1/4(4)+1/4(6)-2 = 3Also, E(X) = 5 and V(X) = 4. Therefore, the strategy that splits between the securities

has a lower standard deviation of percentage return than investing 2million in the first

security.




6-95 955950462572

29

1

9

1

2 ==⎟ ⎠

⎞⎜⎝

⎛ = ∑∑

==

n x xi

i

i

i

11.761.50

61.5019

9

55950425726

1

29

19

1

2

2

==

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−=

∑∑ =

=

s

n

n

x

x

s

i

i

i

i

Subtract 30 and multiply by 10

9228484002579200

29

1

9

1

2 ==⎟ ⎠

⎞⎜⎝

⎛ = ∑∑

==

n x xi

i

i

i

6-58



14.711.5061

1.506119

9

228484005792002

1

29

19

1

2

2

==

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−=

∑∑ =

=

s

n

n

x

x

s

i

i

i

i

Yes, the rescaling is by a factor of 10. Therefore, s2 and s would be rescaled by multiplying s2 by 102

(resulting in 5061.1) and s by 10 (71.14). Subtracting 30 from each value has no affect on the variance or

standard deviation. This is because .)()( 2 X V abaX V =+

6-962

1

2

1

2 )()()( a xn x xa xn

i

i

n

i

i −+−=− ∑∑==

; The sum written in this form shows that the quantity is

minimized when a = x .

6-97 Of the two quantities ( )∑=

−n

i

i x x1

2and , the quantity(∑

=

−n

i

i x1

2μ ) (∑

=

−n

i

i x x1

2) will be smaller

given that μ ≠ x . This is because x is based on the values of the ’s. The value of μ may be quite

different for this sample.

i x

6-98 ii bxa y +=

n

x

x

n

i

i∑== 1

and xban

xbna

n

bxa

y

n

i

i

n

i

i

+=+

=+

=∑∑== 11

)(

( )

1

1

2

2

−

−=∑=

n

x x

s

n

i

i

x and2

x x ss =

221

22

1

2

1

2

2

1

)(

1

)(

1

)(

x

n

i

i

n

i

i

n

i

i

y sbn

x xb

n

xbbx

n

xbabxa

s =−

−=

−

−=

−

−−+=

∑∑∑===

Therefore, x y bss =

6-99 00.835= x °F °F5.10= xs

The results in °C:

6-59



89.431)00.835(9 / 5329 / 532 =+−=+−= x y °C

028.34)5.10()9 / 5( 22222 === x y sbs °C

6-100. Using the results found in Exercise 6-98 with a =s x− and b = 1/ s, the mean and standard deviation

of the zi are z = 0 and sZ = 1.

6-101. Yes, in this case, since no upper bound on the last electronic component is available, use a

measure of central location that is not dependent on this value. That measure is the median.

Sample Median = hours x x

692

7563

2

)5()4( =+

=+

6-60



6-102 a)11

1

1

1

1

1 +

+=

+=

∑∑=

+

+

=+

n

x x

n

x

x

n

i

ni

n

i

i

n

1

1

1 +

+

=

+

+ n

x xn

x

nn

n

11

1

1 ++

+= +

+n

x x

n

n x n

nn

b)1

2

1

12

1

1

22

1 +

⎟ ⎠

⎞⎜⎝

⎛ +

−+=+

=+

=+

∑∑

n

x x

x xns

n

n

i

i

n

n

i

in

( )[ ]

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]

( )[ ]

[ ]

( )

( )2

1

22

21

22

2

1

22

2

1

2

2

2

1

222

1

2

2

1

222

1

2

1

2

1

1

2

2

2

1

1

1

2

1

2

2

11

1

2

1

1

2

1

2

1)1(

21

)1(

211

)1(

21)1(

)1(

21)1(

)1(

2

11

211

12

11

11

2

1

nnn

nnnnn

nnnn

nnn

i

n

nnn

iiin

i

i

nnn

iiin

i

i

nnn

n

i

i

i

n

i

i

nn

n

i

ni

n

n

i

in

n

i

in

i

ni

x xn

nsn

x x x xn

nsn

x x xn

n

n

xnsn

x x xn

n

nn

xsn

x x xn

n

nn

xn xn

n

x x

x x x

n

n

n

x

n

x

n

x x

x x xn

n

n

x x

n

x

x xn

n x

n

n x

n

x

n

x x

n

x

x x

−+

+−=

+−+

+−=

−+

++

+−=

−+

++

+−=

−+

++

−++−=

−

+

+

+

−

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡−+=

−+

+⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−=

+

⎟ ⎠

⎞⎜⎝

⎛

−+

−+

+=

+−

+−

+

⎟ ⎠

⎞⎜⎝

⎛

−+=

+

+

+

+

+=

+=

++=

=+

=+

+=+

=

=+

∑

∑∑∑∑

∑∑∑

∑

∑ ∑

∑∑

∑∑∑

6-61



c) 65.811=n x inches 641 =+n x

37435.42 == nsn 2.106=ns

098.2

37

)811.6564(137

37435.4)137(

76.65137

64)81.65(37

2

1

1

=

−+

+−=

=+

+=

+

+

n

n

s

x

6-103. The trimmed mean is pulled toward the median by eliminating outliers.

a) 10% Trimmed Mean = 89.29

b) 20% Trimmed Mean = 89.19

Difference is very small

c) No, the differences are very small, due to a very large data set with no significant outliers.

6-104. If nT/100 is not an integer, calculate the two surrounding integer values and interpolate between the two.

For example, if nT/100 = 2/3, one could calculate the mean after trimming 2 and 3 observations from each

end and then interpolate between these two means.

6-62



CHAPTER 6

Section 6-1

6-1. Sample average:

mm0044.748

035.5921 ===∑

=

n

x

x

n

i

i

Sample variance:

( )

2

2

2

1

1

2

2

8

1

2

8

1

)mm(000022414.07

0001569.0

18

8

035.59218031.43813

1

03118.43813

035.592

==

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

n

n

x

x

s

x

x

n

i

in

i

i

i

i

i

i

Sample standard deviation:

mm00473.0000022414.0 ==s

The sample standard deviation could also be found using

( )s

x x

n

ii

n

=

−

−=∑

2

1

1where ( )x xi

i

− ==∑

2

1

8

0 0001569.

Dot Diagram:

.. ...: .

-------+---------+---------+---------+---------+---------diameter

73.9920 74.0000 74.0080 74.0160 74.0240 74.0320

There appears to be a possible outlier in the data set.

6-1




min359.1419

82.272

19

19

1 ===∑

=ii x

x

Sample variance:

( )

2

2

2

1

1

2

2

19

1

2

19

1

(min)47.35618

49.6416

119

19

82.2728964.10333

1

8964.10333

82.272

==

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

n

n

x

x

s

x

x

n

i

in

i

i

i

i

i

i


min88.1847.356 ==s


( )s

x x

n

ii

n

=

−

−=∑

2

1

1

where

( ) 49.641619

1

2 =−∑=i

i x x

6-2




yards x 1.706812

84817==

Sample variance:

( )

( )

2

2

2

1

1

2

2

19

1

2

12

1

265.5130211

92.564324

112

12

84817600057949

1

600057949

84817

yards

n

n

x

x

s

x

x

n

i

in

i

i

i

i

i

i

==

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=


yardss 5.226265.51302 ==


( )

1

1

2

−

−

=∑

=

n

x x

s

n

i

i

where

( ) 92.56432412

1

2 =−∑=i

i x x

Dot Diagram: (rounding was used to create the dot diagram)

. . : .. .. : :

-+---------+---------+---------+---------+---------+-----C1

6750 6900 7050 7200 7350 7500

6-3



6-4. Sample mean:

kN22.12618

2272

18

18

1 ===∑

=ii x

x

Sample variance:

( )

2

2

2

1

1

2

2

18

1

2

18

1

)kN(24.68317

11.11615

118

18

2272298392

1

298392

2272

==

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

n

n

x

x

s

x

x

n

i

in

i

i

i

i

i

i


kN14.2624.683 ==s


( )

1

1

2

−

−

=∑

=

n

x x

s

n

i

i

where

( ) 11.1161518

1

2 =−∑=i

i x x

Dot Diagram:

.

: :: . :: . . : . .

+---------+---------+---------+---------+---------+-------yield

90 105 120 135 150 165

6-4




975.438

8.351== x

Sample variance:

( )

143.1517

998.1057

18

8

8.351043.16528

1

403.16528

8.351

2

2

1

1

2

2

19

1

2

8

1

==

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

n

n

x

x

s

x

x

n

i

in

i

i

i

i

i

i


294.12143.151 ==s


( )

1

1

2

−

−

=∑

=

n

x x

s

n

i

i

where

( ) 998.10578

1

2=−∑

=ii x x

Dot Diagram:

. . .. . .. .

+---------+---------+---------+---------+---------+-------

24.0 32.0 40.0 48.0 56.0 64.0

6-5




2

35

1 / 514.81035

28368

35mwatts

x

x i

i

===∑

=

Sample variance:

( )

22

2

2

1

1

2

2

19

1

2

19

1

) / (61.16465

34

743.559830

135

35

2836823552500

1

23552500

28368

mwatts

n

n

x

x

s

x

x

n

i

in

i

i

i

i

i

i

=

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

Sample standard deviation:2 / 32.12861.16465 mwattss ==


( )

1

1

2

−

−

=∑

=

n

x x

s

n

i

i

where

( ) 743.55983035

1

2 =−∑=i

i x x

Dot Diagram (rounding of the data is used to create the dot diagram)

x

. .

. : .: ... . . .: :. . .:. .:: :::

-----+---------+---------+---------+---------+---------+-x

500 600 700 800 900 1000

The sample mean is the point at which the data would balance if it were on a scale.

6-7. 44.512706905 ==μ ; The value 5.44 is the population mean since the actual physical population

of all flight times during the operation is available.


6-6



mmn

x

x

n

i

i

173.29

56.191 ===∑

=

Sample variance:

( )

2

2

2

1

1

2

2

9

1

2

9

1

)(4303.0

8

443.3

19

9

56.19953.45

1

953.45

56.19

mm

n

n

x

x

s

x

x

n

i

in

i

i

i

i

ii

=

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑∑

=

=

=

=


mms 6560.04303.0 ==

Dot Diagram . . . . . . . ..

-------+---------+---------+---------+---------+---------crack length

1.40 1.75 2.10 2.45 2.80 3.15

6-9. Sample average of exercise group:

89.28712

3454.681 ===∑

=

n

x

x

n

i

i

Sample variance of exercise group:

112

12

)68.3454(54.1118521

1s

1118521.54

68.3454

2

2

1

1

2

2

1

2

1

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

n

n

x

x

x

x

n

i

in

i

i

n

i

i

n

i

i

11268.5211

123953.71==

6-7



Sample standard deviation of exercise group:

15.10652.11268 ==s

Dot Diagram of exercise group:

6 h o u r s o f E x e r c i s e

4 5 04 0 03 5 03 0 02 5 02 0 01 5 0

D o t p l o t o f 6 h o u r s o f E x e r c i s e

Sample average of no exercise group:

325.0108

2600.081 ===∑

=

n

x

x

n

i

i

Sample variance of no exercise group:

18

8

)08.2600(947873.4

1s

947873.4

08.2600

2

2

1

1

2

2

1

2

1

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

=

=

∑∑

∑

∑

=

=

=

=

n

n

x

x

x

x

n

i

in

i

i

n

i

i

n

i

i

14688.777

102821.4==

Sample standard deviation of no exercise group:

20.12177.14688 ==s

Dot Diagram of no exercise group:

6-8



N o E x e r c i s e

5 0 04 5 04 0 03 5 03 0 02 5 02 0 01 5 0

D o t p l o t o f N o E x e r c i s e

6-10. Dot Diagram of CRT data in exercise 6-5 (Data were rounded for the plot)

70605040302010

Dotplot for Exp 1-Exp 2

Exp 1

Exp 2

The data are centered a lot lower in the second experiment. The lower CRT resolution reduces thevisual accommodation.

6-11. 184.78

47.571 ===∑

=

n

x

x

n

i

i

( )

000427.07

00299.0

18

8

47.57853.412

1

2

2

1

1

2

2 ==−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

∑∑ =

=

n

n

x

x

s

n

i

in

i

i

02066.0000427.0 ==s

Examples: repeatability of the test equipment, time lag between samples, during which the pH of

the solution could change, and operator skill in drawing the sample or using the instrument.

6-12. sample mean 11.839

0.7481 ===∑

=

n

x

x

n

i

i

drag counts

6-9



sample variance

( )

2

2

2

1

1

2

2

countsdrag61.508

89.404

19

9

0.74862572

1

==

−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

∑∑ =

=

n

n

x

x

s

n

i

in

i

i

sample standard deviation 11.761.50 ==s drag counts

Dot Diagram

. . . . . . . . .

---+---------+---------+---------+---------+---------+---drag counts

75.0 80.0 85.0 90.0 95.0 100.0

6-13. a) 86.65= x °F

°F 16.12=s

Dot Diagram: :

. . . . .. .: .: . .:..: .. :: .... ..

-+---------+---------+---------+---------+---------+-----temp

30 40 50 60 70 80

b) Removing the smallest observation (31), the sample mean and standard deviation become

86.66= x °F s = 10.74 °F



Section 6-2

6-14.

NOTE: The data set contains one additional measurement of 91.2 that should be included in the

book. Solutions use this additional measurement.

Stem-and-leaf display of octane rating N = 82

Leaf Unit = 0.10 83|4 represents 83.4

1 83|4

3 84|33

4 85|37 86|777

13 87|456789

24 88|23334556679

34 89|0233678899

(13) 90|0111344456789

36 91|00011122256688

22 92|22236777

14 93|023347

8 94|2247

4 95|

4 96|15

2 97|

2 98|8

1 99|

1 100|3

6-10



Median = 90.4, Q1 = 88.6 , and Q3 = 92.2

6-15. Stem-and-leaf display for cycles to failure: unit = 100 1|2 represents 1200

1 0T|3

1 0F|

5 0S|777710 0o|88899

22 1*|000000011111

33 1T|22222223333

(15) 1F|444445555555555

22 1S|66667777777

11 1o|888899

5 2*|011

2 2T|22

Median = 1436.5, Q1 = 1097.8, and Q3 = 1735.0

No, only 5 out of 70 coupons survived beyond 2000 cycles.

6-16. Stem-and-leaf display of percentage of cotton N = 64

Leaf Unit = 0.10 32|1 represents 32.1%

1 32|1

6 32|56789

9 33|114

17 33|56666688

24 34|0111223

(14) 34|55666667777779

26 35|001112344

17 35|56789

12 36|234

9 36|6888

5 37|13

3 37|689

Median = 34.7, Q1 = 33.800 , and Q3 = 35.575

6-17. Stem-and-leaf display for Problem 2-4.yield: unit = 1 1|2 represents 12

1 7o|8

1 8*|

7 8T|223333

21 8F|44444444555555

38 8S|66666666667777777

(11) 8o|88888999999

41 9*|00000000001111

27 9T|22233333

19 9F|444444445555

7 9S|666677

1 9o|8

Median = 89.250, Q1 = 86.100, and Q3 = 93.125

6-18. The data in the 42nd is 90.4 which is median.

The mode is the most frequently occurring data value. There are several data values that occur 3

times. These are: 86.7, 88.3, 90.1, 90.4, 91, 91.1, 91.2, 92.2, and 92.7, so this data set has a

multimodal distribution.

6-11



Sample mean: 534.9083

7514.31 ===∑=

n

x

x

n

i

i

6-19.

Sample median is at( )

2

170 += 35.5th , the median is 1436.5.

Modes are 1102, 1315, and 1750 which are the most frequent data.


982591 ===∑=

n

x

x

n

i

i

6-20.

Sample median is at( )

2

164 += 32.5th

The 32nd is 34.7 and the 33rd is 34.7, so the median is 34.7.

Mode is 34.7 which is the most frequent data.


2227.11 ===∑=

n

x

x

n

i

i

6-21.Stem-and-leaf of Billion of kilowatt hours N = 23

Leaf Unit = 100

(18) 0 000000000000000111

5 0 23

3 0 5

2 0

2 0 9

1 1

1 1

1 1

1 1 6

Sample median is at 12th = 38.43.


8.43981 ===∑=

n

x

x

n

i

i

Sample variance: s2 = 150673.8

Sample standard deviation: s = 388.2

6-12



6-22. sample mean: 65.811= x inches, standard deviation 2.106=s inches, and sample median:

66.000~ = x inches

Stem-and-leaf display of female engineering student heights N = 37

Leaf Unit = 0.10 61|0 represents 61.0 inches

1 61|0

3 62|00

5 63|00

9 64|0000

17 65|00000000

(4) 66|0000

16 67|00000000

8 68|00000

3 69|00

1 70|0

6-23. Stem-and-leaf display for Problem 6-23. Strength: unit = 1.0 1|2 represents 12

1 532|9

1 533|2 534|2

4 535|47

5 536|6

9 537|5678

20 538|12345778888

26 539|016999

37 540|11166677889

46 541|123666688

(13) 542|0011222357899

41 543|011112556

33 544|00012455678

22 545|2334457899

13 546|23569

8 547|357

5 548|11257

5479955.9595100100

5.0is percentilei

i th⇒=⇒=×−

6-24. Stem-and-leaf of concentration, N = 60, Leaf Unit = 1.0, 2|9 represents 29Note: Minitab has dropped the value to the right of the decimal to make this display.

1 2|9

2 3|1

3 3|9

8 4|22223

12 4|5689

20 5|01223444

(13) 5|5666777899999

27 6|11244

22 6|556677789

13 7|022333

7 7|6777

3 8|01

1 8|9

The data have a symmetrical bell-shaped distribution, and therefore may be normally distributed.

6-13



Sample Mean 59.8760

3592.01 ===∑=

n

x

x

n

i

i

Sample Standard Deviation

( )

50.1220.156

and

20.156

59

93.9215

160

60

0.3592224257

1

224257 and 0.3592

2

2

1

1

2

2

60

1

260

1

==

=

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

==

∑∑

∑∑

=

=

==

s

n

n

x

x

s

x x

n

i

in

i

i

i

i

i

i

Sample Median 45.59~ = x

Variable N Median

concentration 60 59.45

85.76905.549010060

5.0is percentilei

i th⇒=⇒=×−

6-14



6-25. Stem-and-leaf display for Problem 6-25. Yard: unit = 1.0Note: Minitab has dropped the value to the right of the decimal to make this display.

1 22 | 6

5 23 | 2334

8 23 | 677

16 24 | 00112444

20 24 | 5578

33 25 | 0111122334444

46 25 | 5555556677899

(15) 26 | 000011123334444

39 26 | 56677888

31 27 | 0000112222233333444

12 27 | 66788999

4 28 | 003

1 28 | 5

Sample Mean yards

x

n

x

xi

i

n

i

i

3.260

100

2.26030

100

100

11 ====∑∑==


( )

yardss

yards

n

n

x

x

s

x x

n

i

in

i

i

i

i

i

i

41.13782.179

and

782.179

99

42.17798

1100

100

2.26030(6793512

1

6793512 and 2.26030

2

2

2

1

1

2

2

100

1

2100

1

==

=

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

==

∑∑

∑∑

=

=

==

Sample Median

Variable N Median

yards 100 260.85

2.277905.9090100100

5.0is percentilei

i th⇒=⇒=×−

6-15



6-26. Stem-and-leaf of speed (in megahertz) N = 120

Leaf Unit = 1.0 63|4 represents 634 megahertz

2 63|47

7 64|24899

16 65|223566899

35 66|0000001233455788899

48 67|0022455567899

(17) 68|00001111233333458

55 69|0000112345555677889

36 70|011223444556

24 71|0057889

17 72|000012234447

5 73|59

3 74|68

1 75|

1 76|3

35/120= 29% exceed 700 megahertz.

Sample Mean mhz

x

x i

i

686.78

120

82413

120

120

1 ===∑=


( )

mhzs

mhz

n

n

x

x

s

x x

n

i

in

i

i

i

i

i

i

67.2585.658

and

85.658

119

925.78402

1120

120

8241356677591

1

56677591 and 82413

2

2

2

1

1

2

2

120

1

2120

1

==

=

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

==

∑∑

∑∑

=

=

==

Sample Median 0.683~ = x mhz

Variable N Median

speed 120 683.00

6-16



6-27 Stem-and-leaf display of Problem 6-27. Rating: unit = 0.10 1|2 represents 1.2

1 83|0

2 84|0

5 85|000

7 86|00

9 87|0012 88|000

18 89|000000

(7) 90|0000000

15 91|0000000

8 92|0000

4 93|0

3 94|0

2 95|00

Sample Mean

45.89

40

3578

40

40

11 ====∑∑== i

i

n

i

i x

n

x

x


( )

8.205.8

and

05.839

9.313

140

40

3578320366

1

320366 and 3578

2

2

1

1

2

2

40

1

240

1

==

=

=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

==

∑∑

∑∑

=

=

==

s

n

n

x

x

s

x x

n

i

in

i

i

i

i

i

i

Sample Median

Variable N Median

rating 40 90.000

22/40 or 55% of the taste testers considered this particular Pinot Noir truly exceptional.

6-17



6-28. Stem-and-leaf diagram of NbOCl3 N = 27

Leaf Unit = 100 0|4 represents 40 gram-mole/liter x 10-3

6 0|444444

7 0|5

(9) 1|001122233

11 1|5679

7 2|

7 2|5677

3 3|124

Sample mean 3

27

1 x10mole/litergram153927

41553

27

−= −===∑i

i x

x


( )

3-

2

2

1

1

2

2

27

1

227

1

10xmole/liter-gram62.95785.917030 and

85.91703026

23842802

127

27

4155387792869

1

87792869 and 41553

==

==−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

==

∑∑

∑∑

=

=

==

s

n

n

x

x

s

x x

n

i

in

i

i

i

i

i

i

Sample Median 3x10mole/litergram1256~ −−= x

Variable N Median

NbOCl3 40 1256

6-29. Stem-and-leaf display for Problem 6-29. Height: unit = 0.10 1|2 represents 1.2

Female Students| Male Students

0|61 1

00|62 3

00|63 5

0000|64 9

00000000|65 17 2 65|00

0000|66 (4) 3 66|0

00000000|67 16 7 67|0000

00000|68 8 17 68|0000000000

00|69 3 (15) 69|000000000000000

0|70 1 18 70|0000000

11 71|00000

6 72|00

4 73|00

2 74|01 75|0

The male engineering students are taller than the female engineering students. Also there is a

slightly wider range in the heights of the male students.



Section 6-3

6-18



6-30.

Solution uses the n = 83 observations from the data set.

Frequency Tabulation for Exercise 6-14.Octane Data

--------------------------------------------------------------------------------

Lower Upper Relative Cumulative Cum. Rel.

Class Limit Limit Midpoint Frequency Frequency Frequency Frequency--------------------------------------------------------------------------------

at or below 81.75 0 .0000 0 .0000

1 81.75 84.25 83.0 1 .0120 1 .0120

2 84.25 86.75 85.5 6 .0723 7 .0843

3 86.75 89.25 88.0 19 .2289 26 .3133

4 89.25 91.75 90.5 33 .3976 59 .7108

5 91.75 94.25 93.0 18 .2169 77 .9277

6 94.25 96.75 95.5 4 .0482 81 .9759

7 96.75 99.25 98.0 1 .0120 82 .9880

8 99.25 101.75 100.5 1 .0120 83 1.0000

above 101.75 0 .0000 83 1.0000

--------------------------------------------------------------------------------

Mean = 90.534 Standard Deviation = 2.888 Median = 90.400

100.598.095.593.090.588.085.583.0

30

20

10

0

octane data

F r e q u e n c y

6-19



6-31.Frequency Tabulation for Exercise 6-15.Cycles

--------------------------------------------------------------------------------


Class Limit Limit Midpoint Frequency Frequency Frequency Frequency

--------------------------------------------------------------------------------

at or below .000 0 .0000 0 .0000

1 .000 266.667 133.333 0 .0000 0 .0000

2 266.667 533.333 400.000 1 .0143 1 .0143

3 533.333 800.000 666.667 4 .0571 5 .0714

4 800.000 1066.667 933.333 11 .1571 16 .2286

5 1066.667 1333.333 1200.000 17 .2429 33 .4714

6 1333.333 1600.000 1466.667 15 .2143 48 .6857

7 1600.000 1866.667 1733.333 12 .1714 60 .8571

8 1866.667 2133.333 2000.000 8 .1143 68 .9714

9 2133.333 2400.000 2266.667 2 .0286 70 1.0000

above 2400.000 0 .0000 70 1.0000

--------------------------------------------------------------------------------


225020001750150012501000750500

15

10

5

0

number of cycles to failure

F r e q u e n c y

6-20



6-32.Frequency Tabulation for Exercise 6-16.Cotton content

--------------------------------------------------------------------------------



--------------------------------------------------------------------------------

at or below 31.0 0 .0000 0 .0000

1 31.0 32.0 31.5 0 .0000 0 .0000

2 32.0 33.0 32.5 6 .0938 6 .0938

3 33.0 34.0 33.5 11 .1719 17 .2656

4 34.0 35.0 34.5 21 .3281 38 .5938

5 35.0 36.0 35.5 14 .2188 52 .8125

6 36.0 37.0 36.5 7 .1094 59 .9219

7 37.0 38.0 37.5 5 .0781 64 1.0000

8 38.0 39.0 38.5 0 .0000 64 1.0000

above 39.0 0 .0000 64 1.0000

--------------------------------------------------------------------------------


31.5 32.5 33.5 34.5 35.5 36.5 37.5 38.5

0

10

20

cotton percentage

F r e q u e n c y

6-21



6-33.Frequency Tabulation for Exercise 6-17.Yield

--------------------------------------------------------------------------------



--------------------------------------------------------------------------------

at or below 77.000 0 .0000 0 .0000

1 77.000 79.400 78.200 1 .0111 1 .0111

2 79.400 81.800 80.600 0 .0000 1 .0111

3 81.800 84.200 83.000 11 .1222 12 .1333

4 84.200 86.600 85.400 18 .2000 30 .3333

5 86.600 89.000 87.800 13 .1444 43 .4778

6 89.000 91.400 90.200 19 .2111 62 .6889

7 91.400 93.800 92.600 9 .1000 71 .7889

8 93.800 96.200 95.000 13 .1444 84 .9333

9 96.200 98.600 97.400 6 .0667 90 1.0000

10 98.600 101.000 99.800 0 .0000 90 1.0000

above 101.000 0 .0000 90 1.0000

--------------------------------------------------------------------------------


6-22



6-34. Solutions uses the n = 83 observations from the data set.

Frequency Tabulation for Exercise 6-14.Octane Data

--------------------------------------------------------------------------------



--------------------------------------------------------------------------------

at or below 83.000 0 .0000 0 .0000

1 83.000 84.125 83.5625 1 .0120 1 .0120

2 84.125 85.250 84.6875 2 .0241 3 .0361

3 85.250 86.375 85.8125 1 .0120 4 .0482

4 86.375 87.500 86.9375 5 .0602 9 .1084

5 87.500 88.625 88.0625 13 .1566 22 .2651

6 88.625 89.750 89.1875 8 .0964 30 .3614

7 89.750 90.875 90.3125 16 .1928 46 .5542

8 90.875 92.000 91.4375 15 .1807 61 .7349

9 92.000 93.125 92.5625 9 .1084 70 .8434

10 93.125 94.250 93.6875 7 .0843 77 .9277

11 94.250 95.375 94.8125 2 .0241 79 .9518

12 95.375 96.500 95.9375 2 .0241 81 .9759

13 96.500 97.625 97.0625 0 .0000 81 .9759

14 97.625 98.750 98.1875 0 .0000 81 .9759

15 98.750 99.875 99.3125 1 .0120 82 .9880

16 99.875 101.000 100.4375 1 .0120 83 1.0000

above 101.000 0 .0000 83 1.0000

--------------------------------------------------------------------------------


Histogram 8 bins:

Fuel

F r e q u e n c y

10299969390878481

40

30

20

10

0

Histogram of Fuel ( 8 Bins)

Histogram 16 Bins:

Fuel

F r e q u e n c y

100.898.496.093.691.288.886.484.0

18

16

14

12

10

8

6

4

2

0

Histogram of Fuel ( 16 Bins)

6-23



Yes, both of them give the similar information.

6-35.

Histogram 8 bins:

Cycles to failur e of aluminum test coupons

F r e q u e n c y

225020001750150012501000750500

18

16

14

12

10

8

6

4

2

0

Histogram of Cycles to f ailure (8 bins)

Histogram 16 bins:

Cycles to failur e of aluminum test coupons

F r e q u e n c y

23002000170014001100800500200

14

12

10

8

6

4

2

0

Histogram of Cycles to fail ure (16 bins)

Yes, both of them give the same similar information

6-36.

Histogram 8 bins:

Percentage of cott on in in mater ial used to manufactur e men's shirt s

F r e q u e n c

y

37.636.836.035.234.433.632.832.0

20

15

10

5

0

Histogram of Percentage of cotton (8 Bins)

6-24



Histogram 16 Bins:

Percent age of cott on in material used to manufacture men's shirts

F r e q u e n c y

37.636.836.035.234.433.632.832.0

12

10

8

6

4

2

0

Histogram of Percentage of cotton (8 Bins)

Yes, both of them give the similar information.

6-37.Histogram

Energy consumption

F r e q u e n c y

40003000200010000

20

15

10

5

0

Histogram of Energy

The data is skewed.

6-25



6-38.Frequency Tabulation for Problem 6-22. Height Data

--------------------------------------------------------------------------------



--------------------------------------------------------------------------------

at or below 60.500 0 .0000 0 .0000

1 60.500 61.500 61.000 1 .0270 1 .0270

2 61.500 62.500 62.000 2 .0541 3 .0811

3 62.500 63.500 63.000 2 .0541 5 .1351

4 63.500 64.500 64.000 4 .1081 9 .2432

5 64.500 65.500 65.000 8 .2162 17 .4595

6 65.500 66.500 66.000 4 .1081 21 .5676

7 66.500 67.500 67.000 8 .2162 29 .7838

8 67.500 68.500 68.000 5 .1351 34 .9189

9 68.500 69.500 69.000 2 .0541 36 .9730

10 69.500 70.500 70.000 1 .0270 37 1.0000

above 70.500 0 .0000 37 1.0000

--------------------------------------------------------------------------------


61 62 63 64 65 66 67 68 69 70

0

1

2

3

4

5

6

7

8

height

F r e q u e n c y

6-39. The histogram for the spot weld shear strength data shows that the data appear to be normally

distributed (the same shape that appears in the stem-leaf-diagram).

5320 5340 5360 5380 5400 5420 5440 5460 5480 5500

0

10

20

shear strength

F r e q u e n c y

6-26



6-40.Frequency Tabulation for exercise 6-24. Concentration data

--------------------------------------------------------------------------------



--------------------------------------------------------------------------------

at or below 29.000 0 .0000 0 .0000

1 29.0000 37.000 33.000 2 .0333 2 .0333

2 37.0000 45.000 41.000 6 .1000 8 .1333

3 45.0000 53.000 49.000 8 .1333 16 .2667

4 53.0000 61.000 57.000 17 .2833 33 .5500

5 61.0000 69.000 65.000 13 .2167 46 .7667

6 69.0000 77.000 73.000 8 .1333 54 .9000

7 77.0000 85.000 81.000 5 .0833 59 .9833

8 85.0000 93.000 89.000 1 .0167 60 1.0000

above 93.0000 0 .0800 60 1.0000

--------------------------------------------------------------------------------


Yes, the histogram shows the same shape as the stem-and-leaf display.

6-41. Yes, the histogram of the distance data shows the same shape as the stem-and-leaf display in

exercise 6-25.

220 228 236 244 252 260 268 276 284 292

0

10

20

distance

F r e q u e n c y

6-42. Histogram for the speed data in exercise 6-26. Yes, the histogram of the speed data shows the

same shape as the stem-and-leaf display in exercise 6-26

6-27



620 670 720 770

0

10

20

speed (megahertz)

F r e q u e n c y

6-43. Yes, the histogram of the wine rating data shows the same shape as the stem-and-leaf display in

exercise 6-27.

85 90 95

0

1

2

3

4

5

6

7

rating

F r e q u e n c y

6-28



6-44.

contour

trim

holes/slots

assembly

lubrication

dents

pits

deburr

0

16.2

32.4

48.6

64.8

81

0

20

40

60

80

00

scores

Pareto Chart forAutomobile Defects

percent of total

1

37.0

63.0

72.880.2

86.491.4

96.3100.0

Roughly 63% of defects are described by parts out of contour and parts under trimmed.



Section 6-4

6-45. Descriptive StatisticsVariable N Mean Median TrMean StDev SE Mean

time 8 2.415 2.440 2.415 0.534 0.189

Variable Minimum Maximum Q1 Q3

time 1.750 3.150 1.912 2.973

a.)Sample Mean: 2.415

Sample Standard Deviation: 0.543

b.) Box Plot – There are no outliers in the data.

T i m e

3.2

3.0

2.8

2.6

2.4

2.2

2.0

1.8

1.6

Boxplot of Time

6-46. Descriptive StatisticsVariable N Mean Median Tr Mean StDev SE MeanPMC 20 4.000 4.100 4.044 0.931 0.208Variable Min Max Q1 Q3PMC 2.000 5.200 3.150 4.800

a) Sample Mean: 4

Sample Variance: 0.867


6-29



b)

P e r c e n t a g e

5.5

5.0

4.5

4.0

3.5

3.0

2.5

2.0

Boxplot of Percentage

6-47. Descriptive StatisticsVariable N Mean Median Tr Mean StDev SE MeanTemperat 9 952.44 953.00 952.44 3.09 1.03Variable Min Max Q1 Q3

Temperat 948.00 957.00 949.50 955.00

a) Sample Mean: 952.44

Sample Variance: 9.53


b) Median: 953; Any increase in the largest temperature measurement will not affect the median.c)

T e m p e r a t u

r e

957

956

955

954

953

952

951

950

949

948

Boxplot of Temperatur e

6-48 Descriptive statisticsVariable N Mean Median TrMean StDev SE Mean

drag coefficients 9 83.11 82.00 83.11 7.11 2.37

Variable Minimum Maximum Q1 Q3drag coefficients 74.00 100.00 79.50 84.50

a) Median: 00.82~= x

Upper quartile: Q1 =79.50

Lower Quartile: Q3=84.50

b)

6-30



D r a g C o e f

100

95

90

85

80

75

Boxplot of Drag Coef

c) Variable N Mean Median TrMean StDev SE Mean

drag coefficients 8 81.00 81.50 81.00 3.46 1.22


drag coefficients 74.00 85.00 79.25 83.75

D r a g C o e f 1

85.0

82.5

80.0

77.5

75.0

Boxplot of Drag Coef1

Removing the largest observation (100) lowers the mean and median. Removing this “outlier

also greatly reduces the variability as seen by the smaller standard deviation and the smaller

difference between the upper and lower quartiles.

6-49. Descriptive Statistics of O-ring joint temperature dataVariable N Mean Median TrMean StDev SE Mean

Temp 36 65.86 67.50 66.66 12.16 2.03


Temp 31.00 84.00 58.50 75.00

a) Median = 67.50

Lower Quartile: Q1=58.50

Upper Quartile: Q3=75.00

b) Data with lowest point removedVariable N Mean Median TrMean StDev SE Mean

6-31



Temp 35 66.86 68.00 67.35 10.74 1.82


Temp 40.00 84.00 60.00 75.00

The mean and median have increased and the standard deviation and difference between the upper

and lower quartile has decreased.

c.)Box Plot - The box plot indicates that there is an outlier in the data.

J o i n t T e m p

90

80

70

60

50

40

30

Boxplot of Joint Temp

6-50. This plot conveys the same basic information as the stem and leaf plot but in a different format. The outliers that were

separated from the main portion of the stem and leaf plot are shown here separated from the whiskers.

F u e l

100

95

90

85

Boxplot of Fuel

6-51 The box plot shows the same basic information as the stem and leaf plot but in a different format. The outliers that

were separated from the main portion of the stem and leaf plot are shown here separated from the whiskers.

6-32



B i l l i o n o f k i l o w a t t h o u r s

1800

1600

1400

1200

1000

800

600

400

200

0

Boxplot of Bill ion of kilow att hours

6-52 The box plot and the stem-leaf-diagram show that the data are very symmetrical about the mean. It also

shows that there are no outliers in the data.

C o n c e n t r a t i o n

90

80

70

60

50

40

30

Boxplot of Concentrat ion

6-53.This plot, as the stem and leaf one, indicates that the data fall mostly in one region and that the measurements toward

the ends of the range are more rare.

6-33



W e l d S t r e n g t h

5500

5450

5400

5350

Boxplot of Weld Strength

6-54 The box plot shows that the data are symmetrical about the mean. It also shows that there is an outlier in

the data. These are the same interpretations seen in the stem-leaf-diagram.

S p e e d

775

750

725

700

675

650

Boxplot of Speed

6-55 We can see that the two distributions seem to be centered at different values.

D a t a

MaleFemale

76

74

72

70

68

66

64

62

60

Boxplot of Female, Male

6-56 The box plot shows that there is a difference between the two formulations. Formulation 2 has a

higher mean cold start ignition time and a larger variability in the values of the start times. The

6-34



first formulation has a lower mean cold start ignition time and is more consistent. Care should be

taken, though since these box plots for formula 1 and formula 2 are made using 8 and 10 data

points respectively. More data should be collected on each formulation to get a better

determination.

D a t a

Formula 2Formula 1

3.6

3.2

2.8

2.4

2.0

Boxplot of Formula 1, Formula 2

6.57. All distributions are centered at about the same value, but have different variances.

D a t a

Control_2Control_1ControlHigh Dose

3000

2500

2000

1500

1000

500

0

Boxplot of High Dose, Contr ol, Contr ol_1 , Contr ol_2

6-35



Section 6-5

6-58. Stem-leaf-plot of viscosity N = 40

Leaf Unit = 0.10

2 42 89

12 43 0000112223

16 43 556616 44

16 44

16 45

16 45

16 46

16 46

17 47 2

(4) 47 5999

19 48 000001113334

7 48 5666689

The stem-leaf-plot shows that there are two “different” sets of data. One set of data is centered about 43 and

the second set is centered about 48. The time series plot shows that the data starts out at the higher level and

then drops down to the lower viscosity level at point 24. Each plot gives us a different set of information.

If the specifications on the product viscosity are 48.0±2, then there is a problem with the process

performance after data point 24. An investigation needs to take place to find out why the location of the

process has dropped from around 48.0 to 43.0. The most recent product is not within specification limits.

Index

V i s c o s

i t y

403632282420161284

49

48

47

46

45

44

43

42

Time Series Plot of Viscosity

6-59. Stem-and-leaf display for Force: unit = 1 1|2 represents 12

3 17|558

6 18|357

14 19|00445589

18 20|1399

(5) 21|00238

17 22|005

14 23|5678

10 24|1555899

3 25|158

6-36



Index

P u

l l - o f f F o

r c e

403632282420161284

260

250

240

230

220

210

200

190

180

170

Time Seri es Plot of Pull-off Force

In the time series plot there appears to be a downward trend beginning after time 30. The stem and leaf

plot does not reveal this.

6-60. Stem-and-leaf of Chem Concentration N = 50 Leaf Unit = 0.10

1 16 1

2 16 3

3 16 5

5 16 67

9 16 8899

18 17 000011111

25 17 2233333

25 17 44444444444445555

8 17 6667

4 17 888

1 18 1

Index

C h e m C o n c e n t r a t i o n

50454035302520151051

18.0

17.5

17.0

16.5

16.0

Time Seri es Plot of Chem Concentr ati on

In the time series plot there appears to be fluctuating. The stem and leaf plot does not reveal this.

6-37



6-61. Stem-and-leaf of Wolfer sunspot N = 100 Leaf Unit = 1.0

17 0 01224445677777888

29 1 001234456667

39 2 0113344488

50 3 00145567789

50 4 011234567788

38 5 0457933 6 0223466778

23 7 147

20 8 2356

16 9 024668

10 10 13

8 11 8

7 12 245

4 13 128

1 14

1 15 4

The data appears to decrease between 1790 and 1835, the stem and leaf plot indicates skewed data.

Index

W o l f e r s u n s p o t

1009080706050403020101

160

140

120

100

80

60

40

20

0

Time Seri es Plot of Wol fer sunspot

6-62. Time Series Plot

Index

M i l e s

F l o w n

80726456484032241681

16

14

12

10

8

6

Time Seri es Plot of Miles Flown

6-38



Each year the miles flown peaks during the summer hours. The number of miles flown increased over the

years 1964 to 1970.

Stem-and-leaf of Miles Flown N = 84Leaf Unit = 0.10

1 6 710 7 246678889

22 8 013334677889

33 9 01223466899

(18) 10 022334456667888889

33 11 012345566

24 12 11222345779

13 13 1245678

6 14 0179

2 15 1

1 16 2

When grouped together, the yearly cycles in the data are not seen. The data in the stem-leaf-

diagram appear to be nearly normally distributed.

6-63. Stem-and-leaf of Number of Earthquakes N = 105 Leaf Unit = 1.0

2 0 67

6 0 8888

11 1 00011

19 1 22333333

35 1 4444455555555555

45 1 6666666777

(12) 1 888888889999

48 2 00001111111

37 2 222222223333

25 2 44455

20 2 66667777

12 2 89

10 3 018 3 22

6 3 45

4 3 66

2 3 9

1 4 1

Time Series Plot

Year

N u m b e r o f E a r t h q u a k e s

20902080207020602050204020302020201020001990

45

40

35

30

25

20

15

10

5

Time Series Plot of Number of Eart hquakes

6-39



6-64. Stem-and-leaf of Petroleum Imports N = 32 Leaf Unit = 100

5 5 00149

11 6 012269

15 7 3468

(8) 8 00346889

9 9 4

8 10 1785 11 458

2 12 2

1 13 1

Stem-and-leaf of Total Petroleum Imports as Perc N = 32

Leaf Unit = 10

4 3 2334

9 3 66778

14 4 00124

(7) 4 5566779

11 5 0014

7 5 5688

3 6 013

Stem-and-leaf of Petroleum Imports from Persian N = 32

Leaf Unit = 10

1 0 6

3 0 89

3 1

5 1 33

6 1 4

11 1 66777

16 1 89999

16 2 000011

10 2 2233

6 2 44452 2 67

Time Series plot:

Year

D a

t a

2 0 0 2

1 9 9 9

1 9 9 6

1 9 9 3

1 9 9 0

1 9 8 7

1 9 8 4

1 9 8 1

1 9 7 8

1 9 7 5

14000

12000

10000

8000

6000

4000

2000

0

Variable

Petroleum Imports from Persian

Petroleum Imports

Total Petroleum Imports as Perc

Time Series Plot of Petroleum I m, Total Petrol, Petr oleum Im

6-40



Section 6-6

6-65.

73.99 74.00 74.01 74.02

1

5

10

20

30

40

5060

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for 6-1

Piston Ring DiameterML Estimates - 95% CI

Mean

StDev

74.0044

0.0043570

ML Estimates

The pattern of the data indicates that the sample may not come from a normally distributed population or that the

largest observation is an outlier. Note the slight bending downward of the sample data at both ends of the graph.

6-66.

-40 -20 0 20 40 60

1

5

10

20

30

40

5060

70

80

90

95

99

Data

P e r c e n t

Exercise 6-2 Data

Normal Probability Plot for time

Mean

StDev

14.3589

18.3769

ML Estimates

It appears that the data do not come from a normal distribution. Very few of the data points fall on the line.

6-41



6-67.

There is no evidence to doubt that data are normally distributed

0 10 20 30 40 50 60 70 80

1

5

10

20

30

40

50

60

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for 6-5

Visual Accomodation DataML Estimates - 95% CI

Mean

StDev

43.975

11.5000

ML Estimates

6-68. The normal probability plot shown below does not seem reasonable for normality.

Solar int ense

P e r c e n t

120011001000900800700600500400

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.021

810.5

StDev 128.3

N 35

A D 0.888

P-Value

Probabilit y Plot of Solar int enseNormal - 95% CI

6-42



6-69.

30 40 50 60 70 80 90 100

1

5

10

20

30

40

50

60

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for temperatureData from exercise 6-13

Mean

StDev

65.8611

11.9888

ML Estimates

The data appear to be normally distributed. Although, there are some departures from the line at the ends

of the distribution.

6-70.

80 90 100

1

5

10

20304050607080

90

95

99

Data

P e r c e n t

Normal Probability Plot for 6-14Octane Rating

ML Estimates - 95% CI

Mean

StDev

90.5256

2.88743

ML Estimates

There are a few points outside the confidence limits, indicating that the sample is not perfectly normal. These

deviations seem to be fairly small though.

6-43



6-71.

0 1000 2000 3000

1

5

10

20304050607080

90

95

99

Data

P e r c e n t

Normal Probability Plot for cycles to failure

Data from exercise 6-15

Mean

StDev

1282.78

539.634

ML Estimates

The data appear to be normally distributed. Although, there are some departures from the line at the ends

of the distribution.

6-72.

9585756555453525

99

95

90

80

7060

50

40

30

20

10

5

1

Data

P e r c e n t

Data from exercise 6-24

Normal Probability Plot for concentration

Mean

StDev

59.8667

12.3932

ML Estimates

The data appear to be normally distributed. Nearly all of the data points fall very close to, or on the line.

6-44



6-73.

Students

Female

Students

Male

75706560

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for 6-22...6-29ML Estimates - 95% CI

Both populations seem to be normally distributed, moreover, the lines seem to be roughly parallel

indicating that the populations may have the same variance and differ only in the value of their mean.

6-74. Yes, it is possible to obtain an estimate of the mean from the 50th percentile value of the normal probability

plot. The fiftieth percentile point is the point at which the sample mean should equal the population mean

and 50% of the data would be above the value and 50% below. An estimate of the standard deviation would

be to subtract the 50th percentile from the 84th percentile These values are based on the values from the z-

table that could be used to estimate the standard deviation.




6-75. Based on the digidot plot and time series plots of these data, in each year the temperature has a

similar distribution. In each year, the temperature will increase until the mid year and then it starts

to decrease.

6-45



Data

16.215.615.014.413.813.212.6

2000

2001

2002

2003

2004

Dotplot of 2000, 2001, 200 2, 2003, 2004

Index

D a t a

121110987654321

16

15

14

13

12

Variable

2002

2003

2004

2000

2001

Time Series Plot of 200 0, 2001 , 2002, 20 03, 200 4

Stem-and-leaf of C7 N = 60

Leaf Unit = 0.10

3 12 244

15 12 555666677778

23 13 12334444

25 13 55

(7) 14 222223428 14 556

25 15 112222444

16 15 5899

12 16 000000111222

Time-series plot by month over 5 years

6-46



Index

C 7

60544842363024181261

16

15

14

13

12

Time Series Plot of C7


6-76. a) Sample Mean = 65.083

The sample mean value is close enough to the target value to accept the solution as

conforming. There is a slight difference due to inherent variability.

b) s2

= 1.86869 s = 1.367

c) A major source of variability might be variability in the reagent material. Furthermore, if the

same setup is used for all measurements it is not expected to affect the variability. However, if each measurement uses a different setup, then setup differences could also be a major source of

variability.

A low variance is desirable because it indicates consistency from measurement to measurement. This

implies the measurement error has low variability.

6-77.

The unemployment rate drops steadily from 1997 to 2000 and then increases again quite quickly.

It reaches it’s peak in the middle of 2003 and then begins to drop steadily again.

6-47



Index

C 1 3

117104917865523926131

6.5

6.0

5.5

5.0

4.5

4.0

Time Series Plot of C13

6-78. a) 6001,62433,10

26

1

6

1

2 ==⎟ ⎠

⎞⎜⎝

⎛ = ∑∑

==

n x x

i

i

i

i

Ω=Ω=

Ω=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

∑∑ =

=

46.49.19

9.1916

6

001,62433,10

1

2

2

26

16

1

2

2

s

n

n

x

x

s

i

i

i

i

b) 61521353

2

6

1

6

1

2 ==⎟ ⎠ ⎞⎜

⎝ ⎛ = ∑∑

==

n x x

i

i

i

i

6-48



Ω=Ω=

Ω=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

∑∑ =

=

46.49.19

9.1916

6

521,1353

1

2

2

26

16

1

2

2

s

n

n

x

x

s

i

i

i

i

Shifting the data from the sample by a constant amount has no effect on the sample variance or standard

deviation.

c) 620010061043300

26

1

6

1

2 ==⎟ ⎠

⎞⎜⎝

⎛ = ∑∑

==

n x x

i

i

i

i

Ω=Ω=

Ω=−

−=

−

⎟ ⎠

⎞⎜⎝

⎛

−

=

∑∑ =

=

61.441990

199016

6

62001001043300

1

2

2

26

16

1

2

2

s

n

n

x

x

s

i

i

i

i

Yes, the rescaling is by a factor of 10. Therefore, s2 and s would be rescaled by multiplying s2 by 102

(resulting in 1990Ω2) and s by 10 (44.6Ω).

6-79 a) Sample 1 Range = 4

Sample 2 Range = 4

Yes, the two appear to exhibit the same variability

b) Sample 1 s = 1.604

Sample 2 s = 1.852

No, sample 2 has a larger standard deviation.

c) The sample range is a relatively crude measure of the sample variability as compared to the sample

standard deviation since the standard deviation uses the information from every data point in the sample

whereas the range uses the information contained in only two data points - the minimum and maximum.

6-80 a.) It appears that the data may shift up and then down over the 80 points.

6-49



10 20 30 40 50 60 70 80

13

14

15

16

17

Index

v i s c o s i t y

b.)It appears that the mean of the second set of 40 data points may be slightly higher than the first set of 40.

c.) Descriptive Statistics: viscosity 1, viscosity 2

Variable N Mean Median TrMean StDev SE Mean

Viscosity1 40 14.875 14.900 14.875 0.948 0.150

Viscosity2 40 14.923 14.850 14.914 1.023 0.162

There is a slight difference in the mean levels and the standard deviations.

6-81.

From the stem-and-leaf diagram, the distribution looks like the uniform distribution. From the

time series plot, there is an increasing trend in energy consumption.

Stem-and-leaf of Energy N = 21

Leaf Unit = 100

4 2 0011

7 2 233

9 2 45

10 2 7

(2) 2 89

9 3 01

7 3 22

5 3 445

2 3 66

6-50



Index

E n e r g y

2018161412108642

3800

3600

3400

3200

3000

2800

2600

2400

2200

2000

Time Series Plot of Energy

6-82

6-76 Second half6-76 First half

17

16

15

14

13

Comparative boxplots for viscosity data

Viscosity

Both sets of data appear to have the same mean although the first half seem to be concentrated a little more tightly.

Two data points appear as outliers in the first half of the data.

6-83

908070605040302010

15

10

5

0

Index

s a l e s

6-51



There appears to be a cyclic variation in the data with the high value of the cycle generally increasing. The

high values are during the winter holiday months.

b) We might draw another cycle, with the peak similar to the last year’s data (1969) at about 12.7 thousand

bottles.

6-84.

Descriptive StatisticsVariable N Mean Median Tr Mean StDev SE Meantemperat 24 48.125 49.000 48.182 2.692 0.549Variable Min Max Q1 Q3temperat 43.000 52.000 46.000 50.000

a) Sample Mean: 48.125

Sample Median: 49

b) Sample Variance: 7.246


c)

52

51

50

49

48

47

46

45

44

43

t e m p e r a t u r

The data appear to be slightly skewed.

6-85 a)Stem-and-leaf display for Problem 2-35: unit = 1 1|2 represents 12

1 0T|3

8 0F|4444555

18 0S|6666777777

(7) 0o|8888999

15 1*|111

12 1T|22233

7 1F|45

5 1S|77

3 1o|899

b) Sample Average = 9.325

Sample Standard Deviation = 4.4858

6-52



c)

40302010

20

15

10

5

Index

s p r i n g s

The time series plot indicates there was an increase in the average number of nonconforming springs

made during the 40 days. In particular, the increase occurs during the last 10 days.

6-86. a.) Stem-and-leaf of errors N = 20

Leaf Unit = 0.10

3 0 000

10 1 0000000

10 2 0000

6 3 00000

1 4 0

b.) Sample Average = 1.700

Sample Standard Deviation = 1.174

c.)

5 10 15 20

0

1

2

3

4

Index

e r r o r s

The time series plot indicates a slight decrease in the number of errors for strings 16 - 20.

6-53



6-87. The golf course yardage data appear to be skewed. Also, there is an outlying data point above 7500 yards.

6800

6900

7000

7100

7200

7300

7400

7500

y a r d a g e

6-88

half

6-76 First

half

6-76 Second

18171615141312

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for 6-76 First h...6-76 SecondML Estimates - 95% CI

Both sets of data appear to be normally distributed and with roughly the same mean value. The difference in slopes for

the two lines indicates that a change in variance might have occurred. This could have been the result of a change in

processing conditions, the quality of the raw material or some other factor.

6-54



6-89

40 45 50 55

1

5

10

20

30

40

5060

70

80

9095

99

Data

P e r c e n t

Normal Probability Plot for Temperature

Mean

StDev

48.125

2.63490

ML Estimates

There appears to be no evidence that the data are not normally distributed. There are some repeat points in

the data that cause some points to fall off the line.

6-90

6-44

6-56

4.53.52.51.50.5

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for 6-44...6-56ML Estimates - 95% CI

Although we do not have sufficient data points to really see a pattern, there seem to be no significant deviations from

normality for either sample. The large difference in slopes indicates that the variances of the populations are very

different.

6-55



285

275

265

255

245

235

225

D i s t a n c e i n y a r d s

6-91.

The plot indicates that most balls will fall somewhere in the 250-275 range. In general, the population is grouped more

toward the high end of the region. This same type of information could have been obtained from the stem and leaf

graph of problem 6-25.

6-92. a)

-3000 -2000 -1000 0 1000 2000 3000 4000 5000

1

5

10

20

30

40

50

60

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for No. of Cycles

Mean

StDev

1051.88

1146.43

ML Estimates

The data do not appear to be normally distributed. There is a curve in the line.

b)

6-56



1 2 3 4

1

5

10

20

30

40

5060

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for y*

Mean

StDev

2.75410

0.499916

ML Estimates

After the transformation y*=log(y), the normal probability plot shows no evidence that the data are not normallydistributed.

Trial 1 Trial 2 Trial 3 Trial 4 Trial 5

600

700

800

900

1000

1100

6-93

- There is a difference in the variability of the measurements in the trials. Trial 1 has the most variability

in the measurements. Trial 3 has a small amount of variability in the main group of measurements, but

there are four outliers. Trial 5 appears to have the least variability without any outliers.

- All of the trials except Trial 1 appear to be centered around 850. Trial 1 has a higher mean value

- All five trials appear to have measurements that are greater than the “true” value of 734.5.

- The difference in the measurements in Trial 1 may indicate a “start-up” effect in the data.

There could be some bias in the measurements that is centering the data above the “true” value.

6-57



6-94. a) Descriptive Statistics Variable N N* Mean SE Mean StDev Variance

Density 29 0 5.4197 0.0629 0.3389 0.1148

Variable Minimum Q1 Median Q3 MaximumDensity 4.0700 5.2950 5.4600 5.6150 5.8600

C8

P e r c e n t

6.56.05.55.04.54.0

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

<0.005

5.420

StDev 0.3389

N 2

AD 1.355

P-Value

Probabili ty Plot of C8Normal - 95% CI

9

b) There does appear to be a low outlier in the data.

c) Due to the very low data point at 4.07, the mean may be lower than it should be. Therefore, the

median would be a better estimate of the density of the earth. The median is not affected by a few

outliers.




8-95 a.) α χ λ χ α α −=<<−

1)2( 2

2,2

2

2,2

1 r r r T P

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

<<=

−

r r T T P

r r

22

2

2,2

2

2,2

1α α χ

λ

χ

Then a confidence interval for λ

μ 1

= is ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

2

2,2

1

2

2,2

2,

2

r r

r r T T

α α χ χ

b) n = 20 , r = 10 , and the observed value of Tr is 199 + 10(29) = 489.

A 95% confidence interval for λ

1is )98.101,62.28(

59.9

)489(2,

17.34

)489(2=⎟

⎠

⎞⎜⎝

⎛

8-96 dxe

z

dxe

z

x x

2

21

2

2

1

2

11

2

11

−−∫∞−

−=∫∞

=π π

α α

α

Therefore, )(111 α α zΦ=− .

To minimize L we need to minimize subject to)1()1( 21

1 α α −Φ+−Φ −α α α =+ 21

.

Therefore, we need to minimize .)1()1( 11

1 α α α +−Φ+−Φ −

2

2

1

2

2

1

2)1(

2)1(

11

1

1

1

1

α α

α

π α α ∂α ∂

π α ∂α

∂

−

=+−Φ

−=−Φ

−

−

z

z

e

e

Upon setting the sum of the two derivatives equal to zero, we obtain 2

2

12

2

1 α α α z z

ee =−

. This is solved

by . Consequently, andz zα α α1= − 1

α α α α α =−= 111 2,221α α α == .

8.97 a) n=1/2+(1.9/.1)(9.4877/4)

n=46

b) (10-.5)/(9.4877/4)=(1+p)/(1-p)

p=0.6004 between 10.19 and 10.41.

8-98 a)

n

i

n

i

i

allX P

allX P

X P

)2/1()~(

)2/1()~(

2/1)~(

=≥

=≤

=≤

μ

μ

μ

8-39



1

2

1

2

12

2

1

2

1

)()()()(

−

⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ +⎟ ⎠

⎞⎜⎝

⎛ =

∩−+=∪nnnn

B AP BP AP B AP

n

ii

X X P B AP

⎟ ⎠

⎞

⎜⎝

⎛ −=<<=∪−2

11))max(~)(min()(1 μ

b) α μ −=<< 1))max(~)(min( ii X X P

The confidence interval is min( X i), max( X i)

8-99 We would expect that 950 of the confidence intervals would include the value of μ. This is due to

9the definition of a confidence interval.

Let X bet the number of intervals that contain the true mean (μ). We can use the large sample

approximation to determine the probability that P(930 < X < 970).

Let 950.01000

950== p 930.0

1000

9301 == p and 970.0

1000

9702 == p

The variance is estimated by1000

)050.0(950.0)1(=

−n

p p

9963.0)90.2()90.2(006892.0

02.0

006892.0

02.0

1000

)050.0(950.0

)950.0930.0(

1000

)050.0(950.0

)950.0970.0()970.0930.0(

=−<−<=⎟ ⎠

⎞⎜⎝

⎛ −<−⎟

⎠

⎞⎜⎝

⎛ <=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

<−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

<=<<

Z P Z P Z P Z P

Z P Z P pP

8-40



CHAPTER 8

Section 8-2

8-1 a) The confidence level for n xn x /14.2/14.2 σ μ σ +≤≤− is determined by the

value of z0 which is 2.14. From Table III, Φ(2.14) = P(Z<2.14) = 0.9838 and the

confidence level is 2(0.9838-0.5) = 96.76%.

b) The confidence level for n xn x /49.2/49.2 σ μ σ +≤≤− is determined by the

by the value of z0 which is 2.14. From Table III, Φ(2.49) = P(Z<2.49) = 0.9936 and the

confidence level is is 2(0.9936-0.5) = 98.72%.

c) The confidence level for n xn x /85.1/85.1 σ μ σ +≤≤− is determined by the

by the value of z0 which is 2.14. From Table III, Φ(1.85) = P(Z<1.85) = 0.9678 and the

confidence level is 93.56%.

8-2 a) A zα = 2.33 would give result in a 98% two-sided confidence interval.

b) A zα = 1.29 would give result in a 80% two-sided confidence interval.

c) A zα = 1.15 would give result in a 75% two-sided confidence interval.

8-3 a) A zα = 1.29 would give result in a 90% one-sided confidence interval.

b) A zα = 1.65 would give result in a 95% one-sided confidence interval.

c) A zα = 2.33 would give result in a 99% one-sided confidence interval.

8-4 a) 95% CI for 96.1,100020,10 , ==== z xn σ μ

4.10126.987

)10/20(96.11000)10/20(96.11000

//

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

b) .95% CI for 96.1,100020,25 , ====z xn

σ μ

8.10072.992

)25/20(96.11000)25/20(96.11000

//

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

c) 99% CI for 58.2,100020,10 , ==== z xn σ μ

3.10167.983

)10/20(58.21000)10/20(58.21000

//

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

d) 99% CI for 58.2,100020,25 , ==== z xn σ μ

3.10107.989

)25/20(58.21000)25/20(58.21000

//

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

e) When n is larger, the CI will get narrower. The higher the confidence level, the wider the CI.

8-1



8-5. a) Find n for the length of the 95% CI to be 40. Za/2 = 1.96

84.320

2.39

202.39

20/)20)(96.1(length1/2

2

=⎟ ⎠

⎞⎜⎝

⎛ =

=

==

n

n

n

Therefore, n = 4. b) Find n for the length of the 99% CI to be 40. Za/2 = 2.58

66.620

6.51

206.51

20/)20)(58.2(length1/2

2

=⎟ ⎠

⎞⎜⎝

⎛ =

=

==

n

n

n

Therefore, n = 7.

8-6 Interval (1): 7.32159.3124 ≤≤ μ and Interval (2): 1.32305.3110 ≤≤ μ

Interval (1): half-length =90.8/2=45.4 and Interval (2): half-length =119.6/2=59.8

a) 3.31704.459.31241 =+= x

3.31708.595.31102 =+= x The sample means are the same.

b) Interval (1): 7.32159.3124 ≤≤ μ was calculated with 95% Confidence because it has a

smaller half-length, and therefore a smaller confidence interval. The 99% confidence level will

make the interval larger.

8-7 a) The 99% CI on the mean calcium concentration would be longer.

b) No, that is not the correct interpretation of a confidence interval. The probability that μ is

between 0.49 and 0.82 is either 0 or 1.

c) Yes, this is the correct interpretation of a confidence interval. The upper and lower limits of the

confidence limits are random variables.

8-8 95% Two-sided CI on the breaking strength of yarn: where ⎯ x = 98, σ = 2 , n=9 and z0.025 = 1.96

3.997.96

9/)2(96.1989/)2(96.198

// 025.0025.0

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

8-9 95% Two-sided CI on the true mean yield: where ⎯ x = 90.480, σ = 3 , n=5 and z0.025 = 1.96

11.9385.87

5/)3(96.1480.905/)3(96.1480.90

// 025.0025.0

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

8-10 99% Two-sided CI on the diameter cable harness holes: where ⎯ x =1.5045 , σ = 0.01 , n=10 and

z0.005 = 2.58

5127.14963.1

10/)01.0(58.25045.110/)01.0(58.25045.1

// 005.0005.0

≤≤

+≤≤−

+≤≤−

μ

μ

σ μ σ n z xn z x

8-2



8-11 a) 99% Two-sided CI on the true mean piston ring diameter

For α = 0.01, zα/2 = z0.005 = 2.58 , and ⎯ x = 74.036, σ = 0.001, n=15

x zn

x zn

−⎛

⎝ ⎜

⎞

⎠⎟ ≤ ≤ +

⎛

⎝ ⎜

⎞

⎠⎟0 005 0 005. .

σμ

σ

74 036 2 580 001

15

74 036 2 580001

15

. ..

. ..

−⎛

⎝ ⎜

⎞

⎠⎟ ≤ ≤ +

⎛

⎝ ⎜

⎞

⎠⎟μ

74.0353 ≤ μ ≤ 74.0367

b) 99% One-sided CI on the true mean piston ring diameter

For α = 0.01, zα = z0.01 =2.33 and ⎯ x = 74.036, σ = 0.001, n=15

μ

μ σ

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

≤−

15

001.033.2036.74

01.0n

z x

74.0354≤ μ

The lower bound of the one sided confidence interval is less than the lower bound of the two-

sided confidence. This is because the Type I probability of 99% one sided confidence interval (or

α = 0.01) in the left tail (or in the lower bound) is greater than Type I probability of 99% two-

sided confidence interval (or α/2 = 0.005) in the left tail.

8-12 a) 95% Two-sided CI on the true mean life of a 75-watt light bulb

For α = 0.05, zα/2 = z0.025 = 1.96 , and ⎯ x = 1014, σ =25 , n=20

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n z x

n z x

σ μ

σ

025.0025.0

10251003

20

2596.11014

20

2596.11014

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

b) 95% One-sided CI on the true mean piston ring diameter

For α = 0.05, zα = z0.05 =1.65 and ⎯ x = 1014, σ =25 , n=20

μ

μ

μ σ

≤

≤⎟ ⎠

⎞⎜⎝

⎛ −

≤−

1005

20

2565.11014

05.0n

z x

The lower bound of the one sided confidence interval is lower than the lower bound of the two-

sided confidence interval even though the level of significance is the same. This is because all of

the Type I probability (or α) is in the left tail (or in the lower bound).

8-3



8-13 a) 95% two sided CI on the mean compressive strength

zα/2 = z0.025 = 1.96, and ⎯ x = 3250, σ2 = 1000, n=12

x zn

x zn

−⎛

⎝ ⎜

⎞

⎠⎟ ≤ ≤ +

⎛

⎝ ⎜

⎞

⎠⎟0 025 0 025. .

σμ

σ

89.32673232.1112

62.3196.13250

12

62.3196.13250

≤≤

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

μ

μ

b) 99% Two-sided CI on the true mean compressive strength

zα/2 = z0.005 = 2.58

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

n z x

n z x

σ μ

σ

005.0005.0

6.32733226.412

62.3158.23250

12

62.3158.23250

≤≤

⎟

⎠

⎞⎜

⎝

⎛ +≤≤⎟

⎠

⎞⎜

⎝

⎛ −

μ

μ

The 99% CI is wider than the 95% CI

8-14 95% Confident that the error of estimating the true mean life of a 75-watt light bulb is less than 5

hours.

For α = 0.05, zα/2 = z0.025 = 1.96 , and ⎯σ =25 , E=5

04.965

)25(96.122

2/ =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =

E

zn

aσ

Always round up to the next number, therefore n = 97

8-15 Set the width to 6 hours with σ = 25, z0.025 = 1.96 solve for n.

78.2663

49

349

3/)25)(96.1(width1/2

2

=⎟ ⎠

⎞⎜⎝

⎛ =

=

==

n

n

n

Therefore, n = 267.

8-16 99% Confident that the error of estimating the true compressive strength is less than 15 psi

For α = 0.01, zα/2 = z0.005 = 2.58 , and ⎯σ =31.62 , E=15

306.2915

)62.31(58.222

2/ ≅=⎟ ⎠

⎞

⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =

E

zn

aσ

Therefore, n=30

8-4



8-17 To decrease the length of the CI by one half, the sample size must be increased by 4 times (22).

ln z 5.0/2/ =σ α

Now, to decrease by half, divide both sides by 2.

4/)2/(

4/)2/(

2/)2/(2/)/(

2

2/

2/

2/

ln z

ln z

ln z

==

=

σ

σ

σ

α

α

α

Therefore, the sample size must be increased by 22.

8-18 If n is doubled in Eq 8-7: ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n z x

n z x

σ μ

σ

α α 2/2/

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ===

n

z

n

z

n

z

n

z σ σ σ σ α α α α 2/2/2/2/

414.1

1

414.1414.12

The interval is reduced by 0.293 29.3%

If n is increased by a factor of 4 Eq 8-7:

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ===

n

z

n

z

n

z

n

z σ σ σ σ α α α α 2/2/2/2/

2

1

224

The interval is reduced by 0.5 or ½.

8-19 a) 99% two sided CI on the mean temperature

zα/2 = z0.005 = 2.57, and x = 13.77, σ = 0.5, n=11

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n z x

n z x

σ μ

σ

005.0005.0

157.1413.383

11

5.057.277.13

11

5.057.277.13

≤≤

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

μ

μ

b) 95% lower-confidence bound on the mean temperature

For α = 0.05, zα = z0.05 =1.65 and x = 13.77, σ = 0.5, n =11

μ

μ

μ σ

≤

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

≤−

521.13

11

5.065.177.13

05.0n

z x

c) 95% confidence that the error of estimating the mean temperature for wheat grown is

less than 2 degrees Celsius.

For α = 0.05, zα/2 = z0.025 = 1.96, and σ = 0.5, E = 2

2401.02

)5.0(96.122

2/ =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =

E

zn

aσ

Always round up to the next number, therefore n = 1.

8-5



d) Set the width to 1.5 degrees Celsius with σ = 0.5, z0.025 = 1.96 solve for n.

707.175.0

98.0

75.098.0

75.0/)5.0)(96.1(width1/2

2

=⎟ ⎠

⎞⎜⎝

⎛ =

=

==

n

n

n

Therefore, n = 2.



Section 8-3

8-20 131.215,025.0 =t 812.110,05.0 =t 325.120,10.0 =t

787.225,005.0 =t 385.330,001.0 =t

8-21 a) b)179.212,025.0 =t 064.224,025.0 =t c) 012.313,005.0 =t

d) 073.415,0005.0 =t

8-22 a) b)761.114,05.0 =t 539.219,01.0 =t c) 467.324,001.0 =t

8-23 95% confidence interval on mean tire life

94.36457.139,6016 === s xn 131.215,025.0 =t

07.6208233.58197

16

94.3645131.27.60139

16

94.3645131.27.60139

15,025.015,025.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-24 99% lower confidence bound on mean Izod impact strength

25.025.120 === s xn 539.219,01.0 =t

μ

μ

μ

≤

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

108.1

20

25.0539.225.1

19,01.0n

st x

8-6



8-25 x = 1.10 s = 0.015 n = 25

95% CI on the mean volume of syrup dispensed

For α = 0.05 and n = 25, tα/2,n-1 = t0.025,24 = 2.064

106.11.094

25

015.0064.210.1

25

015.0064.210.1

24,025.024,025.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-26 95% confidence interval on mean peak power

163157 === s xn 447.26,025.0 =t

798.329202.300

7315447.2315

716447.2315

6,025.06,025.0

≤≤

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ +≤≤⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-27 99% upper confidence interval on mean SBP

9.93.11814 === s xn 650.213,01.0 =t

312.125

14

9.9650.23.118

13,005.0

≤

⎟ ⎠

⎞⎜⎝

⎛ +≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤

μ

μ

μ

n

st x

8-28 90% CI on the mean frequency of a beam subjected to loads

132.25,n1.53,s231.67, 4,05.1-n/2, ===== t t xα

1.233230.2

5

53.1132.267.231

5

53.1132.267.231

4,05.04,05.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-7



By examining the normal probability plot, it appears that the data are normally distributed. There

does not appear to be enough evidence to reject the hypothesis that the frequencies are normally

distributed.

23 623 122 6

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Norm al Probabil ity Plot for frequen ciesML Est ima tes - 95% CI

Mean

StDev

231.67

1.36944

ML Est imates

8-29 The data appear to be normally distributed based on examination of the normal probability plot

below. Therefore, there is evidence to support that the annual rainfall is normally distributed.

Rainfall

P e r c e n t

800700600500400300200

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.581

485.9

StDev 90.30

N 20

A D 0.288

P-Value

Probabili ty Plot of RainfallNormal - 95% CI

95% confidence interval on mean annual rainfall

34.908.48520 === s xn 093.219,025.0 =t

080.528520.443

20

34.90093.28.48520

34.90093.28.485

19,025.019,025.0

≤≤

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ +≤≤⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-8




below. Therefore, there is evidence to support that the solar energy is normally distributed.

Solar

P e r c e n t

80757065605550

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.349

65.58

StDev 4.225

N 16

A D 0.386

P-Value

Probabili ty Plot of SolarNormal - 95% CI

95% confidence interval on mean solar energy consumed

225.458.6516 === s xn 131.215,025.0 =t

831.67329.63

16

225.4131.258.65

16

225.4131.258.65

15,025.015,025.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-31 99% confidence interval on mean current required

Assume that the data are a random sample from a normal distribution.

7.152.31710 === s xn 250.39,005.0 =t

34.33306.301

10

7.15250.32.317

10

7.15250.32.317

9,005.09,005.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-9



8-32 a) The data appear to be normally distributed based on examination of the normal probability plot

below. Therefore, there is evidence to support that the level of polyunsaturated fatty acid is normally

distributed.

16 17 18

1

5

10

20

30

40

5060

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for 8-25ML Estimates - 95% CI

b) 99% CI on the mean level of polyunsaturated fatty acid.

For α = 0.01, tα/2,n-1 = t0.005,5 = 4.032

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n

st x

n

st x 5,005.05,005.0 μ

505.17455.16

6

319.0032.498.16

6

319.0032.498.16

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

The 99% confidence for the mean polyunsaturated fat is (16.455, 17.505). There is high

confidence that the true mean is in this interval

8-33 a) The data appear to be normally distributed based on examination of the normal probability

plot below.

235022502150

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for StrengthML Estimates - 95% CI

Mean

StDev

2259.92

34.0550

ML Estimates

8-10



b) 95% two-sided confidence interval on mean comprehensive strength

35.62259.912 === s xn 201.211,025.0 =t

5.22823.2237

126.35201.29.2259

126.35201.29.2259

11,025.011,025.0

≤≤

⎟ ⎠ ⎞⎜

⎝ ⎛ +≤≤⎟

⎠ ⎞⎜

⎝ ⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

c) 95% lower-confidence bound on mean strength

μ

μ

μ

≤

≤⎟ ⎠

⎞⎜⎝

⎛ −

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

4.2241

12

6.35796.19.2259

11,05.0n

st x

8-34 a) According to the normal probability plot there does not seem to be a severe deviation from

normality for this data. This is due to the fact that the data appears to fall along a straight line.

8.308.258.208.15

99

95

90

8070

6050

40

30

20

10

5

1

Data

P e r c e n t

Norm al Probab ility Plot for 8-27ML Est imates - 95% C I

b) 95% two-sided confidence interval on mean rod diameter

For α = 0.05 and n = 15, tα/2,n-1 = t0.025,14 = 2.145

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n

st x

n

st x 14,025.014,025.0 μ

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

15

025.0145.223.8

15

025.0145.223.8 μ

8.216 ≤ μ ≤ 8.244

8-11



c) 95% upper confidence bound on mean rod diameter t0.05,14 = 1.761

241.815

025.0761.123.8

14,025.0

≤

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤

μ

μ

μ

n

st x


below. Therefore, there is evidence to support that the speed-up of CNN is normally distributed.

Speed

P e r c e n t

6.05.55.04.54.03.53.0

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.745

4.313

StDev 0.4328

N 1

A D 0.233

P-Value

Probabilit y Plot of SpeedNormal - 95% CI

3

b) 95% confidence interval on mean speed-up

4328.0313.413 === s xn 179.212,025.0 =t

575.4051.4

13

4328.0179.2313.4

13

4328.0179.2313.4

12,025.012,025.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ +≤≤⎟⎟

⎠ ⎞

⎜⎜⎝ ⎛ −

μ

μ

μ

n

st x

n

st x

c) 95% lower confidence bound on mean speed-up

4328.0313.413 === s xn 782.112,05.0 =t

μ

μ

μ

≤

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

≤⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −

099.4

13

4328.0782.1313.4

12,05.0 n

st x

8-12



8-36 95% lower bound confidence for the mean wall thickness

given x = 4.05 s = 0.08 n = 25

tα,n-1 = t0.05,24 = 1.711

μ ≤⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −

n

st x 24,05.0

μ ≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

25

08.0711.105.4

4.023 ≤ μ

There is high confidence that the true mean wall thickness is greater than 4.023 mm.

3.23.13.02.92.82.72.6

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for Percent EnrichmentML Estimates - 95% CI

Mean

StDev

2.90167

0.0951169

ML Estimates

8-37 a) The data appear to be normally distributed. There is not strong evidence that the percentage of

enrichment deviates from normality.

b) 99% two-sided confidence interval on mean percentage enrichment

For α = 0.01 and n = 12, tα/2,n-1 = t0.005,11 = 3.106, 0.0993s2.9017 == x

991.2813.2

12

0.0993106.3902.2

12

0.0993106.3902.2

11,005.011,005.0

≤≤

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-13



Section 8-4

8-38 31.182

10,05.0 = χ 49.272

15,025.0 = χ 22.262

12,01.0 = χ

93.462

25,005.0 = χ 85.102

20,95.0 = χ 01.72

18,99.0 = χ 14.52

16,995.0 = χ

8-39 99% lower confidence bound for σ2

For α = 0.01 and n = 15, 29.14=−

2

1,nα χ =

2

14,01.0 χ

2

22

00003075.0

14.29

)008.0(14

σ

σ

≤

≤

8-40 95% two sided confidence interval for σ

8.410 == sn

02.192

9,025.0

2

1,2/ ==−

χ χ α n

and 70.22

9,975.0

2

1,2/1 ==−−

χ χ α n

76.830.3

80.7690.10

70.2

)8.4(9

02.19

)8.4(9

2

22

2

<<

≤≤

≤≤

σ

σ

σ

8-41 95% confidence interval for σ: given n = 51, s = 0.37

First find the confidence interval for σ2 :

For α = 0.05 and n = 51, 71.42 and 32.36χα / ,2 12

n− = χ0 025 502. , = χ α1 2 1

2− − =/ ,n χ0 975 50

2. , =

36.32

)37.0(50

42.71

)37.0(50 22

2

≤≤ σ

0.096 ≤ σ2 ≤ 0.2115

Taking the square root of the endpoints of this interval we obtain,

0.31 < σ < 0.46

8-42 95% confidence interval for σ

09.017 == sn

85.282

16,025.0

2

1,2/ ==−

χ χ α n

and 91.62

16,975.0

2

1,2/1 ==−−

χ χ α n

137.0067.00188.00045.0

91.6

)09.0(16

85.28

)09.0(16

2

22

2

<<≤≤

≤≤

σ

σ

σ

8-14




below. Therefore, there is evidence to support that the mean temperature is normally distributed.

Mean Temp

P e r c e n t

252423222120191817

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.367

21.41StDev 0.9463

N 8

AD 0.352

P-Value

Probabil it y Plot of Mean TempNormal - 95% CI

95% confidence interval for σ

9463.08 == sn

01.162

7,025.0

2

1,2/ ==−

χ χ α n

and 69.12

7,975.0

2

1,2/1 ==−−

χ χ α n

926.1626.0

709.3392.0

69.1

)9463.0(7

01.16

)9463.0(7

2

22

2

<<

≤≤

≤≤

σ

σ

σ

8-44 95% confidence interval for σ

99.1541 == sn

34.592

40,025.0

2

1,2/ ==−

χ χ α n

and 43.242

40,975.0

2

1,2/1 ==−−

χ χ α n

46.2013.13

633.41835.172

43.24

)99.15(40

34.59

)99.15(40

2

22

2

<<

≤≤

≤≤

σ

σ

σ

The data don’t appear to be normally distributed based on examination of the normal probability plot

below. Therefore, there is not enough evidence to support that the time of tumor appearance is normallydistributed. So the 95% confidence interval for σ is invalid.

8-15



TimeOfTumor

P e r c e n t

140120100806040

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

<0.005

88.78

S tDev 15.99

N 4

AD 1.631

P-Value

Probability Plot of TimeOfTumorNormal - 95% CI

1

8-45 95% confidence interval for σ 00831.015 == sn

12.262

14,025.0

2

1,2/ ==−

χ χ α n

and 53.62

14,95.0

2

1,1 ==−−

χ χ α n

0122.0

000148.0

53.6

)00831.0(14

2

22

≤

≤

≤

σ

σ

σ

The data do not appear to be normally distributed based on an examination of the normal probability plot

below. Therefore, the 95% confidence interval for σ is not valid.

Gauge Cap

P e r c e n t

3.503.493.483.473.463.453.44

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.018

3.472

StDev 0.008307

N 1

AD 0.878

P-Value

Probability Plot of Gauge CapNormal - 95% CI

5

8-46 a) 99% two-sided confidence interval on σ2

8-16



and913.110 == sn 59.232

9,005.0 = χ 73.12

9,995.0 = χ

038.19396.1

73.1

)913.1(9

59.23

)913.1(9

2

22

2

≤≤

≤≤

σ

σ

b) 99% lower confidence bound for σ2

For α = 0.01 and n = 10, 21.67=−

2

1,nα χ =

2

9,01.0 χ

2

22

5199.1

67.21

)913.1(9

σ

σ

≤

≤

c) 90% lower confidence bound for σ2

For α = 0.1 and n = 10, 14.68=−

2

1,nα χ =

2

9,1.0 χ

σ

σ

σ

≤

≤

≤

498.1

2436.268.14

)913.1(9

2

22

d) The lower confidence bound of the 99% two-sided interval is less than the one-sided interval.

The lower confidence bound for σ2 is in part (c) is greater because the confidence is lower.



Section 8-5

8-47 a) 95% Confidence Interval on the fraction defective produced with this tool.

04333.0300

13ˆ == p 300=n 96.12/ =

α z

06637.002029.0

300

)95667.0(04333.096.104333.0

300

)95667.0(04333.096.104333.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p

α α

b) 95% upper confidence bound 65.105.0 == z zα

06273.0

300

)95667.0(04333.0650.104333.0

)ˆ1(ˆˆ

2/

≤

+≤

−+≤

p

p

n

p p z p p

α

8-17



8-48 a) 95% Confidence Interval on the proportion of such tears that will heal.

676.0ˆ = p 37=n 96.12/ =α

z

827.05245.0

37

)324.0(676.096.1676.0

37

)324.0(676.096.1676.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p

α α

b) 95% lower confidence bound on the proportion of such tears that will heal.

p

p

pn

p p z p

≤

≤−

≤−

−

549.0

37

)33.0(676.064.1676.0

)ˆ1(ˆˆ

α

8-49 a) 95% confidence interval for the proportion of college graduates in Ohio that voted for

George Bush.

536.0768

412ˆ == p 768=n 96.12/ =

α z

571.0501.0768

)464.0(536.096.1536.0

768

)464.0(536.096.1536.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p

α α

b) 95% lower confidence bound on the proportion of college graduates in Ohio that voted for

George Bush.

p

p

pn

p p z p

≤

≤−

≤−

−

506.0

768

)464.0(536.064.1536.0

)ˆ1(ˆˆ

α

8-50 a) 95% Confidence Interval on the death rate from lung cancer.

823.01000

823ˆ == p 1000=n 96.12/ =

α z

8-18



8467.07993.0

1000

)177.0(823.096.1823.0

1000

)177.0(823.096.1823.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p

α α

b) E = 0.03, α = 0.05, zα/2 = z0.025 = 1.96 and p = 0.823 as the initial estimate of p,

79.621)823.01(823.003.0

96.1)ˆ1(ˆ

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p

E

zn α

,

n ≅ 622.

c) E = 0.03, α = 0.05, zα/2 = z0.025 = 1.96 at least 95% confident

11.1067)25.0(03.0

96.1)25.0(

22

2/ =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =

E

zn α

,

n ≅ 1068.

8-51 a) 95% Confidence Interval on the proportion of rats that are under-weight.

4.030

12ˆ == p 30=n 96.12/ =

α z

575.0225.0

30

)6.0(4.096.14.0

30

)6.0(4.096.14.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p

α α

b) E = 0.02, α = 0.05, zα/2 = z0.025 = 1.96 and p = 0.4as the initial estimate of p,

96.2304)4.01(4.002.0

96.1)ˆ1(ˆ

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p

E

zn α

,

n ≅ 2305.

c) E = 0.02, α = 0.05, zα/2 = z0.025 = 1.96 at least 95% confident

2401)25.0(02.0

96.1)25.0(

22

2/ =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =

E

zn α

.

8-52 a) 95% Confidence Interval on the true proportion of helmets showing damage

36.050

18ˆ == p 50=n 96.12/ =

α z

8-19



493.0227.0

50

)64.0(36.096.136.0

50

)64.0(36.096.136.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p

α α

b) 76.2212)36.01(36.002.0

96.1)1(

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p

E

zn α

2213≅n

c) 2401)5.01(5.002.0

96.1)1(

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p

E

zn α

8-53 The worst case would be for p = 0.5, thus with E = 0.05 and α = 0.01, zα/2 = z0.005 = 2.58 we

obtain a sample size of:

64.665)5.01(5.005.0

58.2)1(

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p E

zn α

, n ≅ 666

8-54 E = 0.017, α = 0.01, zα/2 = z0.005 = 2.58

13.5758)5.01(5.0017.0

58.2)1(

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p E

zn α

, n ≅ 5759



Section 8-7

8-55 95% prediction interval on the life of the next tire

given x = 60139.7 s = 3645.94 n = 16

for α=0.05 tα/2,n-1 = t0.025,15 = 2.131

3.681481.52131

16

11)94.3645(131.27.60139

16

11)94.3645(131.27.60139

11

11

1

1

15,025.0115,025.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

The prediction interval is considerably wider than the 95% confidence interval (58,197.3 ≤ μ ≤

62,082.07). This is expected because the prediction interval needs to include the variability in the

parameter estimates as well as the variability in a future observation.

8-56 99% prediction interval on the Izod impact data

8-20



25.025.120 === s xn 861.219,005.0 =t

983.1517.0

2011)25.0(861.225.1

2011)25.0(861.225.1

11

11

1

1

19,005.0119,005.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

The lower bound of the 99% prediction interval is considerably lower than the 99% confidence

interval (1.108 ≤ μ ≤ ∞). This is expected because the prediction interval needs to include the

variability in the parameter estimates as well as the variability in a future observation.

8-57 95% Prediction Interval on the volume of syrup of the next beverage dispensed

x = 1.10 s = 0.015 n = 25 tα/2,n-1 = t0.025,24 = 2.064

13.1068.1

25

11)015.0(064.210.1

25

11)015.0(064.210.1

11

11

1

1

24,025.0124,025.0

≤≤

+−≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x xnst x

The prediction interval is wider than the confidence interval: 106.11.094 ≤≤ μ

8-58 90% prediction interval the value of the natural frequency of the next beam of this type that will

be tested. given⎯ x = 231.67, s =1.53 For α = 0.10 and n = 5, tα/2,n-1 = t0.05,4 = 2.132

2.2351.228

5

11)53.1(132.267.231

5

11)53.1(132.267.231

11

11

1

1

4,05.014,05.0

≤≤

+−≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x xnst x

The 90% prediction in interval is greater than the 90% CI.

8-59 95% Prediction Interval on the volume of syrup of the next beverage dispensed

34.908.48520 === s xn tα/2,n-1 = t0.025,19 = 2.093

551.679049.292

20

11)34.90(093.28.485

20

11)34.90(093.28.485

1111

1

1

19,025.0119,025.0

≤≤

+−≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

The 95% prediction interval is wider than the 95% confidence interval.

8-60 99% prediction interval on the polyunsaturated fat

8-21



319.098.166 === s xn 032.45,005.0 =t

37.1859.15

611)319.0(032.498.16

611)319.0(032.498.16

11

11

1

1

5,005.015,005.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

The length of the prediction interval is much longer than the width of the confidence interval

505.17455.16 ≤≤ μ .

8-61 Given x = 317.2 s = 15.7 n = 10 for α=0.05 tα/2,n-1 = t0.005,9 = 3.250

7.3707.263

10

11)7.15(250.32.317

10

11)7.15(250.32.317

11

11

1

1

9,005.019,005.0

≤≤

+−≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

The length of the prediction interval is longer.

8-62 95% prediction interval on the next rod diameter tested

025.023.815 === s xn 145.214,025.0 =t

29.817.8

15

11)025.0(145.223.8

15

11)025.0(145.223.8

11

11

1

1

14,025.0114,025.0

≤≤

+−≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

95% two-sided confidence interval on mean rod diameter is 8.216 ≤ μ ≤ 8.244

8-63 90% prediction interval on the next specimen of concrete tested

given x = 2260 s = 35.57 n = 12 for α = 0.05 and n = 12, tα/2,n-1 = t0.05,11 = 1.796

5.23265.2193

12

11)57.35(796.12260

12

11)57.35(796.12260

11

11

1

1

11,05.0111,05.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

8-64 90% prediction interval on wall thickness on the next bottle tested.

8-22



given x = 4.05 s = 0.08 n = 25 for tα/2,n-1 = t0.05,24 = 1.711

19.491.325

11)08.0(711.105.4

25

11)08.0(711.105.4

11

11

1

1

24,05.0124,05.0

≤≤

+−≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

8-65 90% prediction interval for enrichment data given x= 2.9 s = 0.099 n = 12 for α = 0.10

and n = 12, tα/2,n-1 = t0.05,11 = 1.796

09.371.2

12

11)099.0(796.19.2

12

11)099.0(796.19.2

11

11

1

1

12,05.0112,05.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

The 90% confidence interval is

95.285.2

12

1)099.0(796.19.2

12

1)099.0(796.19.2

1112,05.012,05.0

≤≤

−≤≤−

+≤≤−

μ

μ

μ n

st xn

st x

The prediction interval is wider than the CI on the population mean with the same confidence.

The 99% confidence interval is

99.281.2

12

1)099.0(106.39.2

12

1)099.0(106.39.2

1112,005.012,005.0

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

st xn

st x

The prediction interval is even wider than the CI on the population mean with greater confidence.

8-66 To obtain a one sided prediction interval, use t α,n-1 instead of tα/2,n-1

Since we want a 95% one sided prediction interval, tα/2,n-1 = t0.05,24 = 1.711

and x = 4.05 s = 0.08 n = 25

1

1

124,05.0

91.3

25

11)08.0(711.105.4

11

+

+

+

≤

≤+−

≤+−

n

n

n

x

x

xn

st x

The prediction interval bound is much lower than the confidence interval bound of

8-23



4.023 mm



Section 8-7

8-67 95% tolerance interval on the life of the tires that has a 95% CL

given ⎯ x = 60139.7 s = 3645.94 n = 16 we find k =2.903

( ) ( ))86.70723,54.49555(

94.3645903.27.60139,94.3645903.27.60139

,

+−

+− ks xks x

95% confidence interval (58,197.3 ≤ μ ≤ 62,082.07) is shorter than the 95%tolerance interval.

8-68 99% tolerance interval on the Izod impact strength PVC pipe that has a 90% CL

given ⎯ x=1.25, s=0.25 and n=20 we find k =3.368

( ) ( ))092.2,408.0(

25.0368.325.1,25.0368.325.1

,

+−

+− ks xks x

The 99% tolerance interval is much wider than the 99% confidence interval on the population

mean (1.090 ≤ μ ≤ 1.410).

8-69 95% tolerance interval on the syrup volume that has 90% confidence levelx = 1.10 s = 0.015 n = 25 and k=2.474

( ) ( ))14.1,06.1(

015.0474.210.1,015.0474.210.1

,

+−

+− ks xks x

8-70 99% tolerance interval on the polyunsaturated fatty acid in this type of margarine that has a

confidence level of 95% ⎯ x = 16.98 s = 0.319 n=6 and k = 5.775

( ) ( )

)82.18,14.15(

319.0775.598.16,319.0775.598.16

,

+−

+− ks xks x


mean (16.46 ≤ μ ≤ 17.51).

8-71 95% tolerance interval on the rainfall that has a confidence level of 95%

34.908.48520 === s xn 752.2=k

( ) ( ))416.734,184.237(

34.90752.28.485,34.90752.28.485

,

+−

+− ks xks x

8-24




mean ( 08.52852.443 ≤≤ μ ).

8-72 95% tolerance interval on the diameter of the rods in exercise 8-27 that has a 90% confidence

level

⎯ x = 8.23 s = 0.0.25 n=15 and k=2.713

( ) ( )

)30.8,16.8(

025.0713.223.8,025.0713.223.8

,

+−

+− ks xks x

The 95% tolerance interval is wider than the 95% confidence interval on the population mean

(8.216 ≤ μ ≤ 8.244).

8-73 99% tolerance interval on the brightness of television tubes that has a 95% CL

given ⎯ x = 317.2 s = 15.7 n = 10 we find k =4.433

( ) (

)80.386,60.247(

7.15433.42.317,7.15433.42.317,

+−+− ks xks x

)


mean

34.33306.301 ≤≤ μ .

8-74 90% tolerance interval on the comprehensive strength of concrete that has a 90% CL

given x = 2260 s = 35.57 n = 12 we find k =2.404

( ) ( )

)5.2345,5.2174(

57.35404.22260,57.35404.22260

,

+−

+− ks xks x


mean 5.22823.2237 ≤≤ μ .

8-75 99% tolerance interval on rod enrichment data that have a 95% CL

given x= 2.9 s = 0.099 n = 12 we find k =4.150

( ) ( )

)31.3,49.2(

099.0150.49.2,099.0150.49.2

,

+−

+− ks xks x

The 99% tolerance interval is much wider than the 95% CI on the population mean (2.84 ≤ μ ≤

2.96).

8-25



8-76 a) 90% tolerance interval on wall thickness measurements that have a 90% CL

given x = 4.05 s = 0.08 n = 25 we find k =2.077

( ) ( )

)22.4,88.3(

08.0077.205.4,08.0077.205.4

,

+−

+− ks xks x

The lower bound of the 90% tolerance interval is much lower than the lower bound on the 95%

confidence interval on the population mean (4.023 ≤ μ ≤ ∞)

b) 90% lower tolerance bound on bottle wall thickness that has confidence level 90%.

given x = 4.05 s = 0.08 n = 25 and k = 1.702

( )91.3

08.0702.105.4 −

− ks x

The lower tolerance bound is of interest if we want to make sure the wall thickness is at least a

certain value so that the bottle will not break.

8-26




8-77 Where α α α =+ 21 . Let 05.0=α

Interval for 025.02/21 === α α α

The confidence level for n xn x /96.1/96.1 σ μ σ +≤≤− is determined by the by the

value of z0 which is 1.96. From Table III, we find Φ(1.96) = P(Z<1.96) = 0.975 and theconfidence level is 95%.

Interval for 04.0,01.0 21 == α α

The confidence interval is n xn x /75.1/33.2 σ μ σ +≤≤− , the confidence level is the

same since 05.0=α . The symmetric interval does not affect the level of significance; however,

it does affect the length. The symmetric interval is shorter in length.

8-78 μ = 50 σ unknown

a) n = 16 x = 52 s = 1.5

116/8

5052=

−=ot

The P-value for t0 = 1, degrees of freedom = 15, is between 0.1 and 0.25. Thus we would

conclude that the results are not very unusual.

b) n = 30

37.130/8

5052=

−=ot

The P-value for t0 = 1.37, degrees of freedom = 29, is between 0.05 and 0.1. Thus we conclude

that the results are somewhat unusual.

c) n = 100 (with n > 30, the standard normal table can be used for this problem)

5.2

100/8

5052=

−=o z

The P-value for z0 = 2.5, is 0.00621. Thus we conclude that the results are very unusual.

d) For constant values of x and s, increasing only the sample size, we see that the standard error

of X decreases and consequently a sample mean value of 52 when the true mean is 50 is more

unusual for the larger sample sizes.

8-79 5,50 2 == σ μ

a) For find or 16=n )44.7( 2 ≥sP )56.2( 2 ≤sP

( ) 10.032.2205.05

)44.7(15)44.7( 2

152

2

15

2 ≤≥≤=⎟ ⎠

⎞⎜⎝

⎛ ≥=≥ χ χ PPSP

Using Minitab =0.0997)44.7( 2 ≥SP

( )≤≤≤=⎟ ⎠

⎞⎜⎝

⎛ ≤=≤ 68.705.05

)56.2(15)56.2( 2

15

2

15

2 χ χ PPSP 0.10

Using Minitab =0.064)56.2( 2 ≤SP

8-27



b) For find or 30=n )44.7( 2 ≥SP )56.2( 2 ≤SP

( ) 05.015.43025.05

)44.7(29)44.7( 2

29

2

29

2 ≤≥≤=⎟ ⎠

⎞⎜⎝


Using Minitab = 0.044)44.7( 2 ≥SP

( ) 025.085.1401.05

)56.2(29)56.2( 2

29

2

29

2 ≤≤≤=⎟ ⎠ ⎞⎜

⎝ ⎛ ≤=≤ χ χ PPSP

Using Minitab = 0.014.)56.2( 2 ≤SP

c) For or 71=n )44.7( 2 ≥sP )56.2( 2 ≤sP

( ) 01.016.104005.05

)44.7(70)44.7( 2

70

2

70

2 ≤≥≤=⎟ ⎠

⎞⎜⎝


Using Minitab =0.0051)44.7( 2 ≥SP

( ) 005.084.35

5

)56.2(70)56.2( 2

70

2

70

2 ≤≤=⎟

⎠

⎞⎜

⎝

⎛ ≤=≤ χ χ PPSP

Using Minitab < 0.001)56.2( 2 ≤SP

d) The probabilities get smaller as n increases. As n increases, the sample variance should

approach the population variance; therefore, the likelihood of obtaining a sample variance much

larger than the population variance will decrease.

e) The probabilities get smaller as n increases. As n increases, the sample variance should

approach the population variance; therefore, the likelihood of obtaining a sample variance much

smaller than the population variance will decrease.

8-80 a) The data appear to follow a normal distribution based on the normal probability plot since the

data fall along a straight line.

b) It is important to check for normality of the distribution underlying the sample data since the

confidence intervals to be constructed should have the assumption of normality for the results to

be reliable (especially since the sample size is less than 30 and the central limit theorem does not

apply).

c) No, with 95% confidence, we can not infer that the true mean could be 14.05 since this value is

not contained within the given 95% confidence interval.

d) As with part b, to construct a confidence interval on the variance, the normality assumption

must hold for the results to be reliable.

e) Yes, it is reasonable to infer that the variance could be 0.35 since the 95% confidence interval

on the variance contains this value.

f) i) & ii) No, doctors and children would represent two completely different populations not

represented by the population of Canadian Olympic hockey players. Because neither doctors nor

children were the target of this study or part of the sample taken, the results should not be

extended to these groups.

8-81 a) The probability plot shows that the data appear to be normally distributed. Therefore, there is

no evidence conclude that the comprehensive strength data are normally distributed.

8-28



b) 99% lower confidence bound on the mean 98.42,s25.12, === n x

896.28,01.0 =t

μ

μ

μ

≤

≤⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ −

≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

99.16

9

42.8896.212.25

8,01.0n

st x

The lower bound on the 99% confidence interval shows that the mean comprehensive strength is

most likely be greater than 16.99 Megapascals.

c) 98% two-sided confidence interval on the mean 98.42,s25.12, === n x

896.28,01.0 =t

25.3399.16

9

42.8896.212.25

9

42.8896.212.25

8,01.08,01.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −

μ

μ

μ

n

st x

n

st x

The bounds on the 98% two-sided confidence interval shows that the mean comprehensive

strength will most likely be greater than 16.99 Megapascals and less than 33.25 Megapascals.

The lower bound of the 99% one sided CI is the same as the lower bound of the 98% two-sided CI

(this is because of the value of α)

d) 99% one-sided upper bound on the confidence interval on σ2 comprehensive strength

90.70,42.8 2 == ss 65.12

8,99.0 = χ

74.343

65.1

)42.8(8

2

22

≤

≤

σ

σ

The upper bound on the 99% confidence interval on the variance shows that the variance of the

comprehensive strength is most likely less than 343.74 Megapascals2.

e) 98% two-sided confidence interval on σ2 of comprehensive strength

90.70,42.8 2 == ss 09.202

9,01.0 = χ 65.12

8,99.0 = χ

74.34323.28

65.1

)42.8(8

09.20

)42.8(8

2

22

2

≤≤

≤≤

σ

σ

The bounds on the 98% two-sided confidence-interval on the variance shows that the variance of

the comprehensive strength is most likely less than 343.74 Megapascals2 and greater than 28.23

Megapascals2.

The upper bound of the 99% one-sided CI is the same as the upper bound of the 98% two-sided

CI because value of α for the one-sided example is one-half the value for the two-sided example.

8-29



f) 98% two-sided confidence interval on the mean 96.31,s23, === n x

896.28,01.0 =t

09.2991.16

9

31.6896.223

9

31.6896.223

8,01.08,01.0

≤≤

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ +≤≤⎟⎟

⎠ ⎞

⎜⎜⎝ ⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

98% two-sided confidence interval on σ2 comprehensive strength

8.39,31.6 2 == ss 09.202

9,01.0 = χ 65.12

8,99.0 = χ

97.19285.15

65.1

)8.39(8

09.20

)8.39(8

2

2

≤≤

≤≤

σ

σ

Fixing the mistake decreased the values of the sample mean and the sample standard deviation.

Because the sample standard deviation was decreased the widths of the confidence intervals were

also decreased.

g) The exercise provides s = 8.41 (instead of the sample variance). A 98% two-sided confidence

interval on the mean 9,41.8s25, === n x

896.28,01.0 =t

12.3388.16

9

41.8896.225

9

41.8896.225

8,01.08,01.0

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

98% two-sided confidence interval on σ2 of comprehensive strength

73.70,41.8 2 == ss 09.202

9,01.0 = χ 65.12

8,99.0 = χ

94.34216.28

65.1

)41.8(8

09.20

)41.8(8

2

22

2

≤≤

≤≤

σ

σ

Fixing the mistake did not have an affect on the sample mean or the sample standard deviation.

They are very close to the original values. The widths of the confidence intervals are also very

similar.

h) When a mistaken value is near the sample mean, the mistake will not affect the sample mean,standard deviation or confidence intervals greatly. However, when the mistake is not near the

sample mean, the value can greatly affect the sample mean, standard deviation and confidence

intervals. The farther from the mean, the greater the effect.

8-82

With σ = 8, the 95% confidence interval on the mean has length of at most 5; the error is then E =

2.5.

8-30



a) 34.39645.2

96.18

5.2

2

2

2

025.0 =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =z

n = 40

b) 13.22365.2

96.16

5.2

2

2

2

025.0 =⎟ ⎠

⎞⎜⎝

⎛ =⎟ ⎠

⎞⎜⎝

⎛ =z

n = 23

As the standard deviation decreases, with all other values held constant, the sample size necessary

to maintain the acceptable level of confidence and the length of the interval, decreases.

8-83 15.33= x 0.62=s 20=n 564.2=k

a) 95% Tolerance Interval of hemoglobin values with 90% confidence

( ) ( )

)92.16,74.13(

62.0564.233.15,62.0564.233.15

,

+−

+− ks xks x

b) 99% Tolerance Interval of hemoglobin values with 90% confidence 368.3=k

( ) ( )

)42.17,24.13(

62.0368.333.15,62.0368.333.15

,

+−

+− ks xks x

8-84 95% prediction interval for the next sample of concrete that will be tested.

given x = 25.12 s = 8.42 n = 9 for α = 0.05 and n = 9, tα/2,n-1 = t0.025,8 = 2.306

59.4565.4

9

11)42.8(306.212.25

9

11)42.8(306.212.25

1

1

1

1

1

1

8,025.018,025.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x xnst x

8-85 a) There is no evidence to reject the assumption that the data are normally distributed.

228218208198188178

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for foam heightML Estimates - 95% CI

Mean

StDev

203.2

7.11056

ML Estimates

8-31



b) 95% confidence interval on the mean 10,5.7s203.20, === n x

262.29,025.0 =t

56.20884.197

1050.7262.22.203

1050.7262.22.203

9,025.09,025.0

≤≤

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ +≤≤⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

c) 95% prediction interval on a future sample

99.22041.185

10

11)50.7(262.22.203

10

11)50.7(262.22.203

11

11 9,025.09,025.0

≤≤

++≤≤+−

+−≤≤+−

μ

μ

μ n

st xn

st x

d) 95% tolerance interval on foam height with 99% confidence 265.4=k

( ) ( )

)19.235,21.171(

5.7265.42.203,5.7265.42.203

,

+−

+− ks xks x

e) The 95% CI on the population mean is the narrowest interval. For the CI, 95% of such intervals

contain the population mean. For the prediction interval, 95% of such intervals will cover a future

data value. This interval is quite a bit wider than the CI on the mean. The tolerance interval is the

widest interval of all. For the tolerance interval, 99% of such intervals will include 95% of the true

distribution of foam height.

8-86 a) Normal probability plot for the coefficient of restitution.

There is no evidence to reject the assumption that the data are normally distributed.

0.660.650.640.630.620.610.600.59

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t

b) 99% CI on the true mean coefficient of restitution

⎯ x = 0.624, s = 0.013, n = 40 ta/2, n-1 = t0.005, 39 = 2.7079

8-32



630.0618.0

40

013.07079.2624.0

40

013.07079.2624.0

39,005.039,005.0

≤≤

+≤≤−

+≤≤−

μ

μ

μ

n

st x

n

st x

c) 99% prediction interval on the coefficient of restitution for the next baseball that will be

tested.

660.0588.0

40

11)013.0(7079.2624.0

40

11)013.0(7079.2624.0

11

11

1

1

39,005.0139,005.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

d) 99% tolerance interval on the coefficient of restitution with a 95% level of confidence

)666.0,582.0(

))013.0(213.3624.0),013.0(213.3624.0(

),(

+−

+− ks xks x

e) The confidence interval in part (b) is for the population mean and we may interpret this to

imply that 99% of such intervals will cover the true population mean. For the prediction

interval, 99% of such intervals will cover a future baseball’s coefficient of restitution. For the

tolerance interval, 95% of such intervals will cover 99% of the true distribution.

8-87 95% Confidence Interval on the proportion of baseballs with a coefficient of restitution that

exceeds 0.635.

2.0

40

8ˆ == p 40=n 65.1=

α z

p

p

pn

p p z p

≤

≤−

≤−

−

0956.0

40

)8.0(2.065.12.0

)ˆ1(ˆˆ

α

8-88 a) The normal probability shows that the data are mostly follow the straight line, however, there

are some points that deviate from the line near the middle. It is probably safe to assume that the

data are normal.

8-33



0 5 10

1

5

10

20

30

4050

60

70

80

90

95

99

Data

P e r c e n t

b) 95% CI on the mean dissolved oxygen concentration

⎯ x = 3.265, s = 2.127, n = 20 ta/2, n-1 = t0.025, 19 = 2.093

260.4270.2

20

127.2093.2265.3

20

127.2093.2265.3

19,025.019,025.0

≤≤

+≤≤−

+≤≤−

μ

μ

μ

n

st x

n

st x

c) 95% prediction interval on the oxygen concentration for the next stream in the system that

will be tested..

827.7297.120

11)127.2(093.2265.3

20

11)127.2(093.2265.3

11

11

1

1

19,025.0119,025.0

≤≤−

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

d) 95% tolerance interval on the values of the dissolved oxygen concentration with a 99% level

of confidence

)003.10,473.3(

))127.2(168.3265.3),127.2(168.3265.3(

),(

−

+−

+− ks xks x

e) The confidence interval in part (b) is for the population mean and we may interpret this to

imply that 95% of such intervals will cover the true population mean. For the prediction

interval, 95% of such intervals will cover a future oxygen concentration. For the tolerance

interval, 99% of such intervals will cover 95% of the true distribution

8-34



8-89 a) There is no evidence to support that the data are not normally distributed. The data points

appear to fall along the normal probability line.

1.4 1.5 1.6 1.7

1

5

10

20

30

40

5060

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for tar contentML Estimates - 95% CI

Mean

StDev

1.529

0.0556117

ML Estimates

b) 99% CI on the mean tar content

⎯ x = 1.529, s = 0.0566, n = 30 ta/2, n-1 = t0.005, 29 = 2.756

557.1501.1

30

0566.0756.2529.1

30

0566.0756.2529.1

29,005.029,005.0

≤≤

+≤≤−

+≤≤−

μ

μ

μ

n

st x

n

st x

c) 99% prediction interval on the tar content for the next sample that will be tested..

688.1370.1

30

11)0566.0(756.2529.1

30

11)0566.0(756.2529.1

11

11

1

1

19,005.0119,005.0

≤≤

++≤≤+−

++≤≤+−

+

+

+

n

n

n

x

x

nst x x

nst x

d) 99% tolerance interval on the values of the tar content with a 95% level of confidence

8-35



)719.1,339.1(

))0566.0(350.3529.1),0566.0(350.3529.1(

),(

+−

+− ks xks x

e) The confidence interval in part (b) is for the population mean and we may interpret this toimply that 95% of such intervals will cover the true population mean. For the prediction

interval, 95% of such intervals will cover a future observed tar content. For the tolerance

interval, 99% of such intervals will cover 95% of the true distribution

8-90 a) 95% Confidence Interval on the population proportion

n=1200 x=8 z0067.0ˆ = p α/2=z0.025=1.96

0113.00021.0

1200)0067.01(0067.096.10067.0

1200)0067.01(0067.096.10067.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

−+≤≤−−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p aa

b) No, there is not sufficient evidence to support the claim that the fraction of defective units

produced is one percent or less at α = 0.05. This is because the upper limit of the control limit is

greater than 0.01.

8-91 a) 99% Confidence Interval on the population proportion

n=1600 x=8 z005.0ˆ = p α/2=z0.005=2.58

009549.00004505.0

1600

)005.01(005.058.2005.0

1600

)005.01(005.058.2005.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

−+≤≤

−−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p aa

b) E = 0.008, α = 0.01, zα/2 = z0.005 = 2.58

43.517)005.01(005.0008.0

58.2)1(

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p

E

zn α

, n ≅ 518

c) E = 0.008, α = 0.01, zα/2 = z0.005 = 2.58

56.26001)5.01(5.0008.0

58.2)1(

22

2/ =−⎟ ⎠

⎞⎜⎝

⎛ =−⎟ ⎠

⎞⎜⎝

⎛ = p p

E

zn α

, n ≅ 26002

d) A bound on the true population proportion reduces the required sample size by a substantial

amount. A sample size of 518 is much more reasonable than a sample size of over 26,000.

8-92 242.0484

117ˆ == p

a) 90% confidence interval; 645.12/ =α

z

8-36



274.0210.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

−+≤≤

−−

p

n

p p z p p

n

p p z p

α α

With 90% confidence, the true proportion of new engineering graduates who were planning to

continue studying for an advanced degree is between 0.210 and 0.274.

b) 95% confidence interval; zα / .2 196=

280.0204.0

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

−+≤≤

−−

p

n

p p z p p

n

p p z p

α α

With 95% confidence, we believe the true proportion of new engineering graduates who were

planning to continue studying for an advanced degree lies between 0.204 and 0.280.

c) Comparison of parts (a) and (b):

The 95% confidence interval is larger than the 90% confidence interval. Higher confidencealways yields larger intervals, all other values held constant.

d) Yes, since both intervals contain the value 0.25, thus there in not enough evidence to determine

that the true proportion is not actually 0.25.

8-93 a) The data appear to follow a normal distribution based on the normal probability plot

since the data fall along a straight line.

b) It is important to check for normality of the distribution underlying the sample data

since the confidence intervals to be constructed should have the assumption of

normality for the results to be reliable (especially since the sample size is less than 30

and the central limit theorem does not apply).

c) 95% confidence interval for the mean

33.673.2211 === s xn 228.210,025.0 =t

982.26478.18

11

33.6228.273.22

11

33.6228.273.22

10,025.010,025.0

≤≤

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

d) As with part b, to construct a confidence interval on the variance, the normality

assumption must hold for the results to be reliable.

e) 95% confidence interval for variance

33.611 == sn

48.202

10,025.0

2

1,2/ ==− χ χ α n

and 25.32

10,975.0

2

1,2/1 ==−− χ χ α n

8-37



289.123565.19

25.3

)33.6(10

48.20

)33.6(10

2

22

2

≤≤

≤≤

σ

σ


below. Therefore, there is evidence to support that the energy intake is normally distributed.

Energy

P e r c e n t

87654

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.011

5.884

StDev 0.5645

N 1

AD 0.928

P-Value

Probabili ty Plot of EnergyNormal - 95% CI

0

b) 99% upper confidence interval on mean energy (BMR)

5645.0884.510 === s xn 250.39,005.0 =t

464.6304.510

5645.0250.3884.5

10

5645.0250.3884.5

9,005.09,005.0

≤≤⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ +≤≤

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

μ

n

st x

n

st x

8-38



9-65


9-126 The parameter of interest is the true,μ.

H0 : μ = μ0

H1 μ ≠ μ0

9-127 a) Reject H0 if z0 < -zα-ε or z0 > zε

)|//

()| //

( 000

000 μ μ

σ

μ

σ

μ μ μ

σ

μ

σ

μ =

−>

−+==

−−<

−

n

X

n

X P

n

X

n

X P P

α ε ε α

ε ε α ε ε α

=−−+−=

Φ−+−Φ=>+−< −−

))1(1())((

)(1)()()( 00 z z z zP z zP

b) β = P(zε ≤ X ≤ zε when d += 01 μ μ )

or β μ μ δα ε ε= − < < = +−P z Z z( | )0 1 0

β μ μ δα εμ

σε

α εδ

σε

δ

σ

εδ

σ

α εδ

σ

= − < < = +

= − − < < −

= − − − −

−−

−

−

P z z

P z Z z

z z

x

n

n n

n n

( | )

( )

( ) ( )

/

/ /

/ /

0

2

2 2

2 2

1 0

Φ Φ

9-128 1) The parameter of interest is the true mean number of open circuits, λ.

2) H0 : λ = 2

3) H1 : λ > 2

4) α = 0.05

5) Since n>30 we can use the normal distribution

z0 =

n

X

/λ

λ −

6) Reject H0 if z0 > zα where z0.05 =1.65

7) x= 1038/500=2.076 n = 500



9-66

z0 = 202.1500/2

2076.2=

−

8) Because 1.202 < 1.65, do not reject the null hypothesis and conclude there is insufficient evidence to

indicate that the true mean number of open circuits is greater than 2 at α = 0.01

9-129 a) 1) The parameter of interest is the true standard deviation of the golf ball distance σ.

2) H0: σ = 10

3) H1: σ < 10

4) α=0.05

5) Because n > 30 we can use the normal distribution

z0 =

)2/(2

0

0

n

S

σ

σ −

6) Reject H0 if z0 < zα where z0.05 =-1.65

7) s= 13.41 n = 100

z0 = 82.4)200/(10

1041.132

=−

8) Since 4.82 > -1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate

that the true standard deviation is less than 10 at α = 0.05

b) 95% percentile: σ μ θ 645.1+=

95% percentile estimator: S X 645.1ˆ +=θ

From the independence

)2/(645.1/)ˆ( 222nnSE σ σ θ +≅

The statistic S can be used as an estimator for σ in the standard error formula.

c) 1) The parameter of interest is the true 95th percentile of the golf ball distance θ.

2) H0: θ = 285

3) H1: θ < 285

4) α = 0.05

5) Since n > 30 we can use the normal distribution

z0 =

)ˆ(ˆ

ˆ0

θ

θ θ

E S

−

6) Reject H0 if z0 < -1.65

7) θ ̂= 282.36 , s = 13.41, n = 100

z0 = 283.1

200/41.13645.1100/41.13

28536.282

222−=

+

−

8) Because -1.283 > -1.65, do not reject the null hypothesis. There is not sufficient evidence to indicate that

the true θ is less than 285 at α = 0.05

9-130 1) The parameter of interest is the true mean number of open circuits, λ.

2) H0 : λ = λ0

3) H1 : λ ≠ λ0

4) α = 0.05

5) test statistic



9-67

∑

∑

=

=

−

=n

i

i

n

i

i

X

X

1

0

12

0

2

2

λ

λ λ

χ

6) Reject H0 if 2

2,2/

2

0 na χ χ > or 2

2,2/1

2

0 na−< χ χ

7) compute ∑=

n

i

i X

1

2λ and plug into

∑

∑

=

=

−

=n

i

i

n

i

i

X

X

1

0

12

0

2

2

λ

λ λ

χ

8) make conclusions

alternative hypotheses1) H0 : λ = λ0

H1 : λ > λ0

Reject H0 if 2

2,

2

0 na χ χ >

2) H0 : λ = λ0

H1 : λ < λ0

Reject H0 if 2

2,

2

0 na χ χ <



9-1

CHAPTER 9

Section 9-1

9-1 a) 25:,25: 10 ≠= μ μ H H Yes, because the hypothesis is stated in terms of the parameter of

interest, inequality is in the alternative hypothesis, and the value in the null and alternativehypotheses matches.

b) 10:,10: 10 => σ σ H H No, because the inequality is in the null hypothesis.

c) 50:,50: 10 ≠= x H x H No, because the hypothesis is stated in terms of the statistic rather

than the parameter.

d) 3.0:,1.0: 10 == p H p H No, the values in the null and alternative hypotheses do not match and

both of the hypotheses are equality statements.

e) 30:,30: 10 >= s H s H No, because the hypothesis is stated in terms of the statistic rather than the

parameter.

9-2 The conclusion does not provide strong evidence that the critical dimension mean equals 100nm.The conclusion is that we don’t have enough evidence to reject the null hypothesis.

9-3 a) nm H nm H 20:,20: 10 <= σ σ

b) This result does not provide strong evidence that the standard deviation has not been reduced.

This result says that we do not have enough evidence to reject the null hypothesis. It is not

support for the null hypothesis.

9-4 a) Newtons H Newtons H 25:,25: 10 <= μ μ

b) No, this results only means that we do not have enough evidence to support H 1

9-5 a) α = P(reject H0 when H0 is true)

= P( X ≤ 11.5 when μ = 12) = ⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ −≤−

4/5.0125.11

/ n

X Pσ

μ = P(Z ≤ −2)

= 0.02275.

The probability of rejecting the null hypothesis when it is true is 0.02275.

b) β = P(accept H0 when μ = 11.25) = ( )25.115.11 => μ when X P

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −>

−

4/5.0

25.115.11

/ n

X P

σ

μ = P(Z > 1.0)

= 1 − P(Z ≤ 1.0) = 1 − 0.84134 = 0.15866

The probability of accepting the null hypothesis when it is false is 0.15866.

c) β = P(accept H0 when μ = 11.25) =

= ( )5.115.11 => μ when X P = ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −>

−

4/5.0

5.115.11

/ n

X P

σ

μ

= P(Z > 0) = 1 − P(Z ≤ 0) = 1 − 0.5 = 0.5

The probability of accepting the null hypothesis when it is false is 0.5



9-2

9-6 a) α = P( X ≤ 11.5 | μ = 12) = ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

16/5.0

125.11

/ n

X P

σ

μ = P(Z ≤ −4) = 0.

The probability of rejecting the null, when the null is true, is approximately 0 with a sample size

of 16.

b) β = P( X > 11.5 | μ =11.25) = ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −>

−

16/5.0

25.115.11

/ n

X P σ

μ

= P(Z > 2) = 1 − P(Z ≤ 2)= 1− 0.97725 = 0.02275.


c) β = P( X > 11.5 | μ =11.5) = ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −>

−

16/5.0

5.115.11

/ n

X P

σ

μ

= P(Z > 0) = 1 − P(Z ≤ 0) = 1− 0.5 = 0.5


9-7 The critical value for the one-sided test is

n z X /5.012 α −≤

a) α=0.01, n=4, from Table III -2.33 = zα and X ≤ 11.42

b) α=0.05, n=4, from Table III -1.65 = zα and X ≤ 11.59

c) α=0.01, n=16, from Table III -2.33 = zα and X ≤ 11.71

d) α=0.05, n=16, from Table III -1.65 = zα and X ≤ 11.95

9-8 a) β= P( X >11.59|µ=11.5)=P(Z>0.36)=1-0.6406=0.3594

b) β= P( X >11.79|µ=11.5)=P(Z>2.32)=1-0.9898=0.0102

c) Notice that the value of β decreases as n increases

9-9 a) x =11.25, then p-value= 00135.0)3(4/5.0

1225.11=−≤=⎟

⎠

⎞⎜⎝

⎛ −≤ Z p Z P

b) x =11.0, then p-value= 000033.0)4(4/5.0

120.11≤−≤=⎟

⎠

⎞⎜⎝

⎛ −≤ Z p Z P

c) x =11.75, then p-value= 158655.0)1(4/5.0

1275.11=−≤=⎟

⎠

⎞⎜⎝

⎛ −≤ Z p Z P

9-10 a) α = P( X ≤ 98.5) + P( X > 101.5)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

9/2

1005.98

9/2

100 X P + ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −>

−

9/2

1005.101

9/2

100 X P

= P(Z ≤ −2.25) + P(Z > 2.25)

= (P(Z ≤- 2.25)) + (1 − P(Z ≤ 2.25))

= 0.01222 + 1 − 0.98778

= 0.01222 + 0.01222 = 0.02444

b) β = P(98.5 ≤ X ≤ 101.5 when μ = 103)



9-3

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−≤

−

9/2

1035.101

9/2

103

9/2

1035.98 X P

= P(−6.75 ≤ Z ≤ −2.25)

= P(Z ≤ −2.25) − P(Z ≤ −6.75)

= 0.01222 − 0 = 0.01222

c) β = P(98.5 ≤ X ≤ 101.5 when μ = 105)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−≤

−

9/2

1055.101

9/2

105

9/2

1055.98 X P

= P(−9.75≤ Z ≤ −5.25)

= P(Z ≤ −5.25) − P(Z ≤ −9.75)

= 0 − 0

= 0.

The probability of accepting the null hypothesis when it is actually false is smaller in part c since the true

mean, μ = 105, is further from the acceptance region. A larger difference exists.

9-11 Use n = 5, everything else held constant (from the values in exercise 9-6):

a) P( X ≤ 98.5) + P( X >101.5)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

5/2

1005.98

5/2

100 X P + ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −>

−

5/2

1005.101

5/2

100 X P

= P(Z ≤ −1.68) + P(Z > 1.68)

= P(Z ≤ −1.68) + (1 − P(Z ≤ 1.68))

= 0.04648 + (1 − 0.95352)

= 0.09296

b) β = P(98.5 ≤ X ≤ 101.5 when μ = 103)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−≤

−

5/2

1035.101

5/2

103

5/2

1035.98 X P

= P(−5.03 ≤ Z ≤ −1.68)

= P(Z ≤ −1.68) − P(Z ≤ −5.03)= 0.04648 − 0

= 0.04648

c) β = P(98.5 ≤ x ≤ 101.5 when μ = 105)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−≤

−

5/2

1055.101

5/2

105

5/2

1055.98 X P

= P(−7.27≤ Z ≤ −3.91)

= P(Z ≤ −3.91) − P(Z ≤ −7.27)

= 0.00005 − 0

= 0.00005

It is smaller, because it is not likely to accept the product when the true mean is as high as 105.

9-12 ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n z X

n z

σ μ

σ μ α α 2/02/0 , where σ =2

a) α=0.01, n=9, then 2/α z =2,57, then 71.101,29.98

b) α=0.05, n=9, then 2/α z =1.96, then 31.101,69.98



9-4

c) α=0.01, n=5, then 2/α z =2.57, then 30.102,70.97

d) α=0.05, n=5, then 2/α z =1.96, then 75.101,25.98

9-13 δ =103-100=3

δ >0 then ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ=

σ

δ β α

n z 2/ , where σ =2

a) β= P(98.69< X <101.31|µ=103)=P(-6.47<Z<-2.54)=0.0055

b) β= P(98.25< X <101.75|µ=103)=P(-5.31<Z<-1.40)=0.0808

a) As n increases, β decreases

9-14 a) p-value=2(1- )( 0 Z Φ )=2(1- )9/2

10098

(

−Φ )=2(1- )3(Φ )=2(1-.99865)=0.0027

b) p-value=2(1- )( 0 Z Φ )=2(1- )9/2

100101(

−Φ )=2(1- )5.1(Φ )=2(1-.93319)=0.13362

c) p-value=2(1- )( 0 Z Φ )=2(1- )9/2

100102(

−Φ )=2(1- )3(Φ )=2(1-.99865)=0.0027

9-15 a) α = P( X > 185 when μ = 175)

=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −>

−

10/20

175185

10/20

175 X P

= P(Z > 1.58)

= 1 − P(Z ≤ 1.58)

= 1 − 0.94295

= 0.05705

b) β = P( X ≤ 185 when μ = 185)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

10/20

185185

10/20

185 X P

= P(Z ≤ 0)

= 0.5

c) β = P( X ≤ 185 when μ = 195)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

10/20

195185

10/20

195 X P

= P(Z ≤ −1.58)

= 0.05705



9-5

9-16 Using n = 16:

a) α = P( X > 185 when μ = 175)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −>

−

16/20

175185

16/20

175 X P

= P(Z > 2)

= 1 − P(Z ≤ 2) = 1 − 0.97725 = 0.02275

b) β = P( X ≤ 185 when μ = 185)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

16/20

185185

16/20

195 X P

= P(Z ≤ 0) = 0.5.

c) β = P( X ≤ 185 when μ = 195)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

16/20

195185

16/20

195 X P

= P(Z ≤ −2) = 0.02275

9-17 ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≥

n Z X

20175 α

a) α=0.01, n=10, then α z =2.32 and critical value is 189.67

b) α=0.05, n=10, then α z =1.64 and critical value is 185.93

c) α=0.01, n=16, then α z = 2.32 and critical value is 186.6

d) α=0.05, n=16, then α z =1.64 and critical value is 183.2

9-18 a) α=0.05, n=10, then the critical value 185.93 (from 9-17 part (b))

β = P( X ≤ 185.37 when μ = 185)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

10/20

18593.185

10/20

185 X P

= P(Z ≤ 0.147)

= 0.5584

b) α=0.05, n=16, then the critical value 183.2 (from 9-17(d)), then

β = P( X ≤ 183.2 when μ = 185)

= ⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ −≤−

16/20

1852.183

16/20

185 X P

= P(Z ≤ -0.36)

= 0.3594

c) as n increases, β decreases



9-6

9-19 p-value=1- )( 0 Z Φ ) wheren

X Z

/

00

σ

μ −=

a) X =180 then 79.010/20

1751800 =

−= Z

p-value=1- 2148.07852.01)79.0( =−=Φ

b) X =190 then 37.210/20

1751900 =

−= Z

p-value=1- 008894.0991106.01)37.2( =−=Φ

c) X =170 then 79.010/20

1751700 −=

−= Z

p-value=1- 785236.0214764.01)79.0( =−=−Φ

9-20 a) α = P( X ≤ 4.85 when μ = 5) + P( X > 5.15 when μ = 5)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤−8/25.0

585.4

8/25.0

5 X P + ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −>−8/25.0

515.5

8/25.0

5 X P

= P(Z ≤ −1.7) + P(Z > 1.7)

= P(Z ≤ −1.7) + (1 − P(Z ≤ 1.7)

= 0.04457 + (1 − 0.95543)

= 0.08914

b) Power = 1 − β

β = P(4.85 ≤ X ≤ 5.15 when μ = 5.1)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−≤

−

8/25.0

1.515.5

8/25.0

1.5

8/25.0

1.585.4 X P

= P(−2.83 ≤ Z ≤ 0.566)

= P(Z ≤ 0.566) − P(Z ≤ −2.83)

= 0.71566 − 0.00233

= 0.71333

1 − β = 0.2867

9-21 Using n = 16:

a) α = P( X ≤ 4.85 | μ = 5) + P( X > 5.15 | μ = 5)

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−

16/25.0

585.4

16/25.0

5 X P + ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −>

−

16/25.0

515.5

16/25.0

5 X P

= P(Z ≤ −2.4) + P(Z > 2.4)

= P(Z ≤ −2.4) +(1 − P(Z ≤ 2.4))= 2(1 − P(Z ≤ 2.4))

= 2(1 − 0.99180)

= 2(0.0082)

= 0.0164.

b) β = P(4.85 ≤ X ≤ 5.15 | μ = 5.1)



9-7

= ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤

−≤

−

16/25.0

1.515.5

16/25.0

1.5

16/25.0

1.585.4 X P

= P(−4 ≤ Z ≤ 0.8) = P(Z ≤ 0.8) − P(Z ≤ −4)

= 0.78814 − 0 = 0.78814

1 − β = 0.21186

c) With larger sample size, the value of α decreased from approximately 0.089 to 0.016. The

power declined modestly from 0.287 to 0.211 while the value for α declined substantially. If the

test with n = 16 were conducted at the α value of 0.089, then it would have greater power than the

test with n = 8.

9-22 5,25.0 0 == μ σ

a) α=0.01, n=8 then

a= n z /2/0 σ μ α + =5+2.57*.25/ 8 =5.22 and

b= n z /2/0 σ μ α − =5-2.57*.25/ 8 =4.77

b) α=0.05, n=8 then

a= n Z /*2/0 σ μ α + =5+1.96*.25/ 8 =5.1732 and

b= n z /2/0 σ μ α − =5-1.96*.25/ 8 =4.8267

c) α=0.01, n=16 then

a= n z /2/0 σ μ α + =5+2.57*.25/ 16 =5.1606 and

b= n z /2/0 σ μ α − =5-2.57*.25/ 16 =4.8393

d) α=0.05, n=16 then

a= n z /2/0 σ μ α + =5+1.96*.25/ 16 =5.1225 and

b= n z /2/0 σ μ α − =5-1.96*.25/ 16 =4.8775

9-23 p-value=2(1 - )( 0 Z Φ ) wheren

x z

/

00

σ

μ −=

a) x =5.2 then 26.28/25.

52.50 =

−= z

p-value=2(1- 0238.0)988089.01(2))26.2( =−=Φ

b) x =4.7 then 39.38/25.

57.40 −=

−= z

p-value=2(1- 0007.0)99965.01(2))39.3( =−=Φ

c) x =5.1 then 1313.18/25.

51.50 =

−= z

p-value=2(1- 2585.0)870762.01(2))1313.1( =−=Φ



9-8

9-24 a) β= P(4.845< X <5.155|µ=5.05)=P(-2.59<Z<1.33)=0.9034

b) β= P(4.8775< X <5.1225|µ=5.05)=P(-2.76<Z<1.16)=0.8741

c) As the n increases, β decreases

9-25 X ~ bin(15, 0.4) H 0: p = 0.4 and H 1: p ≠ 0.4

p1= 4/15 = 0.267 p2 = 8/15 = 0.533

Accept Region: 533.0ˆ267.0 ≤≤ p

Reject Region: 267.0ˆ < p or 533.0ˆ > p

Use the normal approximation for parts a) and b)

a) When p = 0.4, )533.0ˆ()267.0ˆ( >+<= pP pPα

29372.0

14686.014686.0

))05.1(1()05.1(

)05.1()05.1(

15

)6.0(4.0

4.0533.0

15

)6.0(4.0

4.0267.0

=

+=

<−+−<=

>+−<=

⎟⎟

⎟⎟

⎠

⎞

⎜⎜

⎜⎜

⎝

⎛ −

>+

⎟⎟

⎟⎟

⎠

⎞

⎜⎜

⎜⎜

⎝

⎛ −

<=

Z P Z P

Z P Z P

Z P Z P

b) When p = 0.2,

⎟⎟⎟⎟

⎟

⎠

⎞

⎜⎜⎜⎜

⎜

⎝

⎛

−≤≤−=≤≤=

15

)8.0(2.02.0533.0

15

)8.0(2.02.0267.0)533.0ˆ267.0( Z P pP β

2572.0

74215.099936.0

)65.0()22.3(

)22.365.0(

=

−=

≤−≤=

≤≤=

Z P Z P

Z P

9-26 X ~ bin(10, 0.3) Implicitly, H 0: p = 0.3 and H1: p < 0.3

n = 10

Accept region: 1.0ˆ > p

Reject region: 1.0ˆ ≤ p

Use the normal approximation for parts a), b) and c):

a) When p =0.3 α = ( )⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

≤=<

10

)7.0(3.0

3.01.01.0ˆ Z P pP



9-9

08379.0

)38.1(

=

−≤= Z P

b) When p = 0.2 β = ( ) ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

>=>10

)8.0(2.0

2.01.0

1.0ˆ Z P pP

78524.0

)79.0(1

)79.0(

=

−<−=

−>=

Z P

Z P

c) Power = 1 − β = 1 − 078524 = 0.21476

9-27 The problem statement implies H 0: p = 0.6, H 1: p > 0.6 and defines an acceptance region as

80.0500

400ˆ =≤ p and rejection region as 80.0ˆ > p

a) α=P( p̂ >0.80 | p=0.60) = P

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

>

500

)4.0(6.0

60.080.0 Z

= P(Z>9.13)=1-P(Z≤ 9.13) ≈ 0

b) β = P( p̂ ≤ 0.8 when p=0.75) = P(Z ≤ 2.58)=0.99506

9-28 a) Operating characteristic curve:

x

P Zx

P Z

=

= ≤−⎛

⎝ ⎜⎞

⎠⎟ = ≤−⎛

⎝ ⎜⎞

⎠⎟

185

20 10

185

20 10βμ μ

/ /

μ P Z ≤−⎛

⎝ ⎜

⎞

⎠⎟

185

20 10

μ

/= β 1 − β

178 P(Z ≤ 1.11) = 0.8665 0.1335

181 P(Z ≤ 0.63) = 0.7357 0.2643

184 P(Z ≤ 0.16) = 0.5636 0.4364

187 P(Z ≤ −0.32) = 0.3745 0.6255

190 P(Z ≤ −0.79) = 0.2148 0.7852

193 P(Z ≤ −1.26) = 0.1038 0.8962

196 P(Z ≤ −1.74) = 0.0409 0.9591

199 P(Z ≤ −2.21) = 0.0136 0.9864



175 180 185 190 195 200

μ

0

0.2

0.4

0.6

0.8

1

β

Operating Characteristic Curve

b)

175 180 185 190 195 200

μ

0

0.2

0.4

0.6

0.8

1

1 − β

Power Function Curve



9-10

Section 9-2

9-29 a) 10:,10: 10 >= μ μ H H

b) 7:,7: 10 ≠= μ μ H H

c) 5:,5: 10 <= μ μ H H

9-30 a) α=0.01, then a= 2/α z =2.57 and b=- 2/α z =-2.57

b) α=0.05, then a= 2/α z =1.96 and b=- 2/α z =-1.96

c) α=0.1, then a=2/α

z =1.65 and b=-2/α

z =-1.65

9-31 a) α=0.01, then a= α z ≅ 2.33

b) α=0.05, then a= α z ≅ 1.64



9-11

c) α=0.1, then a= α z ≅ 1.29

9-32 a) α=0.01, then a= α −1 z ≅ -2.33

b) α=0.05, then a= α −1 z ≅ -1.64

c) α=0.1, then a=α −1

z ≅ -1.29

9-33 a) p-value=2(1- )( 0 Z Φ )=2(1- )05.2(Φ ) ≅ 0.04

b) p-value=2(1- )( 0 Z Φ )=2(1- )84.1(Φ ) ≅ 0.066

c) p-value=2(1- )( 0 Z Φ )=2(1- )4.0(Φ ) ≅ 0.69

9-34 a) p-value=1- )( 0 Z Φ =1- )05.2(Φ ≅ 0.02

b) p-value=1- )( 0 Z Φ =1- )84.1(−Φ ≅ 0.97

c) p-value=1- )( 0 Z Φ =1- )4.0(Φ ≅ 0.34

9-35 a) p-value= )( 0 Z Φ = )05.2(Φ ≅ 0.98

b) p-value= )( 0 Z Φ = )84.1(−Φ ≅ 0.03

c) p-value= )( 0 Z Φ = )4.0(Φ ≅ 0.65

9-36 a.) 1) The parameter of interest is the true mean water temperature, μ.

2) H0 : μ = 100

3) H1 : μ > 100

4) α = 0.05

5) zx

n0 =

− μ

σ /

6) Reject H0 if z0 > zα where z0.05 = 1.65

7) 98= x , σ = 2

0.39/2

100980 −=

−= z

8) Since -3.0 < 1.65 do not reject H0 and conclude the water temperature is not significantly different

greater than 100 at α = 0.05.

b) P-value = 99865.000135.01)0.3(1 =−=−Φ−

c) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ=

9/2

10410005.0 z β

= Φ(1.65 + −6)

= Φ(-4.35)

≅0

9-37 a) 1) The parameter of interest is the true mean crankshaft wear, μ.

2) H0 : μ = 3

3) H1 : μ ≠ 3

4) α = 0.05

5) zx

n0 =

− μ

σ /

6) ) Reject H0 if z0 < −z α/2 where −z0.025 = −1.96 or z0 > zα/2 where z0.025 = 1.96



9-12

7) x = 2.78, σ = 0.9

95.015/9.0

378.20 −=

−= z

8) Since –0.95 > -1.96, do not reject the null hypothesis and conclude there is not sufficient evidence

to support the claim the mean crankshaft wear is not equal to 3 at α = 0.05.

b) ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ −+−Φ−⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ −+Φ= 15/9.025.33

15/9.025.33 025.0025.0 z z β

= Φ(1.96 + −1.08) − Φ(−1.96 + −1.08)

= Φ(0.88) − Φ(-3.04)

= 0.81057 − (0.00118)

= 0.80939

c)( ) ( )

,21.15)75.0(

)9.0()29.196.1(

)375.3( 2

22

2

22

10.0025.0

2

22

2/ =+

=−

+=

+=

σ

δ

σ β α z z z zn 16≅n

9-38 a) 1) The parameter of interest is the true mean melting point, μ.

2) H0 : μ = 155

3) H1 : μ ≠ 155

4) α = 0.01

5) zx

n0 =

− μ

σ /

6) Reject H0 if z0 < −z α/2 where −z0.005 = −2.58 or z0 > zα/2 where z0.005 = 2.58

7) x = 154.2, σ = 1.5

69.110/5.1

1552.1540 −=

−= z

8) Since –1.69 > -2.58, do not reject the null hypothesis and conclude there is not sufficient evidence

to support the claim the mean melting point is not equal to 155 °F at α = 0.01.

b) P-value = 2*P(Z <- 1.69) =2* 0.045514 = 0.091028

c)

⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−−Φ−⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −−Φ=

⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−Φ−⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ=

5.1

10)150155(58.2

5.1

10)150155(58.2

005.0005.0σ

δ

σ

δ β

n z

n z

= Φ(-7.96)- Φ(-13.12) = 0 – 0 = 0

d)

( ) ( ),35.1

)5(

)5.1()29.158.2(

)155150( 2

22

2

22

10.0005.0

2

22

2/ =+

=−

+=

+=

σ

δ

σ β α z z z zn

n ≅ 2.

9-39 a.) 1) The parameter of interest is the true mean battery life in hours, μ.

2) H0 : μ = 40

3) H1 : μ > 40

4) α = 0.05

5) zx

n0 =

− μ

σ /



9-13


7) 5.40= x , σ = 1.25

26.110/25.1

405.400 =

−= z

8) Since 1.26 < 1.65 do not reject H0 and conclude the battery life is not significantly different greater than

40 at α = 0.05.

b) P-value = 1038.08962.01)26.1(1 =−=Φ−

c) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ=

10/25.1

424005.0 z β

= Φ(1.65 + −5.06)

= Φ(-3.41)

≅ 0.000325

d)( ) ( )

,844.0)4(

)25.1()29.165.1(

)4440( 2

22

2

22

10.005.0

2

22

=+

=−

+=

+=

σ

δ

σ β α z z z zn 1≅n

e) 95% Confidence Interval

μ

μ

μ σ

≤

≤+≤+

85.39

10/)25.1(65.15.40

/05.0 n z x

The lower bound of the 90 % confidence interval must be greater than 40 to verify that the true mean exceeds

40 hours.

9-40 a)

1) The parameter of interest is the true mean tensile strength, μ.

2) H0 : μ = 3500

3) H1 : μ ≠ 3500

4) α = 0.01

5) z

x

n0 =

− μ

σ /

6) Reject H0 if z0 < −zα/2 where −z0.005 = −2.58 or z0 > zα/2

where z0.005 = 2.58

7) 3450= x , σ = 60

89.212/60

350034500 −=

−= z

8) Since −-2.89 < −2.58, reject the null hypothesis and conclude the true mean tensile strength

is significantly different from 3500 at α = 0.01.

b) Smallest level of significance =

P-value = 2[1 − Φ (2.89) ]= 2[1 − .998074] = 0.004The smallest level of significance at which we are willing to reject the null hypothesis is 0.004.

c) δ = 3470 – 3500 = -30



9-14

0.005 0.005

(3470 3500) 12 (3470 3500) 122.58 2.58

60 60

n n z z

δ δ β

σ σ

⎛ ⎞ ⎛ ⎞= Φ − − Φ − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞− −= Φ − − Φ − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= Φ(4.312)- Φ(-0.848) = 1 – 0.1982 = 0.8018

d) zα/2 = z0.005 = 2.58

0.005 0.005 x z x zn n

σ σ μ

⎛ ⎞ ⎛ ⎞− ≤ ≤ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

60 603450 2.58 3450 2.58

12 12μ

⎛ ⎞ ⎛ ⎞− ≤ ≤ +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3405.313≤ μ ≤ 3494.687

With 99% confidence, we believe the true mean tensile strength is between 3405.313 psi and

3494.687 psi. We can test the hypotheses that the true mean tensile strength is not equal to 3500

by noting that the value is not within the confidence interval. Hence we reject the null hypothesis.

9-41 a) 1) The parameter of interest is the true mean speed, μ.

2) H0 : μ = 100

3) H1 : μ < 100

4) α = 0.05

5) zx

n0 =

− μ

σ /

6) Reject H0 if z0 < −zα where −z0.05 = −1.65

7) 2.102= x

, σ = 4

56.18/4

1002.1020 =

−= z

8) Since 1.56> −1.65, do not reject the null hypothesis and conclude the there is insufficient

evidence to conclude that the true speed strength is less than 100 at α = 0.05.

b) 56.10 = z , then p-value= 94.0)( 0 ≅Φ z

c) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−−Φ−=

4

8)10095(1 05.0 z β = 1-Φ(-1.65 - −3.54) = 1-Φ(1.89) = 0.02938

Power = 1-β = 1-0.02938 = 0.97062

d) n =( ) ( )

,597.4)5(

)4()03.165.1(

)10095( 2

22

2

22

15.005.0

2

22

=+

=−

+=

+ σ

δ

σ β α z z z zn ≅ 5

e) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤

n z x

σ μ 05.0



9-15

53.104

8

465.12.102

≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤

μ

μ

Because 100 is included in the CI then we don’t have enough confidence to reject the null

hypothesis.

9-42 a) 1) The parameter of interest is the true mean hole diameter, μ.

2) H0 : μ = 1.50

3) H1 : μ ≠ 1.50

4) α = 0.01

5) zx

n0 =

− μ

σ /

6) Reject H0 if z0 < −zα/2 where −z0.005 = −2.58 or z0 > zα/2

where z0.005 = 2.58

7) 4975.1= x , σ = 0.01

25.125/01.0

50.14975.10 −=

−

= z

8) Since −2.58 < -1.25 < 2.58, do not reject the null hypothesis and conclude the true mean

hole diameter is not significantly different from 1.5 in. at α = 0.01.

b) p-value=2(1- )( 0 Z Φ )=2(1- )25.1(Φ ) ≅ 0.21

c)

⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−−Φ−⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −−Φ=

⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−Φ−⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ=

01.0

25)5.1495.1(58.2

01.0

25)5.1495.1(58.2

005.0005.0σ

δ

σ

δ β

n z

n z

= Φ(5.08)- Φ(-0.08) = 1 – .46812 = 0.53188

power=1-β=0.46812.

d) Set β = 1 − 0.90 = 0.10

n =2

22

2/ )(

δ

σ β α z z +=

2

22

10.0005.0

)50.1495.1(

)(

−

+ σ z z ≅

2

22

)005.0(

)01.0()29.158.2(

−

+= 59.908,

n ≅ 60.

e) For α = 0.01, zα/2 = z0.005 = 2.58

⎟ ⎠

⎞

⎜⎝

⎛

+≤≤⎟ ⎠

⎞

⎜⎝

⎛

− n z xn z x

σ

μ

σ 005.0005.0

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

25

01.058.24975.1

25

01.058.24975.1 μ

1.4923 ≤ μ ≤ 1.5027

The confidence interval constructed contains the value 1.5, thus the true mean hole diameter

could possibly be 1.5 in. using a 99% level of confidence. Since a two-sided 99% confidence



interval is equivalent to a two-sided hypothesis test at α = 0.01, the conclusions necessarily must

be consistent.

9-43 a) 1) The parameter of interest is the true average battery life, μ.

2) H0 : μ = 4

3) H1 : μ > 4

4) α = 0.05

5) zx

n0 =

− μ

σ /


7) 05.4= x , σ = 0.2

77.150/2.0

405.40 =

−= z

8) Since 1.77>1.65, reject the null hypothesis and conclude that there is sufficient evidence

to conclude that the true average battery life exceeds 4 hours at α = 0.05.

b) p-value=1- )( 0 Z

Φ =1- )77.1(Φ ≅ 0.04

c) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−Φ=

2.0

50)45.4(05.0 z β = Φ(1.65 – 17.68) = Φ(-16.03) = 0

Power = 1-β = 1-0 = 1

d) n =( ) ( )

,38.1)5.0(

)2.0()29.165.1(

)45.4( 2

22

2

22

1.005.0

2

22

=+

=−

+=

+ σ

δ

σ β α z z z z

n ≅ 2

e) μ

σ

≤⎟ ⎠

⎞

⎜⎝

⎛

− n z x 05.0

μ

μ

≤

≤⎟ ⎠

⎞⎜⎝

⎛ −

003.4

50

2.065.105.4

Since the lower limit of the CI is just slightly above 4, we conclude that average life is greater

than 4 hours at α=0.05.



9-16

Section 9-3

9-44 a) α=0.01, n=20, the critical values are 861.2±

b) α=0.05, n=12, the critical values are 201.2±

c) α=0.1, n=15, the critical values are 761.1±

9-45 a) α=0.01, n=20, the critical value = 2.539

b) α=0.05, n=12, the critical value = 1.796

c) α=0.1, n=15, the critical value = 1.345

9-46 a) α=0.01, n=20, the critical value = -2.539

b)α

=0.05, n=12, the critical value = -1.796c) α=0.1, n=15, the critical value = -1.345



9-17

9-47 a) 05.0*2025.0*2 ≤≤ p then 1.005.0 ≤≤ p

b) 05.0*2025.0*2 ≤≤ p then 1.005.0 ≤≤ p

c) 4.0*225.0*2 ≤≤ p then 8.05.0 ≤≤ p

9-48 a) 05.0025.0 ≤≤ p b) 025.0105.01 −≤≤− p then 975.095.0 ≤≤ p

c) 4.025.0 ≤≤ p

9-49 a) 025.0105.01 −≤≤− p then 975.095.0 ≤≤ p

b) 05.0025.0 ≤≤ p

c) 25.014.01 −≤≤− p then 75.06.0 ≤≤ p

9-50 a. 1) The parameter of interest is the true mean interior temperature life, μ.

2) H0 : μ = 22.53) H1 : μ ≠ 22.5

4) α = 0.05

5)ns

xt

/0

μ −=

6) Reject H0 if |t0| > tα/2,n-1 where tα/2,n-1 = 2.776

7) 22.496= x , s = 0.378 n=5

00237.05/378.0

5.22496.220 −=

−=t

8) Since –0.00237 >- 2.776, we cannot reject the null hypothesis. There is not sufficient evidence to

conclude that the true mean interior temperature is not equal to 22.5 °C at α = 0.05.

2*0.4 <P-value < 2* 0.5 ; 0.8 < P-value <1.0 b.) The points on the normal probability plot fall along the line. Therefore, there is no evidence to

conclude that the interior temperature data is not normally distributed.

21.5 22.5 23.5

1

5

10

20

30

40

50

60

70

80

90

95

99

Data

P e r c e n t

Normal Probability Plot for tempML Estimates - 95% CI

Mean

StDev

22.496

0.338384

ML Estimates



9-18

c.) d = 66.0378.0

|5.2275.22||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.66, and n = 5, we get β ≅ 0.8 and

power of 1−0.8 = 0.2.

d) d = 66.0378.0

|5.2275.22||| 0 =−=−=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.66, and β ≅ 0.1 (Power=0.9),

40=n .

e) 95% two sided confidence interval

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n

st x

n

st x 4,025.04,025.0 μ

965.22027.22

5

378.0776.2496.22

5

378.0776.2496.22

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

We cannot conclude that the mean interior temperature is not equal to 22.5 since the value is included

inside the confidence interval.

9-51 a. 1) The parameter of interest is the true mean female body temperature, μ.

2) H0 : μ = 98.6

3) H1 : μ ≠ 98.6

4) α = 0.05

5)ns

xt

/0

μ −=


7) 264.98= x , s = 0.4821 n=25

48.325/4821.0

6.98264.980 −=

−=t

8) Since 3.48 > 2.064, reject the null hypothesis and conclude that the there is sufficient evidence to

conclude that the true mean female body temperature is not equal to 98.6 °F at α = 0.05.

P-value = 2* 0.001 = 0.002

b)



9-19

Data appear to be normally distributed.

c) d = 24.14821.0

|6.9898||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 1.24, and n = 25, we get β ≅ 0 and

power of 1−0 ≅ 1.

d) d = 83.04821.0

|6.982.98||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.83, and β ≅ 0.1 (Power=0.9), 20=n .


⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

n

st x

n

st x 24,025.024,025.0 μ

463.98065.9825

4821.0064.2264.98

25

4821.0064.2264.98

≤≤

⎟

⎠

⎞⎜

⎝

⎛ +≤≤⎟

⎠

⎞⎜

⎝

⎛ −

μ

μ

We can conclude that the mean female body temperature is not equal to 98.6 since the value is not included

inside the confidence interval.

9-52 a) 1) The parameter of interest is the true mean rainfall, μ.

2) H0 : μ = 25

3) H1 : μ > 25

4) α = 0.01

5) t0 =

ns

x

/

μ −

6) Reject H0 if t0 > tα,n-1 where t0.01,19 = 2.5397) x = 26.04 s = 4.78 n = 20

t0 = 97.020/78.4

2504.26=

−

8) Since 0.97 < 2.539, do not reject the null hypothesis and conclude there is insufficient evidence to indicate

that the true mean rainfall is greater than 25 acre-feet at α = 0.01. The 0.10 < P-value < 0.25.

b) the data on the normal probability plot fall along the straight line. Therefore there is evidence

that the data are normally distributed.

97 98 99

1

5

10

20

30

4050

60

70

80

90

95

99

Data

P e r c

e n t




9-20

c) d = 42.078.4

|2527||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII h) for α = 0.01, d = 0.42, and n = 20, we get β ≅ 0.7 and

power of 1−0.7 = 0.3.

d) d = 52.078.4

|255.27||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII h) for α = 0.05, d = 0.42, and β ≅ 0.1 (Power=0.9),

75=n .

e) 99% lower confidence bound on the mean diameter

0.01,19

s x t

nμ

⎛ ⎞− ≤⎜ ⎟

⎝ ⎠

4.7826.04 2.539

20

23.326

μ

μ

⎛ ⎞− ≤⎜ ⎟

⎝ ⎠≤

Since the lower limit of the CI is less than 25, we conclude that there is insufficient evidence to indicate that

the true mean rainfall is greater than 25 acre-feet at α = 0.01.

9-53 a)

1) The parameter of interest is the true mean sodium content, μ.

2) H0 : μ = 130

3) H1 : μ ≠ 130

4) α = 0.05

5)ns

xt

/0

μ −=

403020

99

95

90

80

70

6050

40

30

20

10

5

1

Data

P e r c e n t

Normal Probability Plot for rainfallML Estimates - 95% CI

Mean

StDev

26.035

4.66361

ML Estimates



9-21


7) 753.129= x , s = 0.929 n=30

456.130/929.0

130753.1290 −=

−=t

8) Since 1.456 < 2.064, do not reject the null hypothesis and conclude that the there is not

sufficient evidence that the true mean sodium content is different from 130mg at α = 0.05.

From table V the t0 value is found between the values of 0.05 and 0.1 with 29 degrees of freedom,

so 2*0.05<P-value < 2* 0.1 Therefore, 0.1< P-value < 0.2.

b) The assumption of normality appears to be reasonable.

c) d = 538.0929.0

|1305.130||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.53, and n = 30, we get β ≅ 0.2 and power

of 1−0.20 = 0.80

d) d = 11.0929.0

|1301.130||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.11, and β ≅ 0.25 (Power=0.75), 100=n .


⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

n

st x

n

st x 29,025.029,025.0 μ

100.130406.129

30

929.0045.2753.129

30

929.0045.2753.129

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

132131130129128127

99

9590

80

70

60

50

40

30

20

10

5

1

Data

P e r c e n t




9-22

There is no evidence that the mean differs from 130 because that value is inside the confidence

interval.

9-54 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution

is normal.

1) The parameter of interest is the true mean coefficient of restitution, μ.

2) H0 : μ = 0.6353) H1 : μ > 0.635

4) α = 0.05

5) t0 =ns

x

/

μ −

6) Reject H0 if t0 > tα,n-1 where t0.05,39 = 1.685

7) x = 0.624 s = 0.013 n = 40

t0 = 35.540/013.0

635.0624.0−=

−

8) Since –5.25 < 1.685, do not reject the null hypothesis and conclude that there is not

sufficient evidence to indicate that the true mean coefficient of restitution is greater than

0.635 at α = 0.05.

The area to right of -5.35 under the t distribution is greater than 0.9995 from table V.

Minitab gives P-value = 1.

b) From the normal probability plot, the normality assumption seems reasonable:

Baseball Coeff of Restit ut ion

P e r c e n t

0.660.650.640.630.620.610.600.59

99

95

90

80

70

60

50

40

30

20

10

5

1

Probabili ty Plot of Baseball Coeff of Restit utionNormal

c) d = 38.0013.0

|635.064.0||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.38, and n = 40, we get β ≅ 0.25 and power

of 1−0.25 = 0.75.

d) d = 23.0013.0

|635.0638.0||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.23, and β ≅ 0.25 (Power=0.75),

40=n .



9-23

e) Lower confidence bound is 6205.01, =⎟ ⎠

⎞⎜⎝

⎛ − −

n

st x nα

Since 0.635 > 0.6205, then we fail to reject the null hypothesis.


is normal.1) The parameter of interest is the true mean oxygen concentration, μ.

2) H0 : μ = 4

3) H1 : μ ≠ 4

4) α = 0.01

5) t0 =ns

x

/

μ −

6) Reject H0 if |t0 |>tα/2, n-1 = t0.005, 19 = 2.861

7) ⎯ x = 3.265, s = 2.127, n = 20

t0 = 55.120/127.2

4265.3−=

−

8) Because -2.861<-1.55 do not reject the null hypothesis and conclude that there isinsufficient evidence to indicate that the true mean oxygen differs from 4 at α = 0.01.

P-Value: 2*0.05<P-value<2*0.10 therefore 0.10< P-value<0.20


O2 concentr ation

P e r c

e n t

7.55.02.50.0

99

95

90

80

70

60

50

40

30

20

10

5

1

Probabili ty Plot of O2 concentrationNormal

c.) d = 47.0127.2

|43||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII f) for α = 0.01, d = 0.47, and n = 20, we get β ≅ 0.70 and power of 1−0.70 = 0.30.

d) d = 71.0127.2

|45.2||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII f) for α = 0.01, d = 0.71, and β ≅ 0.10 (Power=0.90), 40=n .

e) The 95% confidence interval is:



9-24

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ − −−

n

st x

n

st x nn 1,2/1,2/ α α μ = 62.49.1 ≤≤ μ

Because 4 is within the confidence interval, we fail to reject the null hypothesis.

9-56 a)

1) The parameter of interest is the true mean sodium content, μ.

2) H0 : μ = 300

3) H1 : μ > 300

4) α = 0.05

5)ns

xt

/0

μ −=

6) Reject H0 if t0 > tα,n-1 where tα,n-1 = 1.943

7) 315= x , s = 16 n=7

48.27/16

3003150 =

−=t

8) Since 2.48>1.943, reject the null hypothesis and conclude that there is sufficient evidence that

the leg strength exceeds 300 watts at α = 0.05.

The p-value is between .01 and .025

b) d = 3125.016

|300305||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.3125, and n = 7,

β ≅ 0.9 and power = 1−0.9 = 0.1.

c) if 1-β>0.9 then β<0.1 and n is approximately 100

d) Lower confidence bound is 2.3031, =⎟ ⎠

⎞

⎜⎝

⎛

− − n

s

t x nα

because 300< 303.2 reject the null hypothesis

9-57 a.)1) The parameter of interest is the true mean tire life, μ.

2) H0 : μ = 60000

3) H1 : μ > 60000

4) α = 0.05

5) t0 =

ns

x

/

μ −

6) Reject H0 if t0 > tα,n-1 where 753.115,05.0 =t

7) 94.36457.139,6016 === s xn

t0 = 15.016/94.3645

600007.60139=−

8) Since 0.15 < 1.753., do not reject the null hypothesis and conclude there is insufficient evidence to indicate

that the true mean tire life is greater than 60,000 kilometers at α = 0.05. The P-value > 0.40.

b.) d = 27.094.3645

|6000061000||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.27, and β ≅ 0.1 (Power=0.9),



9-25

4=n .

Yes, the sample size of 16 was sufficient.

9-58 In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean impact strength, μ.

2) H0 : μ = 1.0

3) H1 : μ > 1.0

4) α = 0.05

5) t0 =

ns

x

/

μ −


7) x = 1.25 s = 0.25 n = 20

t0 = 47.420/25.0

0.125.1=

−

8) Since 4.47 > 1.729, reject the null hypothesis and conclude there is sufficient evidence to indicate that the

true mean impact strength is greater than 1.0 ft-lb/in at α = 0.05. The P-value < 0.0005

9-59 In order to use t statistic in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean current, μ.2) H0 : μ = 300

3) H1 : μ > 300

4) α = 0.05

5) t0 =

ns

x

/

μ −

6) Reject H0 if t0 > tα,n-1 where 833.19,05.0 =t

7) 7.152.31710 === s xn

t0 = 46.310/7.15

3002.317=

−

8) Since 3.46 > 1.833, reject the null hypothesis and conclude there is sufficient evidence to indicate that the

true mean current is greater than 300 microamps at α = 0.05. The 0.0025 <P-value < 0.005

9-60 a)

1) The parameter of interest is the true mean height of female engineering students, μ.

2) H0 : μ = 65

3) H1 : μ > 65

4) α = 0.05

5) t0 =ns

x

/

μ −

6) Reject H0 if t0 > tα,n-1 where t0.05,36 =1.68

7) 65.811= x inches 2.106=s inches n = 37

t0 = 34.237/11.2

65811.65=

−

8) Since 2.34 > 1.68, reject the null hypothesis and conclude that there is sufficient evidence to

indicate that the true mean height of female engineering students is not equal to 65 at α = 0.05.

P-value: 0.01<P-value<0.025.

b.) From the normal probability plot, the normality assumption seems reasonable:



9-26

Female height s

P e r c e n t

72706866646260

99

95

90

80

70

60

50

40

30

20

10

5

1

Probabilit y Plot of Female heightsNormal

c) 42.111.2

6562 =−=d , n=37 so, from the OC Chart VII g) for α = 0.05, we find that β≅0.

Therefore, the power ≅ 1.

d.) 47.011.2

6564=

−=d so, from the OC Chart VII g) for α = 0.05, and β≅0.2 (Power=0.8).

30* =n .

9-61 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying

distribution is normal.

1) The parameter of interest is the true mean distance, μ.

2) H0 : μ = 2803) H1 : μ > 280

4) α = 0.05

5) t0 =ns

x

/

μ −

6) Reject H0 if t0 > tα,n-1 where t0.05,99 =1.6604

7) x = 260.3 s = 13.41 n = 100

t0 = 69.14100/41.13

2803.260−=

−

8) Since –14.69 < 1.6604, do not reject the null hypothesis and conclude that there is

insufficient evidence to indicate that the true mean distance is greater than 280 at α = 0.05.

From table V the t0 value in absolute value is greater than the value corresponding to

0.0005. Therefore, the P-value is greater than 0.9995.




9-27

Dist ance for g olf balls

P e r c e n t

310300290280270260250240230220

99.9

99

95

90

80

70

60

50

40

30

20

10

5

1

0.1

Probabil it y Plot of Distance for golf ballsNormal

c) d = 75.041.13

|280290||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.75, and n = 100, β ≅ 0 and power of 1−0 =

1.

d) d = 75.041.13

|280290||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.75, and β ≅ 0.20 (Power=0.80), 15=n .

9-62 a) In order to use t statistics in hypothesis testing, we need to assume that the underlyingdistribution is normal.

1) The parameter of interest is the true mean concentration of suspended solids, μ.

2) H0 : μ = 55

3) H1 : μ ≠ 55

4) α = 0.05

5) t0 =ns

x

/

μ −

6) Reject H0 if |t0 | > tα/2,n-1 where t0.025,59 =2.000

7) x = 59.87 s = 12.50 n = 60

t0 = 018.3

60/50.12

5587.59=

−

8) Since 3.018 > 2.000, reject the null hypothesis and conclude that there is sufficient evidence to

indicate that the true mean concentration of suspended solids is not equal to 55 at α = 0.05.

From table V the t0 value is between the values of 0.001 and 0.0025 with 59 degrees of freedom.

Therefore 2*0.001<P-value < 2* 0.0025 and 0.002< P-value<0.005. Minitab gives a P-value of

0.0038.




9-28

Concentr ation of solids

P e r c e n t

1009080706050403020

99.9

99

95

90

80

70

60

50

40

30

20

10

5

1

0.1

Probabil it y Plot of Concentrati on of solidsNormal

d) 4.050.12

5550

=−

=d , n=60 so, from the OC Chart VII e) for α = 0.05, d= 0.4 and n=60 we

find that β≅0.2. Therefore, the power = 1-0.2 = 0.8.

e) From the same OC chart, and for the specified power, we would need approximately 75

observations.

4.050.12

5550=

−=d Using the OC Chart VII e) for α = 0.05, d = 0.4, and β ≅ 0.10 so

that power=0.90, 75=n .

.



9-29

Section 9-4

9-63 a) α=0.01, n=20, from table V we find the following critical values 6.84 and 38.58

b) α=0.05, n=12, from table V we find the following critical values 3.82 and 21.92

c) α=0.10, n=15, from table V we find the following critical values 6.57 and 23.68

9-64 a) α=0.01, n=20, from table V we find =−

2

1,nα χ 36.19

b) α=0.05, n=12, from table V we find =−

2

1,nα χ 19.68

c) α=0.10, n=15, from table V we find =−

2

1,nα χ 21.06

9-65 a) α=0.01, n=20, from table V we find =−−

2

1,1 nα χ 7.63

b) α=0.05, n=12, from table V we find =−−

2

1,1 nα χ 4.57

c) α=0.10, n=15, from table V we find =−−

2

1,1 nα χ 7.79

9-66 a) 2(0.1)<P-value<2(0.5), then 0.2<P-value<1 b) 2(0.1)<P-value<2(0.5), then 0.2<P-value<1

c) 2(0.05)<P-value<2(0.1), then 0.1<P-value<0.2

9-67 a) 0.1<1-P<0.5 then 0.5<P-value<0.9

b) 0.1<1-P<0.5 then 0.5<P-value<0.9

c) 0.99<1-P<0.995 then 0.005<P-value<0.01

9-68 a) 0.1<P-value<0.5

b) 0.1<P-value<0.5

c) 0.99<P-value<0.995

9-69

a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we needto assume that the underlying distribution is normal.

1) The parameter of interest is the true standard deviation of performance time σ. However,

the answer can be found by performing a hypothesis test on σ2.

2) H0 : σ2 = .752

3) H1 : σ2 >.752

4) α = 0.05

5) χ02 =

( )n s− 1 2

2σ

6) Reject H0 if 2

1,

2

0 −>nα χ χ where 30.262

16,05.0 = χ

7) n = 17, s = 0.09

χ02 = 23.0

75.)09.0(16)1(

2

2

2

2

==−σ

sn

8) Because 0.23 < 26.30 do not reject H0 and conclude there is insufficient evidence to indicate

the true variance of performance time content exceeds 0.752 at α = 0.05.

P-value: Because χ02 =0.23 the P-value>0.995

b) The 95% one sided confidence interval given below, includes the value 0.75. Therefore, we are

not be able to conclude that the standard deviation is greater than 0.75.



9-30

σ

σ

≤

≤

07.0

3.26

)09(.16 22

9-70

a) In order to use the χ2

statistic in hypothesis testing and confidence interval construction, we needto assume that the underlying distribution is normal.

1) The parameter of interest is the true measurement standard deviation σ. However, the

answer can be found by performing a hypothesis test on σ2.

2) H0 : σ2 = .012

3) H1 : σ2 ≠ .012

4) α = 0.05

5) χ02 =

( )n s− 12

2σ

6) Reject H0 if χ χ α02

1 2 12< − −/ ,n where 63.52

14,975.0 = χ or χ χα02

2 12> −, ,n where

12.262

14,025.0 = χ

7) n = 15, s = 0.0083

χ02 = 6446.9

01.

)0083(.14)1(2

2

2

2

==−

σ

sn

8) Since 5.63 < 9.64 < 26.12 do not reject H0

P-value: 0.1<P-value/2<0.5. Therefore, 0.2<P-value<1

b) The 95% confidence interval includes the value 0.01. Therefore, there is not enough evidence

to reject the null hypothesis.

013.000607.0

63.5

)0083(.14

12.26

)0083(.14

2

22

2

≤≤

≤≤

σ

σ

9-71

a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we need

to assume that the underlying distribution is normal.

1) The parameter of interest is the true standard deviation of titanium percentage, σ. However,

the answer can be found by performing a hypothesis test on σ2.

2) H0 : σ2 = (0.25)2

3) H1 : σ2 ≠ (0.25)2

4) α = 0.05

5) χ02 =

( )n s− 1 2

2σ


1 2 12

< − −/ ,n where χ0 995 502. , = 32.36 or χ χα0

22 1

2

> −, ,n where

χ0 005 502. , = 71.42

7) n = 51, s = 0.37

χ02 =

( ) ( . )

( . ).

n s−= =

1 50 0 37

0 25109 52

2

2

2

2σ

8) Since 109.52 > 71.42 we reject H0 and conclude there is sufficient evidence to indicate the

true standard deviation of titanium percentage is significantly different from 0.25 at α = 0.01.



9-31

P-value: p/2<.005 then p<0.01

b) 95% confidence interval for σ:

First find the confidence interval for σ2 :

For α = 0.05 and n = 51, χα / ,2 12

n− = χ0 025 502. , = 71.42 and χ α1 2 1

2− − =/ ,n χ0 975 50

2. , = 32.36

36.32)37.0(50

42.71)37.0(50

2

2

2

≤≤ σ

0.096 ≤ σ2 ≤ 0.2115

Taking the square root of the endpoints of this interval we obtain,

0.31 < σ < 0.46

Since 0.25 falls below the lower confidence bound we would conclude that the population

standard deviation is not equal to 0.25.

9-72

a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we need to

assume that the underlying distribution is normal.

1) The parameter of interest is the true standard deviation of Izod impact strength, σ. However, the


2) H0 : σ2

= (0.10)2

3) H1 : σ

2 ≠ (0.10)2

4) α = 0.01

5) χ02 =

( )n s− 1 2

2σ


1 2 12< − −/ ,n where 84.62

19,995.0 = χ 27 or χ χα02

2 12> −, ,n where 58.382

19,005.0 = χ

7) n = 20, s = 0.25

χ02 = 75.118

)10.0(

)25.0(19)1(2

2

2

2

==−

σ

sn

8) Since 118.75 > 38.58 reject H0 and conclude there is sufficient evidence to indicate the true

standard deviation of Izod impact strength is significantly different from 0.10 at α = 0.01.

b.)P

-value: The P-value<0.005

c.) 99% confidence interval for σ:

First find the confidence interval for σ2

:

For α = 0.01 and n = 20, χα / ,2 12

n− = 84.62

19,995.0 = χ and χ α1 2 12

− − =/ ,n 58.382

19,005.0 = χ

1736.003078.0

84.6

)25.0(19

58.38

)25.0(19

2

22

2

≤≤

≤≤

σ

σ

Since 0.01 falls below the lower confidence bound we would conclude that the population standard deviation

is not equal to 0.01.

9-73a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we need

to assume that the underlying distribution is normal.

1) The parameter of interest is the standard deviation of tire life, σ. However, the answer can

be found by performing a hypothesis test on σ2.

2) H0 : σ2 = 40002

3) H1 : σ2 <40002

4) α = 0.05



9-32

5) χ02 =

2

2)1(

σ

sn −

6) Reject H0 if 2

1,1

2

0 −−<nα χ χ where =2

15,95.0 χ 7.26

7) n = 16, s2 = (3645.94)2

χ02 = 46.124000

)94.3645(15)1(2

2

2

2

==−

σ

sn

8) Since 12.46 > 7.26, fail to reject H0 and conclude there is not sufficient evidence to indicate

the true standard deviation of tire life is less than 4000 km at α = 0.05.

P-value = P(χ2 <12.46) for 15 degrees of freedom 0.5<1-P-value < 0.9

Then 0.1<P-value<0.5

b) The 95% one sided confidence interval below, includes the value 4000, therefore, we could not

be able to conclude that the variance was not equal to 40002.

5240

26.7

)94.3645(15 22

≤

≤

σ

σ

9-74



1) The parameter of interest is the true standard deviation of the diameter,σ. However, the answer can

be found by performing a hypothesis test on σ2.

2) H0 : σ2 = 0.0001

3) H1 : σ2 > 0.0001

4) α = 0.01

5) χ02

=( )n s− 1 2

2σ

6) Reject H0 if χ χα02

12> −,n where χ0 0114

2. , = 29.14

7) n = 15, s

2

= 0.008

χ02 =

( ) ( . )

..

n s−= =

1 14 0 008

000018 96

2

2

2

σ

8) Since 8.96 < 29.14 do not reject H0 and conclude there is insufficient evidence to indicate the true

standard deviation of the diameter exceeds 0.01 at α = 0.01.

P-value = P(χ2 > 8.96) for 14 degrees of freedom: 0.5 < P-value < 0.9

b) Using the chart in the Appendix, with0.015

1.50.01

λ = = and n = 15 we find β = 0.50.

c) 25.101.0

0125.0

0

===σ

σ λ power = 0.8, β=0.2

using chart VII k) the required sample size is 50

9-75



1) The parameter of interest is the true variance of sugar content, σ2. However, the




9-33

2) H0 : σ2

= 18

3) H1 : σ2 ≠ 18

4) α = 0.05

5) χ02 =

( )n s− 1 2

2σ


1 2 12< − −/ ,n where 70.22

9,975.0 = χ or χ χα02

2 12> −, ,n where 02.192

9,025.0 = χ

7) n = 10, s = 4.8

χ02 = 52.11

18

)8.4(9)1( 2

2

2

==−

σ

sn

8) Since 11.52 < 19.02 do not reject H0 and conclude there is insufficient evidence to indicate the true

variance of sugar content is significantly different from 18 at α = 0.01.

P-value: The χ02 is between 0.10 and 0.50. Therefore, 0.2<P-value<1

b) Using the chart in the Appendix, with 2λ = and n = 10 we find β = 0.45.

c) Using the chart in the Appendix, with 49.118

40==λ and β = 0.10, n = 30.



9-34

Section 9-59-76 a)

1) The parameter of interest is the true fraction of satisfied customers.

2) H0 : p = 0.9

3) H1 : p ≠ 0.9

4) α = 0.05

5)( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−

−= ; Either approach will yield the

same conclusion

6) Reject H0 if z0 < − zα/2 where −zα/2 = −z0.025 = −1.96 or z0 > zα/2 where zα/2 = z0.025

= 1.96

7) x = 850 n = 1000 85.01000

850ˆ == p

( )

27.5

)1.0)(9.0(1000

)9.0(1000850

1 00

00 −=

−=

−

−=

pnp

np x z

8) Because -5.27<-1.96 reject the null hypothesis and conclude the true fraction of

satisfied customers is significantly different from 0.9 at α = 0.05.

The P-value: 2(1-Φ(5.27)) ≤ 2(1-1) ≈ 0

b) The 95% confidence interval for the fraction of surveyed customers is:

87.0827.01000

)15.0(85.096.185.

1000

)15.0(85.096.185.

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p α α

Because 0.9 is not included in the confidence interval, we reject the null hypothesis at α

= 0.05.

9-77 a)

1) The parameter of interest is the true fraction of rejected parts

2) H0 : p = 0.03

3) H1 : p < 0.03

4) α = 0.05

5)( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−

−= ; Either approach will yield the same

conclusion

6) Reject H0 if z0 < − zα where −zα = −z0.05 = −1.65

7) x = 10 n = 500 02.0500

10ˆ == p

( )31.1

)97.0)(03.0(500

)03.0(50010

1 00

00 −=

−=

−

−=

pnp

np x z



9-35

8) Because −1.31 > −1.65, do not reject the null hypothesis. There is not enough

evidence to conclude that the true fraction of rejected parts is significantly less than 0.03

at α = 0.05.

P-value = Φ(-1.31) = 0.095

b)

The upper one-sided 95% confidence interval for the fraction of rejected parts is:

0303.0

500

)98.0(02.065.102.

)ˆ1(ˆˆ

≤

+≤

−−≤

p

p

n

p p z p p α

Because 0.03<0.0303 the we fail to reject the null hypothesis

9-78 a)1) The parameter of interest is the true fraction defective integrated circuits

2) H0 : p = 0.05

3) H1 : p ≠ 0.05

4) α = 0.05

5)( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−


conclusion

6) Reject H0 if z0 < − zα/2 where −zα/2 = −z0.025 = −1.96 or z0 > zα/2 where zα/2 = z0.025 = 1.96

7) x = 13 n = 300 . p = =13

300

0043

( )53.0

)95.0)(05.0(300

)05.0(30013

1 00

00 −=

−=

−

−=

pnp

np x z

8) Since −0.53 > −1.65, do not reject null hypothesis and conclude the true fraction of defective

integrated circuits is not significantly less than 0.05, at α = 0.05.

The P-value: 2(1-Φ(0.53))=2(1-0.70194)=0.59612

b) The 95% confidence interval is:

065.002004.0

300

)957.0(043.096.1043

300

)957.0(043.096.1043.

)ˆ1(ˆˆ

)ˆ1(ˆˆ

2/2/

≤≤

+≤≤−

−+≤≤

−−

p

p

n

p p z p p

n

p p z p α α

Since 065.005.002004.0 ≤=≤ p , then we fail to reject the null hypothesis.

9-79 a)

1) The parameter of interest is the true success rate



9-36

2) H0 : p = 0.78

3) H1 : p > 0.78

4) The value for α is not given. We assume α = 0.05.

5)( )00

00

1 pnp

np x z

−

−= or

( )

n

p p

p p z

00

00

1

ˆ

−


conclusion

6) Reject H0 if z0 > zα where zα = z0.05 = 1.65

7) x = 289 n = 350 83.0350

289ˆ ≅= p

( )06.2

)22.0)(78.0(350

)78.0(350289

1 00

00 =

−=

−

−=

pnp

np x z

8) Because 2.06 > 1.65, reject the null hypothesis and conclude the true success rate is

significantly greater than 0.78, at α = 0.05.

P-value=1-0.9803=0.0197

b) The 95% lower confidence interval:

p

p

pn

p p z p

≤

≤−

≤−

−

7969.0

350

)17.0(83.065.183.

)ˆ1(ˆˆ

α

Because 0.7969>0.78, reject the null hypothesis

9-80 a)

1) The parameter of interest is the true percentage of polished lenses that contain surface defects,

p.

2) H0 : p = 0.02

3) H1 : p < 0.02

4) α = 0.05

5)( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−

−= ; Either approach will yield the

same conclusion

6) Reject H0 if z0 < − zα where −zα = −z0.05 = −1.65

7) x = 6 n = 250 024.0250

6ˆ == p

( )452.0

250

)02.01(02.0

02.0024.0

1

ˆ

00

00 =

−

−=

−

−=

n

p p

p p z

8) Since 0.452 > −1.65 do not reject the null hypothesis and conclude the machine cannot be

qualified at the 0.05 level of significance.



9-37

P-value = Φ(0.452) = 0.67364

b) The upper 95% confidence interval is:

0264.0

250

)976.0(024.065.1024.

)ˆ1(ˆ

ˆ

≤

+≤

−

+≤

p

p

n

p p

z p p α

Since 0264.002.0 ≤= p then we fail to reject the null hypothesis.

9-81 a)

1) The parameter of interest is the true percentage of football helmets that contain flaws, p.

2) H0 : p = 0.1

3) H1 : p > 0.14) α = 0.01

5)( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−


conclusion


7) x = 16 n = 200 08.0200

16ˆ == p

( )94.0

200

)10.01(10.0

10.008.0

1

ˆ

00

0

0 −=−

−=

−

−=

n

p p

p p z

8) Because -0.94 < 2.33 do not reject the null hypothesis. There is not enough evidence that the

proportion of football helmets with flaws exceeds 10%.

P-value = 1- Φ(-0.94) =0.8264

b) The 99% lower confidence interval:

p

p

pn

p p z p

≤

≤−

≤−

−

035.0200

)92.0(08.033.208.

)ˆ1(ˆˆ

α

Because 1.0035.0 =≤ p then we fail to reject the null hypothesis.

9-82 a) 1) The parameter of interest is the true proportion of engineering students planning graduate studies

2) H0 : p = 0.50

3) H1 : p ≠ 0.50

4) α = 0.05



9-38

5)

( )00

0

01 pnp

np x z

−

−= or

( )n

p p

p p z

00

0

01

ˆ

−

−= ; Either approach will yield the same conclusion

6) Reject H0 if z0 < − zα/2 where −zα/2 = −z0.025 = −1.96 or z0 > zα/2 where zα/2 = z0.025 = 1.96

7) x = 117 n = 484 2423.0484

117

ˆ == p

( )36.11

)5.0)(5.0(484

)5.0(484117

1 00

0

0 −=−

=−

−=

pnp

np x z

8) Since −11.36 > −1.65, reject the null hypothesis and conclude the true proportion of engineering students

planning graduate studies is significantly different from 0.5, at α = 0.05.

P-value =2[1 − Φ(11.36)] ≅ 0

b.) 242.0484

117ˆ == p

280.0204.0

484

)758.0(242.096.1242.0

484

)758.0(242.096.1242.0

)ˆ1(ˆˆ)ˆ1(ˆˆ 2/2/

≤≤

−≤≤−

−+≤≤−−

p

p

n p p z p p

n p p z p α α

Since the 95% confidence interval does not contain the value 0.5, then conclude that the true proportion of

engineering students planning graduate studies is significantly different from 0.5.

9-83 1) The parameter of interest is the true proportion of batteries that fail before 48 hours, p.

2) H0 : p = 0.002

3) H1 : p < 0.002

4) α = 0.01

5)( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−


6) Reject H0 if z0 < -zα where -zα = -z0.01 = -2.33

7) x = 15 n = 5000 003.05000

15ˆ == p

( )58.1

5000

)998.01(002.0

002.0003.0

1

ˆ

00

00 =

−

−=

−

−=

n

p p

p p z

8) Since 1.58 > -2.33 do not reject the null hypothesis and conclude the proportion of proportion

of cell phone batteries that fail is not less than 0.2% at α=0.01.



9-39

9-84. The problem statement implies that H0: p = 0.6, H1: p > 0.6 and defines an acceptance region as

. p ≤ =315

5000 63 and rejection region as . p > 0 63

a) The probability of a type 1 error is

( ) ( ) 08535.0)37.1(137.1

500)4.0(6.0

6.063.06.0|63.0ˆ =<−=≥=

⎟⎟

⎟⎟

⎠

⎞

⎜⎜

⎜⎜

⎝

⎛ −

≥==≥= Z P Z P Z P p pPα .

b) β = P( P ≤ 0.63 | p = 0.75) = P(Z ≤ −6.196) = 0.

9-85 a) The parameter of interest is the true proportion of engine crankshaft bearings exhibiting surface roughness.

2) H0 : p = 0.10

3) H1 : p > 0.10

4) α = 0.05

5)

( )00

00

1 pnp

np x z

−

−= or

( )n

p p

p p z

00

00

1

ˆ

−



7) x = 10 n = 85 118.085

10ˆ == p

( )54.0

)90.0)(10.0(85

)10.0(8510

1 00

00 =

−=

−

−=

pnp

np x z

8) Because 0.54 < 1.65, do not reject the null hypothesis. There is not enough evidence to conclude that

the true proportion of crankshaft bearings exhibiting surface roughness exceeds 0.10, at α = 0.05.

P-value = 1 – Φ(0.54) = 0.295

b) p= 0.15, p0=0.10, n=85, and zα/2=1.96

639.00016.06406.0)94.2()36.0(

85/)15.01(15.0

85/)10.01(10.096.115.010.0

85/)15.01(15.0

85/)10.01(10.096.115.010.0

/)1(

/)1(

/)1(

/)1( 002/0002/0

=−=−Φ−Φ=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−Φ−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−+−Φ=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−

−−−

Φ−⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−

−+−

Φ= n p p

n p p z p p

n p p

n p p z p p α α

β

11863.117)85.10(

10.015.0

)15.01(15.028.1)10.01(10.096.1

)1()1(

2

2

2

0

002/

≅==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−=

p p

p p z p p zn

β α



9-40

Section 9-7

9-86 Expected Frequency is found by using the Poisson distribution

!)(

x

e x X P

xλ λ −

== where 41.1100/)]4(4)11(3)31(2)30(1)24(0[ =++++=λ

Value 0 1 2 3 4

Observed Frequency 24 30 31 11 4

Expected Frequency 30.12 36.14 21.69 8.67 2.60

Since value 4 has an expected frequency less than 3, combine this category with the previous category:

Value 0 1 2 3-4

Observed Frequency 24 30 31 15

Expected Frequency 30.12 36.14 21.69 11.67

The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3

a) 1) The variable of interest is the form of the distribution for X.

2) H0: The form of the distribution is Poisson

3) H1: The form of the distribution is not Poisson4) α = 0.05

5) The test statistic is

( )∑

=

−=

k

i i

ii

E

E O

1

2

2

0 χ

6) Reject H0 if χ χo2

0 0 5 32 7 81> =. , .

7)( ) ( ) ( ) ( )

23.767.11

67.1115

69.21

69.2131

14.36

14.3630

12.30

12.30242222

2

0 =−

+−

+−

+−

= χ

8) Since 7.23 < 7.81 do not reject H0. We are unable to reject the null hypothesis that the distribution of

X is Poisson.

b) The P-value is between 0.05 and 0.1 using Table IV. From Minitab the P-value = 0.0649.



9-41

9-87 Expected Frequency is found by using the Poisson distribution

!)(

x

e x X P

xλ λ −

== where 907.475/)]9(8)10(7)11(2)1(1[ =++++= …λ

Estimated mean = 4.907

Value 1 2 3 4 5 6 7 8

Observed Frequency 1 11 8 13 11 12 10 9

Expected Frequency 2.7214 6.6770 10.9213 13.3977 13.1485 10.7533 7.5381 4.6237

Since the first category has an expected frequency less than 3, combine it with the next category:

Value 1-2 3 4 5 6 7 8

Observed Frequency 12 8 13 11 12 10 9

Expected Frequency 9.3984 10.9213 13.3977 13.1485 10.7533 7.5381 4.6237


a) 1) The variable of interest is the form of the distribution for the number of flaws.


3) H1: The form of the distribution is not Poisson

4) α = 0.01


( )χ0

22

1

=−

=∑

O E

E

i i

ii

k


00152

1509> =. , .

7)

( ) ( )χ0

22 2

12 9 3984

9 3984

9 4 6237

4 6237=

−+ +

−=

.

.

.

. 6.955

8) Since 6.955 < 15.09 do not reject H0. We are unable to reject the null hypothesis that the distribution

of the number of flaws is Poisson.

b) P-value = 0.2237 (found using Minitab)



9-42

9-88 Estimated mean = 10.131

Value 5 6 8 9 10 11 12 13 14 15

Rel. Freq 0.067 0.067 0.100 0.133 0.200 0.133 0.133 0.067 0.033 0.067

Observed

(Days)

2 2 3 4 6 4 4 2 1 2

Expected

(Days)

1.0626 1.7942 3.2884 3.7016 3.7501 3.4538 2.9159 2.2724 1.6444 1.1106

Since there are several cells with expected frequencies less than 3, the revised table would be:

Value 5-8 9 10 11 12-15

Observed

(Days)

7 4 6 4 9

Expected

(Days)

6.1452 3.7016 3.7501 3.4538 7.9433


a) 1) The variable of interest is the form of the distribution for the number of calls arriving to a switchboard

from noon to 1pm during business days.2) H0: The form of the distribution is Poisson


4) α = 0.05


( )χ0

22

1

=−

=∑

O E

E

i i

ii

k


0 0 5 32 7 81> =. , .

7)

( ) ( ) ( ) ( ) ( )1.72

9433.7

9433.79

4538.3

4538.34

7501.3

7501.36

7016.3

7016.34

1452.6

1452.6722222

2

0 =−

+−

+−

+−

+−

= χ


for the number of calls is Poisson.

b) The P-value is between 0.9 and 0.5 using Table IV. P-value = 0.6325 (found using Minitab)



9-43

9-89 Use the binomial distribution to get the expected frequencies with the mean = np = 6(0.25) = 1.5

Value 0 1 2 3 4

Observed 4 21 10 13 2

Expected 8.8989 17.7979 14.8315 6.5918 1.6479

The expected frequency for value 4 is less than 3. Combine this cell with value 3:

Value 0 1 2 3-4

Observed 4 21 10 15

Expected 8.8989 17.7979 14.8315 8.2397


a) 1) The variable of interest is the form of the distribution for the random variable X.

2) H0: The form of the distribution is binomial with n = 6 and p = 0.25

3) H1: The form of the distribution is not binomial with n = 6 and p = 0.25

4) α = 0.05


( )χ0

22

1

=−

=∑

O E

E

i i

ii

k


0 0 5 32 7 81> =. , .

7)

( ) ( )χ0

22 2

4 88989

88989

15 8 2397

82397=

−+ +

−=

.

.

.

. 10.39

8) Since 10.39 > 7.81 reject H0. We can conclude that the distribution is not binomial with n = 6 and p =

0.25 at α = 0.05.

b) P-value = 0.0155 (from Minitab)

9-90 The value of p must be estimated. Let the estimate be denoted by psample

sample mean =+ + +

=0 39 1 23 2 12 3 1

750 6667

( ) ( ) ( ) ( ).

02778.024

6667.0ˆ ===

n

meansample psample

Value 0 1 2 3

Observed 39 23 12 1

Expected 38.1426 26.1571 8.5952 1.8010

Since value 3 has an expected frequency less than 3, combine this category with that of value 2:

Value 0 1 2-3

Observed 39 23 13

Expected 38.1426 26.1571 10.3962


a) 1) The variable of interest is the form of the distribution for the number of underfilled cartons, X.

2) H0: The form of the distribution is binomial

3) H1: The form of the distribution is not binomial

4) α = 0.05




9-44

( )χ0

22

1

=−

=∑

O E

E

i i

ii

k


00512 384> =. , .

7)( ) ( )

χ02

2 2 239 38 1426

381426

23 26 1571

261571

13 10 3962

1039=

−+

−+

−=

.

.

( . )

.

.

.1.053


of the number of underfilled cartons is binomial at α = 0.05.

b) The P-value is between 0.5 and 0.1 using Table IV. From Minitab the P-value = 0.3048.

9-91 Estimated mean = 49.6741 use Poisson distribution with λ=49.674

All expected frequencies are greater than 3.


a) 1) The variable of interest is the form of the distribution for the number of cars passing through the

intersection.



4) α = 0.055) The test statistic is

( )χ0

22

1

=−

=∑

O E

E

i i

ii

k


005242 3642> =. , .

7) Estimated mean = 49.6741

χ02 769 57= .

8) Since 769.57 >>> 36.42, reject H0. We can conclude that the distribution is not Poisson at α = 0.05.

b) P-value = 0 (found using Minitab)

9-92 The Expected Frequency is found by using the Poisson distribution

!)(

xe x X P

x

λ

λ −

== where 514.19105/)]1(41)1(39)1(7)1(6[ =++++= …λ

Estimated mean = 19.514

Number ofEarthquakes

ObservedFrequency

ExpectedFrequency

6 1 0.03

7 1 0.08

8 4 0.18

10 3 0.78

11 2 1.38

12 2 2.24

13 6 3.36

14 5 4.69

15 11 6.10

16 7 7.43

17 3 8.53

18 8 9.25

19 4 9.50

20 4 9.27



9-45

21 7 8.62

22 8 7.64

23 4 6.48

24 3 5.27

25 2 4.12

26 4 3.0927 4 2.23

28 1 1.56

29 1 1.05

30 1 0.68

31 1 0.43

32 2 0.26

34 1 0.09

35 1 0.05

36 2 0.03

39 1 0.00

41 1 0.00

After combining categories with frequencies less than 3, we get the following table:

Number ofEarthquakes

ObservedFrequency

ExpectedFrequency

6-7-8-10-11-12 13.00 4.68

13 6 3.36

14 5 4.69

15 11 6.10

16 7 7.4317 3 8.53

18 8 9.25

19 4 9.50

20 4 9.27

21 7 8.62

22 8 7.64

23 4 6.48

24 3 5.27

25 2 4.12

26 4 3.09

27-28-29-30-31-32-34-35-36-39-41 16 6.38




a)

1) The variable of interest is the form of the distribution for the number of earthquakes per year

of magnitude 7.0 and greater..



4) α = 0.05

5) The test statistic is( )

χ02

2

1

=−

=∑

O E

E

i i

ii

k

6) Reject H0 if 68.232

14,05.0

2 => χ χ o

7)

( ) ( )48.89

38.6

38.616

86.4

86.41322

2

0 =−

++−

= χ

8) Since 48.89 > 23.68 reject H0. The form of the distribution of the number of earthquakes is

not Poisson.

b) P-value < 0.005



9-46

Section 9-8

9-93 1. The variable of interest is breakdowns among shift.

2. H0: Breakdowns are independent of shift.

3. H1: Breakdowns are not independent of shift.

4. α = 0.05

5. The test statistic is:

( )∑∑

= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0 χ

6. The critical value is 592.122

6,05.= χ

7. The calculated test statistic is 65.112

0

= χ

8. χ χ02

00562/>. ,

, do not reject H0 and conclude that the data provide insufficient evidence to claim that

machine breakdown and shift are dependent at α = 0.05.

P-value = 0.070 (using Minitab)

9-94 1. The variable of interest is calls by surgical-medical patients.

2. H0:Calls by surgical-medical patients are independent of Medicare status.

3. H1:Calls by surgical-medical patients are not independent of Medicare status.

4. α = 0.01


( )

∑∑= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0

χ


1,01.= χ


0= χ

8.2

1,01.0

2

0 χ χ >/ , do not reject H0 and conclude that the evidence is not sufficient to claim that surgical-

medical patients and Medicare status are dependent. P-value = 0.85



9-47

9-95. 1. The variable of interest is statistics grades and OR grades.

2. H0: Statistics grades are independent of OR grades.

3. H1: Statistics and OR grades are not independent.

4. α = 0.01


( )

∑∑= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0

χ


9,01.= χ


0= χ

8.2

9,01.0

2

0 χ χ > Therefore, reject H0 and conclude that the grades are not independent at α = 0.01.

P-value = 0.002

9-96 1. The variable of interest is characteristic among deflections and ranges.

2. H0: Deflection and range are independent.

3. H1: Deflection and range are not independent.

4. α = 0.05


( )∑∑

= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0 χ


4,05.0= χ


0= χ

8.2

4,05.0

2

0 χ χ >/ , do not reject H0 and conclude that the evidence is not sufficient to claim that the data are

not independent at α = 0.05. P-value = 0.652

9-97. 1. The variable of interest is failures of an electronic component.2. H0: Type of failure is independent of mounting position.

3. H1: Type of failure is not independent of mounting position.

4. α = 0.01


( )∑∑

= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0 χ


3,01.0= χ


0= χ

8.2

3,01.0

2

0 χ χ >/ , do not reject H0 and conclude that the evidence is not sufficient to claim that the type of

failure is not independent of the mounting position at α = 0.01. P-value = 0.013

9-98 1. The variable of interest is opinion on core curriculum change.

2. H0: Opinion of the change is independent of the class standing.

3. H1: Opinion of the change is not independent of the class standing.

4. α = 0.05




( )∑∑

= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0 χ


3,05.0= χ


0.= χ .

8. 23,05.0

20 χ χ >>> , reject H0 and conclude that the opinions on the change are not independent of class

standing. P-value ≈ 0

9-99 a)

1. The variable of interest is successes.

2. H0: successes are independent of size of stone.

3. H1: successes are not independent of size of stone.

4. α = 0.05


( )∑∑

= =

−=

r

i

c

j ij

ijij

E

E O

1 1

2

2

0 χ

6. The critical value is 84.321,05. = χ

7. The calculated test statistic2

0 χ = 13.766 with details below.

8.2

1,05.0

2

0 χ χ > , reject H0 and conclude that there is enough evidence to claim that number of

successes and the stone size are not independent.1 2 All

1 55 25 80

66.06 13.94 80.00

2 234 36 270

222.94 47.06 270.00

All 289 61 350

289.00 61.00 350.00

Cell Contents: Count

Expected count

Pearson Chi-Square = 13.766, DF = 1, P-Value = 0.000

b) P-value < 0.005



9-48


9-100 α=0.01

a.) n=25 9783.0)02.2()31.033.2(25/16

868501.0 =Φ=−Φ=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

n=100 9554.0)70.1()63.033.2(100/16

868501.0 =Φ=−Φ=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

n=400 8599.0)08.1()25.133.2(400/16

868501.0 =Φ=−Φ=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

n=2500 2119.0)80.0()13.333.2(2500/16868501.0 =−Φ=−Φ=⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ −+Φ= z β



9-49

b) n=25 31.025/16

85860 =

−= z P-value: 3783.06217.01)31.0(1 =−=Φ−

n=100 63.0100/16

85860 =

−= z P-value: 2643.07357.01)63.0(1 =−=Φ−

n=400 25.1400/16

85860 =

−= z P-value: 1056.08944.01)25.1(1 =−=Φ−

n=2500 13.32500/16

85860 =

−= z P-value: 0009.09991.01)13.3(1 =−=Φ−

The data would be statistically significant when n=2500 at α=0.01

9-101 Sample Mean = p Sample Variance =( ) p p

n

1−

Sample Size, n Sampling Distribution Sample Mean Sample Variance

a. 50 Normal p p p( )1

50

−

b. 80 Normal p p p( )180−

c. 100 Normal p p p( )1

100

−

d) As the sample size increases, the variance of the sampling distribution decreases.

9-102

n Test statistic P-value conclusion

a. 50

12.050/)10.01(10.0

10.0095.00 −=−

−= z

0.4522 Do not reject H 0

b. 10015.0

100/)10.01(10.0

10.0095.00 −=

−

−= z


c. 50037.0

500/)10.01(10.0

10.0095.00 −=

−

−= z


d.

100053.0

1000/)10.01(10.0

10.0095.00 −=

−

−= z


e. The P-value decreases as the sample size increases.

9-103. σ = 12, δ = 205 − 200 = 5,α

20025= . , z0.025 = 1.96,

a) n = 20: 564.0)163.0(12

20596.1 =Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ= β



9-50

b) n = 50: 161.0839.01)986.0(1)986.0(12

50596.1 =−=Φ−=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ= β

c) n = 100: 0116.09884.01)207.2(1)207.2(12

100596.1 =−=Φ−=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ= β

d) β , which is the probability of a Type II error, decreases as the sample size increases because the variance

of the sample mean decreases. Consequently, the probability of observing a sample mean in the

acceptance region centered about the incorrect value of 200 ml/h decreases with larger n.

9-104 σ = 14, δ = 205 − 200 = 5,α2

0025= . , z0.025 = 1.96,

a) n = 20: 6406.0)362.0(14

20596.1 =Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ= β

b) n = 50: 2877.07123.01)565.0(1)565.0(

14

50596.1 =−=Φ−=−Φ=⎟

⎟

⎠

⎞⎜⎜

⎝

⎛ −Φ= β

c) n = 100: 0537.09463.01)611.1(1)611.1(14

100596.1 =−=Φ−=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −Φ= β

d) The probability of a Type II error increases with an increase in the standard deviation.

9-105 σ = 8, δ = 204 − 200 = 4,α

20025= . , z0.025 = 1.96.

a) n = 20: β = −⎛

⎝ ⎜⎜

⎞

⎠⎟⎟ = − =Φ Φ196

4 20

80 28. ( . ) 1 − Φ(0.28) = 1 − 0.61026 = 0.38974

Therefore, power = 1 − β = 0.61026

b) n = 50: β = −⎛

⎝ ⎜⎜

⎞

⎠⎟⎟ = − =Φ Φ196

4 50

82 58. ( . ) 1 − Φ(2.58) = 1 − 0.99506 = 0.00494


c) n = 100: β = −⎛

⎝ ⎜⎜

⎞

⎠⎟⎟ = − =Φ Φ196

4 100

83 04. ( . ) 1 − Φ(3.04) = 1 − 0.99882 = 0.00118


d) As sample size increases, and all other values are held constant, the power increases because the

variance of the sample mean decreases. Consequently, the probability of a Type II error decreases,

which implies the power increases.

9-106 a) α=0.05

n=100 3632.0)35.0()0.265.1(100/)5.0(5.0

6.05.005.0 =−Φ=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

6368.03632.011 =−=−= β Power

n=150 2119.0)8.0()45.265.1(100/)5.0(5.0

6.05.005.0 =−Φ=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ= z β



9-51

7881.02119.011 =−=−= β Power

n=300 03515.0)81.1()46.365.1(300/)5.0(5.0

6.05.005.0 =−Φ=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

96485.003515.011 =−=−= β Power

b) α=0.01

n=100 6293.0)33.0()0.233.2(100/)5.0(5.0

6.05.001.0 =Φ=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

3707.06293.011 =−=−= β Power

n=150 4522.0)12.0()45.233.2(100/)5.0(5.0

6.05.001.0 =−Φ=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

5478.04522.011 =−=−= β Power

n=300 1292.0)13.1()46.333.2(

300/)5.0(5.0

6.05.001.0 =−Φ=−Φ=

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ −+Φ= z β

8702.01292.011 =−=−= β Power

Decreasing the value of α decreases the power of the test for the different sample sizes.

c) α=0.05

n=100 0.0)35.4()0.665.1(100/)5.0(5.0

8.05.005.0 ≅−Φ=−Φ=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −+Φ= z β

1011 ≅−=−= β Power

The true value of p has a large effect on the power. The further p is away from p0 the larger the power of the

test.

d)

242.23)82.4(5.06.0

)6.01(6.065.1)50.01(5.058.2

)1()1(

2

2

2

0

002/

≅==⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−=

p p

p p z p p zn

β α

541.4)1.2(5.08.0

)8.01(8.065.1)50.01(5.058.2

)1()1(

2

2

2

0

002/

≅==⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−−−=

p p

p p z p p zn

β α

The true value of p has a large effect on the sample size. The further p is away from p0 the smaller the

sample size that is required.



9-52

9-107 a) Rejecting a null hypothesis provides a stronger conclusion than failing to reject a null hypothesis.

Therefore, place what we are trying to demonstrate in the alternative hypothesis.

Assume that the data follow a normal distribution.

b) 1) the parameter of interest is the mean weld strength, μ.

2) H0 : μ = 150

3) H1 : μ > 1504) Not given

5) The test statistic is:

ns

xt

/

0

0

μ −=

6) Since no critical value is given, we will calculate the P-value

7) x = 1537. , s= 11.3, n=20

46.120/3.11

1507.1530 =

−=t

P-value = ( ) 10.005.046.1 <−<=≥ valuePt P

8) There is some modest evidence to support the claim that the weld strength exceeds 150 psi.

If we used α = 0.01 or 0.05, we would not reject the null hypothesis, thus the claim would not be

supported. If we used α = 0.10, we would reject the null in favor of the alternative and conclude the

weld strength exceeds 150 psi.

9-108

a)

d = 21

|7573||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve for α = 0.05, d = 2, and n = 10, we get β ≅ 0.0 and

power of 1−0.0 ≅ 1.

d = 31

|7572||| 0 =−

=−

=σ

μ μ

σ

δ


power of 1−0.0 ≅ 1.

b) d = 21

|7573||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 2, and β ≅ 0.1 (Power=0.9),

5* =n . Therefore, 3

2

15

2

1*

=+

=+

=n

n

d = 31

|7572||| 0 =−

=−

=σ

μ μ

σ

δ



13

2

1*

=+

=+

=n

n



9-53

c) 2=σ .

d = 12

|7573||| 0 =−

=−

=σ

μ μ

σ

δ


power of 1−0.10 ≅ 0.90.

d = 5.12

|7572||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve for α = 0.05, d = 1.5, and n = 10, we get β ≅ 0.04 and

power of 1−0.04 ≅ 0.96.

d = 12

|7573||| 0 =−

=−

=σ

μ μ

σ

δ


10* =n . Therefore, 5.52

110

2

1*

=+

=+

=n

n n ≅ 6

d = 5.12

|7572||| 0 =−

=−

=σ

μ μ

σ

δ



17

2

1*

=+

=+

=n

n

Increasing the standard deviation lowers the power of the test and increases the sample size

required to obtain a certain power.

9-109 Assume the data follow a normal distribution.

a) 1) The parameter of interest is the standard deviation, σ.

2) H0 : σ2 = (0.00002)2

3) H1 : σ2< (0.00002)

2

4) α = 0.01

5) The test statistic is: χσ

02

2

2

1=

−( )n s

6) χ0 99 72 124. , .= reject H0 if χ0

2 124< .

7) s = 0.00001 and α = 0.01

75.1)00002.0(

)00001.0(72

22

0 == χ

1.75 > 1.24, do not reject the null hypothesis; that is, there is insufficient evidence to conclude the

standard deviation is at most 0.00002 mm.

b) Although the sample standard deviation is less than the hypothesized value of 0.00002, it is not

significantly less (when α = 0.01) than 0.00002. The value of 0.00001 could have occurred as a result of

sampling variation.

9-110 Assume the data follow a normal distribution.

1) The parameter of interest is the standard deviation of the concentration, σ.

2) H0 : σ2 =42

3) H1 : σ2 < 42



9-54

4) not given


02

2

2

1=

−( )n s

6) will be determined based on the P-value

7) s = 0.004 and n = 10

000009.0)4(

)004.0(92

22

0 == χ

P-value = ( );00009.02 < χ P .0≅− valueP

The P-value is approximately 0, therefore we reject the null hypothesis and conclude that the standard

deviation of the concentration is less than 4 grams per liter.

9-111 Create a table for the number of nonconforming coil springs (value) and the observed number of times the

number appeared. One possible table is:

Value 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Obs 0 0 0 1 4 3 4 6 4 3 0 3 3 2 1 1 0 2 1 2

The value of p must be estimated. Let the estimate be denoted by psample

sample mean 9.32540

)2(19)0(2)0(1)0(0=

++++=

1865.050

325.9ˆ ===

n

meansample psample

Value Observed Expected

0 0 0.00165

1 0 0.01889

2 0 0.10608

3 1 0.38911

4 4 1.04816

5 3 2.21073 6 4 3.80118

7 6 5.47765

8 4 6.74985

9 3 7.22141

10 0 6.78777

11 3 5.65869

12 3 4.21619

13 2 2.82541

14 1 1.71190

15 1 0.94191

16 0 0.47237

17 2 0.21659

18 1 0.09103 19 2 0.03515

Since several of the expected values are less than 3, some cells must be combined resulting in the

following table:

Value Observed Expected

0-5 8 3.77462

6 4 3.80118

7 6 5.47765

8 4 6.74985



9-55

9 3 7.22141

10 0 6.78777

11 3 5.65869

12 3 4.21619

≥13 9 6.29436


a) 1) The variable of interest is the form of the distribution for the number of nonconforming coil springs.



4) α = 0.05


( )χ0

22

1

=−

=∑

O E

E

i i

ii

k

6) Reject H0 if χ χ02

0 0 5 72 1407> =. , .

7)

χ02

2 2

377462

4 38 011

38011

9 6 29436

6 29436= +

−

+ +

−

=

(8 - 3.77462)

17.929

2

.

( . . )

.

( . )

.

8) Since 17.929 > 14.07 reject H0. We are able to conclude the distribution of nonconforming springs is

not binomial at α = 0.05.

b) P-value = 0.0123 (from Minitab)

9-112 Create a table for the number of errors in a string of 1000 bits (value) and the observed number of times the

number appeared. One possible table is:

Value 0 1 2 3 4 5

Obs 3 7 4 5 1 0

The value of p must be estimated. Let the estimate be denoted by psample

sample mean 7.1

20

)0(5)1(4)5(3)4(2)7(1)3(0=

+++++=

0017.01000

7.1ˆ ===

n

meansample psample

Value 0 1 2 3 4 5

Observed 3 7 4 5 1 0

Expected 3.64839 6.21282 5.28460 2.99371 1.27067 0.43103

Since several of the expected values are less than 3, some cells must be combined resulting in the

following table:

Value 0 1 2 ≥3

Observed 3 7 4 6

Expected 3.64839 6.21282 5.28460 4.69541


a) 1) The variable of interest is the form of the distribution for the number of errors in a string of 1000 bits.



4) α = 0.05


( )∑

=

−=

k

i i

ii

E

E O

1

2

2

0 χ



9-56

6) Reject H0 if χ χ02

0 0 5 22 5 99> =. , .

7)

( ) ( )0.88971

69541.4

69541.46

64839.3

64839.3322

2

0 =−

++−

= χ


of the number of errors is binomial at α = 0.05. b) P-value = 0.6409 (found using Minitab)

9-113 Divide the real line under a standard normal distribution into eight intervals with equal

probability. These intervals are [0,0.32), [0.32, 0.675), [0.675, 1.15), [1.15, ∞) and their negative

counterparts. The probability for each interval is p = 1/8 = 0.125 so the expected cell frequencies

are E = np = (100) (0.125) = 12.5. The table of ranges and their corresponding frequencies is

completed as follows.

Interval Obs. Frequency. Exp. Frequency.

x ≤ 5332.5 1 12.5

5332.5< x ≤ 5357.5 4 12.5

5357.5< x ≤ 5382.5 7 12.5

5382.5< x ≤ 5407.5 24 12.55407.5< x ≤ 5432.5 30 12.5

5432.5< x ≤ 5457.5 20 12.5

5457.5< x ≤ 5482.5 15 12.5

x ≥ 5482.5 5 12.5

The test statistic is:

36.635.12

)5.125(

5.12

12.5)-(15

5.12

)5.124(

5.12

12.5)-(1

22222

0 =−

+++−

+= χ

and we would reject if this value exceeds 07.115,05.02 = χ . Because

2

5,05.0

2 χ χ >o , reject the

hypothesis that the data are normally distributed


is normal.

1) The parameter of interest is the true mean concentration of suspended solids, μ.

2) H0 : μ = 50

3) H1 : μ < 50

4) α = 0.05

5) Since n>>30 we can use the normal distribution

z0 =

ns

x

/

μ −

6) Reject H0 if z0 <- zα where z0.05 =1.657) x = 59.87 s = 12.50 n = 60

z0 = 12.660/50.12

5087.59=

−

8) Since 6.12>-1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that

the true mean concentration of suspended solids is less than 50 ppm at α = 0.05.

b) The P-value = 1)12.6( ≅Φ .



9-57

c) We can divide the real line under a standard normal distribution into eight intervals with equal

probability. These intervals are [0,.32), [0.32, 0.675), [0.675, 1.15), [1.15, ∞) and their negative

counterparts. The probability for each interval is p = 1/8 = .125 so the expected cell frequencies

are E = np = (60) (0.125) = 7.5. The table of ranges and their corresponding frequencies is

completed as follows.

Interval Obs. Frequency. Exp. Frequency.x ≤ 45.50 9 7.5

45.50< x ≤ 51.43 5 7.5

51.43< x ≤ 55.87 7 7.5

55.87< x ≤ 59.87 11 7.5

59.87< x ≤ 63.87 4 7.5

63.87< x ≤ 68.31 9 7.5

68.31< x ≤ 74.24 8 7.5

x ≥ 74.24 6 7.5


06.55.7

)5.76(5.7

)5.78(5.7

)5.75(5.7

)5.79(2222

2 =−+−++−+−= o χ

and we would reject if this value exceeds 07.115,05.02 = χ . Since it does not, we cannot reject

the hypothesis that the data are normally distributed.

9-115 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distributionis normal.

1) The parameter of interest is the true mean overall distance for this brand of golf ball, μ.

2) H0 : μ = 270

3) H1 : μ < 270

4) α = 0.05

5) Since n>>30 we can use the normal distribution

z0 =

ns

x

/

μ −

6) Reject H0 if z0 <- zα where z0.05 =1.65

7) x = 1.25 s = 0.25 n = 100

z0 = 23.7100/41.13

0.27030.260−=

−

8) Since –7.23<-1.65, reject the null hypothesis and conclude there is sufficient evidence to indicate that the

true mean distance is less than 270 yard at α = 0.05.

b) The P-value ≅ 0

c) We can divide the real line under a standard normal distribution into eight intervals with equal

probability. These intervals are [0,.32), [0.32, 0.675), [0.675, 1.15), [1.15, ∞) and their

negative counterparts. The probability for each interval is p = 1/8 = .125 so the expected cell

frequencies are E = np = (100) (0.125) = 12.5. The table of ranges and their corresponding

frequencies is completed as follows.

Interval Obs. Frequency. Exp. Frequency.

x ≤ 244.88 16 12.5



9-58

244.88< x ≤ 251.25 6 12.5

251.25< x ≤ 256.01 17 12.5

256.01< x ≤ 260.30 9 12.5

260.30< x ≤ 264.59 13 12.5

264.59< x ≤ 269.35 8 12.5

269.35< x ≤ 275.72 19 12.5

x ≥ 275.72 12 12.5


125.12

)5.1212(

5.12

)5.1219(

5.12

)5.126(

5.12

)5.1216( 22222 =

−+

−++

−+

−= o χ

and we would reject if this value exceeds 07.115,05.02 = χ . Because it does, we can reject the

hypothesis that the data are normally distributed.

9-116 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distributionis normal.

1) The parameter of interest is the true mean coefficient of restitution, μ.2) H0 : μ = 0.635

3) H1 : μ > 0.635

4) α = 0.01

5) Since n>30 we can use the normal distribution

z0 =

ns

x

/

μ −

6) Reject H0 if z0 > zα where z0.05 =2.33

7) x = 0.624 s = 0.0131 n = 40

z0 = 31.540/0131.0

635.0624.0−=

−

8) Since –5.31< 2.33, do not reject the null hypothesis and conclude there is insufficient evidence to indicate

that the true mean coefficient of restitution is greater than 0.635 at α = 0.01.

b) The P-value )31.5(Φ ≅ 1

c.) If the lower bound of the CI was above the value 0.635 then we could conclude that the mean coefficient

of restitution was greater than 0.635.

9-117 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distributionis normal. Use the t-test to test the hypothesis that the true mean is 2.5 mg/L.

1) State the parameter of interest: The parameter of interest is the true mean dissolved oxygen level, μ.

2) State the null hypothesis H0 : μ = 2.5

3) State the alternative hypothesis H1 : μ ≠ 2.5

4) Give the significance level α = 0.055) Give the statistic

t0 =

ns

x

/

μ −

6) Reject H0 if |t0 | <tα/2,n-1

7) Sample statistic x = 3.265 s =2.127 n = 20

and calculate the t-statistic t0 =

ns

x

/

μ −



9-59

8) Draw your conclusion and find the P-value.

b) Assume the data are normally distributed.1) The parameter of interest is the true mean dissolved oxygen level, μ.

2) H0 : μ = 2.5

3) H1 : μ ≠ 2.5

4) α = 0.055)Test statistic

t0 =

ns

x

/

μ −

6) Reject H0 if |t0 | >tα/2,n-1 where tα/2,n-1= t0.025,19b =2.093

7) x = 3.265 s =2.127 n = 20

t0 = 608.120/127.2

5.2265.3=

−

8) Since 1.608 < 2.093, do not reject the null hypotheses and conclude that the true mean is not significantly

different from 2.5 mg/L

c) The value of 1.608 is found between the columns of 0.05 and 0.1 of table V. Therefore the P-value is between 0.1 and 0.2. Minitab provides a value of 0.124

d) The confidence interval found in exercise 8-81 b. agrees with the hypothesis test above. The value of 2.5 is

within the 95% confidence limits. The confidence interval shows that the interval is quite wide due to the large

sample standard deviation value.

260.4270.2

20

127.2093.2265.3

20

127.2093.2265.3

19,025.019,025.0

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

st x

n

st x

9-118 a)

1) The parameter of interest is the true mean sugar concentration, μ.

2) H0 : μ = 11.5

3) H1 : μ ≠ 11.5

4) α = 0.05

5)ns

xt

/0

μ −=


7) 47.11= x , s = 0.022 n=20

10.620/022.0

5.1147.11

0

−=−

=t

8) Since 6.10 > 2.093, reject the null hypothesis and conclude that there is sufficient evidence that

the true mean sugar concentration is different from 11.5 at α = 0.05.

From table V the t0 value in absolute value is greater than the value corresponding to 0.0005 with

19 degrees of freedom. Therefore 2*0.0005 = 0.001>P-value



9-60

b) d = 54.4022.0

|5.114.11||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 4.54, and 20=n we find

β ≅ 0 and the Power ≅ 1.

c) d = 27.2022.0

|5.1145.11||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 2.27, and 1-β>0.9 (β<0.1),

We find that n should be at least 5.

d) 95% two sided confidence interval

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

n

st x

n

st x 19,025.019,025.0 μ

48.1146.11

20022.0093.247.11

20022.0093.247.11

≤≤

⎟ ⎠ ⎞⎜⎝ ⎛ +≤≤⎟ ⎠ ⎞⎜⎝ ⎛ −

μ

μ

We can conclude that the mean sugar concentration content is not equal to 11.5 because that value

is not inside the confidence interval.

e) The normality plot below indicates that the normality assumption is reasonable.

Sugar Concentr at ion

P e r c e n

t

11.5211.5011.4811.4611.4411.4211.40

99

95

90

80

70

6050

40

30

20

10

5

1

Probabilit y Plot of Sugar Concentrati onNormal

9-119 a)

1) The parameter of interest is the true mean body weight, μ.

2) H0 : μ = 300

3) H1 : μ ≠ 300

4) α = 0.05

5)ns

xt

/0

μ −=


7) 50.325= x , s = 198.79 n=27



9-61

667.027/79.198

3005.3250 =

−=t

8) Since 0.667< 2.056, we fail to reject the null hypothesis and conclude that there is not sufficient

evidence that the true mean body weight is different from 300 at α = 0.05.

b) From table V the t0 value is found between the values of 0.25 and 0.4 with 26 degrees of

freedom, so 0.5 < P-value < 0.8 The smallest level of significance at which we would willing to

reject the null hypothesis is the P-value

c) 95% two sided confidence interval

⎟ ⎠

⎞⎜⎝

⎛ +≤≤⎟

⎠

⎞⎜⎝

⎛ −

n

st x

n

st x 19,025.019,025.0 μ

16.40484.246

27

79.198056.25.325

27

79.198056.25.325

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

μ

μ

There is not enough evidence to conclude that the mean body weight differs from 300 because 300

is within the 95% confidence interval.

9-120

a)

1) The parameter of interest is the true mean percent protein, μ.

2) H0 : μ = 80

3) H1 : μ > 80

4) α = 0.05

5) t0 =ns

x

/

μ −


7) x = 80.68 s = 7.38 n = 16

t0 = 37.016/38.7

8068.80=

−

8) Since 0.37 < 1.753, do not reject the null hypothesis and conclude that there is not

sufficient evidence to indicate that the true mean percent protein is greater than 80 at α =

0.05.




9-62

percent protein

P e r c e n t

10090807060

99

95

90

80

70

60

50

40

30

20

10

5

1

Probabili ty Plot of percent proteinNormal

c) The 0.25 < P-value < 0.4, based on Table V.

9-121

a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we

need to assume that the underlying distribution is normal.

1) The parameter of interest is the true variance of tissue assay, σ2.

2) H0 : σ2 = 0.6

3) H1 : σ2 ≠ 0.6

4) α = 0.01

5) χ02 =

( )n s− 1 2

2σ


1 2 12< − −/ ,n where =2

11,995.0 χ 2.60 or 2

1,2/

2

0 −> nα χ χ where

=2

11,005.0 χ 26.76

7) n = 12, s = 0.758

χ02 = 53.10

6.0

)758.0(11)1( 2

2

2

==−σ

sn

8) Since 2.6< 53.10 <26.76 we fail to reject H0 and conclude there is not sufficient evidence to

indicate the true variance of tissue assay is significantly different from 0.6 at α = 0.01.

b) P-value: 0.1<p-value/2<0.5 then 0.2<p-value<1

c) 99% confidence interval for σ, first find the confidence interval for σ2

For α = 0.05 and n = 12, =2

11,995.0 χ 2.60 and =2

11,005.0 χ 26.76

60.2

)758.0(11

76.26

)758.0(11 22

2

≤≤σ

0.236≤ σ2 ≤ 2.43

Because 0.6 falls within the 99% confidence bounds we would conclude that we don’t have

enough evidence to conclude that the population variance is different from 0.6

9-122



9-63

a) In order to use the χ2 statistic in hypothesis testing and confidence interval construction, we

need to assume that the underlying distribution is normal.

1) The parameter of interest is the true variance of the ratio between the numbers of

symmetrical and total synapses, σ2.

2) H0 : σ2= 0.02

3) H1 : σ2 ≠ 0.02

4) α = 0.05

5) χ02 =

( )n s− 1 2

2σ


1 2 12< − −/ ,n where 79.162

30,975.0 = χ or 2

1,2/

2

0 −> nα χ χ where

=2

30,025.0 χ 46.98

7) n = 31, s = 0.198

χ02 = 81.58

02.0

)198.0(30)1( 2

2

2

==−σ

sn

8) Because 58.81 > 46.98 reject H0 and conclude that the true variance of the ratio between the

numbers of symmetrical and total synapses is significantly different from 0.02 at α = 0.05.

b) P-value/2<0.005 so that P-value<0.01

9-123 a)

1) The parameter of interest is the true mean of cut-on wave length, μ.

2) H0 : μ = 6.5

3) H1 : μ ≠ 6.5

4) A value for α is not given. We assume that α = 0.05.

5)ns

xt

/0

μ −=


7) 55.6= x , s = 0.35 n=11

47.011/35.0

5.655.60 =

−=t

8) Since 0.47< 2.228, we fail to reject the null hypothesis and conclude that there is not

sufficient

evidence that the true mean of cut-on wave length is different from 6.5 at α = 0.05.

b) From table V the t0 value is found between the values of 0.25 and 0.4 with 10 degrees of

freedom, so 0.5<P-value<0.8

c) d = 71.035.0

|5.625.6||| 0

=−

=−

= σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.71, and 1-β > 0.95 (β<0.05),

We find that n should be at least 30.

d) d = 28.135.0

|5.695.6||| 0 =−

=−

=σ

μ μ

σ

δ

Using the OC curve, Chart VII e) for α = 0.05, n=11, d = 1.28, we find β≈0.1



9-64

9-124 a) 1) the parameter of interest is the variance of fatty acid measurements, σ2

2) H0 : σ2 = 1.0

3) H1 : σ2 ≠ 1.0

4) α=0.01

5) The test statistic is: χ

σ

02

2

2

1=

−( )n s

6) 41.02

5,995.0 = χ reject H0 if 41.02

0 < χ or 75.162

5,005.0 = χ reject H0 if 75.162

0 > χ

7) n = 6, s = 0.319

509.01

)319.0(52

22

0 == χ

P-value: 0.005<P-value/2<0.01 so that 0.01<P-value<0.02

8) Since 0.509>0.41, do not reject the null hypothesis and conclude that there is insufficient evidence to

conclude that the variance is not equal to 1.0. The P-value is greater than any acceptable significance

level, α, therefore we do not reject the null hypothesis.

b) 1) the parameter of interest is the variance of fatty acid measurements, σ2

(now n=51)2) H0 : σ2 = 1.0

3) H1 : σ2 ≠ 1.0

4) α=0.01


02

2

2

1=

−( )n s

6) 99.272

50,995.0 = χ reject H0 if 99.272

0 < χ or 49.792

5,005.0 = χ reject H0 if 49.792

0 > χ

7) n = 51, s = 0.319

09.51

)319.0(502

22

0 == χ

P-value: P-value/2 < 0.005 so that P-value < 0.01

8) Since 5.09<27.99, reject the null hypothesis and conclude that there is sufficient evidence to conclude

that the variance is not equal to 1.0. The P-value is smaller than any acceptable significance level, α,

therefore we do reject the null hypothesis.

c.) The sample size changes the conclusion that is drawn. With a small sample size, the results are

inconclusive. A larger sample size helps to make sure that the correct conclusion is drawn.

9-125 a) 1) the parameter of interest is the standard deviation, σ

2) H0 : σ2 = 400

3) H1 : σ2 < 400

4) Not given

5) The test statistic is: χ

σ02

2

2

1=

−( )n s

6) Since no critical value is given, we will calculate the p-value

7) n = 10, s = 15.7

χ02

29 15 7

4005546= =

( . ).

P-value = ( )P χ2 5546< . ; 01 05. .< − <P value



8) The P-value is greater than any acceptable significance level α. Therefore, we do not reject the null

hypothesis. There is insufficient evidence to support the claim that the standard deviation is less than

20 microamps.

b) 7) n = 51, s = 20

χ0

2250 157

400 3081= =

( . )

.

P-value = ( )P χ2 3081< . ; 0 01 0 025. .< − <P value

8) The P-value is less than 0.05, therefore we reject the null hypothesis and conclude that the standard

deviation is significantly less than 20 microamps.

c) Increasing the sample size increases the test statistic χ02 and therefore decreases the P-value, providing

more evidence against the null hypothesis.



11-59


11-91 The correlation coefficient for the n pairs of data ( xi , zi) will not be near unity. It will be near zero. The data for the

pairs ( xi , zi) where2

ii y z = will not fall along the line ii x y = which has a slope near unity and gives a correlation

coefficient near unity. These data will fall on a line ii x y = that has a slope near zero and gives a much smaller

correlation coefficient.

11-92 a)

xx

xY

S

S=1

ˆ β , xY 10ˆˆ β β −=

xx

xx

i

xx

iii

xx

xY

SV Cov

nS

x x

nS

x xY Y Cov

S

SY CovY Cov

Cov xY CovCov

2

111

2

1

11110

)ˆ()ˆ,ˆ(

0)())(,(),(

)ˆ,(

)ˆ,ˆ()ˆ,()ˆ,ˆ(

σ β β β

σ β

β β β β β

==

=−

=−

==

−=

∑∑∑ . Therefore,

xxS

xCov

2

10 )ˆ,ˆ(σ

β β −=

b) The requested result is shown in part a).

11-93 a)22

)ˆˆ( 22

10

−=

−

−−= ∑∑

n

e

n

xY MS

iii

E

β β

0)ˆ()ˆ()()( 10 =−−= iii x E E Y E e E β β

)](1[)(2)(12

xx

i

S

x x

nieV −+−=σ Therefore,

22

)(12

2

2

]11[

2

)](1[

2

)(

2

)()(

2

σ σ

σ

=−

−−=

−

+−=

−=

−=

∑

∑∑

−

n

n

n

n

eV

n

e E MS E

xx

i

S

x x

n

ii

E

b) Using the fact that SSR = MSR , we obtain

xx

xx

xx

xx xx R

SS

S

E V SS E MS E

2

1

22

1

2

2

11

2

1 )]ˆ([)ˆ()ˆ()(

β σ β σ

β β β

+=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=

+==

11-94

11

1

1ˆ

x x

Y x

S

S= β



11-60

11

21

11

12

11

11

121

11

1

2

11

121

1

12101

1

1

)(

))(()(

)ˆ(

x x

x x

x x

n

i

x x

x x

n

i

x x

n

i

i

SS

S

x x xS

S

x x x x

S

x xY E

E

ii

iiii

β β

β β

β β β

β

+=−+

=

−++=

⎥⎦

⎤⎢⎣

⎡−

=

∑

∑∑

=

==

No, β1 is no longer unbiased.

11-95

xxSV

2

1 )ˆ(σ

β = . To minimize xxSV ),ˆ( 1 β should be maximized. Because ∑=

−=n

i

i xx x xS1

2)( , Sxx is

maximized by choosing approximately half of the observations at each end of the range of x.

From a practical perspective, this allocation assumes the linear model between Y and x holds throughout the range of x

and observing Y at only two x values prohibits verifying the linearity assumption. It is often preferable to obtain some

observations at intermediate values of x.

11-96 One might minimize a weighted some of squares ∑=

−−n

i

iii x yw1

2

10 )( β β in which a Y i with small variance

(wi large) receives greater weight in the sum of squares.

i

n

i

iii

n

i

iii

n

i

iii

n

i

iii

x x yw x yw

x yw x yw

∑∑

∑∑

==

==

−−−=−−

−−−=−−

1

10

1

2

10

1

1

10

1

2

10

0

)(2)(

)(2)(

β β β β β

∂

β β β β β

∂

Setting these derivatives to zero yields

∑∑∑

∑∑∑

=+

=+

iiiiiii

iiiii

y xw xw xw

yw xww

2

10

10

ˆˆ

ˆˆ

β β

β β

and these equations are solved as follows

( )( )( )( ) ( )22

1ˆ

∑∑∑∑∑∑

−

−=

iiiii

iiiiii

xw xww

yww y xw β

10ˆˆ β β

∑∑∑∑−=

i

ii

i

ii

w xw

w yw .

11-97 )(ˆ x xs

sr y y

x

y −+=



11-61

x x x y

x xS

S y

x xSS

x x y yS y

xx

xy

i yy xx

i xy

1011

2

2

ˆˆˆˆ

)(

)(

)()(

β β β β +=−+=

−+=

−

−−+=

∑∑

11-98 a) i

n

i

ii

n

i

ii x x y x y ∑∑==

−−−=−−1

10

1

2

10

1

)(2)( β β β β β

∂

Upon setting the derivative to zero, we obtain

∑∑∑ =+ iiii y x x x2

10 β β

Therefore,

∑

∑

∑

∑ ∑ −=

−=

2

0

2

0

1

)(ˆ

i

ii

i

iii

x

y x

x

x y x β β β

b)

∑∑∑

∑∑ ==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −=

2

2

22

22

2

0

1][

)()ˆ(

ii

i

i

ii

x x

x

x

Y xV V

σ σ β β

c)∑−± 2

2ˆ1,2/1

ˆi xnt σ

α β

This confidence interval is shorter because ∑∑ −≥ 22)( x x x ii . Also, the t value based on n-1 degrees of

freedom is slightly smaller than the corresponding t value based on n-2 degrees of freedom.



11-1

CHAPTER 11

Section 11-2

11-1 a) iii x y ε β β ++= 10

348571.2542.157 1443

2

=−= xxS

057143.5980.169714

)572(43 −=−= xyS

.

..

( . )( ) .

β

β β

1

0 157214

4314

59 057143

253485712 330

2 3298017 48 013

= =−

= −

= − = − − =

S

S

y x

xy

xx

123.22

59143.13771429.159

59.137

)057143.59(3298017.2ˆ1

=

−=

−=

=

−−==

R yy E

xy R

SSSSS

SSS β

8436112

12322

2

2.

.

n

SS MSˆ E

E ==−

==σ

b) x y 10ˆˆˆ β β +=

99.37)3.4(3298017.2012962.48ˆ =−= y

c) 39.39)7.3(3298017.2012962.48ˆ =−= y

d) 71.639.391.46ˆ =−=−= y ye

11-2 a) iii x y ε β β ++= 10

6.339918.14321520

14782

=−= xxS

445.14167.108320

)75.12)(1478( =−= xyS

32999.0))(0041617512.0(ˆ

00416.06.33991

445.141ˆ

201478

2075.12

0

1

=−=

===

β

β xx

xy

S

S

x y 00416.032999.0ˆ +=

00796.018

143275.0

2ˆ 2 ==

−==

n

SS MS E

E σ



11-2

0

0.1

0.20.3

0.4

0.5

0.6

0.7

0.8

-50 0 50 100

x

y

b) 6836.0)85(00416.032999.0ˆ =+= y

c) 7044.0)90(00416.032999.0ˆ =+= y

d) 00416.0ˆ1 = β

11-3 a) The regression equation is Rating Pts = - 5.56 + 12.7 Yds per Att

Predictor Coef SE Coef T P

Constant -5.558 9.159 -0.61 0.549

Yds per Att 12.652 1.243 10.18 0.000

S = 5.71252 R-Sq = 78.7% R-Sq(adj) = 78.0%

Analysis of Variance

Source DF SS MS F P

Regression 1 3378.5 3378.5 103.53 0.000

Residual Error 28 913.7 32.6

Total 29 4292.2

iii x y ε β β ++= 10

106.2130

)55.219(847.1627

2

=−= xxS

037.26730

)2611)(55.219(21.19375 =−= xyS

56.530

55.219)652.12(30

2611ˆ

652.12106.21

037.267ˆ

0

1

−=⎟ ⎠

⎞⎜⎝

⎛ −=

===

β

β xx

xy

S

S

x y 652.1256.5ˆ +−=

6.3228

7.913

2ˆ 2 ==

−==

n

SS MS E

E σ



11-3

Yds per At t

R a t i n g

P t s

98765

120

110

100

90

80

70

60

S 5.71252

R-Sq 78.7%

R-Sq(adj) 78.0%

Fitted Line PlotRating Pts = - 5.558 + 12.65 Yds per Att

b) 33.89)5.7(652.1256.5ˆ =+−= y

c) 652.12ˆ1 −=− β

d) 79.010652.12

1=×

e) 66.85)21.7(652.1256.5ˆ =+−= y

56.766.851.78

ˆ

−=−=

−= y ye

11-4 a)Regression Analysis - Linear model: Y = a+bX

Dependent variable: SalePrice Independent variable: Taxes

--------------------------------------------------------------------------------

Standard T Prob.

Parameter Estimate Error Value Level

Intercept 13.3202 2.57172 5.17948 .00003

Slope 3.32437 0.390276 8.518 .00000

--------------------------------------------------------------------------------


Source Sum of Squares Df Mean Square F-Ratio Prob. Level

Model 636.15569 1 636.15569 72.5563 .00000

Residual 192.89056 22 8.76775

--------------------------------------------------------------------------------

Total (Corr.) 829.04625 23

Correlation Coefficient = 0.875976 R-squared = 76.73 percent

Stnd. Error of Est. = 2.96104

76775.8ˆ 2 =σ

If the calculations were to be done by hand, use Equations (11-7) and (11-8). x y 32437.33202.13ˆ +=

b) 253.38)5.7(32437.33202.13ˆ =+= y

c) 9273.32)8980.5(32437.33202.13ˆ =+= y

0273.29273.329.30ˆ

9273.32ˆ

−=−=−=

=

y ye

y



11-4

d) All the points would lie along a 45 degree line. That is, the regression model would estimate the values exactly. At

this point, the graph of observed vs. predicted indicates that the simple linear regression model provides a reasonable

fit to the data.

25 30 35 40 45 50

Observed

25

30

35

40

45

50

P r e d i c t e d

Plot of Observed values versus predicted



11-5

11-5 a)Regression Analysis - Linear model: Y = a+bX

Dependent variable: Usage Independent variable: Temperature

--------------------------------------------------------------------------------

Standard T Prob.

Parameter Estimate Error Value Level

Intercept -6.3355 1.66765 -3.79906 .00349

Slope 9.20836 0.0337744 272.643 .00000

--------------------------------------------------------------------------------


Source Sum of Squares Df Mean Square F-Ratio Prob. Level

Model 280583.12 1 280583.12 74334.4 .00000

Residual 37.746089 10 3.774609

--------------------------------------------------------------------------------

Total (Corr.) 280620.87 11

Correlation Coefficient = 0.999933 R-squared = 99.99 percent

Stnd. Error of Est. = 1.94284

7746.3ˆ 2 =σ

If the calculations were to be done by hand, use Equations (11-7) and (11-8).

x y 20836.93355.6ˆ +−=

b) 124.500)55(20836.93355.6ˆ =+−= y

c) If monthly temperature increases by 1 F, y increases by 9.20836.

d) 458.426)47(20836.93355.6ˆ =+−= y

.y = 426458

618.1458.42684.424ˆ −=−=−= y ye

11-6 a)The regression equation is MPG = 39.2 - 0.0402 Engine Replacement


Constant 39.156 2.006 19.52 0.000

Engine Replacement -0.040216 0.007671 -5.24 0.000

S = 3.74332 R-Sq = 59.1% R-Sq(adj) = 57.0%


Source DF SS MS F P

Regression 1 385.18 385.18 27.49 0.000


Total 20 651.41

01.14ˆ 2 =σ

x y 0402.02.39ˆ −=

b) 17.33)150(0402.02.39ˆ =−= y

c) 2956.34ˆ = y

0044.72956.343.41ˆ =−=−= y ye



11-6

11-7 a)

1050950850

60

50

40

x

y

Predictor Coef StDev T PConstant -16.509 9.843 -1.68 0.122x 0.06936 0.01045 6.64 0.000S = 2.706 R-Sq = 80.0% R-Sq(adj) = 78.2% Analysis of VarianceSource DF SS MS F PRegression 1 322.50 322.50 44.03 0.000Error 11 80.57 7.32Total 12 403.08

3212.7ˆ 2 =σ

x y 0693554.05093.16ˆ +−=

b) 6041.46ˆ = y 39592.1ˆ =−= y ye

c) 38.49)950(0693554.05093.16ˆ =+−= y

11-8 a)

10090807060

9

8

7

65

4

3

2

1

0

x

y

Yes, a linear regression would seem appropriate, but one or two points might be outliers.


Constant -10.132 1.995 -5.08 0.000

x 0.17429 0.02383 7.31 0.000

S = 1.318 R-Sq = 74.8% R-Sq(adj) = 73.4%


Source DF SS MS F P

Regression 1 92.934 92.934 53.50 0.000


Total 19 124.200

b) 737.1ˆ 2 =σ and x y 17429.0132.10ˆ +−=

c) 68265.4ˆ = y at x = 85



11-7

11-9 a)

403020100

250

200

150

100

x

y

Yes, a linear regression model appears to be plausible.

Predictor Coef StDev T PConstant 234.07 13.75 17.03 0.000x -3.5086 0.4911 -7.14 0.000S = 19.96 R-Sq = 87.9% R-Sq(adj) = 86.2% Analysis of VarianceSource DF SS MS F PRegression 1 20329 20329 51.04 0.000Error 7 2788 398

Total 8 23117

b) 25.398ˆ 2 =σ and x y 50856.3071.234ˆ −=

c) 814.128)30(50856.3071.234ˆ =−= y

d) 1175.15883.156ˆ == e y

11-10 a)

1.81.61.41.21.00.80.60.40.20.0

40

30

20

10

0

x

y

Yes, a simple linear regression model seems appropriate for these data.

Predictor Coef StDev T PConstant 0.470 1.936 0.24 0.811x 20.567 2.142 9.60 0.000S = 3.716 R-Sq = 85.2% R-Sq(adj) = 84.3%


Source DF SS MS F PRegression 1 1273.5 1273.5 92.22 0.000Error 16 220.9 13.8Total 17 1494.5



11-8

b) 81.13ˆ 2 =σ

x y 5673.20470467.0ˆ +=

c) 038.21)1(5673.20470467.0ˆ =+= y

d) 6629.11371.10ˆ == e y

11-11 a)

2520151050

2600

2100

1600

x

y

Yes, a simple linear regression (straight-line) model seems plausible for this situation.

Predictor Coef StDev T PConstant 2625.39 45.35 57.90 0.000x -36.962 2.967 -12.46 0.000

S = 99.05 R-Sq = 89.6% R-Sq(adj) = 89.0%

Analysis of VarianceSource DF SS MS F PRegression 1 1522819 1522819 155.21 0.000Error 18 176602 9811Total 19 1699421

b) 2.9811ˆ 2 =σ

x y 962.3639.2625ˆ −=

c) 15.1886)20(962.3639.2625ˆ =−= y

d) If there were no error, the values would all lie along the 45

line. The plot indicates age is reasonable

regressor variable.

1600 2100 2600

1700

1800

1900

2000

2100

2200

2300

2400

2500

2600

y

F I T S

1



11-9

11-12

a)The regression equation is

Porosity = 55.6 - 0.0342 Temperature


Constant 55.63 32.11 1.73 0.144Temperature -0.03416 0.02569 -1.33 0.241

S = 8.79376 R-Sq = 26.1% R-Sq(adj) = 11.3%


Source DF SS MS F P

Regression 1 136.68 136.68 1.77 0.241


Total 6 523.33

x y 03416.063.55ˆ −=

33.77ˆ 2 =σ

b) 806.7)1400(03416.063.55ˆ =−= y

c) 39.4ˆ = y 012.7=e

d)

Temperature

P o r o s i t y

15001400130012001100

30

25

20

15

10

5

0

Scatter plot of Porosity vs Temperatur e

The simple linear regression model doesn’t seem appropriate because the scatter plot doesn’t show a linear

relationship between the data.

11-13 a) The regression equation is BOD = 0.658 + 0.178 Time

Predictor Coef SE Coef T PConstant 0.6578 0.1657 3.97 0.003

Time 0.17806 0.01400 12.72 0.000

S = 0.287281 R-Sq = 94.7% R-Sq(adj) = 94.1%


Source DF SS MS F P

Regression 1 13.344 13.344 161.69 0.000



11-10


Total 10 14.087

x y 178.0658.0ˆ +=

083.0ˆ 2 =σ

b) 328.3)15(178.0658.0ˆ =+= y

c) 0.178(3) = 0.534

d) 726.1)6(178.0658.0ˆ =+= y

174.0726.19.1ˆ =−=−= y ye

e)

y

y - h a t

4.03.53.02.52.01.51.00.5

4.5

4.0

3.5

3.0

2.5

2.0

1.5

1.0

Scatt erplot of y-hat vs y

All the points would lie along the 45 degree line y y ˆ= . That is, the regression model would estimate the

values exactly. At this point, the graph of observed vs. predicted indicates that the simple linear regression

model provides a reasonable fit to the data.

11-14 a)The regression equation is

Deflection = 32.0 - 0.277 Stress level


Constant 32.049 2.885 11.11 0.000

Stress level -0.27712 0.04361 -6.35 0.000

S = 1.05743 R-Sq = 85.2% R-Sq(adj) = 83.1%


Source DF SS MS F P

Regression 1 45.154 45.154 40.38 0.000

Residual Error 7 7.827 1.118Total 8 52.981

118.1ˆ 2 =σ



11-11

Deflection

S t r e s s

l e v e l

191817161514131211

75

70

65

60

55

50

Graph of Stress level v s Deflection

b) 045.14)65(277.005.32ˆ =−= y

c) (-0.277)(5) = -1.385

d) 61.3277.0

1=

e) ˆ 32.05 0.277(68) 13.214 y = − = ˆ 11.640 13.214 1.574e y y= − = − =

11-15

x

y

35302520151050

2.7

2.6

2.5

2.4

2.3

2.2

2.1

2.0

1.9

1.8

Scatt erplot of y vs x

It’s possible to fit this data with straight-line model, but it’s not a good fit. There is a curvature shown on

the scatter plot.

a)The regression equation is

y = 2.02 + 0.0287 x


x 0.028718 0.003966 7.24 0.000

S = 0.159159 R-Sq = 67.7% R-Sq(adj) = 66.4%


Source DF SS MS F P



11-12

Regression 1 1.3280 1.3280 52.42 0.000


Total 26 1.9613

x y 0287.002.2ˆ +=

0253.0ˆ 2 =σ

b) 3357.2)11(0287.002.2ˆ =+= y

c)

y-hat

y

2.92.82.72.62.52.42.32.22.12.0

2.7

2.6

2.5

2.4

2.3

2.2

2.1

2.0

1.9

1.8

Scatterpl ot of y vs y-hat

If the relationship between length and age was deterministic, the points would fall on the 45 degree

line y y ˆ= . Because the points in this plot vary substantially from this line, it appears that age is not a

reasonable choice for the regressor variable in this model.

11-16 a) )32(0041612.03299892.0ˆ5

9 ++= x y

x y

x y

0074902.04631476.0ˆ

1331584.00074902.03299892.0ˆ

+=++=

b) 00749.0ˆ1 = β

11-17 Let x = engine displacement (cm3) and xold = engine displacement (in3)

a) The old regression equation is y = 39.2 - 0.0402xold

Because 1 in3 = 16.387 cm3, the new regression equation is

x x y 0025.02.39)387.16/(0402.02.39ˆ −=−=

b) 0025.0ˆ1 −= β

11-18 y x x y x =+−=+ 1110 ˆ)ˆ(ˆˆ β β β β

11-19 a) The slopes of both regression models will be the same, but the intercept will be shifted.

b) x y 9618.3641.2132ˆ −=

9618.36ˆ

39.2625ˆ

1

0

−=

=

β

β vs.

9618.36ˆ

41.2132ˆ

1

0

−=

=∗

∗

β

β



11-13

11-20 a) The least squares estimate minimizes ∑ − 2)( ii x y β . Upon setting the derivative equal to zero, we obtain

0][2)()(22 =+−=−− ∑∑∑ iiiiii x x y x x y β β

Therefore,

∑∑=

2ˆ

i

ii

x

x y β .

b) x y 031461.21ˆ = . The model seems very appropriate—an even better fit.

0

5

10

15

20

25

30

35

40

45

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

watershed

c h l o r i d e



11-14

Section 11-4

11-21 a) 1) The parameter of interest is the regressor variable coefficient, β1

2) H0 1 0:β =

3) H1 1 0:β ≠

4) α = 0.05


)2/(

1/0 −

==nSS

SS

MS

MS f

E

R

E

R

6) Reject H0 if f 0 > f α,1,12 where f 0.05,1,12 = 4.75

7) Using results from Exercise 11-1

123.22

59143.13771429.159

59.137

)057143.59(3298017.2ˆ1

=

−=

−=

=

−−==

R yy E

xy R

SSSSS

SSS β

63.7412/123.22

59.1370 == f

8) Since 74.63 > 4.75 reject H 0 and conclude that compressive strength is significant in predicting intrinsic

permeability of concrete at α = 0.05. We can therefore conclude that the model specifies a useful linear

relationship between these two variables.

P value− ≅ 0 000002.

b) 8436112

12322

2

2 ..

n

SS MSˆ E

E ==−

==σ and 2696.03486.25

8436.1ˆ)ˆ(

2

1 === xxS

seσ

β

c) 9043.03486.25

0714.3

14

18436.1

1ˆ)ˆ(

222

0 =⎥

⎦

⎤⎢

⎣

⎡+=⎥

⎦

⎤⎢

⎣

⎡+=

xx

S

x

nse σ β

11-22 a) 1) The parameter of interest is the regressor variable coefficient, β1.

2) H0 1 0:β =

3) H1 1 0:β ≠

4) α = 0.05


)n /(SS

/ SS

MS

MS f

E

R

E

R

2

10 −

==


7) Using the results from Exercise 11-2

SS S

SS S SS

R xy

E yy R

= =

== −

= − −

=

( . )( . )

.

( . ) .

.

.

β1

127520

0 0041612 141445

05886

886 05886

0143275

2

9573181432750

588600 .

/ .

. f ==

8) Since 73.95 > 4.414, reject H 0 and conclude the model specifies a useful relationship at α = 0.05.

P value− ≅ 0 000001.



11-15

b)4

2

1 108391.46.33991

00796.ˆ)ˆ( −=== x

Sse

xx

σ β

04091.06.33991

9.73

20

100796.

1ˆ)ˆ(

22

0=⎥

⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

xxS

x

nse σ β


Rating Pts = - 5.558 + 12.65 Yds per Att

S = 5.71252 R-Sq = 78.7% R-Sq(adj) = 78.0%


Source DF SS MS F P

Regression 1 3378.53 3378.53 103.53 0.000

Error 28 913.72 32.63

Total 29 4292.25

Refer to ANOVA

01.0

0:

0:

11

10

=

≠=

α

β

β

H

H

Because p-value = 0.000 < α = 0.01, reject H0. We can conclude that there is a linear relationship between

these two variables.

b) 63.32ˆ 2 =σ

243.1106.21

63.32ˆ)ˆ(

2

1 === xxS

seσ

β

159.9106.21

318.7

30

163.32

1ˆ)ˆ(

222

0=⎥

⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

xxS

x

nse σ β

c) 1)The parameter of interest is the regressor variable coefficient, β1.

2) H0 1 10:β =

3) H1 1 10:β ≠

4) α = 0.01

5) The test statistic is)ˆ(

ˆ

1

0,11

0 β

β β

set

−=

6) Reject H0 if t0 < −tα/2,n-2 where −t0.005,28 = −2.763 or t0 > t0.005,28 = 2.763


134.2243.1

10652.120 =

−=t

8) Because 2.134 < 2.763, fail to reject H 0 . There is not enough evidence to conclude that the slope

differs from 10 at α = 0.01.

11-24 Refer to ANOVA of Exercise 11-4

a) 1) The parameter of interest is the regressor variable coefficient, β1.

2) H0 1 0:β =



11-16

3) H1 1 0:β ≠

4) α = 0.05, using t-test

5) The test statistic is tse

01

1

=

( )

β

β



t0

332437

0 3902768518= =

.

..

8) Since 8.518 > 2.074 reject H 0 and conclude the model is useful α = 0.05.

b) 1) The parameter of interest is the slope, β1

2) H0 1 0:β =

3) H1 1 0:β ≠

4) α = 0.05

5) The test statistic is f MS

MS

SS

SS n

R

E

R

E0

1

2= =

−

/

/ ( )



f 063615569 1

19289056 2272 5563= =. /

. /.

8) Since 72.5563 > 4.303, reject H 0 and conclude the model is useful at a significance α = 0.05.

The F-statistic is the square of the t-statistic. The F-test is a restricted to a two-sided test, whereas the

t-test could be used for one-sided alternative hypotheses.

c) 39027563157

767582

1 ..

.

S

ˆ )ˆ (se

xx

==σ

=β

57172563157

40496

24

176758

1 22

0 ..

..

S

x

nˆ )ˆ (se

xx

=⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+σ=β

d) 1) The parameter of interest is the intercept, β0.

2) 000 =β: H

3) 001 ≠β: H

4) α = 0.05, using t-test

5) The test statistic is)ˆ (se

ˆ t

0

00 β

β=



179.55717.2

3201.130 ==t

8) Since 5.179 > 2.074 reject H 0 and conclude the intercept is not zero at α = 0.05.



11-17

11-25 Refer to ANOVA for Exercise 10-65

a) 1) The parameter of interest is the regressor variable coefficient, β1.

2) H0 1 0:β =

3) H1 1 0:β ≠

4) α = 0.01

5) The test statistic is )2/(

1/0 −== nSS

SS

MS

MS

f E

R

E

R



4.7433410/746089.37

1/12.2805830 == f

8) Since 74334.4 > 10.049, reject H 0 and conclude the model is useful α = 0.01. P-value < 0.000001

b) se( 1β̂ ) = 0.0337744, se( 0β̂ ) = 1.66765

c) 1) The parameter of interest is the regressor variable coefficient, β1.

2) H0 1 10:β =

3) H1 1 10:β ≠

4) α = 0.01


)ˆ(

ˆ

1

0,11

0

β

β β

set

−=



37.230338.0

1021.90 −=

−=t

8) Since −23.37 < −3.17 reject H 0 and conclude the slope is not 10 at α = 0.01. P-value ≈ 0.

d) H0: β0 = 0 H1: β0 ≠ 0

8.366765.1

03355.60 −=

−−=t

P-value < 0.005. Reject H0 and conclude that the intercept should be included in the model.

11-26 Refer to ANOVA table of Exercise 11-6

a) 010 =β: H

011 ≠β: H α = 0.01

18,1,0

18,1,01.0

0

285.8

53158.4

α f f

f

f

>/

=

=

Therefore, do not reject H0. P-value = 0.04734. Insufficient evidence to conclude that the model is a useful

relationship.

b) 016628101 .)ˆ (se =β

6139620 .)ˆ (se =β



11-18

c) 05010 .: H −=β

05011 .: H −<β

α = 0.01

180

1801

0

5522

87803001662810

05003540

,

,.

t t

.t

..

).(.t

α−</

=

=−−−

=

Therefore, do not reject H0. P-value = 0.804251. Insufficient evidence to conclude that β1 is < -0.05.

d) 0: 00 = β H 001 ≠β: H α = 0.01

18,2/0

18,005.

0

878.2

8291.12

α t t

t

t

>

=

=

Therefore, reject H0. P-value ≅ 0


a) 010 =β: H

011 ≠β: H

α = 0.05

1110

11105

0

844

027944

, ,

, ,.

f f

. f

. f

α>

=

=

Therefore, reject H0. P-value = 0.00004.

b) 010452401 .)ˆ (se =β

8434690 .)ˆ (se =β

c) 000 =β: H

001 ≠β: H

α = 0.05

1120

11025

0

2012

677181

, /

,.

t |t |

.t

.t

α−</

=

−=

Therefore, do not rejectH0. P-value = 0.12166.



11-19


a) 010 =β: H

011 ≠β: H

α = 0.05

18,1,0

18,1,05.

0

414.4

53.50

α f f

f

f

>=

=


b) 025661301 .)ˆ (se =β

1352620 .)ˆ (se =β

c) 000 =β: H

001 ≠β: H

α = 0.05

18,2/0

18,025.

0

||

101.2

5.079-

α t t

t

t

>=

=



a) 010 =β: H

011 ≠β: H

α = 0.01

18,1,0

18,1,01.

0

285.8

2.155

α f f

f

f

>

=

=

Therefore, reject H0. P-value < 0.00001.

b) 3468451 .)ˆ (se =β

9668120 .)ˆ (se =β

c) 3010 −=β: H

3011 −≠β: H

α = 0.01

1820

18005

0

8782

34662966812

30961836

, /

,.

t |t |.t

..

)(.t

α−>/ =

−=−−−

=

Therefore, do not reject H0. P-value = 0.0153(2) = 0.0306.

d) 000 =β: H

001 ≠β: H

α = 0.01



11-20

8782

895757

18005

0

.t

.t

,. =

=

t t0 2 18> α / , , therefore, reject H0. P-value < 0.00001.

e) H0 0 2500:β =

250001 >β: H α = 0.01

552.2

7651.23468.45

250039.2625

18,01.

0

=

=−

=

t

t

18,0 α t t > , therefore reject H0 . P-value = 0.0064.


a) 010 =β: H

011 ≠β: H

α = 0.01

16,1,0

16,1,01.

0

531.8

224.92

α f f

f

f

>

=

=

Therefore, reject H0 .

b) P-value < 0.00001

c) 1416921 .)ˆ (se =β

9359110 .)ˆ (se =β

d) 000 =β: H

001

≠β: H

α = 0.01

1620

16005

0

9212

2430

, /

,.

t t

.t

.t

α>/

=

=

Therefore, do not reject H0. There is not sufficient evidence to conclude that the intercept differs from zero. Based

on this test result, the intercept could be removed from the model.



11-21

11-31 a) Refer to ANOVA of Exercise 11-13

01.0

0:

0:

11

10

=

≠

=

α

β

β

H

H

Because the P-value = 0.000 < α = 0.01, reject H0. There is evidence of a linear relationship between these

two variables.

b) 083.0ˆ 2 =σ

The standard errors for the parameters can be obtained from the computer output or calculated as follows.

014.091.420

083.0ˆ)ˆ(

2

1 === xxS

seσ

β

1657.091.420

09.10

11

1083.0

1ˆ)ˆ(

222

0 =⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

xxS

x

nse σ β

c)1) The parameter of interest is the intercept β0.

2) 0: 00 = β H

3) 0: 01 ≠ β H

4) 01.0=α

5) The test statistic is)( 0

00

β

β

set =

6) Reject H0 if 2,2/0 −−< nt t α where 250.39,005.0 −=− t or 2,2/0 −> nt t α where 250.39,005.0 =t


97.31657.0

6578.00 ==t

8) Because t0 = 3.97 > 3.250 reject H0 and conclude the intercept is not zero at α = 0.01.

11-32 a) Refer to ANOVA of Exercise 11-14

01.0

0:

0:

11

10

=

≠

=

α

β

β

H

H

Because the P-value = 0.000 < α = 0.01, reject H0. There is evidence of a linear relationship between these

two variables.

b) Yes

c) 118.1ˆ 2 =σ

0436.0588

118.1ˆ)ˆ(

2

1 === xxS

seσ

β



11-22

885.2588

67.65

9

1118.1

1ˆ)ˆ(

222

0 =⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

xxS

x

nse σ β

11-33 a)

05.0

0:

0:

11

10

=

≠

=

α

β

β

H

H

Because the P-value = 0.310 > α = 0.05, fail to reject H0. There is not sufficient evidence of a linear

relationship between these two variables.

The regression equation is

BMI = 13.8 + 0.256 Age


Constant 13.820 9.141 1.51 0.174

Age 0.2558 0.2340 1.09 0.310

S = 5.53982 R-Sq = 14.6% R-Sq(adj) = 2.4%


Source DF SS MS F P

Regression 1 36.68 36.68 1.20 0.310


Total 8 251.51

b) 141.9)ˆ(,2340.0)ˆ(,69.30ˆ01

2 === β β σ sese from the computer output

c) 141.9342.560

256.38

9

169.30

1ˆ)ˆ(

222

0 =⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

xxS

x

nse σ β

11-34

xxSt

/ˆ

ˆ2

10

σ

β = After the transformation 11ˆˆ β β

ab=∗ , xx xx SaS 2=∗ , 00

ˆˆ, β β b xa x == ∗∗ , and

σ σ ˆˆ b=∗. Therefore,

022

10

/)ˆ(

/ˆt

Sab

abt

xx

==∗

σ

β .

11-35 3878.0106.21/295.5

|)5.12(10|=

−=d

Assume α = 0.05, from Chart VI and interpolating between the curves for n = 20 and n = 30, 70.0≅ β .

11-36 a)

∑ 2

2ˆ

ˆ

i xσ

β has a t distribution with n-1 degree of freedom.

b) From Exercise 11-17, ,611768.3ˆ,031461.21ˆ == σ β and 7073.142 =∑ i x .

The t-statistic in part a. is 22.3314 and 0: 00 = β H is rejected at usual α values.



11-23

Sections 11-5 and 11-6

11-37 tα/2,n-2 = t0.025,12 = 2.179a) 95% confidence interval on β1 .

.7423.1.9173.2

)2696.0(179.23298.2

)2696.0(3298.2

)ˆ(ˆ

1

12,025.

12,2/1

−≤≤−

±−

±−

± −

β

β β α

t

set n

b) 95% confidence interval on β0 .

.3115.497145.46

)5959.0(179.20130.48

)ˆ(ˆ

0

012,025.0

≤≤

±

±

β

β β set

c) 95% confidence interval on μ when x0 2 5= . .

0477.43ˆ3293.41

)3943.0(179.21885.42

)(844.1)179.2(1885.42

)(ˆˆ

1885.42)5.2(3298.20130.48ˆ

0

2

20

0

0

|

3486.25

)0714.35.2(

14

1

)(12

12,025.|

|

≤≤

±

+±

+±

=−=

−

−

xY

S

x x

n xY

xY

xxt

μ

σ μ

μ

d) 95% on prediction interval when x0 2 5= . .

2513.451257.39

)4056.1(179.21885.42

)1(844.1179.21885.42

)1(ˆˆ

0

348571.25

)0714.35.2(

141

)(12

12,025.0

2

20

≤≤

±

++±

++±

−

−

y

t y xxS

x x

nσ

It is wider because it depends on both the errors associated with the fitted model and the

future observation.

11-38 tα/2,n-2 = t0.005,18 = 2.878

a) ( ) )ˆ(ˆ118,005.01 β β set ±

0055542.00027682.0

)000484.0)(878.2(0041612.0

1 ≤≤

±

β

b) ( ) )ˆ(ˆ018,005.00 β β set ±

447728.0212250.0

)04095.0)(878.2(3299892.0

0 ≤≤± β

c) 99% confidence interval on μ when x F0 85= .



11-24

7431497.0ˆ6242283.0

0594607.0683689.0

)(00796.0)878.2(683689.0

)(ˆˆ

683689.0ˆ

0

2

20

0

0

|

6.33991

)9.7385(

20

1

)(12

18,005.|

|

≤≤

±

+±

+±

=

−

−

xY

S

x x

n xY

xY

xx

t

μ

σ μ

μ

d) 99% prediction interval when F x900 = .

947256.0420122.0

263567.07044949.0

)1(00796.0878.27044949.0

)1(ˆˆ

7044949.0ˆ

0

6.33991

)9.7390(

201

)(12

18,005.0

0

2

20

≤≤

±

++±

++±

=

−

−

y

t y

y

xxS

x x

nσ

11-39 048.228,025.02,2/ ==− t t nα

a) 95% confidence interval on 1ˆ β

198.15ˆ106.10

)243.1(048.2652.12

)243.1(652.12

)ˆ(ˆ

1

5,025.0

12,2/1

≤≤

±

±

± −

β

β β α

t

set n

b) 95% confidence interval on 0 β

200.13ˆ316.24

)159.9(048.2558.5

)159.9(558.5

)ˆ(ˆ

0

5,025.0

02,2/0

≤≤−

±−

±−

± −

β

β β α

t

set n

c) 95% confidence interval for the mean rating when the average yards per attempt is 8.0

656.95)0.8(652.1256.5ˆ =+−=μ

( )

( )

409.98903.92

)344.1(048.2656.95

106.21

318.78

30

16.32048.2656.95

1ˆˆ

2

2

02

28,025.0

≤≤

±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

μ

σ μ xxS

x x

nt

d) 95% prediction interval on 0.80 = x



11-25

( )

( )

67.107642.83

)866.5(048.2656.95106.21

318.78

30

116.32048.2656.95

11ˆˆ

0

2

2

02

28,025.0

≤≤

±

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −++±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −++±

y

S

x x

nt y

xx

σ

Note for Problems 11-40 through 11-41. These computer printouts were obtained from Statgraphics. For Minitab users, the

standard errors are obtained from the Regression subroutine.

11-40 95 percent confidence intervals for coefficient estimates

--------------------------------------------------------------------------------

Estimate Standard error Lower Limit Upper Limit

CONSTANT 13.3202 2.57172 7.98547 18.6549

Taxes 3.32437 0.39028 2.51479 4.13395

--------------------------------------------------------------------------------

a) 2.51479 ≤ β1 ≤ 4.13395.

b) 7.98547≤ β0 ≤ 18.6549.

c) )(76775.8)074.2(253.38563139.57

)40492.65.7(

24

12−+±

7883.39ˆ7177.36

5353.1253.38

0| ≤≤

±

xY μ

d) 38 253 2 074 8 76775 1 124

7 5 6 4 0492

57 563139

2

. ( . ) . ( )( . . )

.± + + −

5832.449228.31

3302.6253.38

0 ≤≤

±

y

11-41 99 percent confidence intervals for coefficient estimates

--------------------------------------------------------------------------------


CONSTANT -6.33550 1.66765 -11.6219 -1.05011

Temperature 9.20836 0.03377 9.10130 9.93154

--------------------------------------------------------------------------------

a) 9.10130 ≤ β1 ≤ 9.31543

b) −11.6219 ≤ β0 ≤ −1.04911

c) 500 124 2 228 3 774609 112

55 46 5

33089994

2

. ( . ) . ( )( . )

.± + −

500124 14037586498 72024 50152776

0

. .

. .|

±≤ ≤μY x

d) 500 124 2 228 3 774609 1 112

55 46 5

3308 9994

2

. ( . ) . ( )( . )

.± + + −

67456.50457344.495

5505644.4124.500

0 ≤≤

±

y

It is wider because the prediction interval includes errors for both the fitted model and for a future observation.



11-26

11-42 a) 00045.007034.0 1 −≤≤− β

b) 027.390417.28 0 ≤≤ β

c) )(39232.13)101.2(225.28256.48436

)3.149150(

201

2−+±

9794.295406.26

7194236.1225.28

0| ≤≤

±

x yμ

d) )1(39232.13)101.2(225.28256.48436

)3.149150(201 2−++±

1386.363814.20

87863.7225.28

0 ≤≤

±

y

11-43 a) 10183.003689.0 1 ≤≤ β

b) 0691.140877.47 0 ≤≤− β

c) )(324951.7)106.3(6041.4697.67045

)939910(

131

2−+±

)1185.490897.44

514401.26041.46

0| ≤≤

±

x yμ

d) )1(324951.7)106.3(6041.4697.67045

)939910(

131

2−++±

3784.558298.37

779266.86041.46

0 ≤≤

±

y

11-44 a) 22541.011756.0 1 ≤≤ β

b) 32598.53002.14 0 −≤≤− β

c) )(982231.1)101.2(76301.42111.3010

)3.8285(

20

12−+±

4403.50857.4

6772655.076301.4

0| ≤≤

±

x yμ

d) )1(982231.1)101.2(76301.4 2111.3010

)3.8285(

20

12−

++±

7976.77284.1

0345765.376301.4

0 ≤≤

±

y

11-45 a) 590.266552.201 1 ≤≤ β

b) 34696.267015.4 0 −≤≤− β

c) )(2804.398)365.2(814.1284214.1651

)5.2430(

9

12−+±

7941.1458339.111

980124.16814.128

0| ≤≤

±

x yμ

11-46 a) 8239.263107.14 1 ≤≤ β

b) 12594.618501.5 0 ≤≤− β

c) )(8092.13)921.2(038.21 01062.3

)806111.01(

181

2−+±

8694.232066.18

8314277.2038.21

0| ≤≤

±

x yμ

d) )1(8092.13)921.2(038.21 01062.3

)806111.01(

181

2−++±

2559.328201.9

217861.11038.21

0 ≤≤

±

y



11-27

11-47 a) 7272.301964.43 1 −≤≤− β

b) 68.272009.2530 0 ≤≤ β

c) )(21.9811)101.2(154.18866618.1114

)3375.1320(

201

2−+±

5247.19487833.1823

370688.62154.1886

0| ≤≤

±

x yμ

d) )1(21.9811)101.2(154.18866618.1114

)3375.1320(

201

2−++±

4067.21039013.1668

25275.217154.1886

0 ≤≤

±

y

11-48 571.25,025.02,2/ ==− t t nα

a) 95% confidence interval on 1ˆ β

033.0ˆ101.0)026.0(571.2034.0

)026.0(034.0

)ˆ(ˆ

1

5,025.0

12,2/1

≤≤−±−

±−

± −

β

β β α

t

set n


15.138ˆ89.26

)11.32(571.263.55

)11.32(63.55

)ˆ(ˆ

0

5,025.0

02,2/0

≤≤−

±

±

± −

β

β β α

t

set n

c) 95% confidence interval for the mean length when x=1500:63.4)1500(034.063.55ˆ =−=μ

( )

( )

63.439.14

)396.7(571.263.4

8.117142

86.12421500

7

133.77571.263.4

1ˆˆ

2

2

02

5,025.0

≤≤−

±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

μ

σ μ xxS

x x

nt

d) 95% prediction interval when 15000 = x



11-28

( )

( )

17.3491.24

)49.11(571.263.48.117142

86.12421500

7

1133.77571.263.4

11ˆˆ

0

2

2

02

5,025.0

≤≤−

±

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −++±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −++±

y

S

x x

nt y

xx

σ

It’s wider because it depends on both the error associated with the fitted model as well as that of the future

observation.

11-49 250.39,005.02,2/ ==− t t nα

a) 99% confidence interval for 1ˆ β

2235.0ˆ1325.0

)014.0(250.3178.0

)014.0(178.0

)ˆ(ˆ

1

9,005.0

12,2/1

≤≤

±

±

± −

β

β β α

t

set n


196.1ˆ119.0

)1657.0(250.36578.0

)1657.0(6578.0

)ˆ(ˆ

0

9,005.0

02,2/0

≤≤

±

±

± −

β

β β α

t

set n

c) 95% confidence interval on μ when 80 = x

( )

( )

29.287.1

91.420

09.108

11

1083.0262.2082.2

1ˆˆ

082.2)8(178.0658.0ˆ

0

0

0

|

2

2

02

9,025.0|

|

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

=+=

x y

xx

x y

x y

S

x x

nt

μ

σ μ

μ



11-29

Section 11-7

11-50 ( ) 8617.071.159

35.25330.2ˆ 22

1

2=−==

YY

XX

S

S R β

The model accounts for 86.17% of the variability in the data.

11-51 a) 787.02= R


b) There is no major departure from the normality assumption in the following graph.

Residual

P e r

c e n t

1050-5-10-15

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probabil i ty Plot of t he Residuals(response is Rating Pts)

c) Assumption of constant variance appears reasonable.

Fitt ed Value

R e s i d u a l

11010090807060

10

5

0

-5

-10

Residuals Versus the Fitted Values(response is Rating Pts)

11-52 Use the Results of exercise 11-5 to answer the following questions.

a) SalePrice Taxes Predicted Residuals

25.9 4.9176 29.6681073 -3.76810726

29.5 5.0208 30.0111824 -0.51118237

27.9 4.5429 28.4224654 -0.52246536

25.9 4.5573 28.4703363 -2.57033630

29.9 5.0597 30.1405004 -0.24050041



11-30

29.9 3.8910 26.2553078 3.64469225

30.9 5.8980 32.9273208 -2.02732082

28.9 5.6039 31.9496232 -3.04962324

35.9 5.8282 32.6952797 3.20472030

31.5 5.3003 30.9403441 0.55965587

31.0 6.2712 34.1679762 -3.16797616

30.9 5.9592 33.1307723 -2.23077234

30.0 5.0500 30.1082540 -0.10825401

36.9 8.2464 40.7342742 -3.83427422

41.9 6.6969 35.5831610 6.3168390140.5 7.7841 39.1974174 1.30258260

43.9 9.0384 43.3671762 0.53282376

37.5 5.9894 33.2311683 4.26883165

37.9 7.5422 38.3932520 -0.49325200

44.5 8.7951 42.5583567 1.94164328

37.9 6.0831 33.5426619 4.35733807

38.9 8.3607 41.1142499 -2.21424985

36.9 8.1400 40.3805611 -3.48056112

45.8 9.1416 43.7102513 2.08974865

b) Assumption of normality does not seem to be violated since the data appear to fall along a straight line.

-4 -2 0 2 4 6 8

Residuals

Normal Probability Plot

0.1

1

5

20

50

80

95

99

99.9

c u m u l a t i v e p e r c e n t

c) There are no serious departures from the assumption of constant variance. This is evident by the random pattern of the residuals.

26 29 32 35 38 41 44

Predicted Values

-4

-2

0

2

4

6

8

R e s i d u a l s

Plot of Residuals versus Predicted

3.8 4.8 5.8 6.8 7.8 8.8 9.8

Taxes

-4

-2

0

2

4

6

8

R e s i d u a l s

Plot of Residuals versus Taxes

d) %73.762≡ R ;

11-53 Use the results of Exercise 11-5 to answer the following questions

a) R 2 99 986%= . ; The proportion of variability explained by the model.

b) Yes, normality seems to be satisfied because the data appear to fall along the straight line.



11-31

-2.6 -0.6 1.4 3.4 5.4

Residuals

Normal Probability Plot

0.1

1

5

20

50

80

95

99

99.9

c u m u l a t i v

e p e r c e n t

c) There might be lower variance at the middle settings of x. However, this data does not indicate a serious departure

from the assumptions.

180 280 380 480 580 680

Predicted Values

-2.6

-0.6

1.4

3.4

5.4

R e s i d u a l s

Plot of Residuals versus Predicted

21 31 41 51 61 71 81

Temperature

-2.6

-0.6

1.4

3.4

5.4

R e s i d u a l s

Plot of Residuals versus Temperature

11-54 a) %1121.202= R

b) These plots might indicate the presence of outliers, but no real problem with assumptions.

300200100

10

0

-10

x

R e s i d u a l

ResidualsVersusx(responsei s y)

313029282726252423

10

0

-10

FittedValue

Residual

Residuals Versusthe Fitted Values

(responseis y)

c) The normality assumption appears marginal.



11-32

210-1-2

10

0

-10

Normal Score

R e s i d u a l

Normal Probability Plot of the Residuals(response is y)

11-55 a) 879397.02= R

b) No departures from constant variance are noted.

403020100

30

20

10

0

-10

-20

-30

x

R e s i d u a l

Residuals Versusx(response is y)

23018013080

30

20

10

0

-10

-20

-30

FittedValue

R e s i d u a l

ResidualsVersusthe Fitted Values(response is y)

c) Normality assumption appears reasonable.

1.51.00.50.0-0.5-1.0-1.5

30

20

10

0

-10

-20

-30

Normal Score

R e s i d u a l


11-56 a) %27.712= R

b) No major departure from normality assumptions.

210-1-2

3

2

1

0

-1

-2

Normal Score

R e s i d u a l


c) Assumption of constant variance appears reasonable.



11-33

10090807060

3

2

1

0

-1

-2

x

R e s i d u a l

Residuals Versus x(response is y)

876543210

3

2

1

0

-1

-2

FittedValue

R e s i d u a l

Residuals Versusthe Fitted Values(response is y)

11-57 a) %22.852= R

b) Assumptions appear reasonable, but there is a suggestion that variability increases slightly with y .

1.81.61.41.21.00.80.60.40.20.0

5

0

-5

x

R e s i d u a l

Residuals Versus x(response is y)

403020100

5

0

-5

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is y)

c) Normality assumption may be questionable. There is some “bending” away from a line in the tails of the normal

probability plot.

210-1-2

5

0

-5

Normal Score

R e s i d u a l



Compressive Strength = - 2150 + 185 Density


Constant -2149.6 332.5 -6.46 0.000

Density 184.55 11.79 15.66 0.000

S = 339.219 R-Sq = 86.0% R-Sq(adj) = 85.6%


Source DF SS MS F P

Regression 1 28209679 28209679 245.15 0.000

Residual Error 40 4602769 115069

Total 41 32812448



11-34

b) Because the P-value = 0.000 < α = 0.05, the model is significant.

c) 115069ˆ 2=σ

d) %97.858597.032812448

2820967912

===−==T

E

T

R

SS

SS

SS

SS R


e)

Residual

P e r c e n t

10005000-500-1000

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probability Plot of t he Residuals(response is Compressive Strength)

No major departure from the normality assumption.

f)

Fitt ed Value

R e s i d

u a l

50004000300020001000

1000

500

0

-500

-1000

Residuals Versus the Fit ted Values(response is Compressive Strength)

Assumption of constant variance appears reasonable.

11-59 a) 896081.02= R 89% of the variability is explained by the model.

b) Yes, the two points with residuals much larger in magnitude than the others seem unusual.



11-35

1000-100-200

2

1

0

-1

-2

N o r m a l S c o r e

Residual

Normal Probability Plot of the Residuals

(response is y)

c) 9573.02

modelew =n R

Larger, because the model is better able to account for the variability in the data with these two outlying data pointsremoved.

d) 21.9811ˆ 2

modeld =olσ

93.4022ˆ 2

modelnew =σ

Yes, reduced more than 50%, because the two removed points accounted for a large amount of the error.

11-60 a)

1050950850

60

50

40

x

y

x y 0521753.0677559.0ˆ +=

b) 0: 10 = β H 0: 11 ≠ β H α = 0.05

12,1,0

12,1,05.

0

75.4

9384.7

α f f

f

f

>

=

=

Reject Ho.

c) 23842.25ˆ 2=σ

d) 502.7ˆ2

=origσ

The new estimate is larger because the new point added additional variance that was not accounted for by the model.

e) ˆ 0.677559 0.0521753(855) 45.287 y = + =

ˆ 59 45.287 13.713e y y= − = − =

Yes, e14 is especially large compared to the other residuals.

f) The one added point is an outlier and the normality assumption is not as valid with the point included.



11-36

210-1-2

10

0

-10

Normal Score

R

e s i d u a l


g) Constant variance assumption appears valid except for the added point.

1050950850

10

0

-10

x

R e s i d u a l


555045

10

0

-10

FittedValue

R e s i d u a l


11-61 Yes, when the residuals are standardized the unusual residuals are easier to identify.

1.0723949 -0.7005724 -0.1382218 0.6488043 -2.3530824 -2.1686819 0.4684192

0.4242137 0.0982116 0.5945823 0.2051244 -0.8806396 0.7921072 0.7242900

-0.4737526 0.9432146 0.1088859 0.3749259 1.0372187 -0.7774419

11-62 For two random variables X1 and X2,

),(2)()()( 212121 X X Cov X V X V X X V ++=+

Then,

[ ][ ] [ ]

[ ])(1

2

2)ˆˆ(

)ˆ,(2)ˆ()()ˆ(

2

22

2

)(12

)(12)(122

)(12

10

2

xx

i

xx

i

xx

i

xx

i

S

x x

n

S

x x

nS

x x

n

S

x x

ni

iiiiii

xV

Y Y CovY V Y V Y Y V

−

−−

−

+−=

+−++=

+−++=

−+=−

σ

σ σ σ

σ β β σ

a) Because ei is divided by an estimate of its standard error (when2σ is estimated by σ

2 ), r i has approximately unit

variance.

b) No, the term in brackets in the denominator is necessary.

c) If xi is near x and n is reasonably large, r i is approximately equal to the standardized residual.

d) If xi is far from x , the standard error of ei is small. Consequently, extreme points are better fit by least squares

regression than points near the middle range of x. Because the studentized residual at any point has variance of

approximately one, the studentized residuals can be used to compare the fit of points to the regression line over therange of x.

11-63 Using ,12

yy

E

S

SS R −=

2

2

0ˆ

)1)(2(

σ

E yy

n

SS

E yy

S

SS

S

SSSSSSSSn

F E

yy

E

yy

E −

=−

=−−

=−

Also,



11-37

22

1

22

1

2

1

22

1

2

1

2

10

)(ˆ

)(ˆ)(

))((ˆ2)(ˆ)(

))(ˆ(

)ˆˆ(

x xSSS

x x y y

x x y y x x y y

x x y y

x ySS

i E yy

ii

iiii

ii

ii E

−=−

−−−=

−−−−+−=

−−−=

−−=

∑

∑∑

∑∑∑

∑

∑

β

β

β β

β

β β

Therefore, 2

02

2

10

/ˆ

ˆt

SF

xx

==σ

β

Because the square of a t random variable with n-2 degrees of freedom is an F random variable with 1 and n-2 degrees

of freedom, the usually t -test that compares || 0t to 2,2/ −nt α is equivalent to comparing2

00 t f = to

2

2,1, 2,2/ −=

− nt f n α α .

a) 2079.01

)23(9.00 =

−= f . Reject 0: 10 = β H .

b) Because 28.423,1,05. = f , 0 H is rejected if 28.4

1

232

2

>

− R

R .

That is, H0 is rejected if

157.0

28.428.27

)1(28.423

2

2

22

>

>

−>

R

R

R R



11-38

Section 11-8

11-64 a) 0:0 = ρ H

0:1 ≠ ρ H α = 0.05

18,025.00

18,025.0

64.01

2208.0

0

||

101.2

657.5

t t

t

t

>

=

==−

−

Reject H0. P-value = (<0.0005)(2) = <0.001

b) 5.0:0 = ρ H

5.0:1 ≠ ρ H α = 0.05

2/0

025.

2/1

0

||

96.1

265.2)17)(0.5)(arctanh0.8)((arctanh

α z z

z

z

>

=

=−=

Reject H0. P-value = (0.012)(2) = 0.024.

c) )+0.8nhtanh(arcta)-0.8nhtanh(arcta1717

025.025. z z≤≤ ρ

where z. .025 196= . 9177.05534.0 ≤≤ ρ .

Because ρ = 0 and ρ = 0.5 are not in the interval, so reject H0.

11-65 a) 0:0 = ρ H

0:1 > ρ H α = 0.05

18,05.00

18,05.0

75.01

22075.0

0

734.1

81.42

t t

t

t

>

=

==−

−

Reject H0. P-value < 0.0005

b) 5.0:0 = ρ H

5.0:1 > ρ H α = 0.05

α z z

z

z

>

=

=−=

0

05.

2/1

0

65.1

7467.1)17)(0.5)(arctanh0.75)((arctanh

Reject H0. P-value = 0.04

c) )0.75nhtanh(arcta17

05.0 z−≥ ρ where 64.105. = z

26.2≥ ρ

Because ρ = 0 and ρ = 0.5 are not in the interval, reject the nul hypotheses from parts (a) and (b).

11-66 n = 25 r = 0.83

a) 0:0 = ρ H

0:1 ≠ ρ H α = 0.05



11-39

23,2/0

23,025.

)83.0(1

2383.0

1

2

0

069.2

137.722

α t t

t

t r

nr

>

=

===−−

−

Reject H0 . P-value = 0.

b) )+0.83nhtanh(arcta)-0.83nhtanh(arcta2222

025.025. z z ≤≤ ρ

where z. .025 196= . 9226.06471.0 ≤≤ ρ .

a) 8.0:0 = ρ H

8.0:1 ≠ ρ H α = 0.05

2/0

025.

2/1

0

96.1

4199.0)22)(8.0arctanh83.0(arctanh

α z z

z

z

>/

=

=−=

Do not reject H0. P-value = (0.3373)(2) = 0.6746.

11-67 n = 50 r = 0.62

a) 0:0 = ρ H

0:1 ≠ ρ H α = 0.01

48,005.00

48,005.

)62.0(1

4862.0

1

2

0

682.2

475.522

t t

t

t r

nr

>

=

===−−

−

Reject H0. P-value ≅ 0

b) )+0.62nhtanh(arcta)-0.62nhtanh(arcta4747

005.005. z z ≤≤ ρ

where 575.2005.0 = z . 8007.03358.0 ≤≤ ρ .

c) Yes.

11-68 a) r = 0.933203

b) 0:0 = ρ H

0:1 ≠ ρ H α = 0.05

15,2/0

15,025.

)8709.0(1

15933203.0

1

2

0

131.2

06.102

α t t

t

t r

nr

>

=

===−−

−

Reject H0.

c) x y 498081.072538.0ˆ +=

0: 10 = β H

0: 11≠ β H α = 0.05

15,1,0

15,1,05.

0

543.416.101

α f f

f f

>>

==

Reject H0. Conclude that the model is significant at α = 0.05. This test and the one in part b) are identical.

d) No problems with model assumptions are noted.



11-40

5040302010

3

2

1

0

-1

-2

x

R e s i d u a l


25155

3

2

1

0

-1

-2

FittedValue

R e s i d u a l

Residuals Versusthe Fitted Values(response is y)

210-1-2

3

2

1

0

-1

-2

Normal Score

R e s i d u a l


11-69 a) x y 990987.00280411.0ˆ +−=

OR

S t a t

10095908580757065

100

90

80

70

60

Scatterplot of Stat vs OR

b) 0: 10 = β H

0: 11 ≠ β H α = 0.05

18,1,0

18,1,05.

0

41.4

838.79

α f f

f

f

>>

=

=

Reject H0 .c) 903.0816.0 ==r

d) 0:0 = ρ H

0:1 ≠ ρ H α = 0.05



11-41

18,2/0

18,025.

20

101.2

9345.8816.01

1890334.0

1

2

α t t

t

R

n Rt

>

=

=−

=−

−=

Reject H0.

e)5.0:0

= ρ H

5.0:1 ≠ ρ H α = 0.05

2/0

025.

0

96.1

879.3

α z z

z

z

>

=

=

Reject H0.

f) )+0.90334nhtanh(arcta)-0.90334nhtanh(arcta1717

025.025. z z ≤≤ ρ where 96.1025.0 = z .

9615.07677.0 ≤≤ ρ .

11-70 a) x y 419415.01044.69ˆ +=

b) 0: 10 = β H

0: 11≠ β H

α= 0.05

24,1,0

24,1,05.

0

260.4

744.35

α f f

f

f

>

=

=

Reject H0.

c) 77349.0=r

d) 0:0 = ρ H

0:1 ≠ ρ H α = 0.05

24,2/0

24,025.

5983.01

2477349.0

0

064.2

9787.5

α t t

t

t

>

=

==−

Reject H0.

e) 6.0:0 = ρ H

6.0:1 ≠ ρ H α = 0.05

2/0

025.

2/1

0

96.1

6105.1)23)(6.0arctanh77349.0(arctanh

α z z

z

z

>/

=

=−=

Do not reject H0 .

f) )+0.77349nhtanh(arcta)-0.77349nhtanh(arcta2323

025.025. z z ≤≤ ρ where z. .025 196= .

8932.05513.0 ≤≤ ρ .

11-71 a)



11-42

Supply Voltage

C u r r e n t W i t h o

u t E l e c t ( m A )

76543210

50

40

30

20

10

Scatt erplot of Current WithoutElect(mA) vs Supply Voltage


Current WithoutElect(mA) = 5.50 + 6.73 Supply Voltage


Constant 5.503 3.104 1.77 0.114

Supply Voltage 6.7342 0.7999 8.42 0.000

S = 4.59061 R-Sq = 89.9% R-Sq(adj) = 88.6%


Source DF SS MS F P

Regression 1 1493.7 1493.7 70.88 0.000


Total 9 1662.3

x y 73.650.5ˆ +=

Yes, because the P-value ≈ 0, the regression model is significant at α = 0.05.

b) 948.0899.0 ==r

c)

306.2425.8

306.2

425.8948.01

210948.0

1

2

0:

0:

8,025.00

8,025.0

220

1

0

=>=

=

=−

−=

−

−=

≠

=

t t

t

r

nr t

H

H

ρ

ρ

Reject H0.

d)



11-43

9879.07898.0

310

6.19948.0harctantanh

310


3harctantanh

3harctantanh 2/2/

≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

ρ

ρ

ρ α α

n

zr

n

zr

11-72 a)

2000

2 0 0 1

1615141312

16

15

14

13

12

Scatterplot of 2001 vs 2000


Y2001 = - 0.014 + 1.01 Y2000


Constant -0.0144 0.3315 -0.04 0.966

Y2000 1.01127 0.02321 43.56 0.000

S = 0.110372 R-Sq = 99.5% R-Sq(adj) = 99.4%


Source DF SS MS F P

Regression 1 23.117 23.117 1897.63 0.000


Total 11 23.239

x y 011.1014.0ˆ +−=

Yes, because the P-value ≈ 0, the regression model is significant at α = 0.05.

b) 9975.0995.0 ==r

c)



11-44

( )

( )

025.00

025.02/

0

2/1

0

2/1

00

1

0

||

96.1

6084.5

312)9.0harctan9975.0h(arctan

3)harctanh(arctan

9.0:

9.0:

z z

z z

z

z

n R z

H

H

>

==

=−−=

−−=

≠

=

α

ρ

ρ

ρ

Reject H0. P-value = (1-1)(2) = 0.000.

d)

9993.09908.0



3harctantanh

3harctantanh 2/2/

≤≤

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛

−+≤≤⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

−−

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

ρ

ρ

ρ α α

n

zr

n

zr

11-73 a) r= 787.0 =0.887

b)

0:

0:

1

0

≠

=

ρ

ρ

H

H

164.10887.01

230887.0

1

2

220=−

−

=−

−

= r

nr

t

28,025.0

28,025. 048.2

t t

t

>

=

Reject H0. The P-value < 0.0005

c)

9452.07741.0

330


330


≤≤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

ρ

ρ

d)



( )

( )

025.00

025.02/

0

2/1

0

2/1

00

1

0

||

96.1

808.2

312)7.0harctan887.0h(arctan

3)harctanh(arctan

7.0:

7.0:

z z

z z

z

z

n R z

H

H

>

==

=

−−=

−−=

≠

=

α

ρ

ρ

ρ

Reject H0. The P-value = 2(0.0025) = 0.005

11-74

x

y

43210-1-2-3-4-5

4

3

2

1

0

-1

-2

-3

-4

-5

Scatt erplot of y vs x

r = 0. The correlation coefficient does not detect the relationship between x and y because the relationshipis not linear as shown on the graph above.



11-45

Section 11-9

11-75 a) Yes, ε β β lnlnlnln 10 ++= x y

b) No

c) Yes, ε β β lnlnlnln 10 ++= x y

d) Yes, ε β β ++=

x y

1110

11-76 a) There is curvature in the data.

38 033 028 0

80 0

70 0

60 0

50 0

40 0

30 0

20 0

10 0

0

Temperature (K)

V a p o r P r e s s u r e ( m m H

g )



11-46

b) y = - 1956.3 + 6.686 x

c)Source DF SS MS F P

Regression 1 491662 491662 35.57 0.000Residual Error 9 124403 13823

Total 10 616065

d) There is a curve in the residuals.

e) The data are linear after the transformation to y* = ln y and x* = 1/ x.

ln y = 20.6 - 5201(1/x)

3500300025002000150010005000

15

10

5

0

x

y

-200 -100 0 100 200 300 400 500 600

-100

0

100

200

Fitted Value

R e s i d u a l

Residuals Versus the Fitted Values

(response is Vapor Pr)

0.00370.00320.0027

7

6

5

4

3

2

1

1/T

L n ( V P )



11-47


Source DF SS MS F PRegression 1 28.511 28.511 66715.47 0.000Residual Error 9 0.004 0.000Total 10 28.515

1 2 3 4 5 6 7

-0.03

-0.02

-0.01

0.00

0.01

0.02

Fitted Value

R e s i d u a l


(response is y*)

There is still curvature in the data, but now the plot is convex instead of concave.

11-77 a)

3500300025002000150010005000

15

10

5

0

x

y

b) x y 00385.08819.0ˆ +−=

c) 0: 10 = β H

0: 11 ≠ β H α = 0.05

48,1,05.00

0 03.122

f f

f

>

=

Reject H0. Conclude that regression model is significant atα

= 0.05d) No, it seems the variance is not constant, there is a funnel shape.



11-48

1050

3

2

1

0

-1

-2

-3

-4

-5

Fitted Value

R e s i d u a l


e) x y 00097.05967.0ˆ +=∗

. Yes, the transformation stabilizes the variance.



11-49


11-78 a) ∑∑∑===

−=−n

i

i

n

i

i

n

i

ii y y y y

111

ˆ)ˆ( and ∑∑ += ii xn y 10ˆˆ β β from the normal equations

Then,

0ˆˆˆˆ

)ˆˆ(ˆˆ

ˆ)ˆˆ(

1

10

1

10

1

10

1

10

1

10

=−−+=

+−+=

−+

∑∑

∑∑

∑∑

==

==

=

n

i

i

n

i

i

n

i

i

n

i

i

i

n

i

i

xn xn

x xn

y xn

β β β β

β β β β

β β

b) i

n

iii

n

iii

n

iii x y x y x y y ∑∑∑

===

−=−111

ˆ)ˆ(

and ∑∑∑===

+=n

ii

n

iii

n

ii x x x y

1

21

10

1

ˆˆ β β from the normal equations. Then,

0ˆˆˆˆ

)ˆˆ(ˆˆ

1

21

10

1

21

10

1011

21

10

=−−+

=+−+

∑∑∑∑

∑∑∑

====

===

n

ii

n

ii

n

ii

n

ii

ii

n

i

n

ii

n

ii

x x x x

x x x x

β β β β

β β β β

c) y yn

n

i

i =∑=1

ˆ1

)ˆˆ(ˆ10 x y β β ∑ ∑ +=

y

x x y

x xn ynn

x x ynn

xnn

xn

yn

i

i

i

i

n

ii

=

+−=

+−=

+−=

+=

+=

∑

∑∑

∑∑=

1

11

11

10

101

ˆˆ

)ˆˆ(1

)ˆ)ˆ((1

)

ˆˆ

(

1

)ˆˆ(1

ˆ1

β β

β β

β β

β β

β β



11-50

11-79 a)

1.1 1.3 1.5 1.7 1.9 2.1

x

0.7

1

1.3

1.6

1.9

2.2

y

Plot of y vs x

Yes, a linear relationship seems plausible.

b) Model fitting results for: y

Independent variable coefficient std. error t-value sig.level

CONSTANT -0.966824 0.004845 -199.5413 0.0000x 1.543758 0.003074 502.2588 0.0000

--------------------------------------------------------------------------------

R-SQ. (ADJ.) = 1.0000 SE= 0.002792 MAE= 0.002063 DurbWat= 2.843

Previously: 0.0000 0.000000 0.000000 0.000

10 observations fitted, forecast(s) computed for 0 missing val. of dep. var.

. .y x= − +0966824 154376

c) Analysis of Variance for the Full Regression

Source Sum of Squares DF Mean Square F-Ratio P-value

Model 1.96613 1 1.96613 252264. .0000

Error 0.0000623515 8 0.00000779394

--------------------------------------------------------------------------------

Total (Corr.) 1.96619 9

R-squared = 0.999968 Stnd. error of est. = 2.79176E-3

R-squared (Adj. for d.f.) = 0.999964 Durbin-Watson statistic = 2.84309

2) H0 1 0:β =

3) H1 1 0:β ≠

4) α = 0.05

5) The test statistic is)/(

/0

pnSS

k SS f

E

R

−=


7) Using the results from the ANOVA table

9.2552638/0000623515.0

1/96613.10 == f

8) Because 2552639 > 5.32 reject H0 and conclude that the regression model is significant at α = 0.05.

P-value < 0.000001

d) 95 percent confidence intervals for coefficient estimates

--------------------------------------------------------------------------------


CONSTANT -0.96682 0.00485 -0.97800 -0.95565

x 1.54376 0.00307 1.53667 1.55085

--------------------------------------------------------------------------------

11.53667 1.55085 β ≤ ≤

e) 2) H0 0 0:β =



11-51

3) H1 0 0:β ≠

4) α = 0.05

5) The test statistic is tse

00

0

=

( )

β

β


7) Using the results from the table above

t00 96682

000485199 34= − = −

.

..

8) Since −199.34 < −2.306 reject H0 and conclude the intercept is significant at α = 0.05.

11-80 a) x y 64.1534.93ˆ +=

b) 0: 10 = β H

0: 11 ≠ β H α = 0.05

14,1,05.00

14,1,05.

0

60.4

872.12

f f

f

f

>

=

=

Reject H0. Conclude that β1 0≠ at α = 0.05.

c) )322.23961.7( 1 ≤≤ β

d) )923.111758.74( 0 ≤≤ β

e) 44.132)5.2(64.1534.93ˆ =+= y

[ ]

91.138ˆ97.125

468.644.132

27.136145.244.132

5.2|

017.7

)325.25.2(

161

0

2

≤≤

±

+±

=

−

xY μ

11-81 x y 5075.02232.1ˆ +=∗where y y /1=∗

. No, the model does not seem reasonable.

The residual plots indicate a possible outlier.

11-82 x y 2047.25755.4ˆ += , r = 0.992, R 2

= 98.40%

The model appears to be an excellent fit. The R 2

is large and both regression coefficients are significant. No, the

existence of a strong correlation does not imply a cause and effect relationship.

11-83 x y 7916.0ˆ =

Even though y should be zero when x is zero, because the regressor variable does not usually assume values near zero, a

model with an intercept fits this data better. Without an intercept, the residuals plots are not satisfactory.

11-84 a)

181716

110

100

90

80

70

60

50

40

30

index

d a y s



11-52

b) The regression equation is

x y 296.15193ˆ +−=

Analysis of VarianceSource DF SS MS F P

Regression 1 1492.6 1492.6 2.64 0.127


Total 15 9419.4

Do not reject Ho. We do not have evidence of a relationship. Therefore, there is not sufficient evidence to conclude that

the seasonal meteorological index ( x) is a reliable predictor of the number of days that the ozone level exceeds 0.20

ppm ( y).

c) 95% CI on β1

504.35912.4

)421.9(145.2296.15

)421.9(296.15

)ˆ(ˆ

1

12,025.

12,2/1

≤≤−

±

±

± −

β

β β α

t

set n

d) The normality plot of the residuals is satisfactory. However, the plot of residuals versus run order exhibits a strong

downward trend. This could indicate that there is another variable should be included in the model and it is one that

changes with time.

11-85 a)

b) x y 29646714.0ˆ −=

161412108642

40

30

20

10

0

-10

-20

-30

-40

Observation Order

R e s i d u a l

403020100-10-20-30-40

2

1

0

-1

-2


Residual

1.00.90.80.70.60.50.40.3

0.7

0.6

0.5

0.4

0.3

0.2

x

y



11-53

c) Analysis of Variance

Source DF SS MS F P

Regression 1 0.03691 0.03691 1.64 0.248


Total 7 0.17189

R 2 = 21.47%

Because the P-value > 0.05, reject the null hypothesis and conclude that the model is significant.

d) There appears to be curvature in the data. There is a dip in the middle of the normal probability plot and the plot of

the residuals versus the fitted values shows curvature.

11-86 a)

b) x y 9636.03.33ˆ +=

c)Predictor Coef SE Coef T P

Constant 66.0 194.2 0.34 0.743

Therm 0.9299 0.2090 4.45 0.002

S = 5.435 R-Sq = 71.2% R-Sq(adj) = 67.6%


Source DF SS MS F P

Regression 1 584.62 584.62 19.79 0.002


Reject the null hypothesis and conclude that the model is significant. Here 77.3% of the variability is explained by the

model.

d) 1: 10 = β H

1: 11 ≠ β H α = 0.05

0.20.10.0-0.1-0.2

1.5

1.0

0.5

0.0

-0.5

-1.0

-1.5


Residual

0.4 0.5 0.6

-0.2

-0.1

0.0

0.1

0.2

Fitted Value

R e s i d u a l

920 930 940

920

930

940

x

y



11-54

3354.02090.0

19299.0

)ˆ(

1ˆ

1

10 −=

−=

−=

β

β

set

306.28,025.2,2/ ==− t t na

Because 2,2/0 −−> nat t , we cannot reject H o and we conclude that there is not enough evidence to reject the claimthat the devices produce different temperature measurements. Therefore, we assume the devices produce equivalent

measurements.

e) The residual plots to not reveal any major problems.

-5 0 5

-1

0

1


Residual


(response is IR)

920 930 940

-5

0

5

Fitted Value

R e s i d u a l


(response is IR)

11-87 a)

2 3 4 5 6 7

1

2

3

4

5

6

7

8

x

y



11-55

b) x y 66.1699.0ˆ +−=

c)Source DF SS MS F P

Regression 1 28.044 28.044 22.75 0.001


Total 9 37.904

Reject the null hypothesis and conclude that the model is significant.

d) x0 = 4.25 257.4ˆ0| = x yμ

)3717.0(306.2257.4

625.20

)75.425.4(

10

12324.1306.2257.4

2

±

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+±

114.5399.3 | ≤≤o x yμ

e) The normal probability plot of the residuals appears linear, but there are some large residuals in the lower fitted

values. There may be some problems with the model.


No. Of Atoms (x 10E9) = - 0.300 + 0.0221 power(mW)


Constant -0.29989 0.02279 -13.16 0.000

power(mW) 0.0221217 0.0006580 33.62 0.000

S = 0.0337423 R-Sq = 98.9% R-Sq(adj) = 98.8%


Source DF SS MS F P

Regression 1 1.2870 1.2870 1130.43 0.000


Total 14 1.3018

8765432

2

1

0

-1

-2

Fitted Value

R e s i d u a l

-2 -1 0 1 2

-1

0

1


Residual



11-56

power(mW)

N o .

O f A t o m s ( x

1 0 E 9 )

5040302010

0.8

0.6

0.4

0.2

0.0

S 0.0337423

R-Sq 98.9%

R-Sq(adj) 98.8%

Fitted Line PlotNo. Of Atoms (x 10E9) = - 0.2999 + 0.02212 power(mW)

b) Yes, there is a significant regression at α = 0.05 because p-value = 0.000 < α.

c) 994.0989.0 ==r

d)

.160.2766.32

160.2

766.32994.1

215994.0

1

2

0:

0:

13,025.00

13,025.0

220

1

0

=>=

=

=−

−=

−

−=

≠

=

t t

t

r

nr t

H

H

ρ

ρ

Reject H0. P-value = 0.000.

e) 95% confidence interval for 1

ˆ β

0234.0ˆ0206.0

)00066.0(160.2022.0

)00066.0(022.0

)ˆ(ˆ

1

13,025.0

12,2/1

≤≤

±

±

± −

β

β β α

t

set n

11-89 a)



11-57

carat

p

r i c e

0.500.450.400.350.30

3500

3000

2500

2000

1500

1000

Scatter plot of pri ce vs carat

The relationship between carat and price is not linear. Yes, there is one outlier, observation number 33.

b) The person obtained a very good price—high carat diamond at low price.

c) All the data


price = - 1696 + 9349 carat


Constant -1696.2 298.3 -5.69 0.000

carat 9349.4 794.1 11.77 0.000

S = 331.921 R-Sq = 78.5% R-Sq(adj) = 77.9%


Source DF SS MS F P

Regression 1 15270545 15270545 138.61 0.000


Total 39 19457057

tα/2,n-2 = t0.025,38 = 2.024

95% confidence interval on β1 .

.26.1095674.7741

)1.794(024.29349

)1.794(9349

)ˆ(ˆ

1

38,025.

12,2/1

≤≤

±

±

± −

β

β β α

t

set n

With unusual data omittedThe regression equation is

price_1 = - 1841 + 9809 carat_1

Predictor Coef SE Coef T PConstant -1841.2 269.9 -6.82 0.000

carat_1 9809.2 722.5 13.58 0.000

S = 296.218 R-Sq = 83.3% R-Sq(adj) = 82.8%


Source DF SS MS F P

Regression 1 16173949 16173949 184.33 0.000



11-58


Total 38 19420517

tα/2,n-2 = t0.025,37 = 2.026

95% confidence interval on β1 .

79.1127222.8345

)5.722(026.29809)5.722(9809

)ˆ(ˆ

1

37,025.

12,2/1

≤≤

±±

± −

β

β β α

t

set n

The width for the outlier removed is narrower than for the first case.

11-90The regression equation is

Population = 3549143 + 651828 Count


Constant 3549143 131986 26.89 0.000

Count 651828 262844 2.48 0.029

S = 183802 R-Sq = 33.9% R-Sq(adj) = 28.4%


Source DF SS MS F P

Regression 1 2.07763E+11 2.07763E+11 6.15 0.029

Residual Error 12 4.05398E+11 33783126799

Total 13 6.13161E+11

x y 6518283549143ˆ +=

Yes, the regression is significant at α = 0.05. Care needs to be taken in making cause and effect statements

based on a regression analysis. In this case, it is surely not the case that an increase in the stork count iscausing the population to increase, in fact, the opposite is most likely the case. However, unless a designed

experiment is performed, cause and effect statements should not be made on regression analysis alone. The

existence of a strong correlation does not imply a cause and effect relationship.




12-98. Because

yy

R

S

SS R =2

and

yy

E

S

SS R =− 21 ,

)1/(

/0 −−

=k nSS

k SSF

E

Rand this is the usual F-test for significance

of regression. Then, 75.33)1420/()9.01(

4/90.00 =

−−−=F and the critical value is . Because

33.75 > 3.06, regression is significant.

f. , , .05 415 3 06=

12-73



12-99. Using n = 20, k = 4, f . Reject H. , , .05 4 15 3 06= 0 if

816.0

)1(

06.315/)1(

4/

2

2

2

2

≥

−

≥−

R

R

R

R

Then, results in a significant regression.449.02 ≥ R

12-100. Because ,Y X X X ')'(ˆ 1−= β Y H I Y X X X X Y X Y e )(')'(ˆ 1 −=−=−= − β

12-101. From Exercise 12-76, ei is ith element of ( I - H )Y . That is,

22

,

2

1,

2

,

2

1,

2

2,

2

1,

,11,,11,22,11,

)...)1(...()(

...)1(...

σ niiiiiiiiii

nniiiiiiiiiiiii

hhhhhheV

and

Y hY hY hY hY hY he

+++−++++=

−−−−+−−−−=

+−

++−−

The expression in parentheses is recognized to be the ith diagonal element of ( I - H )( I - H ') = I - H by matrix

multiplication. Consequently, . Assume that i < j. Now,2

, )1()( σ iii heV −=

nn j j j j j j j j j j j j j

nniiiiiiiiiiiii

Y hY hY hY hY hY he

Y hY hY hY hY hY he

,11,,11,22,11,

,11,,11,22,11,

...)1(...

...)1(...

−−−−+−−−−=−−−−+−−−−=

++−−

++−−

Because the yi‘s are independent,

i jiii jii ji ji ji hhhhhhhheeCov ,,1,1,2,2,1,1, )1(...(),( −++++= −−

2

,,1,1,,,1,1, )...)1(... σ n jni j j ji j j jii jii

hhhhhhhh +++−+++ ++++

The expression in parentheses is recognized as the ijth element of ( I - H )( I - H ') = I - H . Therefore, .2),( σ ij ji heeCov −=

12-102. ε β ε β ε β β R X X X Y X X X +=+=+== − 'X)(X')('X)(X'')'(ˆ -1-11

12-103. a) Min L = )()'( β β X y X y −− subject to :Tβ = c

This is equivalent to Min Z = ( )' ( ) ' (y X y X T c)+ −β λ β2

where λ ],...,2

,1

[' pλ λ λ = is a vector of La Grange multipliers.

∂

∂ββ λ

ZX y X X T= − + +2 2 2' ( ' ) '

( )cT Z

−= β ∂λ

∂ 2 . Set

∂

∂β

∂

∂λ

Z Z= =0 0, .

Then we get

cT

y X T X X

c

c

=

=+

β

λ β

ˆ

''ˆ)'(

where is the constrained estimator. From the first of these equations,βc ( ' ) ( ' ' ) ( ' ) 'λ βc X X X y T X X T= − = −− λ−1 1

From the second,T T or λX X T ( ' ) 'λ− =−1 c = − −[ ( ' ) ' ]T X X T1 1(T - c)β

Then ( ' ) '[ ( ' ) ' ] ( )

( ' ) '[ ( ' ) ' ] ( )

β β β

β β

c X X T T X X T T c

X X T T X X T c T

= − −

= + −

− − −

− − −

1 1 1

1 1 1

12-74



b) This solution would be appropriate in situations where you have tested the hypothesis that Tβ = c and concluded

that this hypothesis cannot be rejected.

12-104. a) For the piecewise linear function to be continuous at x = x , the point-slope formula for a line can be used to

show that

*

⎪⎩

⎪⎨

⎧

>−+

≤−+=

∗ x x x x

x x x x

y

)(

)(

*

20

**

10

β β

β β

where β β are arbitrary constants.β0 1 2, ,

Let .

⎪⎩

⎪⎨

⎧

>

≤=

∗

∗

x x

x x

z

,1

,0

Then, y can be written as . z x x x x y ))(()( 1210

∗∗ −−+−+= β β β β

Let

122

11

00

2

1

)(

β β β

β β

β β

−=

=

=

−=

−=

∗

∗

∗

∗

∗

z x x x

x x x

Then,22110

*** x x y β β β ++= .

b) Should refer to exercise 12-80. If there is a discontinuity at x x= ∗, then a model that can be used is

⎪⎩

⎪⎨

⎧

>+

≤+=

∗ x x x

x x x

y

10

*

10

α α

β β

Let

⎪⎩

⎪⎨

⎧

>

≤=

∗

∗

x x

x x

z

,1

,0

Then, y can be written as y x x z x z= + + − + − = + + + x∗ ∗ ∗ ∗β β α β α β β β β β0 1 0 0 1 1 0 1 1 2 3[( ) ( ) ] 2

where

xz x

x x

=

=

−=

−=

=

=

∗

∗

∗

∗

2

1

113

002

11

00

β α β

β α β

β β

β β

c) Should refer to exercise 12-80. One could estimate x* as a parameter in the model. A simple approach is to refit the

model in Exercise 12-80 with different choices for x* and to select the value for x* that minimizes the residual sum of

squares.

12-75



CHAPTER 12

Sections 12-1

12-1. a) ′ =

⎡

⎣

⎢

⎢⎢

⎤

⎦

⎥

⎥⎥

X X

10 223 553

223 52009 12352

553 12352 31729

.

′ =

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

X y

19160

435508

104736 8

.

.

.

b) so

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

=

126.1

713.3

055.171

ˆ β 21 126.1714.3055.171ˆ x x y −+=

c) 49.189)43(126.1)18(714.3055.171ˆ =−+= y

12-2. a) y X X X ′′=−1

)(ˆ β

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

=

2532.0

0931.0

9122.1

ˆ β

b) 21 2532.00931.09122.1ˆ x x y ++−=

3678.29)50(2532.0)200(0931.09122.1ˆ =++−= y

12-3. a) Model 1 Model 2

x2 2= y x= + +100 2 81 . ( )y x x= + + +95 15 3 2 41 1

x

x

x

y = +108 2 1

.y x= +101 55 1

x2 8= ( )y x= + +100 2 4 81 . ( )y x= + + +95 15 3 8 161 1

y = +132 2 1 .y x= +119 17 5 1

MODEL 1

0

20

40

60

80

100

120

140

160

0 2 4 6 8 1 0

x

y

y = 132 + 2

y = 108 + 2

12-1



MODEL 2

0

50

100

150

200

250

300

350

0 5 10 15

x

yy = 101 + 5.5 x

y = 119 + 17.5 x

The interaction term in model 2 affects the slope of the regression equation. That is, it modifies the

amount of change per unit of onx1 y .

b) x2 5= ( )y x= + +100 2 4 51

y x= +120 2 1

Then, 2 is the expected change on y per unit of .x1

NO, it does not depend on the value of x2, because there is no relationship or interaction between these

two variables in model 1.

c) x2 5= x2 2= x2 8=

. ( ) (y x x= + + +95 15 3 5 2 51 1 )

.y x= +110 115 1

.y x= +101 55 1 .y x= +119 17 5 1

Change per

unit of x 1

11.5 5.5 17.5

Yes, result does depend on the value of x2, because x2 interacts with x1.

12-4. Predictor Coef SE Coef T P Constant -123.1 157.3 -0.78 0.459X1 0.7573 0.2791 2.71 0.030X2 7.519 4.010 1.87 0.103

X3 2.483 1.809 1.37 0.212X4 -0.4811 0.5552 -0.87 0.415

S = 11.79 R-Sq = 85.2% R-Sq(adj) = 76.8%


Source DF SS MS F P

Regression 4 5600.5 1400.1 10.08 0.005Residual Error 7 972.5 138.9Total 11 6572.9

a) 4321 4811.0483.2519.77573.01.123ˆ x x x x y −+++−=

b) 00.139ˆ 2 =σ

c) , , , , and3.157)ˆ( 0 = β se 2791.0)ˆ( 1 = β se 010.4)ˆ( 2 = β se 809.1)ˆ( 3 = β se 5552.0)ˆ( 4 = β se

d) 476.290)98(4811.0)90(483.2)24(519.7)75(7573.01.123ˆ =−+++−= y

12-2



12-5. The regression equation is mpg = 49.9 - 0.0104 cid - 0.0012 rhp - 0.00324 etw + 0.29 cmp - 3.86 axle

+ 0.190 n/v

Predictor Coef SE Coef T PConstant 49.90 19.67 2.54 0.024cid -0.01045 0.02338 -0.45 0.662

rhp -0.00120 0.01631 -0.07 0.942etw -0.0032364 0.0009459 -3.42 0.004cmp 0.292 1.765 0.17 0.871axle -3.855 1.329 -2.90 0.012n/v 0.1897 0.2730 0.69 0.498

S = 2.22830 R-Sq = 89.3% R-Sq(adj) = 84.8%



a) 654321 1897.0855.3292.000324.00012.001045.090.49ˆ x x x x x x y +−+−−−=

where vn xaxle xcmp xetw xrhp xcid x /654321 ======

b) 965.4ˆ 2 =σ

, , , ,67.19)ˆ( 0 = β se 02338.0)ˆ( 1 = β se 01631.0)ˆ( 2 = β se 0009459.0)ˆ( 3 = β se

, and765.1)ˆ( 4 = β se 329.1)ˆ( 5 = β se 273.0)ˆ( 6 = β se

c)

9.30(1897.0)07.3(855.3)9.9(292.0)4500(0032.0)253(0012.0)215(01045.090.49ˆ +−+−−−= y= 29.867

12-6. The regression equation is

y = 7.46 - 0.030 x2 + 0.521 x3 - 0.102 x4 - 2.16 x5Predictor Coef StDev T PConstant 7.458 7.226 1.03 0.320x2 -0.0297 0.2633 -0.11 0.912x3 0.5205 0.1359 3.83 0.002x4 -0.10180 0.05339 -1.91 0.077x5 -2.161 2.395 -0.90 0.382S = 0.8827 R-Sq = 67.2% R-Sq(adj) = 57.8% Analysis of VarianceSource DF SS MS F PRegression 4 22.3119 5.5780 7.16 0.002Error 14 10.9091 0.7792Total 18 33.2211

a) 5432 1606.21018.05205.00297.04578.7ˆ x x x x y −−+−= b) 7792.ˆ 2 =σ

c) , , , and226.7)ˆ( 0 = β se 2633.)ˆ( 2 = β se 1359.)ˆ( 3 = β se 05339.)ˆ( 4 = β se 395.2)ˆ( 5 = β se

d) )0.2(1606.2)90(1018.0)30(5205.0)20(0297.04578.7ˆ −−+−= y 996.8ˆ = y

12-7.

12-3




x1 -9.604 3.723 -2.58 0.020x2 0.4152 0.2261 1.84 0.085x3 18.294 1.323 13.82 0.000 --------------------------------------------------------------------------------

a) 321 294.1844152.0604.982.47ˆ x x x y ++−=

b) 3.12ˆ 2 =σ

c) , , , and94.49)ˆ( 0 = β se 723.3)ˆ( 1 = β se 2261.0)ˆ( 2 = β se 323.1)ˆ( 3 = β se

d) 38.91)5(29.18)220(44152.0)5.14(604.982.47ˆ =++−= y

12-8. Predictor Coef SE Coef T PConstant -0.03023 0.06178 -0.49 0.629temp 0.00002856 0.00003437 0.83 0.414

soaktime 0.0023182 0.0001737 13.35 0.000soakpct -0.003029 0.005844 -0.52 0.609difftime 0.008476 0.001218 6.96 0.000diffpct -0.002363 0.008078 -0.29 0.772

S = 0.002296 R-Sq = 96.8% R-Sq(adj) = 96.2%


Source DF SS MS F PRegression 5 0.00418939 0.00083788 158.92 0.000


a) 54321 002363.0008476.0003029.0002318.0000029.003023.0ˆ x x x x x y −+−++−=

where DIFFPCT x DFTIME xSOAKPCT xSOAKTIME xTEMP x ===== 54321

b) 62 1027.5ˆ −= xσ

c) The standard errors are listed under the StDev column above.

d)

)80.0(002363.0)1(008476.0

)1.1(003029.0)1(002318.0)1650(000029.003023.0ˆ

−+

−++−= y

02247.0ˆ = y

12-9. The regression equation is rads = - 440 + 19.1 mAmps + 68.1 exposure time

Predictor Coef SE Coef T PConstant -440.39 94.20 -4.68 0.000mAmps 19.147 3.460 5.53 0.000exposure time 68.080 5.241 12.99 0.000

S = 235.718 R-Sq = 84.3% R-Sq(adj) = 83.5%


Source DF SS MS F PRegression 2 11076473 5538237 99.67 0.000Residual Error 37 2055837 55563Total 39 13132310

12-4



a) 21 080.68147.1939.440ˆ x x y ++−=

where Time Exposure xmAmps x == 21

b) 55563ˆ 2 =σ

, , and20.94)ˆ( 0 = β se 460.3)ˆ( 1 = β se 241.5)ˆ( 2 = β se

c) 675.186)5(080.68)15(147.1993.440ˆ =++−= y

12-10. The regression equation is ARSNAILS = 0.488 - 0.00077 AGE - 0.0227 DRINKUSE - 0.0415 COOKUSE

+ 13.2 ARSWATER

Predictor Coef SE Coef T PConstant 0.4875 0.4272 1.14 0.271AGE -0.000767 0.003508 -0.22 0.830DRINKUSE -0.02274 0.04747 -0.48 0.638COOKUSE -0.04150 0.08408 -0.49 0.628ARSWATER 13.240 1.679 7.89 0.000

S = 0.236010 R-Sq = 81.2% R-Sq(adj) = 76.5%



a) 4321 240.1304150.002274.0000767.04875.0ˆ x x x x y +−−−=

where ARSWater xCookUse x DrinkUse x AGE x ==== 4321

b) 05570.0ˆ 2 =σ

, , , , and4272.0)ˆ( 0= β se 003508.0)ˆ( 1

= β se 04747.0)ˆ( 2= β se 08408.0)ˆ( 3

= β se 679.1)ˆ( 4 = β se

c) 9307.1)135.0(240.13)5(04150.0)5(02274.0)30(000767.04875.0ˆ =+−−−= y

12-11. The regression equation is density = - 0.110 + 0.407 dielectric constant + 2.11 loss factor

Predictor Coef SE Coef T PConstant -0.1105 0.2501 -0.44 0.670dielectric constant 0.4072 0.1682 2.42 0.042loss factor 2.108 5.834 0.36 0.727

S = 0.00883422 R-Sq = 99.7% R-Sq(adj) = 99.7%



12-5



a) 21 108.24072.01105.0ˆ x x y ++−=

where LossFactor xConst Dielectric x == 21

b) 00008.0ˆ 2 =σ


c) 97074.0)03.0(108.2)5.2(4072.01105.0ˆ =++−= y

12-12. The regression equation is y = - 171 + 7.03 x1 + 12.7 x2

Predictor Coef SE Coef T PConstant -171.26 28.40 -6.03 0.001x1 7.029 1.539 4.57 0.004x2 12.696 1.539 8.25 0.000

S = 3.07827 R-Sq = 93.7% R-Sq(adj) = 91.6%



a) 21 7.1203.7171ˆ x x y ++−=

b) 48.9ˆ 2 =σ


c) )5.12(7.12)5.14(03.7171ˆ ++−= y

= 89.685

12-13. The regression equation is Useful range (ng) = 239 + 0.334 Brightness (%) - 2.72 Contrast (%)

Predictor Coef SE Coef T PConstant 238.56 45.23 5.27 0.002Brightness (%) 0.3339 0.6763 0.49 0.639Contrast (%) -2.7167 0.6887 -3.94 0.008

S = 36.3493 R-Sq = 75.6% R-Sq(adj) = 67.4%



a) 21 7167.23339.056.238ˆ x x y −+=

where Contrast x Brightness x %% 21 ==

b) 1321ˆ 2 =σ

12-6



c) , , and23.45)ˆ( 0 = β se 6763.0)ˆ( 1 = β se 6887.0)ˆ( 2 = β se

d) 5195.61)75(7167.2)80(3339.056.238ˆ =−+= y

12-14. The regression equation is Stack Loss(y) = - 39.9 + 0.716 X1 + 1.30 X2 - 0.152 X3

Predictor Coef SE Coef T PConstant -39.92 11.90 -3.36 0.004X1 0.7156 0.1349 5.31 0.000X2 1.2953 0.3680 3.52 0.003X3 -0.1521 0.1563 -0.97 0.344

S = 3.24336 R-Sq = 91.4% R-Sq(adj) = 89.8%


Source DF SS MS F PRegression 3 1890.41 630.14 59.90 0.000Residual Error 17 178.83 10.52

Total 20 2069.24

a) 321 1521.02953.17156.092.39ˆ x x x y −++−=

b) 52.10ˆ 2 =σ


d) 7653.23)85(1521.0)26(2953.1)60(7156.092.39ˆ =−++−= y

12-15. The regression equation isRating Pts = 3.22 + 1.22 Pct Comp + 4.42 Pct TD - 4.09 Pct Int


Constant 3.220 6.519 0.49 0.626Pct Comp 1.2243 0.1080 11.33 0.000Pct TD 4.4231 0.2799 15.80 0.000Pct Int -4.0914 0.4953 -8.26 0.000

S = 1.92094 R-Sq = 97.8% R-Sq(adj) = 97.5%



a) 1073 091.4423.4224.122.3ˆ x x x y −++=

Where X3 = percentage of completions, X7 = percentage of TDs, and X10 = percentage of interceptions

b) 7.3ˆ 2 =σ


d) 079.82)3(091.4)4(423.4)60(224.122.3ˆ =−++= y

12-16. Predictor Coef SE Coef T P

12-7



Constant -29.46 93.98 -0.31 0.758GF 0.12356 0.05319 2.32 0.035GA -0.15953 0.04458 -3.58 0.003ADV 0.1236 0.2023 0.61 0.550PPGF -0.764 1.269 -0.60 0.556PCTG 3.100 4.406 0.70 0.492PEN 0.2402 0.3453 0.70 0.497BMI 0.0036 0.1151 0.03 0.976

AVG -18.86 28.31 -0.67 0.515SHT -0.03494 0.03419 -1.02 0.323PPGA 0.0526 0.2137 0.25 0.809PKPCT 0.0505 0.7221 0.07 0.945SHGF 0.5833 0.3935 1.48 0.159SHGA 0.3733 0.2712 1.38 0.189FG 0.1578 0.2227 0.71 0.489

S = 3.58133 R-Sq = 90.1% R-Sq(adj) = 80.9%



7654321 0036.02402.01.3764.01236.015953.012356.046.29ˆ x x x x x x x y +++−+−+−=

141312111098 1578.03733.05833.00505.00526.003494.086.18 x x x x x x x +++++−−

where

FG xSHGA xSHGF xPKPCT xPPGA xSHT x AVG x

BMI xPEN xPCTG xPPGF x ADV xGA xGF x

=======

=======

141312111098

7654321

83.12ˆ 2 =σ The standard errors of the coefficients are listed under the SE Coef column above.

12-17. Predictor Coef SE Coef T PConstant 383.80 36.22 10.60 0.002

Xl -3.6381 0.5665 -6.42 0.008X2 -0.11168 0.04338 -2.57 0.082S = 12.35 R-Sq = 98.5% R-Sq(adj) = 97.5%



a) 21 1119.06381.380.383ˆ x x y −−=

b) , , , and0.153ˆ 2 =σ 22.36)ˆ( 0 = β se 5665.0)ˆ( 1 = β se 04338.)ˆ( 2 = β se

c) 95.180)1000(1119.0)25(6381.380.383ˆ =−−= y

d) Predictor Coef SE Coef T P

Constant 484.0 101.3 4.78 0.041Xl -7.656 3.846 -1.99 0.185

X2 -0.2221 0.1129 -1.97 0.188X1*X2 0.004087 0.003871 1.06 0.402S = 12.12 R-Sq = 99.0% R-Sq(adj) = 97.6%Analysis of VarianceSource DF SS MS F PRegression 3 29951.4 9983.8 67.92 0.015


1221 0041.0222.0656.70.484ˆ x x x y −−−=

e) , , , and0.147ˆ 2 =σ 3.101)ˆ( 0 = β se 846.3)ˆ( 1 = β se 113.0)ˆ( 2 = β se 0039.0)ˆ( 12 = β se

12-8



f) 1.173)1000)(25(0041.0)1000(222.0)25(656.70.484ˆ =−−−= y

The predicted value is smaller

12-18. a)2

222111

'

021

'

0 )]()([),,( x x x x y f iii −−−−−= ∑ β β β β β β

))](()([2

)]()([2

11222111

'

0

1

222111'0'

0

x x x x x x y f

x x x x y f

iiii

iii

−−−−−−−=

−−−−−−=

∑

∑

β β β ∂β

∂

β β β ∂β ∂

))](()([2 22222111

'

0

2

x x x x x x y f

iiii −−−−−−−= ∑ β β β ∂β

∂

Setting the derivatives equal to zero yields

)()()()(

)())(()(

222

22222111'0

1122112

2

111

'

0

'

0

x x y x x x x x xn

x x y x x x x x xn

yn

iiiii

iiiii

i

−=−+−−+

−=−−+−+

=

∑∑∑

∑∑∑∑

β β β

β β β

β

b) From the first normal equation, y='

0ˆ β .

c) Substituting y yi − for y in the first normal equation yields .i 0ˆ '

0 = β



Sections 12-2

12-19. a) n = 10, k = 2, p = 3, α = 0.05

0...: 210 ==== k H β β β

0:1 ≠ j H β for at least one j

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

=

=−=

∑∑∑

8.104736

8.43550

1916

'

449010

)1916(6.371595

2

1

2

ii

ii

i

yy

y x

y x

y

y X

S

[ ]

7.833.44064490

3.440610

19169.371511

9.371511

8.104736

8.43550

1916

126.1713.3055.171''ˆ

2

=−=−=

=−=

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

R yy E

R

SSSSS

SS

y X β

7,2,05.00

7,2,05.0

0

74.4

25.1847/7.83

2/3.4406

f f

f

f pn

SS

k

SS

E

R

>

=

===−

12-9



Reject H0 and conclude that the regression model is significant at α = 0.05. P-value = 0.000

b) 957.11ˆ 2 =−

== pn

SS MS E

E σ

229.0)00439.0(957.11ˆ)ˆ(11

2

1 === cse σ β

10199.0)00087.0(957.11ˆ)ˆ( 22

2

2 === cse σ β

0: 10 = β H 02 = β

0: 11 ≠ β H 02 ≠ β

21.16229.0

713.3

)ˆ(

ˆ

1

10

==

= β

β

set

04.1110199.0

126.1

)ˆ(

ˆ

2

20

−=−

=

= β

β

set

365.27,025.07,2/ == t t α

Reject H0, P-value < 0.001 Reject H0, P-value < 0.001

Both regression coefficients significant

12-20. 00.742= yyS

a) 0: 210 == β β H


α = 0.01

( )

45.56

55.685742

55.685

48405548.5525

10

220

996536768

220

2532.00931.09122.1

)(

''ˆ

2

1

2

=

−=

−=

=

−=

−⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

−=∑

=

R yy E

n

i

i

R

SSSSS

n

y

y X SS β

7,2,01.00

7,2,01.0

0

55.9

51.427/45.56

2/55.685

f f

f

f

pn

SS

k

SS

E

R

>

=

===

−

Reject H0 and conclude that the regression model is significant at α = 0.01. P-value = 0.000121

0643.87

45.56ˆ 2 ==

−==

pn

SS MS E

E σ

0254.0)59799.7(0643.8)ˆ( 1 =−= E se β

12-10



b) 0: 10 = β H

0: 11 ≠ β H

67.30254.0

0931.0

)ˆ(

ˆ

1

10

==

= β

β

set

7,005.00

7,005.0

||

499.3

t t

t

>

=

Reject H0 and conclude that 1 β is significant in the model at α = 0.01

P-value = 2 1 0( (− <P ))t t = 2(1 - 0.996018) = 0.007964

c) 1 x is useful as a regressor in the model.

12-21. a) 82.4: 01 =t β P-value = 2(4.08 E-5) = 8.16 E-5

21.8: 02 =t β P-value = 291.91 E-8) = 3.82 E-8

98.0: 03 =t β P-value = 2 (0.1689) = 0.3378

b) 0: 30 = β H

0: 31 ≠ β H

α = 0.05

98.00 =t

23,025.00

22,025.0

||

074.2

t t

t

>/

=

Do not reject H . Here0 3 x does not contribute significantly to the model.

12-22. a) 0: 43210 ==== β β β β H

H1 at least one 0≠ j β

α = 0.05

7,4,05.00

7,4,05.0

0

12.4

08.10

f f

f

f

>

=

=

Reject H P-value = 0.0050 b) α = 0.05

0: 10 = β H 02 = β 03 = β 04 = β

0: 11 ≠ β H 02 ≠ β 03 ≠ β 04 ≠ β

71.20 =t 87.10 =t 37.10 =t 87.00 −=t

365.27,025.0,2/ ==− t t pnα

0257.00 || t t >/ for 2 β , 3 β and 4 β

Reject H0 for 1 β .

12-11



12-23. a) 0: 6543210 ====== β β β β β β H

H1: at least one 0≠ β

14,6,05.00

14,6,05.014,6,

0

848.2

53.19

f f

f f

f

>

==

=

α

Reject H0 and conclude regression model is significant at α = 0.05

b) The t-test statistics for β1 through β6 are -0.45, -0.07, -3.42, 0.17, -2.90, 0.69. Because t0.025,14=-2.14,

the regressors that contribute to the model at α=0.05 are etw and axle.

12-24. a) 0:0 = j H β for all j


14,4,05.00

14,4,05.

0

11.3

16.7

f f

f

f

>

=

=

Reject H0 and conclude regression is significant at α = 0.05. P-value = 0.0023

b) 7792.0ˆ =σ

α = 0.05 t tn pα / , . , .2 025 14 2 145− = =

H 0 2 0:β = β3 0= β4 0= β5 0=

H1 2 0:β ≠ β3 0≠ β4 0≠ β5 0≠

t0 0 113= − . t0 3 83= . t0 1 91= − . t0 0 9= − .

| | / ,t t0 2/> α 14 14 14 | | / ,t t0 2> α | | / ,t t0 2/> α | | / ,t t0 2 14/> α

Do not reject H0 Reject H0 Do not reject H0 Do not reject H0

All variables do not contribute to the model.

12-25. a) 0: 3210 === β β β H

H j1

0:β ≠ for at least one j

16,3,05.00

16,3,05.

0

34.3

828.31

f f

f

f

>

=

=

Reject H0 and conclude regression is significant at α = 0.05

b) 2856.12ˆ 2 =σ

α = 0.05 t tn pα / , . , .2 025 16 2 12− = =

H0 1 0:β = β2 0= β3 0=

H1 1 0:β ≠ β2 0≠ β3 0≠

58.20 −=t 84.10 =t 82.130 =t

16,025.00 || t t > 16,025.00 || t t >/ 16,025.00 || t t >

Reject H0 Do not reject H0 Reject H0

12-26. ARSNAILS = 0.001 + 0.00858 AGE - 0.021 DRINKUSE + 0.010 COOKUSE


Constant 0.0011 0.9067 0.00 0.999

AGE 0.008581 0.007083 1.21 0.242

DRINKUSE -0.0208 0.1018 -0.20 0.841

COOKUSE 0.0097 0.1798 0.05 0.958

12-12



S = 0.506197 R-Sq = 8.1% R-Sq(adj) = 0.0%


Source DF SS MS F P

Regression 3 0.3843 0.1281 0.50 0.687


a)0:

0:

1

3210

≠

===

j H

H

β

β β β

for at least one j; k=4

17,3,05.00

17,3,05.0

0

197.3

50.0

05.0

f f

f

f

<

=

=

=α

Do not reject H0. There is insufficient evidence to conclude that the model is significant at α=0.05. The P-

value =0.687.

b) 0: 10 = β H

0: 11 ≠ β H

α = 0.05

21.1007083.0

008581.0

)ˆ(

ˆ

1

10 ===

β

β

set

11.217,025.0 =t

17,2/0 || α t t < . Fail to reject H , there is not enough evidence to conclude that0 1 β is significant in the

model at α = 0.05.

0: 20 = β H

0: 21 ≠ β H

α = 0.05

2.01018.0

0208.0

)ˆ(

ˆ

2

20 −=

−==

β

β

set

11.217,025.0 =t

17,2/0||

α t t < . Fail to reject H , there is not enough evidence to conclude that

0 2 β is significant in the

model at α = 0.05.

0: 30 = β H

0: 31 ≠ β H

α = 0.05

05.01798.0

0097.0

)ˆ(

ˆ

3

30 ===

β

β

set

12-13



11.217,025.0 =t

17,2/0 || α t t < . Fail to reject H , there is not enough evidence to conclude that0 3 β is significant in the

model at α = 0.05.

12-27. a) 0: 210 == β β H

for at least one:0 H 0≠ j β

α = 0.05

37,2,05.00

37,2,05.0

0

252.3

67.99

f f

f

f

>

=

=

The regression equation israds = - 440 + 19.1 mAmps + 68.1 exposure time


Constant -440.39 94.20 -4.68 0.000

mAmps 19.147 3.460 5.53 0.000

exposure time 68.080 5.241 12.99 0.000

S = 235.718 R-Sq = 84.3% R-Sq(adj) = 83.5%


Source DF SS MS F P

Regression 2 11076473 5538237 99.67 0.000


Total 39 13132310

Reject H0 and conclude regression model is significant at α = 0.05 P-value < 0.000001

b) 55563ˆ 2 == E MSσ

460.3ˆ)ˆ( 2

1 == jjcse σ β

0: 10 = β H

0: 11 ≠ β H

α = 0.05

539.5460.3

147.19

)ˆ(

ˆ

1

10

==

= β

β

set

0262.237,025.0340,025.0 ==− t t

37,2/0 || α t t > , Reject H and conclude that0 1 β is significant in the model at α = 0.05

241.5ˆ)ˆ( 2

2 == jjcse σ β

0: 20 = β H

12-14



0: 21 ≠ β H

α = 0.05

99.12241.5080.68

)ˆ(

ˆ

2

20

==

= β

β

set

0262.237,025.0340,025.0 ==− t t

37,2/0 || α t t > , Reject H conclude that0 2 β is significant in the model at α = 0.05

12-28.The regression equation is

y = - 171 + 7.03 x1 + 12.7 x2


Constant -171.26 28.40 -6.03 0.001

x1 7.029 1.539 4.57 0.004

x2 12.696 1.539 8.25 0.000

S = 3.07827 R-Sq = 93.7% R-Sq(adj) = 91.6%


Source DF SS MS F P

Regression 2 842.37 421.18 44.45 0.000


Total 8 899.22

a) 0: 210 == β β H


α = 0.05

6,2,05.00

6,2,05.0

0

14.5

45.446/85.56

2/37.842

f f

f

f

pn

SS

k

SS

E

R

>

=

===−

Reject H0 and conclude regression model is significant at α = 0.05 P-value ≈ 0

b) 48.9ˆ 2 == E MSσ

539.1ˆ)ˆ( 21 == jjcse σ β

0: 10 = β H

0: 11 ≠ β H

α = 0.05

12-15



568.4539.1

03.7

)ˆ(

ˆ

1

10

==

= β

β

set

447.26,025.039,025.0 ==− t t

6,2/0 || α t t > , Reject H ,0 1 β is significant in the model at α = 0.05

539.1ˆ)ˆ( 2

2 == jjcse σ β

0: 20 = β H

0: 21 ≠ β H

α = 0.05

252.8539.1

7.12

)ˆ(

ˆ

2

20

==

= β

β

set

447.26,025.039,025.0 ==− t t


c) With a smaller sample size, the difference in the estimate from the hypothesized value needs to be

greater to be significant.

12-29.Useful range (ng) = 239 + 0.334 Brightness (%) - 2.72 Contrast (%)


Brightness (%) 0.3339 0.6763 0.49 0.639

Contrast (%) -2.7167 0.6887 -3.94 0.008

S = 36.3493 R-Sq = 75.6% R-Sq(adj) = 67.4%


Source DF SS MS F P

Regression 2 24518 12259 9.28 0.015


Total 8 32446

a) 0: 210 == β β H


α = 0.05

12-16



6,2,05.00

6,2,05.0

0

14.5

28.96/7928

2/24518

f f

f

f pn

SS

k

SS

E

R

>

=

===−

Reject H0 and conclude that the regression model is significant at α = 0.05 P-value =0.015

b) 1321ˆ 2 == E MSσ

6763.0ˆ)ˆ( 2

1 == jjcse σ β

0: 10 = β H

0: 11 ≠ β H

α = 0.05

49.06763.0

3339.0

)ˆ(

ˆ

1

10

==

=

β

β

set

447.26,025.039,025.0 ==− t t

6,2/0 || α t t < , Fail to reject H , there is no enough evidence to conclude that0 1 β is significant in the model at α =

0.05

6887.0ˆ)ˆ( 2

2 == jjcse σ β

0: 20 = β H

0: 21 ≠ β H

α = 0.05

94.36887.0

7167.2

)ˆ(

ˆ

2

20

−=−

=

= β

β

set

447.26,025.039,025.0 ==− t t



Stack Loss(y) = - 39.9 + 0.716 X1 + 1.30 X2 - 0.152 X3


Constant -39.92 11.90 -3.36 0.004

X1 0.7156 0.1349 5.31 0.000

X2 1.2953 0.3680 3.52 0.003

X3 -0.1521 0.1563 -0.97 0.344

S = 3.24336 R-Sq = 91.4% R-Sq(adj) = 89.8%

12-17




Source DF SS MS F P

Regression 3 1890.41 630.14 59.90 0.000


Total 20 2069.24

a)

0: 3210 === β β β H


α = 0.05

17,3,05.00

17,3,05.0

0

20.3

90.5917/83.178

3/41.189

f f

f

f pn

SS

k

SS

E

R

>

=

===−

Reject H0 and conclude that the regression model is significant at α = 0.05 P-value < 0.000001

b) 52.10ˆ 2 == E MSσ

1349.0ˆ)ˆ( 2

1 == jjcse σ β

0: 10 = β H

0: 11 ≠ β H

α = 0.05

31.51349.0

7156.0

)ˆ(

ˆ

1

10

==

= β

β

set

110.217,025.0421,025.0 ==− t t

17,2/0 || α t t > . Reject H and conclude that0 1 β is significant in the model at α = 0.05.

3680.0ˆ)ˆ( 2

2 == jjcse σ β

0: 20 = β H

0: 21 ≠ β H

α = 0.05

52.33680.0

2953.1

)ˆ(

ˆ

2

2

0

==

= β

β

set

110.217,025.0421,025.0 ==− t t

17,2/0 || α t t > . Reject H and conclude that0 2 β is significant in the model at α = 0.05.

12-18



1563.0ˆ)ˆ( 2

3 == jjcse σ β

0: 30 = β H

0: 31 ≠ β H

α = 0.05

97.01563.0

1521.0

)ˆ(

ˆ

3

30

−=−

=

= β

β

set

110.217,025.0421,025.0 ==− t t

17,2/0 || α t t < . Fail to reject H , there is not enough evidence to conclude that0 3 β is significant in the model at α =

0.05.

12-31.

Rating Pts = 3.22 + 1.22 Pct Comp + 4.42 Pct TD - 4.09 Pct Int


Constant 3.220 6.519 0.49 0.626

Pct Comp 1.2243 0.1080 11.33 0.000

Pct TD 4.4231 0.2799 15.80 0.000

Pct Int -4.0914 0.4953 -8.26 0.000

S = 1.92094 R-Sq = 97.8% R-Sq(adj) = 97.5%


Source DF SS MS F P

Regression 3 4196.3 1398.8 379.07 0.000


a) 0: 10730 === β β β H


α = 0.05

26,3,05.00

26,3,05.

0

98.2

07.379

f f

f

f

>>

=

=

Reject H0

and conclude that the regression model is significant at α = 0.05 P-value

≈0

b) All at α = 0.05 056.226,025. =t

0: 30 = β H 0: 70 = β H 0: 100= β H

H1 3 0:β ≠ 0: 71 ≠ β H 0: 101 ≠ β H

33.110 =t 80.150 =t 26.80 −=t

26,2/0 || α t t > 26,2/0 || α t t > 26,2/0 || α t t >

Reject H0 Reject H0 Reject H0

12-19



c)


Rating Pts = - 33.6 + 2.18 Pct Comp - 4.54 Pct Int


Constant -33.60 19.46 -1.73 0.096

Pct Comp 2.1842 0.2855 7.65 0.000Pct Int -4.540 1.580 -2.87 0.008

S = 6.13862 R-Sq = 76.3% R-Sq(adj) = 74.5%


Source DF SS MS F P

Regression 2 3274.8 1637.4 43.45 0.000


Total 29 4292.2

SSR (full model) = 4196.3 SSR (reduced model) = 3274.8

Difference = 4196.3 – 3274.8 =921.5α = 0.05

24,1,05.00

26,1,05.0

0

26.4

1.2497.3

5.9211/9.2558

f f

f

MS f

E

>>

=

===

Reject H0 and conclude that the TD percentage regressor is significant at α = 0.05

P-value 0≈ Note that f 0 = 691.6 = t-statistic

2= 15.80

2= 249.6 (except for rounding error)

12-32. a) H j0 0:β = for all j

H j1 0:β ≠for at least one j

26,5,0

26,5,05.

0

59.2

9902.158

α f f

f

f

>

=

=


P-value < 0.000001

b) α = 0.05 t tn pα / , . , .2 025 26 2 056− = =

0: 10= β H 02 = β 03 = β 04 = β 05 = β

0: 11 ≠ β H 02 ≠ β 03 ≠ β 04 ≠ β 05 ≠ β

83.00 =t 25.120 =t 52.0

0 −=t 96.60 =t 29.00 −=t

Do not Reject H0 Reject H0 Do not reject H0 Reject H0 Do not reject H0

c) 21 009325.0002687.0010889.0ˆ x x y ++=

d) H j0 0:β = for all j

H j1 0:β ≠ for at least one j

29,2,05.00

29,2,05.

0

33.3

455.308

f f

f

f

>

=

=


12-20



α = 0.05 045.229,025.,2/ ==− t t pnα

0: 10= β H 02 = β

0: 11 ≠ β H 02 ≠ β

31.180 =t 37.60 =t

| | / ,t t0 2 2> α 9 | | / ,t t0 2> α 29

Reject H0 for each regressor variable and conclude that both variables are significant at α = 0.05

e) 67.6ˆ)( −= E d part σ . Part c) is smaller, suggesting a better model.

12-33. , Assume no interaction model.22110ˆˆˆˆ x x y β β β ++=

a) 0: 210 == β β H


3,2,05.00

3,2,05.0

0

55.9

59.97

f f

f

f

>

=

=

Reject H0 P-value = 0.002

b) 0: 10= β H 0: 20

= β H

0: 11 ≠ β H 0: 21 ≠ β H

42.60 −=t 57.20 −=t

182.33,025.03,2/ == t t α 182.33,025.03,2/ == t t α

3,025.00 || t t > 3,025.00 || t t >/

Reject H0 for regressor 1 β . Do not reject H0 for regressor 2 β .

c) 1012),|( 012 = β β β RSS

0: 20 = β H

0: 21 ≠ β H

α = 0.05

3,1,05.00

3,1,05.3,1,

0

13.10

629.6

f f

f f

f

>/

==

=

α

Do not reject H0

d) 0:12210

=== β β β H

H 1 at least one 0≠ j β

α = 0.05

2,3,05.00

2,3,05.02,3,

0

16.19

714.7

f f

f f

f

>/

==

=

α

Do not reject H0

12-21



e) 0: 120 = β H

0: 121 ≠ β H

α = 0.05

2,1,05.00

2,1,05.0

0

2112

51.18

11.1147

9.163

9.163297874.29951),|(

f f

f

MSSSR f

SSR

E

>/

=

===

=−= β β β

Do not reject H0

f) 0.147ˆ 2 =σ

σ2 (no interaction term) = 153.0

)ˆ( 2σ E MS was reduced in the model with the interaction term.

12-34. a) 0:0 = j H β for all j


15,14,05.00

15,14,05.

0

424.2

77.9

f f

f

f

>

=

=

Reject H0 and conclude that the regression model is significant at α = 0.05

b) 0:0 = j H β

0:1 ≠ j H β

131.215,025. =t

GF : Reject H32.20 =t 0

GA : Reject H58.30 −=t 0

ADV : Do not reject H61.00 =t 0

PPGF : Do not reject H60.00 −=t 0

PCTG : Do not reject H70.00 =t 0

PEN : Do not reject H70.00 =t 0

BMI : Do not reject H03.00 =t 0

AVG : Do not reject H67.00 −=t 0

SHT : Do not reject H02.10 −=t 0

PPGA : Do not reject H25.00

=t 0

PKPCT : Do not reject H07.00 =t 0

SHGF : Do not reject H48.10 =t 0

SHGA : Do not reject H38.10 =t 0

FG : Do not reject H71.00 =t 0

No, only "GF" (β1 ) and “GA” (β2 ) are significant at α = 0.05

c)

12-22



35.3

10.19

1075.028108.071.17ˆ

27,2,05.

0

41

=

=

−+−=

f

f

x x y

Reject H0

0: 10 = β H 04 = β

0: 11 ≠ β H 04 ≠ β

70.50=t 86.00

−=t

Reject H0 Do not reject H0

Based on the t-test, power play goals for (PPGF) is not a logical choice to add to the model that already contains the

number of goals for.



Sections 12-3 and 12-4

12-35. a)00

2

,2/0ˆˆ ct pnσ β α −±

183.292927.49

128.121055.171

)217.51)(365.2(055.171

)ˆ(055.171

0

07,025.

≤≤

±

±

±

β

β set

11

2

,2/1ˆˆ ct pn σ β α −±

393.7033.0

680.3713.3

)556.1)(365.2(713.3

)ˆ(713.3

1

17,025.

≤≤

±

±

±

β

β set

22

2

,2/2ˆˆ ct

pnσ β α −±

513.0765.2

639.1126.1

)693.0)(365.2(126.1

)ˆ(126.1

2

07,025.

≤≤−

±−

±−

±−

β

β set

b) 181 = x

471.189ˆ

43

0

2

=

=

y

x

127.220815.158

)305065.0(7875.550)365.2(471.189

305065.0)'(

0|

0

1'

0

≤≤

±

=−

xY

X X X X

μ

c) α = 0.05

181 = x

471.189ˆ43

0

2

==

y x

12-23



878.252064.126

)305065.1(7875.550)365.2(471.189

))'(1(ˆˆ

305065.0)'(

0

0

1'

0

2

,2/

0

1'

0

≤≤

±

+±

=

−

−

−

y

X X X X t y

X X X X

pn σ α

12-36.a)00

2,2/0 ˆˆ ct pn σ β α −±

8678.216922.25

78.239122.1

)055.10)(365.2(9122.1

)ˆ(9122.1

0

07,025.

≤≤−

±−

±−

±−

β

β set

11

2

,2/1ˆˆ ct pnσ β α −

±

2887.01025.0

1956.00931.0

)0827.0)(365.2(0931.0

)ˆ(0931.0

1

17,025.

≤≤−

±

±

±

β

β set

22

2


7257.02193.0

4725.02532.0

)1998.0)(365.2(2532.0

)ˆ(2532.0

2

07,025.

≤≤−

±

±

±

β

β set

b) x 1 200=

37.29ˆ50

0

2

==

y x

429.39311.19

059.1037.29

)211088.0(694.85)365.2(37.29

211088.0)'(

0|

0

1'

0

≤≤

±

±

=−

xY

X X X X

μ

c) α = 0.05

x 1 200=

37.29ˆ

50

0

2

=

=

y

x

463.53277.5

093.2437.29

)211088.1(694.85)365.2(37.29

))'(1(ˆˆ

211088.0)'(

0

0

1'

0

2

,2/

0

1'

0

≤≤

±

±

+±

=

−

−

−

y

X X X X t y

X X X X

pn σ α

12-37. a) 269.1477.20 1 ≤≤− β

12-24



159.22428.14

076.1245.0

3

2

≤≤

≤≤−

β

β

b) 91.372ˆ0| = xY μ 721.4)ˆ( 0 = yse 921.216,005. =t

162.105582.77

)721.4(921.2372.91

0 ≤≤

±

yc) 91.372ˆ

0| = xY μ 3.163)ˆ(0| = xY se μ

611.100133.82

)163.3)(921.2(372.91

0| ≤≤

±

xY μ

12.38. a) 95 % CI on coefficients

4172.10973.0 1 ≤≤ β

8319.07941.1

7613.67953.1

0026.179646.1

4

3

2

≤≤−

≤≤−

≤≤−

β

β

β

b) 44.290ˆ0| = xY

μ 61.7)ˆ(0| = xY se μ 365.27,025. =t

44.30844.272

)61.7)(365.2(44.290

)ˆ(ˆ

0

00

|

|,2/|

≤≤

±

± −

xY

xY pn xY set

μ

μ μ α

c) ))(1(ˆˆ0

1

0

2

,2/0 xXXx−

−′

′+± σ α pnt y

64.32325.257

)038.14(365.244.290

0

≤≤

±

y

12-39. a) 3295.09467.6 1 −≤≤− β

1417.03651.0 2 ≤≤− β

b) 5156.308276.45 1 ≤≤− β

04251.003433.0

8984.03426.1

12

2

≤≤−

≤≤−

β

β

These part b. intervals are much wider.

Yes, the addition of this term increased the standard error of the regression coefficient estimators.

12-40. a) 535.0595.0 2 ≤≤− β

9756.22968.7

013.0216.0

812.0229.0

5

4

3

≤≤−

≤≤−

≤≤

β

β

β

b) .|μY x08 99568= se Y x( ) .|μ

00 472445= t. , .025 14 2 145=

12-25



009.10982.7

)472445.0)(145.2(99568.8

)ˆ(ˆ

0

00

|

|,2/|

≤≤

±

± −

xY

xY pn xY set

μ

μ μ α

c) .y0 8 99568= se y( ) .0 100121=

143.118481.6

)00121.1(145.299568.8

0 ≤≤± y

12-41. a)

11

2


1577.261363.12

014458.7147.19

)460.3)(0262.2(147.19

)ˆ(147.19

1

137,025.

≤≤

±

±

±

β

β set

22

2,2/2

ˆˆ ct pn σ β α −±

6993.784607.57

014458.7080.68

)241.5)(0262.2(080.68

)ˆ(080.68

2

237,025.

≤≤

±

±

±

β

β set

b)

1.85ˆ0| −= xY

μ 6.54)ˆ(0| = xY se μ 7154.237,005.0 =t

2.634.233)6.54)(7154.2(1.85

)ˆ(ˆ

0

00

|

|,2/|

≤≤−

±−

± −

xY

xY pn xY set

μ

μ μ α

c) 1.85ˆ0 −= y 95.241)ˆ( 0 = yse

89.57109.742

)95.241(7154.21.85

0 ≤≤−

±−

y

12-42.The regression equation isARSNAILS = 0.001 + 0.00858 AGE - 0.021 DRINKUSE + 0.010 COOKUSE

Predictor Coef SE Coef T PConstant 0.0011 0.9067 0.00 0.999AGE 0.008581 0.007083 1.21 0.242DRINKUSE -0.0208 0.1018 -0.20 0.841COOKUSE 0.0097 0.1798 0.05 0.958

S = 0.506197 R-Sq = 8.1% R-Sq(adj) = 0.0%


12-26




a) 898.217,005.0 =t

531.0511.0

274.0316.0

0291.0012.0

634.2617.2

3

2

1

0

≤≤−

≤≤−

≤≤−

≤≤−

β

β

β

β

b)

214.0ˆ 0| = xY μ 216.0)ˆ( 0| = xY se μ 898.216,005.0 =t

840.0412.0

)216.0)(898.2(214.0

)ˆ(ˆ

0

00

|

|,2/|

≤≤−

±

± −

xY

xY pn xY set

μ

μ μ α

c) 214.0ˆ0 = y 5503.0)ˆ( 0 = yse

809.1381.1

)5503.0(898.2214.0

0 ≤≤−

±

y

12-43. a) 860.18,05.0 =t

-0.576 ≤ β0 ≤ 0.355

959.12743.8

7201.00943.0

2

1

≤≤−

≤≤

β

β

b)

8787.0ˆ0| = xY

μ 00926.0)ˆ(0| = xY se μ 860.116,005.0 =t

89592.086148.0

)00926.0)(860.1(8787.0

)ˆ(ˆ

0

00

|

|,2/|

≤≤

±

± −

xY

xY pn xY set

μ

μ μ α

c) 8787.0ˆ0 = y 0134.0)ˆ( 0 = yse

90250.085490.0

)0134.0(86.18787.0

0 ≤≤

±

y


12-27



y = - 171 + 7.03 x1 + 12.7 x2

Predictor Coef SE Coef T PConstant -171.26 28.40 -6.03 0.001x1 7.029 1.539 4.57 0.004x2 12.696 1.539 8.25 0.000

S = 3.07827 R-Sq = 93.7% R-Sq(adj) = 91.6%



a)

11

2

,2/1ˆˆ ct

pnσ β α −

±

796.10264.3

766.303.7

)539.1)(447.2(03.7

)ˆ(03.7

1

16,025.

≤≤

±

±

±

β

β set

22

2

,2/2ˆˆ ct

pnσ β α −

±

466.16934.8

766.37.12

)539.1)(447.2(7.12

)ˆ(7.12

1

26,025.

≤≤

±

±

±

β

β set

b)

NewObs Fit SE Fit 95% CI 95% PI

1 140.82 6.65 (124.54, 157.11) (122.88, 158.77)XX

82.140ˆ0| =

xY μ 65.6)ˆ(

0| = xY se μ 447.26,025.0 =t

09.15755.124

)65.6)(447.2(82.140

)ˆ(ˆ

0

00

|

|,2/|

≤≤

±

± −

xY

xY pn xY set

μ

μ μ α

c) 82.140ˆ0 = y 33.7)ˆ( 0 = yse

76.15888.122

)33.7(447.282.140

0 ≤≤

±

y

12-28



d) The smaller the sample size, the wider the interval

12-45.

The regression equation isUseful range (ng) = 239 + 0.334 Brightness (%) - 2.72 Contrast (%)

Predictor Coef SE Coef T PConstant 238.56 45.23 5.27 0.002Brightness (%) 0.3339 0.6763 0.49 0.639Contrast (%) -2.7167 0.6887 -3.94 0.008

S = 36.3493 R-Sq = 75.6% R-Sq(adj) = 67.4%


Source DF SS MS F PRegression 2 24518 12259 9.28 0.015Residual Error 6 7928 1321

Total 8 32446

a) 707.36,005.0 =t

164.0270.5

841.2173.2

2

1

−≤≤−

≤≤−

β

β

b)Predicted Values for New Observations


1 44.6 21.9 (-36.7, 125.8) (-112.8, 202.0)

Values of Predictors for New Observations

New ContrastObs Brightness (%) (%)

1 70.0 80.0

6.44ˆ0| =

xY μ 9.21)ˆ(

0| = xY se μ 707.36,005.0 =t

8.1257.36

)9.21)(707.3(6.44

)ˆ(ˆ

0

00

|

|,2/|

≤≤−

±

± −

xY

xY pn xY set

μ

μ μ α

c) 6.44ˆ0 = y 44.42)ˆ( 0 = yse

0.2028.112

)44.42(707.36.44

0 ≤≤−

±

y

d) Predicted Values for New Observations


1 187.3 21.6 (107.4, 267.2) (30.7, 344.0)

12-29



Values of Predictors for New Observations

New ContrastObs Brightness (%) (%)

1 50.0 25.0

CI: 2.2674.1070| ≤≤ xY μ

PI: 0.3447.30 0 ≤≤ y

These intervals are wider because the regressors are set at extreme values in the x space and the standard

errors are greater.

12-46. a) 110.217,025.0 =t

178.0482.0

072.2519.0

00.1431.0

3

2

1

≤≤−

≤≤

≤≤−

β

β

β

b) 023.36ˆ0| =

xY μ 803.1)ˆ(

0| = xY se μ 110.217,025.0 =t

827.39219.32

)803.1)(110.2(023.36

)ˆ(ˆ

0

00

|

|,2/|

≤≤

±

± −

xY

xY pn xY set

μ

μ μ α

c) 023.36ˆ0 = y 698.3)ˆ( 0 = yse

852.43194.28

)698.3(110.2023.36

0 ≤≤

±

y

d) Prediction at x1 = 80, x2 = 19, x3 = 93 is 795.270| = xY μ

CI: 559.34030.210| ≤≤ xY μ

PI: 417.37173.18 0 ≤≤ y

12-47. a) 056.226,025.0 =t

073.3110.5

999.4848.3

446.1002.1

623.16183.10

10

7

3

0

−≤≤−

≤≤

≤≤

≤≤−

β

β

β

β

b) )ˆ(3877.0)'(ˆ0|0

1'

0

2

xY se X X X X μ σ ==−

c) 079.82)3(091.4)4(423.4)60(224.122.3ˆ =−++= y

876.82282.81

7971.0079.82

)3877.0)(056.2(079.82

)ˆ(ˆ

0

00

|

|24,025.|

≤≤

±

±

±

xY

xY xY set

μ

μ μ

12-48. a) 000099.0000042.0 1 ≤≤− β

12-30



01424.001897.0

010979.000597.0

00898.001504.0

00268.000196.0

5

4

3

2

≤≤−

≤≤

≤≤−

≤≤

β

β

β

β

b) 022466.0ˆ 0| = xY μ 000595.0)ˆ( 0| = xY se μ 056.226,025. =t

0237.00212.0

)000595.0)(056.2(0220086.0

0| ≤≤

±

xY μ

c) 0171.0ˆ0| = xY μ 000548.0)ˆ(

0| = xY se μ 045.229,025. =t

0183.00159.0

)000548.0)(045.2(0171.0

0| ≤≤

±

xY μ

d) : width = 2.2 E-3

: width = 2.4 E-3

The interaction model has a shorter confidence interval. Yes, this suggests the interaction model is preferable.

12-49. a) t.005,14=2.977

523.0102.1

101.0811.7

546.5962.4

0006.0

047.005.0

059.008.0

458.108658.8

10

9

8

7

3

2

0

−≤≤

≤≤−

≤≤−

≤≤−

≤≤−

≤≤−

≤≤−

β

β

β

β

β

β

β

b)71.29ˆ 0| = xY μ

395.1)ˆ( 0| = xY se μ

863.33557.25

)395.1)(977.2(71.29

)ˆ(ˆ

0

00

|

|14,005.|

≤≤

±

±

xY

xY xY set

μ

μ μ

c) 972 457.30035.00208.0001.61ˆ x x x y −−−=

429.0485.6001.0006.0

006.0036.0

147.71855.50

898.2

9

7

2

0

17,005.

−≤≤−−≤≤−

−≤≤−

≤≤

=

β β

β

β

t

d) The intervals in part c) are narrower. All of the regressors used in part c) are significant, but not all of

those used in part a) are significant. The model used in part c) is preferable.

12-50. a) 2369.00102.0 1 ≤≤ β

b)125956.0329.19ˆ x y +−=

12-31



c) 3456.01735.0 1 ≤≤ β

d) The simple linear regression model has the shorter interval. Obviously there are extraneous variables in

the model from part a), but there are two good predictors in that model, only one of which is included in

the model of part b). Still the shorter interval is an initial indicator that the original model with all variables

might be improved.



Section 12-5

12-51.

a)The regression equation ismpg = 49.9 - 0.0104 cid - 0.0012 rhp - 0.00324 etw + 0.29 cmp - 3.86 axle

+ 0.190 n/v

Predictor Coef SE Coef T PConstant 49.90 19.67 2.54 0.024cid -0.01045 0.02338 -0.45 0.662rhp -0.00120 0.01631 -0.07 0.942

etw -0.0032364 0.0009459 -3.42 0.004cmp 0.292 1.765 0.17 0.871axle -3.855 1.329 -2.90 0.012n/v 0.1897 0.2730 0.69 0.498

S = 2.22830 R-Sq = 89.3% R-Sq(adj) = 84.8%

b) There appears to be an outlier. Otherwise, the normality assumption is not violated.

Standardized Residual

S c o r e

3210-1-2-3

2

1

0

-1

-2

Norma l Pr obab i li t y P lo t o f the Res iduals(response is mpg)

c) The plots do not show any violations of the assumptions.

12-32



Fitted Value

S t a n d a r d

i z e d

R e s i d u a l

4035302520

2

1

0

-1

-2

-3

Residuals Versus the Fi t ted Values(response is mpg)

cid

S t a n d a r d i z e d

R e s i d u a l

500400300200100

2

1

0

-1

-2

-3

Residuals Versus cid(response is mpg)

et w

S

t a n d a r d i z e d

R e s i d u a l

6000550050004500400035003000

2

1

0

-1

-2

-3

Residuals Versus etw(response is mpg)

cmp


R e s i d u a l

10.009.759.509.259.008.758.50

2

1

0

-1

-2

-3

Residuals Ver sus cmp(response is mpg)

12-33



axle

S t a n d a r d

i z e d

R e s i d u a l

4.254.003.753.503.253.002.752.50

2

1

0

-1

-2

-3

Residuals Versus axle(response is mpg)

n/ v


R e s i d u a l

4035302520

2

1

0

-1

-2

-3

Residuals Versus n/ v(response is mpg)

d)

0.036216, 0.000627, 0.041684, 0.008518, 0.026788, 0.040384, 0.003136,

0.196794, 0.267746, 0.000659, 0.075126, 0.000690, 0.041624, 0.070352,

0.008565, 0.051335, 0.001813, 0.019352, 0.000812, 0.098405, 0.574353

None of the values is greater than 1 so none of the observations are influential.

12-52. a)2 = R 852.0

b) The residual plots look reasonable. There is some increase in variability at the middle of the predicted

values.

c) Normality assumption is reasonable. The residual plots appear reasonable too.

Residual

P e r c e n t

20100-10-20

99

95

90

80

70

60

50

40

30

20

10

5

1


12-34



x1

R

e s i d u a l

9080706050403020

10

5

0

-5

-10

-15

Residuals Versus x1(response is y)

x2

R

e s i d u a l

262524232221

10

5

0

-5

-10

-15


x3

R

e s i d u a l

94939291908988878685

10

5

0

-5

-10

-15


x4

R

e s i d u a l

1101051009590

10

5

0

-5

-10

-15


12-53. a) %8.972 = R b) Assumption of normality appears adequate.

Residual

P e r c e n t

543210-1-2-3-4

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probability Plot of the Residuals(response is Rating Pts)

c) Model appears adequate. Some suggestion of nonconstant variance in the plot of x (percentage of TDs)7

Fitt ed Value

R e s i d u a l

120110100908070

4

3

2

1

0

-1

-2

-3

-4

-5

Residuals Versus the Fitted Values(response is Rating Pts)

Pct Comp

R e s i d u a l

70.067.565.062.560.057.555.0

4

3

2

1

0

-1

-2

-3

-4

-5

Residuals Versus Pct Comp(response is Rating Pts)

12-35



Pct TD

R e s i d u a l

1098765432

4

3

2

1

0

-1

-2

-3

-4

-5

Residuals Versus Pct TD(response is Rating Pts)

Pct Int

R e s i d u a l

4.54.03.53.02.52.01.51.0

4

3

2

1

0

-1

-2

-3

-4

-5

Residuals Versus Pct I nt(response is Rating Pts)

d) No, none of the observations has a Cook’s distance greater than 1.

12-54. a) R2= 0.969

b) Normality is acceptable

210-1-2

0.005

0.000

-0.005

Normal Score

R e s i d u a l

Normal Probability Plot of the Residuals(response is PITCH)

c) Plot is acceptable.

0.070.060.050.040.030.020.01

0.005

0.000

-0.005

Fitted Value

R e s i d u a l

Residual s Versus the Fitted Values(response is PITCH)

d) Cook’s distance values

12-36



0.0191 0.0003 0.0026 0.0009 0.0293 0.1112 0.1014 0.0131 0.0076 0.0004 0.0109

0.0000 0.0140 0.0039 0.0002 0.0003 0.0079 0.0022 4.5975 0.0033 0.0058 0.1412

0.0161 0.0268 0.0609 0.0016 0.0029 0.3391 0.3918 0.0134 0.0088 0.5063

The 19th observation is influential

12-55. a) %3.842 = R

b) Assumption of normality appears adequate.

Residual

P e r c e n t

7505002500-250-500

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probability Plot of the Residuals(response is rads)

c) There are funnel shapes in the graphs, so the assumption of constant variance is violated. The model is

inadequate.

Fitt ed Value

R e s i d u a l

150010005000-500

750

500

250

0

-250

-500

Residuals Versus the Fitted Values(response is rads)

exposure time

R e s i d u a l

20151050

750

500

250

0

-250

-500

Residuals Versus exposure time(response is rads)

mAmps

R e s i d u a l

40353025201510

750

500

250

0

-250

-500

Residuals Versus mAmps(response is rads)


12-37



0.032728 0.029489 0.023724 0.014663 0.008279 0.008611

0.077299 0.3436 0.008489 0.007592 0.006018 0.003612

0.001985 0.002068 0.021386 0.105059 0.000926 0.000823

0.000643 0.000375 0.0002 0.000209 0.002467 0.013062

0.006095 0.005442 0.0043 0.002564 0.0014 0.001459

0.015557 0.077846 0.07828 0.070853 0.057512 0.036157

0.020725 0.021539 0.177299 0.731526 No, none of the observations has a Cook’s distance greater than 1.

12-56. a) %1.82 = R

b) Assumption of normality appears is not adequate.

Residual

P e r c e n t

1.51.00.50.0-0.5-1.0

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probabilit y Plot of the Residuals(response is ARSNAILS)

c) The graphs indicate non-constant variance. Therefore, the model is not adequate.

Fitted Value


R e s i d

u a l

0.70.60.50.40.30.20.10.0

4

3

2

1

0

-1

Residuals Versus the Fi t t ed Values(response is ARSNAILS)

AGE

R e s i d u a l

9080706050403020100

1.5

1.0

0.5

0.0

-0.5

Residuals Versus AGE(response is ARSNAILS)

12-38



DRINKUSE

R e s i d u a

l

54321

1.5

1.0

0.5

0.0

-0.5

Residuals Versus DRI NKUSE(response is ARSNAILS)

COOKUSE

R e s i d u a

l

5.04.54.03.53.02.52.0

1.5

1.0

0.5

0.0

-0.5

Residuals Versus COOKUSE(response is ARSNAILS)


0.0032 0.0035 0.00386 0.05844 0.00139 0.00005 0.00524 0.00154*

0.00496 0.05976 0.37409 0.00105 1.89094 0.68988 0.00035 0.00092 0.0155

0.00008 0.0143 0.00071

There are two influential points with Cook’s distance greater than one. The data shown as (*) indicate thatCook’s distance is very large.

12-57. a) %7.992 = R


Residual

P e r c e n

t

0.020.010.00-0.01-0.02

99

95

90

80

70

6050

40

30

20

10

5

1

Normal Probabili ty Plot of the Residuals(response is density)

c) There is a non-constant variance shown in graphs. Therefore, the model is inadequate.

Fitt ed Value

R e s i d u a l

1.21.11.00.90.80.7

0.010

0.005

0.000

-0.005

-0.010

Residuals Versus the Fitt ed Values(response is density)

12-39



dielect ric constant

R e s i d u a

l

3.02.82.62.42.22.0

0.010

0.005

0.000

-0.005

-0.010

Residuals Versus dielectric constant(response is density)

loss factor

R e s i d u a

l

0.0450.0400.0350.0300.0250.0200.015

0.010

0.005

0.000

-0.005

-0.010

Residuals Versus loss factor(response is density)


0.255007 0.692448 0.008618 0.011784 0.058551 0.077203

0.10971 0.287682 0.001337 0.054084 0.485253

No, none of the observations has a Cook’s distance greater than 1.

12-58. a) %7.932 = Rb) The normal assumption appears inadequate

Residual

P e r c e n t

5.02.50.0-2.5-5.0-7.5

99

95

90

80

70

60

50

40

30

20

10

5

1


c) The constant variance assumption is not invalid.

Fitted Value


R e s i d u a l

10090807060

2.0

1.5

1.0

0.5

0.0

-0.5

-1.0

Residuals Versus the Fi t t ed Values(response is y)

12-40



x1

R e s i d u a l

15.014.514.013.513.0

3

2

1

0

-1

-2

-3

-4


x2

R e s i d u a l

13.012.512.011.511.0

3

2

1

0

-1

-2

-3

-4



1.36736 0.7536 0.7536 1.36736 0.0542 0.01917 0.03646 0.02097 0.00282There are two influential points with Cook’s distances greater than 1.

12-59. a) %6.752 = R


Residual

P

e r c e n t

806040200-20-40-60-80

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probability Plot of the Residuals(response is Useful range (ng))

c) Assumption of constant variance is a possible concern. One point is a concern as a possible outlier.

12-41



Fitt ed Value

R e s i d

u a l

2001751501251007550

80

60

40

20

0

-20

-40

Residuals Versus the Fitted Values(response is Useful range (ng))

Contrast (% )

R e s i d u a l

80706050403020

80

60

40

20

0

-20

-40

Residuals Versus Contrast ( % )(response is Useful range (ng))

Brightness (% )

R e s i d

u a l

1009080706050

80

60

40

20

0

-20

-40

Residuals Versus Brightness (% )(response is Useful range (ng))


0.006827 0.032075 0.045342 0.213024 0.000075

0.154825 0.220637 0.030276 0.859916


12-60. a) %4.912

= R b) Assumption of normality appears adequate.

Residual

P e r c e n t

86420-2-4-6-8

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probability Plot of the Residuals(response is S tack Loss(y))

c) Assumption of constant variance appears reasonable

12-42



Fitt ed Value

R

e s i d u a l

403530252015105

5.0

2.5

0.0

-2.5

-5.0

-7.5

Residuals Versus the Fitted Values(response is S tack Loss(y))

X1

R

e s i d u a l

80757065605550

5.0

2.5

0.0

-2.5

-5.0

-7.5

Residuals Versus X1(response is Stack Loss(y))

X2

R

e s i d u a l

28262422201816

5.0

2.5

0.0

-2.5

-5.0

-7.5

Residuals Versus X2(response is S tack Loss(y))

X3

R

e s i d u a l

959085807570

5.0

2.5

0.0

-2.5

-5.0

-7.5

Residuals Versus X3(response is S tack Loss(y))


0.15371 0.059683 0.126414 0.130542 0.004048 0.019565

0.048802 0.016502 0.044556 0.01193 0.035866 0.065066

0.010765 0.00002 0.038516 0.003379 0.065473 0.001122

0.002179 0.004492 0.692


12-61. a) 985.02 = R

b) 99.02 = R

R2

increases with addition of interaction term. No, adding additional regressor will always increase r 2

12-62. a) . Yes, the R 955.02 = R 2 using these two regressors is nearly as large as the R 2 from the model with

five regressors.

b) Normality is acceptable, but there is some indication of outliers.

0.070.060.050.040.030.020.01

0.005

0.000

-0.005

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is PITCH)

210-1-2

0.005

0.000

-0.005

Normal Score

R e s i d u a l

Normal Probability Plot of the Residuals(response is PITCH)

c) Cook’s distance values

0.0202 0.0008 0.0021 0.0003 0.0050 0.0000 0.0506 0.0175 0.0015 0.0003 0.0087

12-43



0.0001 0.0072 0.0126 0.0004 0.0021 0.0051 0.0007 0.0282 0.0072 0.0004 0.1566

0.0267 0.0006 0.0189 0.0179 0.0055 0.1141 0.1520 0.0001 0.0759 2.3550The last observation is very influential

12-63. a) There is some indication of nonconstant variance since the residuals appear to “fan out” with increasing

values of y.

0 40 80 120 160 200 240

-7

-4

-1

2

5

8

Predicted

R e s i d u a l s

Residual Plot for y

b)

Source Sum of Squares DF Mean Square F-Ratio P-value

Model 30531.5 3 10177.2 840.546 .0000Error 193.725 16 12.1078

Total (Corr.) 30725.2 19

R-squared = 0.993695 Stnd. error of est. = 3.47963


or 99.37 %;R 2

0 9937= .

or 99.25%;R Adj2 0 9925= .

c)Model fitting results for: log(y)

--------------------------------------------------------------------------------

Independent variable coefficient std. error t-value sig.levelCONSTANT 6.22489 1.124522 5.5356 0.0000x1 -0.16647 0.083727 -1.9882 0.0642x2 -0.000228 0.005079 -0.0448 0.9648

x3 0.157312 0.029752 5.2875 0.0001

--------------------------------------------------------------------------------R-SQ. (ADJ.) = 0.9574 SE= 0.078919 MAE= 0.053775 DurbWat= 2.031Previously: 0.0000 0.000000 0.000000 0.00020 observations fitted, forecast(s) computed for 0 missing val. of dep. var.

. . . .y x x∗ = − − +6 22489 016647 0 000228 01573121 2 x3

d)

12-44



4 4.3 4.6 4.9 5.2 5.5

-0.15

-0.1

-0.05

0

0.05

0.1

Predicted

R e s i d u a l s

Residual Plot for log(y)

Plot exhibits curvature

There is curvature in the plot. The plot does not give much more information as to which model is preferable.

e)

3.3 5.3 7.3 9.3 11.3

-0.15

-0.1

-0.05

0

0.05

0.1

x3

R e s i d u a l s


Plot exhibits curvature

Variance does not appear constant. Curvature is evident.

f)Model fitting results for: log(y)

Independent variable coefficient std. error t-value sig.levelCONSTANT 6.222045 0.547157 11.3716 0.0000

x1 -0.198597 0.034022 -5.8374 0.0000x2 0.009724 0.001864 5.2180 0.00011/x3 -4.436229 0.351293 -12.6283 0.0000--------------------------------------------------------------------------------R-SQ. (ADJ.) = 0.9893 SE= 0.039499 MAE= 0.028896 DurbWat= 1.869

Analysis of Variance for the Full RegressionSource Sum of Squares DF Mean Square F-Ratio P-value

Model 2.75054 3 0.916847 587.649 .0000Error 0.0249631 16 0.00156020Total (Corr.) 2.77550 19R-squared = 0.991006 Stnd. error of est. = 0.0394993


12-45



3.8 4.1 4.4 4.7 5 5.3

-0.07

-0.04

-0.01

0.02

0.05

0.08

Predicted

R e s i d u a l s


Using 1/x3

The residual plot indicates better conformance to assumptions.

Curvature is removed when using 1/x3 as the regressor instead of x3 and the log of the response data.

12-64. a)

The regression equation isW = - 19.3 + 0.260 GF

Predictor Coef SE Coef T PConstant -19.329 8.943 -2.16 0.039GF 0.25956 0.04220 6.15 0.000

S = 5.43788 R-Sq = 57.5% R-Sq(adj) = 55.9%

b) R-Sq = 57.5

c) Model appears adequate.

Fitted Value

S t a n d a r d i z e d R e s i d u a l

504540353025

2

1

0

-1

-2

Residuals Versus the Fi t t ed Values(response is W)

d) No, the residuals do not seem to be related to PPGF. Since there is no pattern evident in the plot, it does not

seem that this variable would contribute significantly to the model.

12-46



PPGF

S t a n d a r d

i z e d

R e s i d u a l

8070605040

2

1

0

-1

-2

Residuals Ver sus PPGF(response is W)

12-65. a) p = k + 1 = 2 + 1 = 3

Average size = p/n = 3/25 = 0.12

b) Leverage point criteria:

24.0

)12.0(2

)/(2

>

>

>

ii

ii

ii

h

h

n ph

2929.0

2593.0

18,18

17,17

=

=

h

h

Points 17 and 18 are leverage points.



Sections 12-6

12-66. a)233.0205.18915.26219ˆ x x y −+−=

b) H j0 0:β = for all j


α = 0.05

5,2,05.0

5,2,05.

0

79.5

2045.17

f f

f

f

>

=

=

Reject H and conclude that model is significant at α = 0.050

c) H 0 11 0:β =H1 11 0:β ≠

α = 0.05

571.2||

571.2

45.2

0

5,025.38,025.,

0

>/

===

−=

−−

t

t t t

t

pnα

Do not reject H and conclude insufficient evidence to support value of quadratic term in model at α = 0.050

d) One residual is an outlier

12-47



Normality assumption appears acceptable

Residuals against fitted values are somewhat unusual, but the impact of the outlier should be considered.

-50 -40 -30 -20 -10 0 10 20 30 40 50

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5


Residual


(response is y)

500 600 700 800

-50

-40

-30

-20

-10

0

10

20

30

40

50

Fitted Value

R e s i d u a l


(response is y)

12-67. a)2495.1232.1633.1ˆ x x y −+−=

b) f 0 = 1858613, reject H0

c) t 0 = −601.64, reject H 0

d) Model is acceptable. Observation number 10 has large leverage.

12-48



-0.004 -0.002 0.000 0.002 0.004 0.006

-1

0

1


Residual


(response is y)

0.0 0.5 1.0 1.5 2.0 2.5

-0.004

-0.002

0.000

0.002

0.004

0.006

x

R e s i d u a l

Residuals Versus x

(response is y)

-8 -7 -6 -5 -4 -3 -2 -1

-0.004

-0.002

0.000

0.002

0.004

0.006

Fitted Value

R e s i d u a l


(response is y)

12-68. a)247.138.146.4ˆ x x y ++−=

b) 0:0 = j H β for all j

12-49




α = 0.05

9,2,05.00

9,2,05.

0

26.4

99.1044

f f

f

f

>

=

=

Reject H0 and conclude regression model is significant at α = 0.05

c) 0: 110 = β H

0: 111 ≠ β H α = 0.05

9,025.00

9,025.

0

||

262.2

97.2

t t

t

t

>

=

=

Reject H0 and conclude that β is significant at α = 0.0511

d) Observation number 9 is an extreme outlier.

-5 -4 -3 -2 -1 0 1

-2

-1

0

1

2


Residual


(response is y)

25 35 45 55 65 75 85

-5

-4

-3

-2

-1

0

1

Fitted Value

R e s

i d u a l


(response is y)

12-50



e)32 51.004.701.4836.87ˆ x x x y +−+−=

0: 330 = β H

0: 331 ≠ β H α = 0.05

8,025.00

8,025.

0

||

306.2

91.0

t t

t

t

>/=

=

Do not reject H0 and conclude that cubic term is not significant at α = 0.05

12-69. a) Predictor Coef SE Coef T P

Constant -1.769 1.287 -1.37 0.188xl 0.4208 0.2942 1.43 0.172x2 0.2225 0.1307 1.70 0.108

x3 -0.12800 0.07025 -1.82 0.087x1x2 -0.01988 0.01204 -1.65 0.118x1x3 0.009151 0.007621 1.20 0.247x2x3 0.002576 0.007039 0.37 0.719

x1^2 -0.01932 0.01680 -1.15 0.267x2^2 -0.00745 0.01205 -0.62 0.545x3^3 0.000824 0.001441 0.57 0.575

S = 0.06092 R-Sq = 91.7% R-Sq(adj) = 87.0%


Source DF SS MS F P

Regression 9 0.655671 0.072852 19.63 0.000Residual Error 16 0.059386 0.003712

Total 25 0.715057

1312321 009.002.0128.0222.0421.0769.1ˆ x x x x x y +−−++−=

2

3

2

2

2

123 001.0007.0019.0003.0 x x x x +−−+

b) H0 all 023321 ==== β β β β


16,9,05.00

16,9,05.

0

54.2

628.19

f f

f

f

>

=

=

Reject H0 and conclude that the model is significant at α = 0.05

c) Model is acceptable.

12-51



-0.05 0.00 0.05 0.10

-2

-1

0

1

2


Residual


(response is y)

0.0 0.1 0.2 0.3 0.4 0.5

-0.05

0.00

0.05

0.10

Fitted Value

R e s i d u a l


(response is y)

d) 0: 2313123322110 ====== β β β β β β H

H1 at least one 0≠ β

16,6,05.0

16,6,05.

60359.0

0321231312332211

0

74.2

612.1003712.0

/),,,|,,,,,(

f f

f

MS

r SS f

E

R

>/

=

===β β β β β β β β β β

Do not reject H0

0359.0

619763.065567068.0

)|()|(

)|(

03210321231312332211

0321231312332211

=

−=

−= β β β β β β β β β β β β β β

β β β β β β β β β β

R R

R

SSSS

SS

Reduced Model: 3322110 x x x y β β β β +++=

12-70.

a) Create an indicator variable for sex (e.g. 0 for male, 1 for female) and include this variable in the model.

b)

12-52



The regression equation isARSNAILS = - 0.214 - 0.008 DRINKUSE + 0.028 COOKUSE + 0.00794 AGE + 0.167 SEXID

Predictor Coef SE Coef T PConstant -0.2139 0.9708 -0.22 0.828DRINKUSE -0.0081 0.1050 -0.08 0.940

COOKUSE 0.0276 0.1844 0.15 0.883AGE 0.007937 0.007251 1.09 0.290SEXID 0.1675 0.2398 0.70 0.495

S = 0.514000 R-Sq = 10.8% R-Sq(adj) = 0.0%

where SEXID = 0 for male and 1 for female

c) Since the P-value for testing 0:0 =sex H β against 0:1 ≠sex H β is 0.495, there is no evidence that

the person’s sex affects arsenic in the nails.

12-71. a) Use indicator variable for transmission type.

There are three possible transmission types: L4, L5 and M6. So, two indicator variables could be usedwhere x3=1 if trns=L5, 0 otherwise and x4=1 if trns=M6, 0 otherwise.

b) 4321 179.4138.000525.01457.0677.56ˆ x x x x y −−−−=

c) The P-value for testing 0: 30 = β H is 0.919 which is not significant. However, the P-value for testing

0: 40 = β H is 0.02 which is significant for values of α > 0.02. Thus, it appears that whether or not the

transmission is manual affects mpg, but there is not a significant difference between the types of automatic

transmission.

12-72. 121222110 x x x y β β β β +++=

1221 031.0094.6153.0503.11ˆ x x x y −−+=

where

⎩⎨⎧

=416typefor tool 1

302typefor tool 02 x

Test of different slopes:

0: 120 = β H

0: 121 ≠ β H α = 0.05

16,025.00

16,025.

0

||

12.2

79.1

t t

t

t

>/

=

−=

Do not reject H0. Conclude that data is insufficient to claim that (2) regression models are needed.

Test of different intercepts and slopes using extra sums of squares:

0: 1220 == β β H

H1 at least one is not zero

7508.882

60910.13035995.1013

)|()|,,()|,( 01012210122

=

−=

−= β β β β β β β β β SSSSSS

12-53



40.10874059.0

2/7508.8822/)|,( 01220 ===

E MS

SS f

β β β

Reject H0.

12-73. a) The min C p equation is: x1, x2

and0.3= pC 92.55563= E MS 21 080.68147.1939.440ˆ x x y ++−=

The min equation is the same as the min C E MS p.

b) Same as equation in part (a).

c) Same as equation in part (a).

d) Same as equation in part (a).

e) All methods give the same equation with either min C p or min MS E .

12-74. The default settings for F-to-enter and F-to-remove for Minitab were used. Different settings can change the models

generated by the method.

a) The min MS E equation is: x1, x2, x 3

8.3= pC 6.134= E

MS

321 343.2691.77487.01.162ˆ x x x y +++−=

The min C p equation is: x1, x2,

4.3= pC 7.145= E MS

21 882.95727.092.3ˆ x x y ++=

b) Same as the min C p equation in part (a)

c) Same as part min MS E equation in part (a)

d) Same as part min C p equation in part (a)

e) The minimum MSE and forward models all are the same. Stepwise and backward regressions generate the minimum

C p model. The minimum C p model has fewer regressors and it might be preferred, but MS E has increased.

12-75. a) The min C p equation is: x1

and1.1= pC 0000705.0= E MS

1467864.020052.0ˆ x y +−=The min equation is the same as the min C E MS p.







a) The min MS E equation is: x1, x3, x 4

6.2= pC 6644.0= E MS

431 12338.04790.05530.0419.2ˆ x x x y −++=

The min C p equation is: x3, x4

6.1= pC 7317.0= E MS

43 12418.05113.0656.4ˆ x x y −+=

b) Same as the min C p equation in part (a)

c) Same as the min C p equation in part (a)

d) Same as the min C p equation in part (a)

12-54



e) The minimum MSE and forward models all are the same. Stepwise and backward regressions generate the minimum

C p model. The minimum C p model has fewer regressors and it might be preferred, but MS E has increased.


and2.1= pC 55.1178= E MS

25453.206.253ˆ x y −=The min equation is the same as the min C E MS p.





12-78. a) The min C p equation is: x1 ,x2

and0.3= pC 4759.9= E MS

21 696.12029.7171ˆ x x y ++−=The min equation is the same as the min C E MS p.


c) Same as equation in part (a).d) Same as equation in part (a).


12-79. a) The min C p equation is: x1, x2

and9.2= pC 49.10= E MS

21 30.1671.04.50ˆ x x y ++−=The min equation is the same as the min C E MS p.




e) All methods give the same equation with either min C p or min MS E .f) There are no observations with a Cook’s distance greater than 1 so the results will be the same.



a)

Best Subsets Regression: W versus GF, GA, ...Response is W

PP P P K S S

A P C P B A S P P H HMallows G G D G T E M V H G C G G F

Vars R-Sq R-Sq(adj) C-p S F A V F G N I G T A T F A G1 63.5 62.1 29.5 5.0407 X1 57.5 55.9 38.6 5.4379 X2 86.3 85.3 -3.3 3.1391 X X2 75.1 73.3 13.8 4.2359 X X3 87.2 85.7 -2.5 3.1017 X X X3 86.9 85.4 -2.2 3.1287 X X X4 87.7 85.8 -1.4 3.0886 X X X X4 87.7 85.7 -1.3 3.1010 X X X X5 88.2 85.7 -0.1 3.0958 X X X X X5 88.1 85.6 0.1 3.1122 X X X X X6 88.6 85.7 1.3 3.1028 X X X X X X6 88.4 85.4 1.5 3.1276 X X X X X X7 89.2 85.8 2.4 3.0888 X X X X X X X

12-55



7 89.2 85.8 2.4 3.0926 X X X X X X X8 89.6 85.6 3.8 3.1068 X X X X X X X X8 89.4 85.3 4.1 3.1386 X X X X X X X X9 89.8 85.3 5.4 3.1453 X X X X X X X X X9 89.6 84.9 5.8 3.1793 X X X X X X X X X

10 89.9 84.5 7.4 3.2245 X X X X X X X X X X10 89.8 84.5 7.4 3.2259 X X X X X X X X X X11 90.0 84.0 9.1 3.2823 X X X X X X X X X X X

11 89.9 83.7 9.4 3.3097 X X X X X X X X X X X12 90.1 83.1 11.0 3.3647 X X X X X X X X X X X X12 90.1 83.1 11.1 3.3709 X X X X X X X X X X X X13 90.1 82.1 13.0 3.4677 X X X X X X X X X X X X X13 90.1 82.1 13.0 3.4682 X X X X X X X X X X X X X14 90.1 80.9 15.0 3.5813 X X X X X X X X X X X X X X

Min C p:gagf

x x y 157.0179.081.30ˆ −+=

Min MSE:bmi ppgf gagf x x x x y 8832.008321.016569.015651.0442.30ˆ ++−+=

b) Same as min C p in part (a)

c)bmigagf x x x y 088.0158.0177.016.29ˆ +−+=

d) Same as the min C p model in part (a)

e) The model with minimum C p is a good choice. It’s MSE is not much larger than the minimum MSE

model and it is a simple model with few regressors.



Best Subsets Regression: Rating Pts versus Att, Comp, ...

Response is Rating Pts

Yds

Pc p P

t e P cr c t

C C t LA o o Y A o I It m m d t T T n n n

Vars R-Sq R-Sq(adj) Mallows C-p S t p p s t D D g t t1 78.7 78.0 35767.9 5.7125 X1 76.6 75.8 39318.4 5.9891 X2 91.9 91.3 13596.8 3.5886 X X2 90.0 89.3 16755.2 3.9830 X X3 97.8 97.5 3736.4 1.9209 X X X3 96.6 96.2 5635.3 2.3568 X X X4 100.0 100.0 8.3 0.17007 X X X X4 99.5 99.4 879.8 0.95852 X X X X5 100.0 100.0 8.9 0.16921 X X X X X5 100.0 100.0 9.0 0.16942 X X X X X

6 100.0 100.0 5.4 0.15408 X X X X X X6 100.0 100.0 5.5 0.15463 X X X X X X7 100.0 100.0 5.6 0.15072 X X X X X X X7 100.0 100.0 6.1 0.15258 X X X X X X X8 100.0 100.0 7.2 0.15261 X X X X X X X X8 100.0 100.0 7.2 0.15285 X X X X X X X X9 100.0 100.0 9.0 0.15583 X X X X X X X X X9 100.0 100.0 9.1 0.15597 X X X X X X X X X

10 100.0 100.0 11.0 0.15977 X X X X X X X X X X

12-56



a) Note that R-squared in the models shown is rounded by Minitab because it is near 100%. Because MSE is not zero,

R-squared is not exactly 100%.

Minimum C p=5.4:

int

/

26352.4715.3

08458.0828.300056.082871.02992.3ˆ

pct pcttd

td att yds yds pctcomp

x x

x x x x y

−

+−+++=

Minimum MSE=0.0227:

intint

/

9205.30793.05494.3

05211.008689.483061.0005207.09196.0ˆ

pct pcttd

td att yds pctcompatt

x x x

x x x x y

−−

+−+++=

b) pctcomp pcttd pct att yds x x x x y 83.0358.3209.4048.4158.3ˆint/ ++−+=

c) Same as part (b).

d) Same as min MSE model in part (a)

e) The minimum C p model seems the best because the model contains few variables with a small value for MSE and

high R-squared.


and0.0= pC 2298.0= E MS

100850.0038.0ˆ x y +−=The min model is the same as the min C E MS p model

b) The full model that contains all 3 variables

321 010.0021.000858.0001.0ˆ x x x y −−−=

where CookUse x DrinkUse x AGE x === 321

c) No variables are selected

d) The min C p equation has only intercept term with 5.0−= pC and 2372.0= E MS

The min equation is the same as the min C E MS p in part (a).

e) None of the variables seem to be good predictors of arsenic in nails based on the models above (none of

the variables are significant).

12-83.

This analysis includes the emissions variables hc, co, and co2. It would be reasonable to model without these

variables as regressors.

Best Subsets Regression: mpg versus cid, rhp, ...

Response is mpg

ac r e c x n c

Mallows i h t m l / h c oVars R-Sq R-Sq(adj) C-p S d p w p e v c o 2

1 66.0 64.2 26.5 3.4137 X1 59.1 57.0 35.3 3.7433 X2 81.6 79.5 8.6 2.5813 X X2 78.1 75.7 13.0 2.8151 X X3 88.8 86.8 1.3 2.0718 X X X3 88.8 86.8 1.4 2.0755 X X X4 90.3 87.8 1.5 1.9917 X X X X4 89.9 87.3 2.0 2.0302 X X X X5 90.7 87.6 2.9 2.0057 X X X X X5 90.7 87.6 2.9 2.0064 X X X X X6 91.0 87.2 4.5 2.0442 X X X X X X6 91.0 87.1 4.5 2.0487 X X X X X X

12-57



7 91.3 86.6 6.2 2.0927 X X X X X X X7 91.2 86.4 6.3 2.1039 X X X X X X X8 91.4 85.6 8.0 2.1651 X X X X X X X X8 91.4 85.6 8.1 2.1654 X X X X X X X X9 91.4 84.4 10.0 2.2562 X X X X X X X X X

a)

The minimum C p (1.3) model is:

axleetwcid x x x y 457.300354.002076.0001.61ˆ −−−=

The minimum MSE (4.0228) model is:

020096.0184.329.10034252.0017547.05.49ˆcaxlecmpetwcid x x x x x y −−+−−=

b) 020084.03.300375.00178.031.63ˆcaxleetwcid x x x x y −−−−=

c) Same as equation as the min equation in part (a) E MS

d) vnaxleetw x x x y /385.04.400321.018.45ˆ +−−=

e) The minimum C p model is preferred because it has a very low MSE as well (4.29)f) Only one indicator variable is used for transmission to distinguish the automatic from manual types and two

indicator variables are used for drv:xtrans=0 for automatic (L4, L5) and 1 for manual (M6) and

xdrv1=0 if drv=4 or R and 1 if drv =F; xdrv2=0 if drv = 4 or F and 1 if drv=R.

The minimum C p (4.0) model is the same as the minimum MSE (2.267) model:

21

02

342.26131.3

401.7011118.0216.15936.30038023.010ˆ

drvdrv

transccocmpetw

x x

x x x x x y

+

+−−++−=

Stepwise:

21/ 7.12.35.4271.00044.012.39ˆdrvdrvtrnsvnetw x x x x x y ++−+−=

Forward selection:

21

/

21.2

4.31.2336.000377.012.41ˆ

drvdrv

transaxlevnetw

x x

x x x x y

+

+−−+−=

Backward selection: same as minimum C p and minimum MSE.

Prefer the model giving the minimum C p and minimum MSE.

12-84.

2

2

2

2

1110

331.009.18914.26202ˆ

)125.297(331.0)125.297(607.7395.759ˆ

)'(331.0'607.7395.759ˆ

)'('ˆ

x x y

x x y

x x y

x x y

−+−=

−−−−=

−−=

++= ∗∗∗ β β β

a) , where2)'(166.47'783.90395.759ˆ x x y −−= xx x

Sx

'=−

b) At x = 285 016.19336.11

125.297285' −=

−= x

943.802)106.1(166.47)106.1(783.90395.759ˆ 2 =−−−−= y psi

c) ( ) ( ) . . ...

..

y x x= − −− −759 395 90 783 47166297125119336

297125119336

2

12-58



2

2

331.009.18914.26204ˆ

)125.297(331.0)125.297(607.7395.759ˆ

x x y

x x y

−+−=

−−−−=

d) They are the same.

e)2

)'(440.0'847.0385.0'ˆ x x y −−=where

yS

y y y

−=' and

xS

x x x

−='

The "proportion" of total variability explained is the same for both standardized and un-standardized models.

Therefore, R 2

is the same for both models.

where

yS

y y y

−=' and

xS

x x x

−='

2

1110 )'('' x x y∗∗∗ ++= β β β

2

1110 )'('' x x y∗∗∗ ++= β β β


generated by the method.a)

2

231 004.0031.0083.0304.0ˆ x x x y +−+−=

004.004.4 == E p MSC

b) 2

3321 0008.0042.0022.0078.0256.0ˆ x x x x y +−++−=

004.066.4 == E p MSC

c) The forward selection model in part (a) is more parsimonious with a lower C p and equivalent MSE. Therefore, we

prefer the model in part (a).

12-86. n = 30, k = 9, p =9 + 1 = 10 in full model.

a) 100ˆ 2 == E MSσ 92.02 = R

yy

E

yy

R

S

SS

S

SS

R −== 1

2

2000

)1030(100

)(

=

−=

−= pn MSSS E E

0 92 1

25000

2000. = −

=

S

yy

yy

S

56.25559

23000

23000200025000

===

=−=

−=

k

SS MS

SSSSS

R R

E yy R

20,9,0

20,9,05.

0

39.2

56.25100

56.2555

α f f

f

MS

MS f

E

R

>

=

===

Reject H and conclude at least one0 j

β is significant at α = 0.05.

b) k = 4 p = 5 SSE = 2200

12-59



88530

2200=

−=

−=

pn

SS MS E

E

Yes, MSE is reduced with new model (k = 4).

c) pn pSS

C E p 2

ˆ

)(2

+−=σ

2)5(230100

2200=+−=

pC

Yes, C p is reduced from the full model.12-87. n = 25 k = 7 p = 8 MSE full( ) = 10

a) p =4 SSE = 300

29.14425

300=

−=

−=

pn

SS MS E

E

1385

)4(22510

300

2)(

=+=

+−=

+−= pn MS

SSC

full E

E p

YES, C p > p

b) p = 5 SSE = 275

11530

275=

−=

−=

pn

SS MS E

E 5.12)5(225

10

275=+−= pC

Yes, both MS E and C p are reduced.

12-60




12-88. a)543 6887.1485.21231.04203ˆ x x x y ++−=

b) H0 3 4 5 0:β β β= = =


α = 0.01

38.4

25.1651

36,3,01.

0

=

=

f

f

Reject H0 and conclude that regression is significant. P-value < 0.00001

c) All at α = 0.01 t. , .005 36 2 72=

0: 30 = β H H0 4 0:β = H0 5 0:β =

H1 3 0:β ≠ H1 4 0:β ≠ H1 5 0:β ≠

06.20 −=t 91.220 =t 00.30 =t

| | / ,t t0 2/> α 36 36 36 | | / ,t t0 2> α | | / ,t t0 2> α

Do not reject H0 Reject H0 Reject H0

d) R Adj. R 2 0 993= .2 0 9925= .

e) Normality assumption appears reasonable. However there is a gap in the line.

-100 0 100

-2

-1

0

1

2


Residual


(response is y)

f) Plot is satisfactory.

3000 4000 5000

-100

0

100

Fitted Value

R e s i d

u a l


(response is y)

12-61



g) Slight indication that variance increases as x3 increases.

300002950029000

100

0

-100

x3

R e s i d u a l

Residuals Versus x3

(response is y)

h) 89.3862)1589(6887.1)170(485.21)28900(231.04203ˆ =++−= y

12-89. a) H 0 3 4 5 0:β β β∗

= = =


α = 0.01

36,3,0

36,3,01.

0

38.4

39.1321

α f f

f

f

>>

=

=

Reject H0 and conclude that regression is significant. P-value < 0.00001

b) α = 0.01 t. , .005 36 2 72=

H0 3 0:β ∗ = H0 4 0:β = H0 5 0:β =

H1 3 0:β

∗

≠ H1 4 0:β ≠ H1 5 0:β ≠ 45.1

0 −=t 95.190 =t 53.20 =t

| | / ,t t0 2/> α 36 36 36 | | / ,t t0 2> α | | / ,t t0 2/> α

Do not reject H0 Reject H0 Do not reject H0

c) Curvature is evident in the residuals plots from this model.

Residual

P e r c e n t

0.030.020.010.00-0.01-0.02-0.03

99

95

90

80

70

60

5040

30

20

10

5

1

Normal Probabilit y Plot of t he Residuals(response is y*)

12-62



Fitt ed Value

R e

s i d u a l

8.58.48.38.28.18.0

0.03

0.02

0.01

0.00

-0.01

-0.02

Residuals Versus the Fitted Values(response is y*)

x3*

R e s i d u a l

10.3210.3110.3010.2910.2810.2710.26

0.03

0.02

0.01

0.00

-0.01

-0.02

Residuals Versus x3*(response is y*)

12-90. From data in table 12-6

a)54321 005.0594.0454.02206.0291.086.2ˆ x x x x x y +−++−=

0: 543210 ===== β β β β β H

0oneleastat:1 ≠ j H β

α = 0.01

19,5,0

19,5,01.

0

17.4

81.4

α f f

f

f

>

=

=

Reject H0. P-value = 0.005

b) α = 0.05 093.219,025. =t

0: 10 = β H H0 2 0:β = H0 3 0:β = H0 4 0:β = H0 5 0:β =

H1 1 0:β ≠ H1 2 0:β ≠ H1 3 0:β ≠ H1 4 0:β ≠ H1 5 0:β ≠

47.20

−=t 74.20

=t 42.20 =t 80.20 −=t 26.00 =t

| | / ,t t0 2 1> α | | / ,t t0 2 19> α 19 19 199 | | / ,t t0 2> α | | / ,t t0 2> α | | / ,t t0 2/> α

Reject H0 Reject H0 Reject H0 Reject H0 Do not reject H0

c)4321 609.0455.019919.0290.0148.3ˆ x x x x y −++−=

0: 43210 ==== β β β β H


α = 0.05

20,4,0

20,4,05.

0

87.2

28.6

α f f

f

f

>

=

=

Reject H0.

α = 0.05 t. , .025 20 2 086=

H0 1 0:β = H0 2 0:β = H0 3 0:β = H0 4 0:β =

H1 1 0:β ≠ H1 2 0:β ≠ H1 3 0:β ≠ H1 4 0:β ≠ 53.20 −=t 89.20 =t 49.20 =t 05.30 −=t

| | / ,t t0 2 2> α | | / ,t t0 2 20> α 20 200 | | / ,t t0 2> α | | / ,t t0 2> α

Reject H0 Reject H0 Reject H0 Reject H 0

d) The addition of the 5th

regressor results in a loss of one degree of freedom in the denominator and the reduction in

SS E is not enough to compensate for this loss.

12-63



e) Observation 2 is unusually large. Studentized residuals

-0.80199 -4.99898 -0.39958 2.22883 -0.52268 0.62842 -0.45288 2.21003 1.37196

-0.42875 0.75434 -0.32059 -0.36966 1.91673 0.08151 -0.69908 -0.79465 0.27842

0.59987 0.59609 -0.12247 0.71898 -0.73234 -0.82617 -0.45518

f) R2 for model in part (a): 0.558. R2 for model in part (c): 0.557. R2 for model x1, x2, x3, x4 without obs. #2: 0.804. R2

increased because observation 2 was not fit well by either of the previous models.

g) 0: 43210 ==== β β β β H

0:1 ≠ j H β α = 0.05

19,4,05.00

19,4,05.

0

90.2

53.19

f f

f

f

>

=

=

Reject H0.

α = 0.05 t. , .025 19 2 093=

H0 1 0:β = H0 2 0:β = H0 3 0:β = H0 4 0:β =

H1 1 0:β ≠ H1 2 0:β ≠ H1 3 0:β ≠ H1 4 0:β ≠

96.30 −=t 43.60 =t 64.30 =t 39.30 −=t

19,025.00 || t t > 19,025.00 || t t > 19,025.00 || t t > 19,025.00 || t t >

Reject H0 Reject H0 Reject H0 Reject H0

h) There is some indication of curvature.

x1

R e s i d u a l

1614121086420

4

3

2

1

0

-1

-2


x2

R e s i d u a l

2520151050

4

3

2

1

0

-1

-2


x3

R e s i d u a l

12108642

4

3

2

1

0

-1

-2


x4

R e s i d u a l

12108642

4

3

2

1

0

-1

-2


12-64



12-91. a) *

4

*

3

*

2

*

1 44.156.353.612.687.4ˆ x x x x y −−−+=

b) 0: 43210 ==== β β β β H


α = 0.05

20,4,05.00

20,4,05.

0

87.2

79.21

f f f

f

>=

=

Reject H0 and conclude that regression is significant at α = 0.05.

α = 0.05 086.220,025. =t

H0 1 0:β = H0 2 0:β = H0 3 0:β = H0 4 0:β =

H1 1 0:β ≠ H1 2 0:β ≠ H1 3 0:β ≠ H1 4 0:β ≠

76.50 =t 96.50 −=t 90.20 −=t 99.40 −=t

20,025.00 || t t > 20,025.00 || t t > 20,025.00 || t t > 20,025.00 || t t >

Reject H0 Reject H0 Reject H0 Reject H0

c) The residual plots are more satisfactory than those in Exercise 12-90.

Fitt ed Value

R e s i d u a l

3210-1-2

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0

Residuals Versus the Fitt ed Values(response is y*)

x1*

R e s i d u a l

0.70.60.50.40.30.2

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0


x2*

R e s i d u a l

0.70.60.50.40.30.2

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0


12-65



x3*

R e

s i d u a l

0.70.60.50.40.3

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0


x4*

R e

s i d u a l

3.53.02.52.01.5

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0


12-92. a) 20006.0023.2405.1709ˆ x x y −+−=

b) 0: 1110 == β β H


α = 0.05

7,2,05.00

7,2,05.

0

74.411.300

f f

f f

>>

==

Reject H0.

c) H 0 11 0:β =

H1 11 0:β ≠

α = 0.05

7,1,0

7,1,05.

0

111

0

59.5

01.55

04413.0

1/4276.2/)|(

α

β β

f f

f

f

MS

r SSF

E

R

>>

=

=

==

Reject H0.

d) There is some indication of non-constant variance.

161514131211

0.3

0.2

0.1

0.0

-0.1

-0.2

-0.3

Fitted Value

R e s i d u a l


(response is y)

12-66



e.) Normality assumption is reasonable.

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3

-1

0

1


Residual


12-93. a)54

*

3

* 000418.00055.0404.1068.21ˆ x x x y ++−=

013156.0= E MS 0.4= pC

b) Same as model in part (a)

c) x4, x5 with and1.4= pC 0134.0= E MS

d) The model in part (c) is simpler with values for MS E and C p similar to those in part (a) and (b). The part (c) model is

preferable.

e) 4.52)ˆ( 3 =∗ β VIF

3.9)ˆ( 4 = β VIF

1.29)ˆ( 5 = β VIF

Yes, VIFs for x and x exceed 103*

5

12-94. a) *

4

*

3

*

2

*

1 44.156.353.612.687.4ˆ x x x x y −−−+=

41642.0)( = p MS E Min 0.5= pC

b) Same as part (a)

c) Same as part (a)

d) All models are the same.

12-95. a)21 4.1085.00.300ˆ x x y ++=

8.405)2(4.10)100(85.0300ˆ =++= y

b) S yy = 1230 5. SSE = 120 3.

2.11103.1205.1230 =−=−= E yy R SSSSS

37.55025.10

1.555

025.10315

3.120

1.5552

2.1110

0 ===

=−

=−

=

===

E

R

E

E

R

R

MS

MS f

pn

SS MS

k

SS MS

H0 1 2 0:β β= =


α = 0.05

12-67



12,2,05.00

12,2,05.

0

89.3

37.55

f f

f

f

>

=

=

Reject H0 and conclude that the regression model is significant at α = 0.05

c) 9022.05.12302.11102 ===

yy

R

SSS R or 90.22%

d) k = 3 p = 4 SSE ' .= 117 20

65.1011

2.117'' ==

−=

pn

SS MS E

E

No, MSE increased with the addition of x3 because the reduction in SSE was not enough to compensate for the loss in

one degree of freedom in the error sum of squares. This is why MS E can be used as a model selection criterion.

e) 30.111320.1175.1230 =−=−= E yy R SSSSS

1.3

20.111030.1113

)|,()|(),,|( 01201230123

=

−=

−= β β β β β β β β β β β R R R SSSSSS

0: 30 = β H

0: 31 ≠ β H

α = 0.05

11,1,05.00

11,1,05.

01230

84.4

291.011/2.117

1/1.3

/'

/),,|(

f f

f

pnSS

r SS f

E

R

>/

=

==−

=β β β β

Do not reject H0.

12-96. a)

The model with the minimum C p (-1.3) value is:W = 69.9 + 0.120 HR_B + 0.0737 BB_B - 0.0532 SO_B + 0.0942 SB

The model assumptions are not violated as shown in the graphs below.

Where X1 = AVG, X2 = R, X3 = H, X4 = 2B, X5 = 3B, X6 = HR, X7 = RBI, X8 = BB, X9 = SO, X10 = SB

X11 = GIDP, X12 = LOB and X13 = OBP

The model assumptions appear reasonable.


S c o r e

3210-1-2-3

2

1

0

-1

-2

Normal Probabi l i ty Plot o f the Residuals(response is W)

12-68



Fitted Value

S t a n d a r d

i z e d

R e s i d u a l

10095908580757065

2

1

0

-1

-2


SB

S t a n d a r d

i z e d

R e s i d u a l

16014012010080604020

2

1

0

-1

-2

Residuals Ver sus SB(response is W)

SO_B


R e s i d u a l

1300120011001000900800

2

1

0

-1

-2

Residuals Ver sus SO_B(response is W)

BB_B

S

t a n d a r d i z e d

R e s i d u a l

650600550500450400

2

1

0

-1

-2

Residuals Ver sus BB_B(response is W)

HR_B


R e s i d u a l

260240220200180160140120100

2

1

0

-1

-2

Residuals Ver sus HR_B(response is W)

b)

Minimum C p (1.1) model:W = 96.5 + 0.527 SV - 0.125 HR_P - 0.0847 BB_P + 0.0257 SO_P

12-69



Based on the graphs below, the model assumptions are not violated


S c o r e

3210-1-2-3

2

1

0

-1

-2

Normal Probabi l i ty Plot o f the Residuals(response is W)

Fitted Value


R e s i d u a l

10090807060

2

1

0

-1

-2


SV

S t a n d a r d i z

e d

R e s i d u a l

55504540353025

2

1

0

-1

-2

Residuals Ver sus SV(response is W)

HR_P

S t a n d a

r d i z e d

R e s i d u a l

220200180160140120

2

1

0

-1

-2

Residuals Ver sus HR_P(response is W)

12-70



BB_P

S t a n d a r d

i z e d

R e s i d u a l

650600550500450400350

2

1

0

-1

-2

Residuals Ver sus BB_P(response is W)

SO_P


R e s i d u a l

1300120011001000900

2

1

0

-1

-2

Residuals Ver sus SO_P(response is W)

c) Minimum C p (10.7) model:

pso pbb

phr er eraobplobbso

rbibbbhbavg

x x

x x x x x x

x x x x x y

_ _

_ _

32 _ _

008755.0032314.0

06698.081681.079.1342.87007897.0010207.0

19914.013662.004041.045303.0327749.2ˆ

+−

−+−−+−

+−−−+−=

Every variable in the above model is significant at α=0.10. If α is decreased to 0.05, SO_P is no longer

significant. The residual plots do not show any violations of the model assumptions (only a few plots of residuals vs. the predictors are shown)

12-71




S c o r e

3210-1-2-3

2

1

0

-1

-2

Normal P robab i l i t y P l o t o f t he Resi dua l s(response is W)

Fitted Value

S t a

n d a r d i z e d

R e s i d u a l

1009080706050

2

1

0

-1

-2


SO_B


R e s i d u a l

1300120011001000900800

2

1

0

-1

-2

Residuals Versus SO_B(response is W)

AVG_B


R e s i d u a l

0.2850.2800.2750.2700.2650.2600.2550.250

2

1

0

-1

-2

Residuals Versus AVG_B(response is W)

ERA


R e s i d u a l

5.55.04.54.03.5

2

1

0

-1

-2

Residuals Ver sus ERA(response is W)

12-97. a)

yy

R

S

SS R =2

12-72



03.047.05.0

47.0)50.0(94.0)(2

=−=−=

===

R yy E

yy R

SSSSS

S RSS

0...: 6210 ==== β β β H


α = 0.05

7,6,05.00

7,6,05.

0

87.3

28.187/03.0

6/47.0

/

/

f f

f

pnSS

k SS f

E

R

>

=

==−

=

Reject H0.

b) k = 5 n = 14 p = 6 R2 0 92= .

04.046.05.0''

46.0)50.0(92.0)(' 2

=−=−=

===

R yy E

yy R

SSSSS

S RSS

01.0

46.047.0

)()()|,( 06,,2,1,0

=

−=

−== reduced SS fullSSSS R Rii j R β β ββ …

8,1,05.00

8,1,05.

06,,2,1,

0

32.5

28/04.0

1/01.0

)/('

/)|,(

f f

f

pnSS

r SS f

E

ii j R

>/

=

==−

== β β β

…

Do not reject H0. There is not sufficient evidence that the removed variable is significant at α = 0.05.

c) 005.08

04.0)( ==

−=

pn

SSreduced MS E

E

004.07

03.0)( == full MS E

No, the MS E is larger for the reduced model, although not by much. Generally, if adding a variable to a model

reduces the MS E it is an indication that the variable may be useful in explaining the response variable. Here the

decrease in MS E is not very great because the added variable had no real explanatory power.




13-47.)1(

)(1 1

2

−

−

=∑∑

= =

na

y y

MS

a

i

n

j

iij

E and ijiij a y ε μ ++= . Then .iijiij y y ε ε −=− and

1

)(1

.

−

−∑=

n

n

j

iij ε ε

is recognized to be the sample variance of the independent random variables

inii ε ε ε ,,, 21 … . Therefore,21

2

.

1

)(

σ

ε ε

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=∑

=

n E

n

j

iij

and2

1

2

)( σ σ

== ∑=

a

i

E a

MS E .

The development would not change if the random effects model had been specified because

.iijiij y y ε ε −=− for this model also.

13-48. The two sample t-test rejects equality of means if the statistic

ty y

s

y y

s p n n p n

=−

+=

−| | | | is too large. The ANOVA F-test rejects equality of means if 1 2

1 1

1 2

2

E

2...i

2

1i

MS

)yy(n

F

−

=

∑= is too large.

Now,

n2E

2.2.1

E

2.2.12

n

MS

)yy(

MS

)yy(F

−=

−= and .2

pEsMS =

Consequently, . Also, the distribution of the square of a t random variable with a(n - 1) degrees of

freedom is an F distribution with 1 and a(n - 1) degrees of freedom. Therefore, if the tabulated t value for a

two-sided t-test of size , then the tabulated F value for the F test above is t Therefore, t >

whenever F = and the two tests are identical.

2tF =

0tis α02 . 0t

20

2 tt >

13-49.)1(2

)( 2

.

1

2

1

−

−

=∑∑

==

n

y y

MS

iij

n

ji

E and1

)( 2

.

1

−

−∑=

n

y y iij

n

jis recognized as the sample standard deviation

calculated from the data from population i. Then,

2

2

2

2

1 ss MS E

+= which is the pooled variance estimate used

in the t-test.

13-50. from the independence of Y Y )()( .

2

1

.

1

ii

a

i

ii

a

i

Y V cY cV ∑∑==

= Ya1 2. ., ,..., ..

Also, . Then,2

.)(iii nY V σ =

ii

a

i

ii

a

i

ncY cV 2

1

2

.

1

)( ∑∑==

= σ

13-41



13-51. If b, c, and d are the coefficients of three orthogonal contrasts, it can be shown that

a

y

y

d

yd

c

yc

b

yb i

a

ii

a

ii

a

i

ii

a

i

i

a

i

ii

a

i

i

a

i

ii

i

2

.

12

.

12

1

2

.

1

2

1

2

.

1

2

1

2

.

4

1

)()()()( ∑∑

∑

∑

∑

∑

∑

∑=

=

=

=

=

=

=

= −=++ always holds. Upon dividing both sides

by n, we have N

y

n

yQQQ i

a

i

2..

2

1

2

3

2

2

2

1 −=++ ∑=

which equals . The equation above can be

obtained from a geometrical argument. The square of the distance of any point in four-dimensional space from the zero

point can be expressed as the sum of the squared distance along four orthogonal axes. Let one of the axes be the 45

degree line and let the point be ( ). The three orthogonal contrasts are the other three axes. The square of

the distance of the point from the origin is and this equals the sum of the squared distances along each of the

four axes.

SStreatments

.4.3.2.1 ,,, y y y y

2

.

1

i

a

i

y∑=

13-52. Because2

2

12 )(

σ

μ μ

a

n i

a

i −=Φ ∑=, we only need to shows that

2

1

2

)(2

μ μ −≤ ∑=

i

a

i

D.

Let μ1 and μ2 denote the means that differ by D. Now, is minimized for x equal

to the mean of μ

2

2

2

1 )()( x x −+− μ μ

1 and μ2. Therefore,

2

1

2

2

2

1

2212

2211 )()()()

2

+-()

2

+-( μ μ μ μ μ μ

μ μ μ

μ μ μ −≤−+−≤+ ∑

=i

a

i

Then, 2

1

2222

12

2

21 )(24422

μ μ μ μ μ μ

−≤=+=⎟ ⎠

⎞⎜⎝

⎛ −+⎟

⎠

⎞⎜⎝

⎛ −∑

=i

a

i

D D D.

13-53.a

s

na

y y

MSi

a

i

iij

n

j

a

i

E

2

1

2

.

11

)1(

)(

∑∑∑ === =−

−

= where1

)( 2

.

12

−

−

= ∑=

n

y y

s

iij

n

j

i . Because s is the

sample variance of

i2

,,...,, 21 inii y y y2

2)1(

σ

iSn −has a chi-square distribution with n-1 degrees of

freedom. Then, a n MSE( )− 1

2σis a sum of independent chi-square random variables. Consequently,

a n MSE( )− 1

2σhas a chi-square distribution with a(n - 1) degrees of freedom. Consequently,

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −≤≤

−=

−=≤−

≤

−−−

−−−

2

)1(,2

1

2

)1(,2

2

)1(,2

2

)1(,2

1

)1()1(

1))1(

(

2

2

nana

nana

E E

E

MSna MSnaP

MSnaP

α α

α α

χ σ

χ

α χ σ

χ

Using the fact that a(n - 1) = N - a completes the derivation.

13-42



13-54. From Exercise 13-53,2

)(

σ

E MSa N −has a chi-square distribution with N - a degrees of freedom. Now,

nY V i

22

. )(σ

σ τ

+= and mean square treatment = is n times the sample variance of T MS

.,...,, ..2.1 a y y y Therefore, 222

)1(

)(

)1(2

σ σ σ τ

σ

τ +

−=+

−

n

MSa

n

MSa T

n

T

has a chi-squared distribution with a - 1

degrees of freedom. Using the independence of MST and MSE, we conclude that

⎟ ⎠

⎞⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+ 222/

σ σ σ τ

E T MS

n

MShas an F distribution.a N( ) ,(− −1 a)

Therefore,

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

⎥⎦

⎤

⎢⎣

⎡

−≤≤⎥⎦

⎤

⎢⎣

⎡

−=

−=≤+

≤

−−−−−

−−−−−

1

1

1

1

1)(

,1,2

1,1,2

22

1

2

2

1

,1,22

2

,1,1

E

T

f E

T

f

a N a

E

T

a N a

MS

MS

n MS

MS

nP

f n MS

MS f P

a N aa N aα α

α α

σ

σ

α σ σ

σ

τ

τ

by an algebraic solution for σ

στ2

2

and )(2

2

U LP ≤≤σ

σ τ .

13-55. a) As in Exercise 13-54,22

2

σ σ

σ

τ +n MS

MS

E

T has an F distribution.a N( ) ,(− −1 a)

and

)11

(

)11

111

(

)11

(

)(1

22

2

2

2

2

2

2

2

+≤

+≤

+=

+≤+≤+=

≤≤=

≤≤=−

U

U

L

LP

LU P

LU P

U LP

τ

τ

τ

τ

τ

σ σ

σ

σ

σ

σ

σ

σ

σ α

13-43



b)

)1

1

1

1(

)11(

)11

1(

)(1

22

2

2

22

2

2

2

2

+≤

+≤

+=

+≤+

≤+=

+≤+

≤+=

≤≤=−

LU P

U LP

U LP

U LP

τ

τ

τ

τ

σ σ

σ

σ

σ σ

σ

σ

σ

σ α

Therefore, )1

1,

1

1(

++ LU is a confidence interval for

σ

σ στ

2

2 2+

13-56. 1

)( 2

...

1

−

−

=

∑=

a

y yn

MS

ii

a

i

T and for any random variable X , E X .V X E X( ) ( ) [ ( )]2 2

= +Then,

1

})]([)({

)(

2

......

1

−

−+−=

∑=

a

Y Y E Y Y V n

MS E iii

a

iT

Now,aan N a N n N N n N n N n

Y Y Y Y Y Y Y Y 11

12

121

11

1111

11...1 .........)(...)(

2111−−−−−−−−++−=−

and

constraint thefrom...)()(

)()()(

122

1111

...1

2112

2

11

211...1

1

11

τ τ τ τ

σ σ

=−−−−=−

−=⎟ ⎠

⎞⎜⎝

⎛ −+−=−

aa

N n

N n N n

N n

N nnY Y E

N

n N nY Y V

Then,

2 2 2 21 1

1 1

2

2 1

{( ) } [(1 ) ]

( )1 1

1

i

i

a an

n N N i i

i iT

a

i i

i

n n

E MSa a

n

a

i iσ τ σ

τ

σ

= =

=

− + − += =

− −

= +−

∑ ∑

∑

τ

Because , this does suggest that the null hypothesis is as given in the exercise.2)( σ = E MS E

13-57. a) If A is the accuracy of the interval, then At

n

MS

na

E =−

2

)1(,2

α

Squaring both sides yields222

)1(,2

At n

MS

na

E =−α

As in Exercise 13-48, )1(,1,

2

)1(,2

−−= nana

F t α α . Then,

2

)1(,1,2

A

F MSn

na E −= α

13-44



b) Because n determines one of the degrees of freedom of the tabulated F value on the right-side of the equation in part (a),

some approximation is needed. Because the value for a 95% confidence interval based on a normal distribution is 1.96, w

approximate)1(,

2−na

t α by 2 and we approximate

)1(,1,

2

)1(,2

−−= nana

F t α α by 4.

Then, 84

)4)(4(2 ==n . With n = 8, a(n - 1) = 35 and F .0 05135 4 12. , , .=

The value 4.12 can be used for F in the equation for n and a new value can be computed for n as

824.84

)12.4)(4(2≅==n

Because the solution for n did not change, we can use n = 8. If needed, another iteration could be used to

refine the value of n.

13-45



CHAPTER 13

Section 13-2

13-1. a) Analysis of Variance for STRENGTHSource DF SS MS F P

COTTON 4 475.76 118.94 14.76 0.000Error 20 161.20 8.06Total 24 636.96

Reject H0 and conclude that cotton percentage affects mean breaking strength.

3530252015

25

15

5

COTTON

S T R E N G T H

b) Tensile strength seems to increase up to 30% cotton and declines at 35% cotton.

Cotton Percentage

M e a n S t r e n g t h

3530252015

22

20

18

16

14

12

10

Scatt erplot of Mean Strength vs Cotton Percentage

Individual 95% CIs For MeanBased on Pooled StDev

Level N Mean StDev ------+---------+---------+---------+15 5 9.800 3.347 (-----*----)20 5 15.400 3.130 (----*----)

25 5 17.600 2.074 (----*----)30 5 21.600 2.608 (----*----)35 5 10.800 2.864 (-----*----)

------+---------+---------+---------+Pooled StDev = 2.839 10.0 15.0 20.0 25.0

c) The normal probability plot and the residual plots show that the model assumptions are reasonable.

13-1



201510

6

5

4

3

2

1

0

-1

-2

-3

-4

FittedValue

R e s i d u a l

ResidualsVersus the Fitted Values(response is STRENGTH)

210-1-2

6

5

4

3

2

1

0

-1

-2

-3

-4

Normal Score

R e s i d u a l

Normal Probability Plot of the Residuals(response is STRENGTH)

352515

6

5

4

3

2

1

0

-1

-2

-3

-4

COTTON

R e s i d u a l

Residuals Versus COTTON(response is STRENGTH)

13-2 a) Analysis of Variance for FLOW

Source DF SS MS F PFLOW 2 3.6478 1.8239 3.59 0.053Error 15 7.6300 0.5087Total 17 11.2778

Do not reject H0. There is no evidence that flow rate affects etch uniformity.

200160125

5

4

3

FLOW

U N I F O R M I T

b) Residuals are acceptable.

-1 0 1

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is obs)

13-2



200190180170160150140130120

1

0

-1

FLOW

R

e s i d u a l

Residual s VersusFLOW(response is UNIFORMI)

4.54.03.5

1

0

-1

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is UNIFORMI)

13-3. a) Analysis of Variance for STRENGTHSource DF SS MS F PTECHNIQU 3 489740 163247 12.73 0.000Error 12 153908 12826Total 15 643648

Reject H0. Techniques affect the mean strength of the concrete.

b) P-value ≅ 0

c) Residuals are acceptable

4321

200

100

0

-100

-200

TECHNIQU

R e s i d u a l

Residuals VersusTECHNIQU(response is STRENGTH)

315030502950285027502650

200

100

0

-100

-200

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response i s STRENGTH)

-200 -100 0 100 200

-2

-1

0

1

2


Residual


13-3



13-4 a) Analysis of Variance for CIRCUIT TYPESource DF SS MS F PCIRCUITT 2 260.9 130.5 4.01 0.046Error 12 390.8 32.6Total 14 651.7

Reject H0

b)There is some indication of greater variability in circuit two. There is some curvature in the normal probability plot.

-10 0 10

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is time)

321

10

0

-10

CIRCUITT

R e s i d u a l

ResidualsVersusCIRCUITT(response is RESPONSE)

282318

10

0

-10

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is RESPONSE)

c) 95% Confidence interval on the mean of circuit type 3.

96.2384.12

5

6.32179.24.18

5

6.32179.24.18

1

1

12,025.0312,025.03

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

MSt y

n

MSt y E

i

E

13-5. a) Analysis of Variance for CONDUCTIVITYSource DF SS MS F P

COATINGTYPE 4 1060.5 265.1 16.35 0.000Error 15 243.3 16.2Total 19 1303.8

Reject H0, P-value ≅ 0

13-4



b) There is some indication of that the variability of the response may be increasing as the mean response

increases. There appears to be an outlier on the normal probability plot.

54321

5

0

-5

-10

COATINGT

R e s i d u a l

Residuals Versus COATINGT(response is CONDUCTI)

145140135130

5

0

-5

-10

FittedValue

R e s i d u a l

ResidualsVersusthe Fitted Values(response is CONDUCTI)

-10 -5 0 5

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is CONDUCTI)

c) 95% Confidence interval on the mean of coating type 1

29.14971.140

4

2.16131.200.145

4

2.16131.200.145

1

1

15,015.0115,025.01

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

MS

t yn

MS

t y

E

i

E

99% confidence interval on the difference between the means of coating types 1 and 4.

14.2436.7

4

)2.16(2947.2)25.12900.145(4

)2.16(2947.2)25.12900.145(

22

41

41

15,005.0414115,005.041

≤−≤

−−≤−≤−−

+−≤−≤−−

μ μ

μ μ

μ μ n

MSt y y

n

MSt y y E E

13-5



13-6 a) Analysis of Variance for ORIFICE Source DF SS MS F PORIFICE 5 1133.37 226.67 30.85 0.000Error 18 132.25 7.35Total 23 1265.63

Reject H0

b) P-value ≅ 0

c)

2.01.51.00.5

4

3

2

1

0

-1

-2

-3

-4

-5

ORIFICE

R e s i d u a l

Residuals VersusORIFICE(response is RADON)

807060

4

3

2

1

0

-1

-2

-3

-4

-5

FittedValue

R e s i d u a l

Residuals Versusthe Fitted Values(response is RADON)

-5 -4 -3 -2 -1 0 1 2 3 4

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is percent)

d) 95% CI on the mean radon released when diameter is 1.40

84.6715.62

4

35.7101.265

4

35.7101.265

1

1

18,015.0518,025.05

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

MSt y

n

MSt y E

i

E

13-6



13-7. a) Analysis of Variance for STRENGTHSource DF SS MS F PRODDING 3 28633 9544 1.87 0.214Error 8 40933 5117Total 11 69567

Do not reject H0

b) P-value = 0.214

c) The residual plot suggests nonconstant variance. The normal probability plot looks acceptable.

25201510

100

0

-100

RODDING

R e s i d u a l

Residuals Versus RODDING(response is STRENGTH)

160015501500

100

0

-100

FittedValue

R e s i d u a l

ResidualsVersusthe Fitted Values(response is STRENGTH)

-100 0 100

-2

-1

0

1

2


Residual


13-8 a) Analysis of Variance of PREPARATION METHODSource DF SS MS F PPREPMETH 3 22.124 7.375 14.85 0.000Error 16 7.948 0.497Total 19 30.072

Reject H0

b) P-value ≅ 0

c) There are some differences in the amount variability at the different preparation methods and there is some

curvature in the normal probability plot. There are also some potential problems with the constant variance

assumption apparent in the fitted value plot.

13-7



4321

1

0

-1

-2

PREPMETH

R e s i d u a l

Residuals Versus PREPMETH(response is TEMPERAT)

15141312

1

0

-1

-2

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is TEMPERAT)

-2 -1 0 1

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is temp)

d) 95% Confidence interval on the mean of temperature for preparation method 1

47.1513.14

5

497.0120.28.14

5

497.0120.28.14

1

3

16,015.0116,025.01

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

MSt y

n

MSt y E

i

E

13-9. a) Analysis of Variance for STRENGTHSource DF SS MS F PAIRVOIDS 2 1230.3 615.1 8.30 0.002Error 21 1555.8 74.1Total 23 2786.0

Reject H0

b) P-value = 0.002

c) The residual plots indicate that the constant variance assumption is reasonable. The normal probability

plot has some curvature in the tails but appears reasonable.

13-8



958575

10

0

-10

FittedValue

R e s i d u a l

Residuals Versusthe Fitted Values(response is STRENGTH)

100-10

2

1

0

-1

-2


Residual


321

10

0

-10

AIRVOIDS

R e s i d u a l

Residuals Versus AIRVOIDS(response is STRENGTH)

d) 95% Confidence interval on the mean of retained strength where there is a high level of air voids

83.8117.69

8

1.74080.25.75

8

1.74080.25.75

1

3

21,015.0321,025.03

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

MSt y

n

MSt y E

i

E

e) 95% confidence interval on the difference between the means of retained strength at the high level and thelow levels of air voids.

33.2642.8

8

)1.74(2080.2)5.75875.92(

8

)1.74(2080.2)5.75875.92(

22

41

41

21,025.0313121,025.031

≤−≤

−−≤−≤−−

+−≤−≤−−

μ μ

μ μ

μ μ n

MSt y y

n

MSt y y E E

13-9



13-10. a)

ANOVA

Source DF SS MS F PFactor 5 2.5858 0.5172 18.88 0.000Error 30 0.8217 0.0274Total 35 3.4075

D a t a

10.50-0.5-0.75Crosslinker Level -1

9.2

9.0

8.8

8.6

8.4

8.2

8.0

Boxplot of Crosslinker Level -1, -0.75 , -0.5, 0, 0.5 , 1

Yes, the box plot and ANOVA show that there is a difference in the cross-linker level.

b) Anova table in part (a) showed the p-value = 0.000 < α = 0.05. Therefore there is at least one

level of cross-linker is different. The variability due to random error is SSE = 0.8217

c) Domain spacing seems to increase up to the 0.5 cross-linker level and declines once cross-

linker level reaches 1.

Cross-linker lev el

M e a n d o m a i n s

p a c i n g

1.00.50.0-0.5-1.0

8.9

8.8

8.7

8.6

8.5

8.4

8.3

8.2

8.1

8.0

Scatterplot of Mean domain spacing vs Cross-linker level

13-10



d) The normal probability plot and the residual plots show that the model assumptions are

reasonable.

Residual

P e r c e n t

0.40.20.0-0.2-0.4

99

90

50

10

1

Fitted Value

R e s i d u a l

8.88.68.48.28.0

0.2

0.0

-0.2

-0.4

Residual

F r e q u e

n c y

0.30.20.10.0-0.1-0.2-0.3

6.0

4.5

3.0

1.5

0.0

Observ ation Order

R e s i d u a l

35302520151051

0.2

0.0

-0.2

-0.4

Normal Probab ilit y Plo t o f the Residuals Residuals Versus t he Fit ted Values

Hist o gram of t he Residuals Residuals Versus t h e Order of t he Dat a

Residual Plots for Domain spacing

13-11. a) No, the diet does not affect the protein content of cow’s milk.

Comparative boxplots

C4

C 3

lupinsBarley+lupinsBarley

4.5

4.0

3.5

3.0

2.5

Boxp lo t o f C3 by C4

ANOVA

Source DF SS MS F PC4 2 0.235 0.118 0.72 0.489

Error 76 12.364 0.163

Total 78 12.599

S = 0.4033 R-Sq = 1.87% R-Sq(adj) = 0.00%

b) P-value = 0.489. The variability due to random error is SSE = 0.146.

13-11



c) The Barley diet has the highest average protein content and lupins the lowest.

Diet type

m

e a n p r o t e i n

LupinsBarley+LupinsBarley

3.900

3.875

3.850

3.825

3.800

3.775

3.750

Sca t te rp lo t o f m ean p ro te in vs d ie t type

d) Based on the residual plots, there is no violation of the ANOVA assumptions.

Residual

S c o r e

1.51.00.50.0-0.5-1.0

3

2

1

0

-1

-2

-3

Normal Probabi l i ty Plo t o f the Residuals

(response i s protein content)

Fitted Value

R e s i d u a l

3.9003.8753.8503.8253.8003.7753.750

0.5

0.0

-0.5

-1.0

Residuals Versus the Fi t t ed Values(response i s protein content)

Diet

R E S I 1

LupinsBarley+LupinsBarley

0.5

0.0

-0.5

-1.0

Scatter p lo t o f RESI 1 v s Diet

13-12



13-12. a) From the analysis of variance shown below, 07.48,3,05.0 =F > F 0 = 3.43 there is no

difference in the spoilage percentage when using different AO solutions.

ANOVA

Source DF SS MS F PAO solutions 3 3364 1121 3.43 0.073Error 8 2617 327Total 11 5981

b) From the table above, the P-value = 0.073 and the variability due to random error is SSE =

2617.

c) A 400ppm AO solution should be used because it produces the lowest average spoilage

percentage.

AO solut ion

A v g .

S p o i l a g e

4003002001000

70

60

50

40

30

20

Scatterplot of Avg. Spoilage vs AO solution


reasonable.

Residual

P e r c e n t

40200-20-40

99

90

50

10

1

Fitted Value

R e s i d u a

l

7060504030

20

0

-20

Residual

F r e q u e n c y

3020100-10-20

4

3

2

1

0

Observat ion Order

R e s i d u a l

121110987654321

20

0

-20

No rm al Prob ab ilit y Plo t o f t h e Resid u als Resid u als Versu s t h e Fit t ed Valu es

Hist ogram of t he Residuals Residuals Versus t he Order of t he Dat a

Residual Plots f or % Spoilage

13-13 a) Analysis of Variance for TEMPERATURESource DF SS MS F PTEMPERAT 3 0.1391 0.0464 2.62 0.083Error 18 0.3191 0.0177

13-13



Total 21 0.4582

Do not reject H0

b) P-value = 0.083

c) Residuals are acceptable

-0.2 -0.1 0.0 0.1 0.2

-2

-1

0

1

2


Residual


(response is density)

180170160150140130120110100

0.2

0.1

0.0

-0.1

-0.2

TEMPERAT

R e s i d u a l

Residual sVersusTEMPERAT(response is DENSITY)

21.7521.6521.55

0.2

0.1

0.0

-0.1

-0.2

FittedValue

R e s i d u a l

Residual s Versus the Fitted Values(response is DENSITY)

13-14 Fisher's pairwise comparisonsFamily error rate = 0.117Individual error rate = 0.0500Critical value = 2.131

Intervals for (column level mean) - (row level mean)125 160

160 -1.9775-0.2225

250 -1.4942 -0.39420.2608 1.3608

There are significant differences between levels 125 and 160.

13-15. Fisher's pairwise comparisonsFamily error rate = 0.264

Individual error rate = 0.0500Critical value = 2.086Intervals for (column level mean) - (row level mean)

15 20 25 3020 -9.346

-1.85425 -11.546 -5.946

-4.054 1.54630 -15.546 -9.946 -7.746

-8.054 -2.454 -0.25435 -4.746 0.854 3.054 7.054

2.746 8.346 10.546 14.546

13-14



Significant differences between levels 15 and 20, 15 and 25, 15 and 30, 20 and 30, 20 and 35, 25 and 30, 25

and 35, and 30 and 35.

13-16. Fisher's pairwise comparisons

Family error rate = 0.0251Individual error rate = 0.0100

Critical value = 3.055

Intervals for (column level mean) - (row level mean)

1 2

2 -18.4263.626

3 -8.626 -1.22613.426 20.826

No significant differences at α = 0.01.

13-17. Fisher's pairwise comparisons Family error rate = 0.184Individual error rate = 0.0500Critical value = 2.179Intervals for (column level mean) - (row level mean)

1 2 32 -360

-113 -137 48

212 3974 130 316 93

479 664 442

Significance differences between levels 1 and 2, 1 and 4, 2 and 3, 2 and4, and 3 and 4.

13-18 Fisher's pairwise comparisonsFamily error rate = 0.330Individual error rate = 0.0500Critical value = 2.101Intervals for (column level mean) - (row level mean)

0.37 0.51 0.71 1.02 1.400.51 1.723

9.7770.71 3.723 -2.027

11.777 6.0271.02 6.973 1.223 -0.777

15.027 9.277 7.2771.40 13.723 7.973 5.973 2.723

21.777 16.027 14.027 10.7771.99 15.973 10.223 8.223 4.973 -1.777

24.027 18.277 16.277 13.027 6.277

Significant differences between levels 0.37 and all other levels; 0.51and 1.02, 1.40, and 1.99; 0.71 and 1.40 and 1.99; 1.02 and 1.40 and 1.99

13-15



13-19 Fisher's pairwise comparisons

Family error rate = 0.0649Individual error rate = 0.0100Critical value = 2.947Intervals for (column level mean) - (row level mean)

1 2 3 4

2 -8.6428.142

3 5.108 5.35821.892 22.142

4 7.358 7.608 -6.14224.142 24.392 10.642

5 -8.642 -8.392 -22.142 -24.3928.142 8.392 -5.358 -7.608

Significant differences b/t 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 5,4 and 5.

13-20 Fisher's pairwise comparisonsFamily error rate = 0.189Individual error rate = 0.0500Critical value = 2.120Intervals for (column level mean) - (row level mean)

1 2 32 -0.9450

0.94503 1.5550 1.5550

3.4450 3.44504 0.4750 0.4750 -2.0250

2.3650 2.3650 -0.1350

There are significant differences between levels 1 and 3, 4; 2 and 3, 4;and 3 and 4.

13-21. Fisher's pairwise comparisonsFamily error rate = 0.118Individual error rate = 0.0500Critical value = 2.080Intervals for (column level mean) - (row level mean)

1 22 1.799

19.7013 8.424 -2.326

26.326 15.576

Significant differences between levels 1 and 2; and 1 and 3.

13-22.

a) 1952.06

0274.02042.2

225,025.0 =

×==

b

MSt LSD E

Fisher 95% Individual Confidence IntervalsAll Pairwise Comparisons among Levels of Cross-linker level

Simultaneous confidence level = 65.64%

Cross-linker level = -0.5 subtracted from:

Cross-linker

13-16



level Lower Center Upper-0.75 -0.5451 -0.3500 -0.1549-1 -0.7451 -0.5500 -0.35490 -0.1118 0.0833 0.27850.5 0.0215 0.2167 0.41181 -0.1451 0.0500 0.2451

Cross-linker

level ---------+---------+---------+---------+-0.75 (---*---)-1 (---*---)0 (---*---)0.5 (---*---)1 (---*---)

---------+---------+---------+---------+-0.50 0.00 0.50 1.00

Cross-linker level = -0.75 subtracted from:

Cross-linkerlevel Lower Center Upper-1 -0.3951 -0.2000 -0.00490 0.2382 0.4333 0.6285

0.5 0.3715 0.5667 0.76181 0.2049 0.4000 0.5951

Cross-linkerlevel ---------+---------+---------+---------+-1 (---*---)0 (---*---)0.5 (---*---)1 (---*---)

---------+---------+---------+---------+-0.50 0.00 0.50 1.00

Cross-linker level = -1 subtracted from:

Cross-linker

level Lower Center Upper ---------+---------+---------+---------+0 0.4382 0.6333 0.8285 (---*---)0.5 0.5715 0.7667 0.9618 (---*---)1 0.4049 0.6000 0.7951 (---*---)

---------+---------+---------+---------+-0.50 0.00 0.50 1.00

Cross-linker level = 0 subtracted from:

Cross-linkerlevel Lower Center Upper0.5 -0.0618 0.1333 0.32851 -0.2285 -0.0333 0.1618

Cross-linker

level ---------+---------+---------+---------+0.5 (---*---)1 (---*---)

---------+---------+---------+---------+-0.50 0.00 0.50 1.00

Cross-linker level = 0.5 subtracted from:

Cross-linkerlevel Lower Center Upper

13-17



1 -0.3618 -0.1667 0.0285

Cross-linkerlevel ---------+---------+---------+---------+1 (---*---)

---------+---------+---------+---------+-0.50 0.00 0.50 1.00

Cross-linker levels -0.5, 0, 0.5 and 1 do not differ. Cross-linker levels -0.75 and -1 do not differ from one other but both are significantly different to the others.

b) The mean values are

8.0667, 8.2667, 8.6167, 8.7, 8.8333, 8.6667

0676.06

0274.0ˆ ===

b

MS E

X σ

The width of a scaled normal distribution is 6(0.0676) = 0.405

0

0.2

0.4

0.6

0.8

1

8 8.2 8.4 8.6 8.8 9

With a scaled normal distribution over this plot, the conclusions are similar to those from the LSD

method.

13-23.

a) There is no significant difference in protein content between the three diet types.

Fisher 99% Individual Confidence IntervalsAll Pairwise Comparisons among Levels of C4

Simultaneous confidence level = 97.33%

C4 = Barley subtracted from:

C4 Lower Center UpperBarley+lupins -0.3207 -0.0249 0.2709lupins -0.4218 -0.1260 0.1698

C4 -------+---------+---------+---------+--

Barley+lupins (-----------*-----------)lupins (-----------*-----------)

-------+---------+---------+---------+---0.25 0.00 0.25 0.50

C4 = Barley+lupins subtracted from:

C4 Lower Center Upper -------+---------+---------+---------+--lupins -0.3911 -0.1011 0.1889 (-----------*-----------)

-------+---------+---------+---------+--

13-18



-0.25 0.00 0.25 0.50

b) The mean values are: 3.886, 3.8611, 3.76 (barley, b+l, lupins)

From the ANOVA the estimate of σ can be obtainedSource DF SS MS F P

C4 2 0.235 0.118 0.72 0.489 Error 76 12.364 0.163 Total 78 12.599

S = 0.4033 R-Sq = 1.87% R-Sq(adj) = 0.00%The minimum sample size could be used to calculate the standard error ofa sample mean

081.025

163.0ˆ ===

b

MS E

X σ

The graph would not show any differences between the diets.

13-24. 55=μ , τ1 = -5, τ2 = 5, τ3 = -5, τ4 = 5.

)1(4)1(31,)25(4

)100(2

−=−=−==Φ nnaan

n

Various choices for n yield:

n Φ2

Φ a(n-1) Power=1-β

4 4 2 12 0.80

5 5 2.24 16 0.90

Therefore, n = 5 is needed.

13-25 188=μ , τ1 = -13, τ2 = 2, τ3 = -28, τ4 = 12, τ5 = 27.

)1(5)1(41,)100(5

)1830(2 −=−=−==Φ nnaann

Various choices for n yield:n Φ

2Φ a(n-1) Power=1-β

2 7.32 2.7 5 0.55

3 10.98 3.13 10 0.95

Therefore, n = 3 is needed.



, .Section 13-3

13-26 a) Analysis of Variance for UNIFORMITYSource DF SS MS F PWAFERPOS 3 16.220 5.407 8.29 0.008Error 8 5.217 0.652Total 11 21.437

Reject H0, and conclude that there are significant differences among wafer positions.

b) 585.13

652.0407.5ˆ 2 =

−=

−=

n

MS MS E Treatments

τ σ

c) 652.0ˆ 2 == E MSσ

d) Greater variability at wafer position 1. There is some slight curvature in the normal probability plot.

13-19



4321

1

0

-1

WAFERPOS

R e s i d u a l

Residuals Versus WAFERPOS(response is UNIFORMI)

4321

1

0

-1

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is UNIFORMI)

-1 0 1

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is uniformi)

13-27 a) Analysis of Variance for OUTPUT

Source DF SS MS F P

LOOM 4 0.3416 0.0854 5.77 0.003

Error 20 0.2960 0.0148

Total 24 0.6376

Reject H0, there are significant differences among the looms.

b) 01412.05

0148.00854.0ˆ 2 =−=−=n

MS MS E Treatments

τ σ

c) 0148.0ˆ 2 == E MSσ

d) Residuals are acceptable

54321

0.2

0.1

0.0

-0.1

-0.2

LOOM

R e s i d u a l

ResidualsVersusLOOM(response is OUTPUT)

4.14.03.93.8

0.2

0.1

0.0

-0.1

-0.2

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is OUTPUT)

13-20



-0.2 -0.1 0.0 0.1 0.2

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is OUTPUT)

13-28. a) Yes, the different batches of raw material significantly affect mean yield at α = 0.01 because p-

value is small.

Source DF SS MS F PBatch 5 56358 11272 4.60 0.004

Error 24 58830 2451

Total 29 115188

b) Variability between batches

2.17645

245111272ˆ 2 =

−=

−=

n

MS MS E Treatmentsτ

σ

c) Variability within batches 2451ˆ 2 == MSE σ


reasonable.

13-21



Residual

P e r c e

n t

100500-50-100

99

90

50

10

1

Fitted Value

R e s i d u

a l

160015501500

100

50

0

-50

-100

Residual

F r

e q u e n c y

1007550250-25-50-75

8

6

4

2

0

Observation Order

R

e s i d u a l

30282624222018161412108642

100

50

0

-50

-100

No rm al Pr ob ab ilit y Plo t o f t h e Resid uals Resid uals Ver su s t h e Fit t ed Valu es


Residual Plots for Yield

13-29. a) Analysis of Variance for BRIGHTNENESS

Source DF SS MS F PCHEMICAL 3 54.0 18.0 0.75 0.538Error 16 384.0 24.0Total 19 438.0

Do not reject H0, there is no significant difference among the chemical types.

b) 2.15

0.240.18ˆ 2 −=−=τ σ set equal to 0

c) 0.24ˆ 2 =σ

d) Variability is smaller in chemical 4. There is some curvature in the normal probability plot.

4321

10

5

0

-5

CHEMICAL

R e s i d u a l

Residuals Versus CHEMICAL(response is BRIGHTNE)

8281807978

10

5

0

-5

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is BRIGHTNE)

13-22



1050-5

2

1

0

-1

-2

N o r m a

l S c o r e

Residual

Normal Probability Plot of the Residuals(response is BRIGHTNE)

13-30 a) 237.2ˆˆˆ 222 =+= σ σ σ positiontotal

b) 709.0ˆ

ˆ

2

2

=total

position

σ

σ

c) It could be reduced to 0.6522. This is a reduction of approximately 71%.

13-31.

a) Instead of testing the hypothesis that the individual treatment effects are zero, we are testing

whether there is variability in protein content between all diets.

0:

0:

2

1

2

0

≠

=

τ

τ

σ

σ

H

H

b) The statistical model is

⎩⎨⎧

=

=++=

n j

ai y iji

,...,2,1

,...,2,1ε τ μ

and),0(~ 2σ ε N i ),0(~ 2τ

σ τ N i

c)

The last TWO observations were omitted from two diets to generate equal sample sizes with n = 25.

ANOVA: Protein versus DietTypeAnalysis of Variance for Protein

Source DF SS MS F P

DietType 2 0.2689 0.1345 0.82 0.445

Error 72 11.8169 0.1641

Total 74 12.0858

S = 0.405122 R-Sq = 2.23% R-Sq(adj) = 0.00%

1641.02 == E MSσ

001184.025

1641.01345.02 −=−

=−

=n

MS MS E tr

τ σ

13-23



Section 13-4

13-32.

a) Analysis of variance for GlucoseSource DF SS MS F P

Time 1 36.13 36.125 0.06 0.819

Min 1 128.00 128.000 0.21 0.669

Error 5 3108.75 621.750Total 7 3272.88

No, there is no effect of exercise time on the average blood glucose.

b) P-value = 0.819

c) The normal probability plot and the residual plots show that the model assumptions are

reasonable.

Residual

P e r c e n t

50250-25-50

99

90

50

10

1

Fitted Value

R e s i d u a l

110105100

20

0

-20

-40

Residual

F r e q u e n c y

20100-10-20-30-40

3

2

1

0

Observation Order

R e s i d u a l

87654321

20

0

-20

-40


Hist o gram of t h e Resid uals Resid uals Versus t h e Order of t h e Dat a

Residual Plot s for Glucose

13-33. a) Analysis of Variance for SHAPE

Source DF SS MS F PNOZZLE 4 0.102180 0.025545 8.92 0.000VELOCITY 5 0.062867 0.012573 4.39 0.007

Error 20 0.057300 0.002865Total 29 0.222347

Reject H0, nozzle type affects shape measurement.

13-24



28.7423.4620.4316.5914.3711.73

1.15

1.05

0.95

0.85

0.75

VELOCITY

S

H A P E

54321

1.15

1.05

0.95

0.85

0.75

NOZZLE

S H A P E

b) Fisher's pairwise comparisons

Family error rate = 0.268

Individual error rate = 0.0500



1 2 3 4

2 -0.15412

0.01079

3 -0.20246 -0.13079

-0.03754 0.034124 -0.24412 -0.17246 -0.12412

-0.07921 -0.00754 0.04079

5 -0.11412 -0.04246 0.00588 0.04754

0.05079 0.12246 0.17079 0.21246

There are significant differences between levels 1 and 3, 4; 2 and 4; 3

and 5; and 4 and 5.

c) The residual analysis shows that there is some inequality of variance. The normal probability plot is

acceptable.

302010

0.1

0.0

-0.1

VELOCITY

R e s i d u a l

Residuals Versus VELOCITY(response is SHAPE)

54321

0.1

0.0

-0.1

NOZZLE

R e s i d u a l

Residuals Versus NOZZLE(response is SHAPE)

13-25



1.00.90.80.7

0.1

0.0

-0.1

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is SHAPE)

0.10.0-0.1

2

1

0

-1

-2


Residual

Normal Probability Plot of the Residuals(response is SHAPE)

13-34 a) Analysis of Variance of HARDNESSSource DF SS MS F PTIPTYPE 3 0.38500 0.12833 14.44 0.001SPECIMEN 3 0.82500 0.27500 30.94 0.000Error 9 0.08000 0.00889

Total 15 1.29000

Reject H0, and conclude that there are significant differences in hardness measurements between the tips.

b)Fisher's pairwise comparisons





1 2 3

2 -0.4481

0.3981

3 -0.2981 -0.2731

0.5481 0.5731

4 -0.7231 -0.6981 -0.8481

0.1231 0.1481 -0.0019

Significant difference between tip types 3 and 4

13-26



c) Residuals are acceptable.

4321

0.15

0.10

0.05

0.00

-0.05

-0.10

SPECIMEN

R e s i d u a l

Residuals Versus SPECIMEN(response is HARDNESS)

4321

0.15

0.10

0.05

0.00

-0.05

-0.10

TIPTYPE

R e s i d u a l

Residuals Versus TIPTYPE(response is HARDNESS)

10.210.110.09.99.89.79.69.59.49.39.2

0.15

0.10

0.05

0.00

-0.05

-0.10

FittedValue

R e s i d u a l

Residual s Versus the Fitted Values(response is HARDNESS)

-0.10 -0.05 0.00 0.05 0.10 0.15

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is hardness)

13-35. a) Analysis of Variance for ARSENIC

Source DF SS MS F PTEST 2 0.0014000 0.0007000 3.00 0.125SUBJECT 3 0.0212250 0.0070750 30.32 0.001

Error 6 0.0014000 0.0002333Total 11 0.0240250

Do not reject H0, there is no evidence of differences between the tests.

13-27



b) Some indication of variability increasing with the magnitude of the response.

4321

0.02

0.01

0.00

-0.01

-0.02

-0.03

SUBJECT

R e s i d u a l

Residuals Versus SUBJECT(response is ARSENIC)

321

0.02

0.01

0.00

-0.01

-0.02

-0.03

TEST

R e s i d u a l

Residuals Versus TEST(response is ARSENIC))

0.150.100.05

0.02

0.01

0.00

-0.01

-0.02

-0.03

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is ARSENIC)

-0.03 -0.02 -0.01 0.00 0.01 0.02

-2

-1

0

1

2

N o r m a

l S c o r e

Residual

Normal Probability Plot of the Residuals(response is ARSENIC)

13-28



13-36. a) Analysis of Variance of PROPECTINSource DF SS MS F PSTORAGE 3 1972652 657551 4.33 0.014LOT 8 1980499 247562 1.63 0.169Error 24 3647150 151965Total 35 7600300

Reject H0, and conclude that the storage times affect the mean level of propectin.

b) P-value = 0.014

c)Fisher's pairwise comparisons





0 7 14

7 -171

634

14 -214 -445

592 360

21 239 8 50

1045 813 856

There are differences between 0 and 21 days; 7 and 21 days; and 14 and 21 days. The propectin levels are

significantly different at 21 days from the other storage times so there is evidence that the mean level of

propectin decreases with storage time. However, differences such as between 0 and 7 days and 7 and 14 days

were not significant so that the level is not simply a linear function of storage days.

d) Observations from lot 3 at 14 days appear unusual. Otherwise, the residuals are acceptable.

10005000-500

2

1

0

-1

-2


Residual

Normal Probability Plot of the Residuals(response is propecti)

9876543210

1000

500

0

-500

LOT

R e s i d u a l

Residuals Versus LOT(response is PROTOPE)

20100

1000

500

0

-500

STORAGE

R e s i d u a l

ResidualsVersusSTORAGE(response is PROTOPEC)

13-29



150010005000

1000

500

0

-500

Fitted Value

R e s i d u a l

Residual s Versus the Fitted Va lues(response is PROTOPEC)

13-37. A version of the electronic data file has the reading for length 4 and width 5 as 2. It should be 20.

a) Analysis of Variance for LEAKAGE

Source DF SS MS F PLENGTH 3 72.66 24.22 1.61 0.240WIDTH 4 90.52 22.63 1.50 0.263Error 12 180.83 15.07Total 19 344.01

Do not reject H0, mean leakage voltage does not depend on the channel length.

b) One unusual observation in width 5, length 4. There are some problems with the normal probability plot,

including the unusual observation.

54321

10

5

0

WIDTH

R e s i d u a l

Residuals Versus WIDTH(response is LEAKAGE)

1050

10

5

0

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is LEAKAGE)

0 5 10

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is LEAKAGE)

4321

10

5

0

LENGTH

R e s i d u a l

Residuals Versus LENGTH(response is LEAKAGE)

13-30



c) Analysis of Variance for LEAKAGE VOLTAGESource DF SS MS F PLENGTH 3 8.1775 2.7258 6.16 0.009WIDTH 4 6.8380 1.7095 3.86 0.031Error 12 5.3100 0.4425Total 19 20.3255

Reject H0. And conclude that the mean leakage voltage does depend on channel length. By removing the

data point that was erroneous, the analysis results in a conclusion. The erroneous data point that was anobvious outlier had a strong effect the results of the experiment.




13-38 a) Analysis of Variance for SURFACE ROUGNESSAnalysis of Variance for y

Source DF SS MS F P

Material 3 0.2402 0.0801 4.96 0.020

Error 11 0.1775 0.0161

Total 14 0.4177

Reject H0

b) One observation is an outlier.

1 2 3 4

-0.2

-0.1

0.0

0.1

0.2

0.3

Material

R e s i d u a l

Residuals Versus Material

(response is Surf Rou)

13-31



-0.2 -0.1 0.0 0.1 0.2 0.3

-2

-1

0

1

2


Residual


(response is Surf Rou)

c) There appears to be a problem with constant variance. This may be due to the outlier in the data.

0.1 0.2 0.3 0.4 0.5

-0.2

-0.1

0.0

0.1

0.2

0.3

Fitted Value

R e s i d u a l

Residuals Versus the Fitted Values(response is Surf Rou)

d) 95% confidence interval on the difference in the means of EC10 and EC1

602.0118.0

2

)0161.0(

4

)0161.0(201.2)130.0490.0(

2

)0161.0(

4

)0161.0(201.2)130.0490.0(

41

41

21

11,,025.04131

21

11,,025.041

≤−≤

++−≤−≤+−−

++−≤−≤+−−

μ μ

μ μ

μ μ n

MS

n

MSt y y

n

MS

n

MSt y y E E E E

13-39. a) Analysis of Variance for RESISTANCESource DF SS MS F P ALLOY 2 10941.8 5470.9 76.09 0.000Error 27 1941.4 71.9Total 29 12883.2

Reject H0, the type of alloy has a significant effect on mean contact resistance.

13-32



b) Fisher's pairwise comparisonsFamily error rate = 0.119




1 2

2 -13.58

1.98

3 -50.88 -45.08-35.32 -29.52

There are differences in the mean resistance for alloy types 1 and 3; and types 2 and 3.

c) 99% confidence interval on the mean contact resistance for alloy 3

83.14797.132

10

9.71771.24.140

10

9.71771.24.140

1

3

27,005.0327,005.03

≤≤

+≤≤−

+≤≤−

μ

μ

μ n

MSt y

n

MSt y E

i

E

d) Variability of the residuals increases with the response. The normal probability plot has some

curvature in the tails, indicating a problem with the normality assumption. A transformation of the

response should be conducted.

321

30

20

10

0

-10

-20

ALLOY

R e s i d u a l

Residuals VersusALLOY(response is RESISTAN))

140130120110100

30

20

10

0

-10

-20

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is RESISTAN)

-20 -10 0 10 20 30

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is RESISTAN)

13-33



13-40 a) Analysis of Variance for SCORESource DF SS MS F PMETHOD 2 13.55 6.78 1.68 0.211Error 21 84.77 4.04Total 23 98.32

Do not reject H0

b) P-value = 0.211

c) There is some curvature in the normal probability plot. There appears to be some differences in the

variability for the different methods. The variability for method one is larger than the variability for method

3.

3210-1-2-3-4-5

2

1

0

-1

-2


Residual

Normal Probability Plot of the Residuals(response is score)

321

3

2

1

0

-1

-2

-3

-4

-5

METHOD

R e s i d u a l

Residuals Versus METHOD(response is SCORE)

d.) 342.08

04.478.6ˆ 2 =

−=

−=

n

MS MS E Treatments

τ σ

04.4ˆ 2 == E MSσ

13-41. a) Analysis of Variance for VOLUMESource DF SS MS F PTEMPERATURE 2 16480 8240 7.84 0.007Error 12 12610 1051

Total 14 29090Reject H0.

b) P-value = 0.007

c) Fisher's pairwise comparisonsFamily error rate = 0.116




70 75

13-34



75 -16.7

72.7

80 35.3 7.3

124.7 96.7

There are significant differences in the mean volume for temperature levels 70 and 80; and 75 and 80. The

highest temperature results in the smallest mean volume.

d) There are some relatively small differences in the variability at the different levels of temperature. The

variability decreases with the fitted values. There is an unusual observation on the normal probability plot.

-50 0 50

-2

-1

0

1

2


Residual

Normal Probability Plot of the Residuals(response is vol)

807570

50

0

-50

TEMPERAT

R e s i d u a l

Residuals Versus TEMPERAT(response is VOLUME)

125012401230122012101200119011801170

50

0

-50

FittedValue

R e s i d u a l

Residuals Versus the Fitted Values(response is VOLUME)

13-42. a)Analysis of Variance of Weight GainSource DF SS MS F PMEANWEIG 2 0.2227 0.1113 1.48 0.273 AIRTEMP 5 10.1852 2.0370 27.13 0.000Error 10 0.7509 0.0751Total 17 11.1588

Reject H0 and conclude that the air temperature has an effect on the mean weight gain.

b) Fisher's pairwise comparisonsFamily error rate = 0.314


Critical value = 2.179Intervals for (column level mean) - (row level mean)

50 60 70 80 90

60 -0.9101

0.1034

70 -1.2901 -0.8868

-0.2766 0.1268

80 -0.9834 -0.5801 -0.2001

0.0301 0.4334 0.8134

90 -0.3034 0.0999 0.4799 0.1732

0.7101 1.1134 1.4934 1.1868

100 1.0266 1.4299 1.8099 1.5032 0.8232

13-35



2.0401 2.4434 2.8234 2.5168 1.8368

There are significant differences in the mean air temperature levels 50

and 70, 100; 60 and 90, 100; 70 and 90, 100; 80 and 90, 100; and 90 and

100. The mean of temperature level 100 is different from all the other

temperatures.

c) There appears to be some problems with the assumption of constant variance.

200150100

0.5

0.0

-0.5

MEANWEIG

R e s i d u a l

Residuals Versus MEANWEIG(response is WEIGHTGA)

1009080706050

0.5

0.0

-0.5

AIRTEMP

R e s i d u a l

Residuals Versus AIRTEMP(response is WEIGHTGA)

210

0.5

0.0

-0.5

FittedValue

R e s i d u a l

Residuals Versusthe Fitted Values(response is WEIGHTGA)

0.50.0-0.5

2

1

0

-1

-2


Residual

Normal Probability Plot of the Residuals(response is wt gain)

13-43. a) Analysis of Variance for PCTERROR

Source DF SS MS F P ALGORITH 5 2825746 565149 6.23 0.000PROJECT 7 2710323 387189 4.27 0.002Error 35 3175290 90723Total 47 8711358

13-36



Reject H0 , the algorithms are significantly different.

b) The residuals look acceptable, except there is one unusual point.

87654321

1000

500

0

-500

PROJECT

R e s i d u a l

Residuals Versus PROJECT(response is PCTERROR)

654321

1000

500

0

-500

ALGORITH

R e s i d u a l

Residuals Versus ALGORITH(response is PCTERROR)

10005000

1000

500

0

-500

FittedValue

R e s i d u a l

Residuals Versus the Fit ted Values(response is PCTERROR)

10005000-500

2

1

0

-1

-2


Residual

Normal Probability Plot of the Residuals(response is PCTERROR)

c) The best choice is algorithm 5 because it has the smallest mean and a low variability.

13-44. a) The normal probability plot shows that the normality assumption is not reasonable.

13-37



Obs

P e r c e n t

150010005000-500

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

<0.005

293.8

StDev 309.4

N 3

A D 2.380

P-Value

Probabilit y Plot of ObsNormal - 95% CI

0

b) The normal probability plot shows that the normality assumption is reasonable.

log(obs)

P e r c e n t

4.03.53.02.52.01.51.00.5

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.066

2.209StDev 0.5060

N 30

A D 0.687

P-Value

Probabilit y Plot of l og(obs)Normal - 95% CI

Source DF SS MS F P

Treatments 2 1.188 0.594 2.57 0.095Error 27 6.237 0.231

Total 29 7.425

There is evidence to support the claim that the treatment means differ at α = 0.1 for the

transformed data since the P-value = 0.095.

c) The normal probability plot and the residual plots show that the model assumptions are

reasonable.

13-38



Residual

P e r c e

n t

1.00.50.0-0.5-1.0

99

90

50

10

1

Fitted Value

R e s i d u

a l

2.52.42.32.22.1

1.0

0.5

0.0

-0.5

-1.0

Residual

F r

e q u e n c y

0.80.40.0-0.4-0.8

8

6

4

2

0

Observation Order

R

e s i d u a l

30282624222018161412108642

1.0

0.5

0.0

-0.5

-1.0



Residual Plots for log( obs)

13-45. a) μ=1.6, =0.284, =0.5333 2Φ Φ

Numerator degrees of freedom = 141 ν ==−a

Denominator degrees of freedom =215)1( ν ==−na

From Chart Figure 13-6, β ≈0.8 and the power = 1 - β = 0.2

b)

n Φ2 Φ a(n-1) β Power = 1-β

50 3.56 1.89 245 0.05 0.95

The sample size should be approximately n = 50.

13-46. a) μ = (1+5+8+4)/4 = 4.5 and

5.2

25.6)4(4

])5.44()5.48()5.45()5.41[(4 22222

=Φ

=−+−+−+−

=Φ

Numerator degrees of freedom = 131 ν ==−a

Denominator degrees of freedom =212)1( ν ==−na

From Figure 13-6, β = 0.05 and the power = 1 - β = 0.95

b)

n Φ2

Φ a(n-1) β Power = 1-β

4 6.25 2.5 12 0.05 0.95

3 4.6875 2.165 8 0.25 0.75

The sample size should be approximately n = 4.

13-39



13-40




15-43 Under the null hypothesis each rank has probability of 0.5 of being either positive or negative. Define the

random variable Xi as

⎩⎨⎧

=negativeisrank

positiveisrank X i

0

1

Then,

24

)12)(1()( Then,

4

1

4

1

2

1)]([

2

10

2

11)(

ceindependen by )()( where

2

1)(since

4

)1(

2

1)()( and

222

2

11

++=

=−=−+=

=

=+

====

+

+

=

+

=

+

∑

∑∑∑

nnn RV

X E X V

i xV RV

x E nn

ii x E R E i x R

ii

i

i

n

i

i

n

i

i

15-44 a) 5

4

)( 5=>= + ii X X P p There appears to be an upward trend.

b) V is the number of values for which . The probability distribution of V is binomial with n = 5

and p = 0.5.

ii X X >+5

c) V = 4

1. The parameter of interest is number of values of i for which .ii X X >+5

trendupwardanisthere:3

trendnoisthere:.2

1

0

H

H

4. α = 0.05

5. The test statistic is the observed number of values where or V = 4.ii X X >+5

6. We reject H 0 if the P-value corresponding to V = 4 is less than or equal to α = 0.05.

7. Using the binomial distribution with n = 5 and p = 0.5, P-value = P( V ≥ 4| p = 0.5) = 0.18758. Conclusion: do not reject H 0. There is not enough evidence to suspect an upward trend in the wheel

opening dimensions at α=0.05.

15-45 a) 32 sequences are possible

b) Because each sequence has probability 1/32 under H0, the distribution of W*

is obtained by counting the

sequences that result in each value of W*

w* 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Prob*1/32 1 1 1 2 2 3 3 3 3 3 3 2 2 1 1 1

c) P( W* > 13) = 2/32

d) By enumerating the possible sequences, the probability that W* exceeds any value can be calculated under

the null hypothesis as in part (c). This approach can be used to determine the critical values for the

test.

15-19



CHAPTER 15

Section 15-2

15-1 a)

1. The parameter of interest is median of pH.

0.7~:3

0.7~:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05

5. The test statistic is the observed number of plus differences or r + = 8.

6. We reject H 0 if the P-value corresponding to r + = 8 is less than or equal to α = 0.05.

7. Using the binomial distribution with n = 10 and p = 0.5, P-value = 2P(R + ≥ 8 | p = 0.5) = 0.1

8. Conclusion: we cannot reject H 0. There is not enough evidence to reject the manufacturer’s claim that the

median of the pH is 7.0

b)

1. The parameter of interest is median of pH.

0.7~:3

0.7~:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05

5. The test statistic isn

nr z

5.0

5.0*

0−=

6. We reject H 0 if |Z0|>1.96 for α=0.05.

7. r*=8 and 90.1105.0

)10(5.08

5.0

5.0*

0 =−

=−

=n

nr z

8. Conclusion: we cannot reject H 0. There is not enough evidence to reject the manufacturer’s claim that

the median of the pH is 7.0

P-value = 2[1 - P(|Z0| < 1.90)] = 2(0.0287) = 0.0574

15-2 a)

1. The parameter of interest is median titanium content.

5.8~:3

5.8~:.2

1

0

≠=

μ

μ

H

H

4. α=0.05

5. The test statistic is the observed number of plus differences or r +

= 7.

6. We reject H 0 if the P-value corresponding to r + = 7 is less than or equal to α=0.05.

7. Using the binomial distribution with n=10 and p=0.5, P-value = 2P(R *≤7|p=0.5)=0.359

8. Conclusion: we cannot reject H 0. There is not enough evidence to reject the manufacturer’s claim that

the median of the titanium content is 8.5.

b)

1. Parameter of interest is the median titanium content

5.8~:.3

5.8~:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05

5. Test statistic isn

nr z

5.0

5.00

−=

+

6. We reject H 0 if the | Z 0| > Z 0.025 = 1.96

7. Computation: 34.1205.0

)20(5.070 −=

−= z

8. Conclusion, cannot reject H 0. There is not enough evidence to conclude that the median titanium content

differs from 8.5. The P-value = 2*P( | Z | > 1.34) = 0.1802.

15-1



15-3 a)

1. Parameter of interest is the median impurity level.

5.2~:.3

5.2~:.2

1

0

<

=

μ

μ

H

H

4. α=0.05


6. We reject H 0 if the P-value corresponding to r +

= 2 is less than or equal to α = 0.05.7. Using the binomial distribution with n = 22 and p = 0.5, P-value = P(R +≤ 2 | p = 0.5) = 0.0002

8. Conclusion, reject H 0. The data supports the claim that the median is impurity level is less than 2.5.

b)

1. Parameter of interest is the median impurity level

5.2~:.3

5.2~:.2

1

0

<

=

μ

μ

H

H

4. α=0.05

5. Test statistic isn

nr z

5.0

5.00

−=

+

6. We reject H 0 if the Z 0 < Z 0.05 = -1.65

7. Computation:84.3225.0

)22(5.020 −=

−

= z

8. Conclusion, reject H 0 and conclude that the median impurity level is less than 2.5. The P-value = P( Z <-

3.84) = 0.000062

15-4 a)

1. Parameter of interest is the median margarine fat content

0.17~:.3

0.17~:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05


= 3.

6. We reject H 0 if the P-value corresponding to r + = 7 is less than or equal to α=0.05.

7. Using the binomial distribution with n=6 and p=0.5, P-value = 2*P(R *≥3|p=0.5,n=6)=1.

8. Conclusion, cannot reject H 0. There is not enough evidence to conclude that the median fat content differs

from 17.0.

b)

1. Parameter of interest is the median margarine fat content

0.17~:.3

0.17~:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05

5. Test statistic is

n

nr z

5.0

5.00

−=

+


7. Computation: 065.0

)6(5.030 =

−= z

8. Conclusion, cannot reject H 0. The P-value = 1-Φ(0) = 1 – 0.5 = 0.5. There is not enough evidence toconclude that the median fat content differs from 17.0.

15-2



15-5 a)

1. Parameter of interest is the median compressive strength

2250~:.3

2250~:.2

1

0

>

=

μ

μ

H

H

4. α = 0.05


6. We reject H 0 if the P-value corresponding to r +

= 7 is less than or equal to α=0.05.7. Using the binomial distribution with n = 12 and p = 0.5, P-value = P( R + ≥ 7 | p = 0.5) = 0.3872

8. Conclusion, cannot reject H 0. There is not enough evidence to conclude that the median compressive

strength is greater than 2250.

b)

1. Parameter of interest is the median compressive strength

2250~:.3

2250~:.2

1

0

>

=

μ

μ

H

H

4. α=0.05


n

nr z

5.0

5.00

−=

+



)12(5.070 =−= z

8. Conclusion, cannot reject H 0. The P-value = 1-Φ(0.58) = 1 - 0.7190 = 0.281. There is not enough

evidence to conclude that the median compressive strength is greater than 2250.

15-6 a)

1. Parameters of interest are the median caliper measurements

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.05

5. The test statistic is r = min( r + , r -).

6. Because α =0.05 and n = 12, Appendix A, Table VIII gives the critical value of = 2. We reject*

05.0r

H 0 in favor of H 1 if r ≤ 2.7. There are four ties that are ignored so that r + = 6 and r - = 2 and r = min(6,2) = 2

8. Conclusion, reject H 0. There is a significant difference in the median measurements of the two calipers at

α = 0.05.

b)

1. Parameters of interest are the median caliper measurements

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.05

5. Test statisticn

nr z

5.0

5.00

−=

+



)8(5.060 =−= z

8. Conclusion, do not reject H 0. There is not a significant difference in the median measurements of the two

calipers at α = 0.05.

The P-value = 2[1-P( | Z 0| < 1.41) = 0.159.

15-7 a)

15-3



1. Parameters of interest are the median hardness readings for the two tips

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.05

5. The test statistic is r = min( r +, r -).

6. Because α = 0.05 and n = 8, Appendix A, Table VIII gives the critical value of = 0. We reject*

05.0r

H 0 in favor of H 1 if r ≤ 0.

7. r + = 6 and r - = 2 and r = min(6,2) = 2

8. Conclusion, cannot reject H 0. There is not a significant difference in the tips.

b) P-value = 2P(R − ≥ 6 | p = 0.5) = 0.289. The P-value is greater than 0.05, therefore we cannot reject H0

and conclude that there is no significant difference between median hardness readings for the two tips. This

result agrees with the result found in Exercise 15-9.

15-8 a)

1. Parameters of interest are the median drying times for the two formulations

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.01

5. The test statistic is r = min( r +

, r -

).


01.0r


7. There are two ties that are ignored so that r +

= 3 and r - = 15 and r = min(3,15) = 3

8. Conclusion, reject H 0. There is a significant difference in the drying times of the two formulations at α =

0.01.

b)

1. Parameters of interest are the median drying times of the two formulations

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.01

5. Test statisticnnr z

5.05.00 −=

+



)18(5.030 −=

−= z

8. Conclusion, reject H 0 and conclude that there is a significant difference between the drying times for the

two formulations at α = 0.01.

The P-value = 2[1-P( | Z 0| < 2.83) = 0.005.

c) P-value = 2P(R − ≤ 3 | p = 0.5) = 0.0075. The exact P-value computed here agrees with the normal

approximation in Exercise 15-11 in the sense that both calculations would lead to the rejection of H0.

15-9 a) for x > 0. Because 3.5 is the median . Solving for λ, we find λ = 0.198

and E( X ) = 1/λ = 5.05

xe x f λ λ −=)( 5.05.3

0

=−∫xe λ λ

b) Because r * = 3 > 1, do not reject H0

c) β = P(Type II error)=P(R * > 1| =4.5)~μ

15-4



For ~μ = 4.5, E( X ) = 6.49 and P(X < 3.5| =4.5) = 0.4167. Therefore, with~μ =−∫ xe

154.0

5.3

0

154.0

n =10, p = 0.4167, β = P(Type II error)=P(R * > 1| =4.5) = 0.963~μ

15-10 a)

1. Parameters of interest are the median blood cholesterol levels

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.05

5. The test statistic r = min( r + , r -).


05.0r


7. r + = 14 and r - = 1 and r = min(14,1) = 1

8. Conclusion, reject H 0. There a significant difference in the blood cholesterol levels atα = 0.05.

b)

1. Parameter of interest is the difference in blood cholesterol levels

0~:.30

~:.2

1

0

≠=

D

D

H H

μ μ

4. α=0.05

5. Test statisticn

nr z

5.0

5.00

−=

+



)15(5.0140 =

−= z

8. Conclusion, reject H 0. There is a difference between the blood cholesterol levels.

The P-value = 2*P (| Z | > 3.36) = 0.0008.

15-11 a) α = P(Z > 1.96) = 0.025

b) 115.0)20.1(1|96.1/

=−<=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ === Z P

n

X P μ

σ β

c) The sign test that rejects if R − ≤ 1 has α = 0.011 based on the binomial distribution.

d) ( ) 1587.0)1(1|2 =>==≥= − Z P RP μ β . Therefore, R − has a binomial distribution with p =

0.1587 and n = 10 when μ = 1. Then β = 0.487. The value of β is greater for the sign test than for the normal

test because the Z-test was designed for the normal distribution.

15-5



Section 15-3

Note to Instructors: When an observation equals the hypothesized mean the observation is dropped and not

considered in the analysis.

15-12 a)

1. Parameter of interest is the mean titanium content

5.8:.3

5.8:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05

5. The test statistic is w = min( w + , w -).

6. We reject H 0 if w ≤ = 52, because α = 0.05 and n = 20, Appendix A, Table IX gives the critical

value.

*

05.0w

7. w+ = 80.5 and w - = 109.5 and w = min(80.5,109.5) = 80.5

8. Conclusion, because 80.5 > 52, we cannot reject H 0. The mean titanium content is not significantly

different from 8.5 at α = 0.05.

b)

1. Parameter of interest is the mean titanium content

5.8:.3

5.8:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05

5. The test statistic is

24/)12)(1(

4/)1(0

++

+−=

+

nnn

nnW Z

6. We reject H 0 if the | Z 0| > Z 0.025 = 1.96, at α = 0.05

7. w+ = 80.5 and w - = 109.5 and

58.024/)39)(20(19

4/)20(195.800

−=−

= Z

8. Conclusion, because 0.58 < 1.96, we cannot reject H 0. The mean titanium content is not significantly

different from 8.5 at α=0.05.

15-13 1. Parameter of interest is the mean pH

0.7:.3

0.7:.2

1

0

≠

=

μ

μ

H

H

4. α = 0.05


6. We reject H 0 if w ≤ = 8, because α = 0.05 and n = 10, Appendix A, Table IX gives the critical

value.

*

05.0w

7. w+ = 50.5 and w - = 4.5 and w = min(50.5, 4.5) = 4.5

8. Conclusion, because 4.5 < 8, we reject H 0 and conclude that the mean pH is not equal to 7.

15-14 1. Parameter of interest is the difference in the mean caliper measurements

0:.3

0:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.05


6. We reject H 0 if w ≤ = 3 because α = 0.05 and n = 8 Appendix A, Table IX gives the critical value.*

05.0w

7. w+ = 21.5 and w - = 14.5 and w = min(21.5,14.5) = 14.5

8. Conclusion, do not reject H 0 because 14.5 > 13. There is a not a significant difference in the mean

measurements of the two calipers at α = 0.05.

15-6



15-15 1. Parameter of interest is the mean impurity level

5.2:.3

5.2:.2

1

0

<

=

μ

μ

H

H

4. α = 0.05

5. The test statistic is w+

6. We reject H 0 if w+ ≤ = 60 because α = 0.05 and n = 20 Appendix A, Table IX gives the critical

value.

* 05.0w

7. w+

= 5 and w-= 205

8. Conclusion, because 5 < 60, we reject H 0 and conclude that the mean impurity level is less than 2.5 ppm.

15-16 1. Parameter of interest is the difference in mean drying times

0:.3

0:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.01

5. The test statistic is w = min( w+

, w-).

6. We reject H 0 if w ≤ = 27 because α = 0.01 and n = 18 Appendix A, Table IX gives the critical

value.

*

05.0w

7. w+

= 141.5 and w-

= 29.5 and w = min(141.5, 29.5) = 29.58. Conclusion, not enough evidence to reject H 0.

15-17 1. Parameter of interest is the difference in mean hardness readings for the two tips

0:.3

0:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α = 0.05


6 We reject H 0 if w ≤ = 3, because α = 0.05 and n = 8 Appendix A, Table IX gives the critical value.*

05.0w

7. w+ = 24.5 and w - = 11.5 and w = min(24.5, 11.5) = 11.5

8. Conclusion, because 11.5 > 3 cannot reject H 0. There is not a significant difference in the tips.



Section 15-4

15-18 a)

1. The parameters of interest are the mean current (note: set circuit 1 equal to sample 2 so that Table X can

be used. Thereforeμ1=mean of circuit 2 and μ2=mean of circuit 1)

211

210

:.3

:.2

μ μ

μ μ

>

=

H

H

4. α = 0.025

5. The test statistic is1

21212

2

)1)((w

nnnnw −

+++=

6. We reject H 0 if w2 ≤ = 51, because α=0.025 and n*

025.0w 1=8 and n2=9, Appendix A, Table X gives the

critical value.

7. w1 = 78and w2 = 75 and because 75 is less than 51, do not reject H 08. Conclusion, do not reject H 0. There is not enough evidence to conclude that the mean of circuit 1 exceeds

the mean of circuit 2.

b)

1. The parameters of interest are the mean image brightness of the two tubes

211

210

:.3

:.2

μ μ

μ μ

>

=

H

H

4. α = 0.025


1

11

0

w

wW

zσ

μ −=

15-7



6. We reject H 0 if Z 0 > Z 0.025 = 1.96

7. w1 = 78,1w

μ =72 and =1082

1wσ

58.039.10

72780 =

−= z

Because Z 0 < 1.96, cannot reject H 08. Conclusion, fail to reject H 0. There is not a significant difference in the heat gain for the heating units at α

= 0.05. P-value =2[1 - P( Z < 0.58 )] = 0.5619

15-19 1. The parameters of interest are the mean flight delays

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α=0.01


21212

2

)1)((w

nnnnw −

+++=

6. We reject H 0 if w ≤ = 23, because α=0.01 and n*

01.0w 1=6 and n2=6, Appendix A, Table X gives the

critical value.

7. w1 = 40 and w2 = 38 and because 40 and 38 are greater than 23, we cannot reject H 08. Conclusion, do not reject H 0. There is no significant difference in the flight delays at α = 0.01.

15-20 a)

1. The parameters of interest are the mean heat gains for heating units

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α = 0.05


21212

2

)1)((w

nnnnw −

+++=

6. We reject H 0 if w ≤ = 78, because α = 0.01 and n*

01.0w 1 = 10 and n2 = 10, Appendix A, Table X gives the

critical value.

7. w1 = 77 and w2 = 133 and because 77 is less than 78, we can reject H 08. Conclusion, reject H 0 and conclude that there is a significant difference in the heating units at α = 0.05.

b)1. The parameters of interest are the mean heat gain for heating units

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α=0.05


1

11

0

w

wW

zσ

μ −=

6. We reject H 0 if | Z 0| > Z 0.025 = 1.96

7. w1 = 77,1w

μ =105 and =1752

1wσ

12.223.13

105770 −=

−= z

Because | Z 0 | > 1.96, reject H 0

8. Conclusion, reject H 0 and conclude that there is a difference in the heat gain for the heating units atα=0.05. P-value =2[1 - P( Z < 2.19 )] = 0.034

15-21 a)

1. The parameters of interest are the mean etch rates

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α = 0.05

15-8




21212

2

)1)((w

nnnnw −

+++=


05.0w 1 = 10 and n2 = 10, Appendix A, Table X gives

the critical value.

7. w1 = 73 and w2 = 137 and because 73 is less than 78, we reject H 08. Conclusion, reject H0 and conclude that there is a significant difference in the mean etch rate at α = 0.05.

b)

1. The parameters of interest are the mean temperatures

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α = 0.05


1

11

0

w

wW

zσ

μ −=

6. We reject H 0 if | Z 0| > Z 0.025 = 1.96

7. w1 = 55 ,1w

μ =232.5 and =581.252

1wσ

06.111.24

5.2322580 =

−= z

Because | Z 0| < 1.96, do not reject H 08. Conclusion, do not reject H 0. There is not a significant difference in the pipe deflection temperatures at α

= 0.05. P-value =2[1 - P( Z < 1.06 )] = 0.2891

15-22 a)

1. The parameters of interest are the mean temperatures

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α = 0.05


21212

2

)1)((w

nnnnw −

+++=



the critical value.

7. w1 = 258 and w2 = 207 and because both 258 and 207 are greater than 185, we cannot reject H 08. Conclusion, do not reject H0. There is not significant difference in the pipe deflection temperature at α =

0.05.

b)

1. The parameters of interest are the mean etch rates

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α=0.05


1

11

0

w

wW

zσ

μ −=

6. We reject H 0 if | Z 0| > Z 0.025 = 1.96

7. w1 = 73,1w

μ =105 and =1752

1wσ

42.223.13

105730 −=

−= z

Because | Z 0| > 1.96, reject H 08. Conclusion, reject H 0 and conclude that there is a significant difference between the mean etch rates. P-

value = 0.0155

15-9



15-23 a) The data are analyzed in ascending order and ranked as follows:

Group Distance Rank

2 244 1

2 258 2

2 260 3

1 263 4.5

2 263 4.5

2 265 6

1 267 7

2 268 8

2 270 9

1 271 11

1 271 11

2 271 11

2 273 13

1 275 14.5

1 275 14.5

1 279 16

2 281 17

1 283 18

1 286 19

1 287 20

The sum of the ranks for group 1 is w1=135.5 and for group 2, w2=74.5

Since is less than2

w0.05 78w = , we reject the null hypothesis that both groups have

the same mean.

b) When the sample sizes are equal it does not matter which group we select for w1

31.2175

1055.135

17512

)11010(10*10

1052

)11010(10

0

2

1

1

=−

=

=++

=

=++=

Z

W

W

σ

μ

Because z0 > z0.025 = 1.96, we reject H0 and conclude that the sample means are different

for the two groups.

When z0 = 2.31 the P-value = 0.021



Section 15-5

15-24 a) Kruskal-Wallis Test on strength

methods N Median Ave Rank Z

1 5 550.0 8.7 0.43

2 5 553.0 10.7 1.65

3 5 528.0 4.6 -2.08

Overall 15 8.0

15-10



H = 4.83 DF = 2 P = 0.089

H = 4.84 DF = 2 P = 0.089 (adjusted for ties)

Do not reject H0. The conditioning method does not have an effect on the breaking strength at α = 0.05.

b) The P-value is about 0.089 (use the chi-square distribution)

15-25 a) Kruskal-Wallis Test on strengthmixingte N Median Ave Rank Z1 4 2945 9.6 0.552 4 3075 12.9 2.123 4 2942 9.0 0.244 4 2650 2.5 -2.91Overall 16 8.5H = 10.00 DF = 3 P = 0.019H = 10.03 DF = 3 P = 0.018 (adjusted for ties)* NOTE * One or more small samples

Reject H0 and conclude that the mixing technique has an effect on the strength.

b) P-value = 0.018 (use the chi-square distribution)

15-26 a) Kruskal-Wallis Test on angle

manufact N Median Ave Rank Z1 5 39.00 7.6 -1.272 5 44.00 12.4 0.833 5 48.00 15.8 2.314 5 30.00 6.2 -1.88Overall 20 10.5H = 8.37 DF = 3 P = 0.039

Do not reject H0. There is not a significant difference between the manufacturers at α = 0.01.

b) The P-value is about 0.039 (use the chi-square distribution)

15-27 a) Following is the data and ranks for the conductivity test.

Type r_i.

Coating Type 1 141 143 146 150

10 12.5 15 19 56.5

Coating Type 2 137 143 149 152

9 12.5 18 20 59.5

Coating Type 3 127 132 133 134

1.5 5.5 7 8 22

Coating Type 4 127 129 129 132

1.5 3.5 3.5 5.5 14

Coating Type 5 142 144 147 148

11 14 16 17 58Since there is a fairly large number of ties, we use Equation 15-12 as the test statistic.

22 1 20(21)

[2868 ]20 1 4

34.89474

s = −−=

and the test statistic is

02.144

)21(20125.2694

89474.34

1 2

=⎥⎦

⎤⎢⎣

⎡−=h

15-11



b) An approximate P-value can be obtained from the approximate chi-squared distribution for H.

Here a = 5 and each n = 4 so that the guidelines for the adequacy of the approximation are not

quite satisfied. The guideline specifies each n > 4.

But if we apply the chi-squared result we use a – 1 = 4 degrees of freedom and determine

0.005 < P-value < 0.01

From Minitab we obtain the following results that agree with the calculated values.

Kruskal-Wallis Test on C1

C2 N Median Ave Rank Z

1 4 144.5 14.1 1.37

2 4 146.0 14.9 1.65

3 4 132.5 5.5 -1.89

4 4 129.0 3.5 -2.65

5 4 145.5 14.5 1.51

Overall 20 10.5

H = 13.98 DF = 4 P = 0.007


Since , we would reject the null hypothesis and conclude that the treatments

differ. Minitab provides the P-value = 0.007

2

0.05,4 9.49h χ > =

15-28 a) Kruskal-Wallis Test on uniformity

flow

rate N Median Ave Rank Z

125 6 3.100 5.8 -2.06

160 6 4.400 13.0 1.97

250 6 3.800 9.7 0.09

Overall 18 9.5

H = 5.42 DF = 2 P = 0.067


Do not reject H0. There is not a significant difference between the flow rates on the uniformity at α = 0.05.

b) P-value = 0.066 (use the chi-square distribution)

15-29 a) D A A D D A A A D D D A

The total number of runs is r = 6.

The observed value of 6 is below the mean of 7. For a sample of size n = 12 with equal samples

from each group, the number of runs can range from 2 to 12. Therefore, the P-value can be

computed as the probability of 6 or fewer or 8 or greater runs. This will be greater usual levels of

significance so that the null hypothesis is not rejected.

The normal approximation can be used for a simpler calculation

6055.0727.2

76

727.2)112(12

)12)6)(6(2(662

)1(

)2(2

712

12)6)(6(22

22

21212

21

−=−

=−

=

=−

−×××=

−

−=

=+

=+

=

R

R

R

R

r z

N N

N nnnn

N

N nn

σ

μ

σ

μ

At α = 0.05 reject H0 when 96.1≥ z . Because 6055.0= z < 1.96, fail to reject H0. There is

not sufficient evidence to conclude that there is a difference in distributions.

15-12



Or, Minitab output:Runs test for C5

Runs above and below K = 1.5

The observed number of runs = 6

The expected number of runs = 7

6 observations above K, 6 below

* N is small, so the following approximation may be invalid.

P-value = 0.545

b) The sum of the ranks for group 1 is w1=40 and for group 2, w2=38

Since both w1 and are both greater than w2w 0.05 =26, there is insufficient evidence to reject the

null hypothesis that both groups have the same mean.

In part (a) we test whether the two samples come from the same distribution. In part (b) we test

whether their means are equal.

15-30

a) The sample numbers of the ranked data are:12112112221121122

The number of runs is r =10.

2665.0955.3

47.910

955.3)117(17

)17)8)(9(2(892

)1(

)2(2

47.917

17)8)(9(22

22

21212

21

=−

=−

=

=−

−×××=

−

−=

=+

=+

=

R

R

R

R

r z

N N

N nnnn

N

N nn

σ

μ

σ

μ

At α = 0.05 reject H0 when . Because96.1|| ≥ z 2665.0= z < 1.96, we fail to reject H0. There is

not enough evidence to conclude that there is a difference in distributions.The P-value ≈ 0.79

Or, using Minitab:Runs test for C5

Runs above and below K = 1.47059

The observed number of runs = 10

The expected number of runs = 9.47059

8 observations above K, 9 below

* N is small, so the following approximation may be invalid.

P-value = 0.790

b)

The sum of the ranks for group 2 (which has the smaller sample size) is 78. Since w0.05=51, wefail to reject H0. There is insufficient evidence to conclude that the groups have the same mean.

Yes, the hypotheses differ. The one from part (a) test whether the samples came from the same

distribution. Here we test whether the populations have the same mean.

15-31

The sample numbers of the ranked data are: 222111112221211112211222; thus, r = 9

15-13



67.1739.5

139

739.5)124(24

)24)12)(12(2(12122

)1(

)2(2

1324

24)12)(12(22

22

21212

21

−=−

=−

=

=−

−×××=

−

−=

=+

=+

=

R

R

R

R

r z

N N

N nnnn

N

N nn

σ

μ

σ

μ

At α = 0.05, fail to reject H0 since z>z0=-1.96. There is not enough evidence to conclude

that the distributions differ.

15-32 Let 0 be the data below 2000 psi and 1 be the data above 2000 psi

1 0 1 1 1 0 0 1 1 1 1 1 0 1 0 1 1 1 1 0 The total number of runs is r = 10.

3317.0272.3

4.910

272.3)120(20

)20)14)(6(2(1462

)1(

)2(2

4.920

20)14)(6(22

22

21212

21

=−

=−

=

=−

−×××

=−

−

=

=+

=+

=

R

R

R

R

r z

N N

N nnnn

N

N nn

σ

μ

σ

μ

At α = 0.05 reject H0 when 96.1≥ z . Because 3317.0= z < 1.96, we fail to reject H0

15-14




15-33 a)

1. Parameter of interest is median surface finish

0.10~:3

0.10~:.2

1

0

≠

=

μ

μ

H

H

4. α = 0.055. The test statistic is the observed number of plus differences or r + = 5.

6. We reject H 0 if the P-value corresponding to r + = 5 is less than or equal to α = 0.05.

7. Using the binomial distribution with n = 10 and p = 0.5, P-value = 2P( R * ≥ 5| p = 0.5) =1.0

8. Conclusion, we cannot reject H 0. We cannot reject the claim that the median is 10 μin.

b)

1. Parameter of interest is median of surface finish

0.10~:3

0.10~:.2

1

0

≠

=

μ

μ

H

H

4. α=0.05


n

nr z

5.0

5.00

−=

+

6. We reject H 0 if the | Z 0| > Z 0.025 = 1.967. Computation: 0

105.0

)10(5.050 =

−= z

8. Conclusion, cannot reject H 0. There is not sufficient evidence that the median surface finish differs from

10 microinches.

The P-value = 2[1-Φ(0)] = 1

15-34 a)

1. The parameter of interest is the median fluoride emissions.

6~:.3

6~:.2

1

0

<

=

μ

μ

H

H

4. α=0.05


= 4.6. We reject H 0 if the P-value corresponding to r + = 4 is less than or equal to α = 0.05.

7. Using the binomial distribution with n = 15 and p = 0.5, P-value = P(R + ≤ 4 | p = 0.5) = 0.1334

8. Conclusion, do not reject H 0. The data does not support the claim that the median fluoride impurity level

is less than 6.

Using Minitab (Sign Rank Test)Sign test of median = 6.000 versus < 6.000

N Below Equal Above P Mediany 15 9 2 4 0.1334 4.000

Do not reject H0

b)

1. Parameter of interest is median fluoride impurity level

0.6~:3

0.6~:.2

1

0

<

=

μ

μ

H

H

4. α=0.05


n

nr z

5.0

5.00

−=

+

6. We reject H 0 if the Z 0 < -Z 0.05 = -1.65


)13(5.040 −=

−= z

15-15



8. Conclusion, do not reject H 0. There is not enough evidence to conclude that the median fluoride impurity

level is less than 6.0 ppm. The P-value = 2[1-Φ(1.39)] = 0.1645.

15-35 1. Parameters of interest are the mean weights

0:.3

0:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α=0.055. The test statistic is w = min( w +, w -).

6. We reject H 0 if w ≤ = 8 because α = 0.05 and n = 10 Appendix A, Table IX gives the critical value.*

05.0w

7. w+

= 55 and w -= 0 and w = min(55, 0) = 0

8. Conclusion, because 0 < 8, we reject H 0 and conclude that there is a difference in the weights due to the

diet modification experiment.

15-36 1. Parameter of interest is the difference between the impurity levels.

0~:.3

0~:.2

1

0

≠

=

D

D

H

H

μ

μ

4. α = 0.05

5. The test statistic r = min(r +, r -).

6. Because α = 0.05 and n = 7, Appendix A, Table VIII gives the critical value of = 0. We reject*05.0r


7. r + = 1 and r - = 6 and r = min(1,6) = 1

8. Conclusion, cannot reject H 0. There is not a significant difference in the impurity levels.

15-37 a)

1. The parameters of interest are the mean mileages for a Volkswagen and Mercedes.

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α = 0.01


21212

2

)1)((w

nnnnw −

+++=


01.0w 1 = 10 and n2 = 10, Appendix A, Table X gives the

critical value.

7. w1 = 55 and w2 = 155 and because 55 is less than 71, we reject H 08. Conclusion, reject H0 and conclude that there is a difference in the mean mileages of the two different

vehicles.

b)

1. The parameters of interest are the mean mileages for a Volkswagen and Mercedes

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α = 0.01


1

11

0

w

wW

z

σ

μ −=

6. We reject H 0 if | Z 0| > Z 0.005 = 2.58

7. w1 = 55,1w

μ =105 and =1752

1wσ

78.323.13

105550 −=

−= z

Because | Z 0| > 2.58, reject H 08. Conclusion, reject H 0 and conclude that there is a significant difference in the mean mileages at α = 0.01.

P-value =2[1 - P(Z < 3.78 )] = 0.00016

15-16



15-38 a)

1. The parameters of interest are the in mean fill volumes

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α=0.05

5. The test statistic1

21212

2

)1)((w

nnnnw −

+++=

6. We reject if H 0 w ≤ = 78, because α = 0.05 and n*


the critical value.

7. w1 = 58 and w2 = 152 and because 58 is less than 78, we reject H 08. Conclusion, reject H0 and conclude that there is a significant difference in the mean fill volumes at α =

0.05.

b)

1. The parameters of interest are the mean mileages for a Volkswagen and Mercedes

211

210

:.3

:.2

μ μ

μ μ

≠

=

H

H

4. α=0.05


1

11

0

w

wW

z σ

μ −

=

6. We reject H 0 if | Z 0| > Z 0.025 = 1.96

7. w1 = 58,1w

μ = 105 and = 1752

1wσ

55.323.13

105580 −=

−= z

Because | Z 0| > 1.96, reject H 08. Conclusion, reject H 0 and conclude that there is a significant difference in the mean mileages at α = 0.05.

P-value =2[1 - P(Z < 3.55 )] ≅ 0.000385

15-39 Kruskal-Wallis Test on RESISTAN ALLOY N Median Ave Rank Z1 10 98.00 5.7 -4.312 10 102.50 15.3 -0.09

3 10 138.50 25.5 4.40Overall 30 15.5

H = 25.30 DF = 2 P = 0.000H = 25.45 DF = 2 P = 0.000 (adjusted for ties)

Reject H0, P-value ≅ 0

15-40 Kruskal-Wallis Test on TIME

time N Median Ave Rank Z

0 9 1106.0 24.8 2.06

7 9 672.2 19.7 0.38

14 9 646.2 20.9 0.79

21 9 377.6 8.7 -3.23

Overall 36 18.5

H = 11.61 DF = 3 P = 0.009

Reject H0 at α = 0.05

15-41 Kruskal-Wallis Test on VOLUMETEMPERAT N Median Ave Rank Z70 5 1245 12.4 2.6975 5 1220 7.9 -0.0680 5 1170 3.7 -2.63Overall 15 8.0H = 9.46 DF = 2 P = 0.009

15-17




Reject H0 at α = 0.05

15-42 Kruskal-Wallis Test on C4

C5 N Median Ave Rank Z1 10 68.55 13.2 -1.01

2 10 379.20 20.4 2.16

3 10 149.75 12.9 -1.14

Overall 30 15.5

H = 4.65 DF = 2 P = 0.098

Since P=0.098<0.1, we would reject the null hypothesis and conclude that the treatments

differ.

15-18




16-81 a) According to Table 16-10, if there is no shift, ARL=500.

If the shift in mean is 1 X

σ , ARL=17.5.

b) According to Table 16-10, if there is no shift, ARL=500.

If the shift in mean is 2 X

σ , ARL=3.63.

16-82 a) I-MR chart for Refining percentage

Observat ion

I n d i v i d u a l V a l u e

635649423528211471

30

20

10

_

X=15.65

UC L=25.26

LCL=6.04

Observat ion

M

o v i n g

R a n g e

635649423528211471

15

10

5

0

__

MR=3.61

UC L=11.81

LCL=0

1

1

1

11

I -MR Chart of r efine

I-MR chart for Dist& Marketing percentage

Observat ion


635649423528211471

25

20

15

10

5

_ X=12.45

UC L=23.34

LCL=1.56

Observat ion

M

o v i n g

R a n g e

635649423528211471

16

12

8

4

0

__

MR=4.09

UC L=13.38

LCL=0

1

1

I -MR Chart of distr ibuti on

16-58



b) I-MR chart for crude oil percentage of retail price. It appears there have been a couple of shifts

in the process around observations 16 and 56.

Observat ion

I n d i v i d u a l V a

l u e

635649423528211471

55

50

45

40

35

_

X=44.63

UC L=50.45

LCL=38.82

Observat ion

M

o v i n g

R a n g e

635649423528211471

8

6

4

2

0

__ MR=2.185

UC L=7.139

LCL=0

11

1

11

1

1

1

1

111

1111

1

1

I -MR Chart of cr ude

c)

Observat ion


635649423528211471

60

40

20

_ X=25.41

UC L=41.85

LCL=8.97

Observat ion

M

o v i n g

R a n g e

635649423528211471

24

18

12

6

0

__ MR=6.18

UC L=20.20

LCL=0

1

1

1

1

1

1

11

I -MR Chart of ref ine_v

16-59



Observat ion


635649423528211471

40

30

20

10

0

_ X=19.56

UC L=37.02

LCL=2.09

Observat ion

M

o v i n g

R a n g e

635649423528211471

24

18

12

6

0

__ MR=6.57

UC L=21.46

LCL=0

11

1

I -MR Chart of distr ibuti on_v

16-83 Let p denote the probability that a point plots outside of the control limits when the mean has shifted from μ0 to

μ = μ0 + 1.5σ. Then:

( )

( )5.0)06(

4when5.135.13

3/

5.1

/3

/

5.1

5.1|33

5.1|1

000

0

=<<−=

=−+<<−−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

−<

−<−

−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=+<<−=

+=<<=−

Z P

nn Z nP

nn

X

nP

n X

nP

UCL X LCLP p

σ

σ

σ

μ

σ

σ

σ μ μ σ

μ σ

μ

σ μ μ

Therefore, the probability the shift is undetected for three consecutive samples is (1 - p)3 = 0.53 =0. 125.

If 2-sigma control limits were used, then

( )

1587.001587.0)15(

4when2/

5.1

/2

/

5.1

5.1|225.1|1 0000

=−=−<<−=

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

−<

−<−

−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=+<<−=+=<<=−

Z P

nnn

X

nP

n X

nPUCL X LCLP p

σ

σ

σ

μ

σ

σ

σ μ μ σ

μ σ

μ σ μ μ

Therefore, the probability the shift is undetected for three consecutive samples is (1 - p)3 = 0.15873 = 0.004.

16-84

nk UCL

CLnk LCL

/

/

0

0

0

σ μ

μ σ μ

+=

= −=

a)

16-60



( )

]()([1

)(1

///11

|1

|11

000

0

nk nk

nk Z nk P

nk

n

X

nk P

nk X

nk P

UCL X LCLP p

δ δ

δ δ

σ

δσ

σ

μ

σ

δσ

δσ μ μ σ

μ σ

μ

δσ μ μ

−Φ−−Φ−=

−<<−−−=⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −<

−<−−−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=+<<−−=

+=<<−=−

where Φ(Z) is the standard normal cumulative distribution function.

16-85

nk UCL

CL

nk LCL

/

/

0

0

0

σ μ

μ

σ μ

+=

=

−=

a) ARL = 1/ p where p is the probability a point plots outside of the control limits. Then,

( )

1)(2)()()(

|/

|1 00

0

−Φ=−Φ−Φ=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ <

−<−=<<=−

k k k k Z k P

k n

X k PUCL X LCLP p μ

σ

μ μ

where Φ(Z) is the standard normal cumulative distribution function. Therefore, p = 2 − 2Φ(k) and ARL =

1/[2−2Φ(k)]. The mean time until a false alarm is 1/ p hours.

b) ARL = 1/ p where

( )

( )

)()(1

)()(

//|1 01

δ δ

δ δ

δ δ

σ

δσ

σ

δσ δσ μ μ

nk nk p

and

nk nk

nk Z nk P

nk X

nk PUCL X LCLP p

−−Φ+−Φ−=

−−Φ−−Φ=−<<−−=

⎟ ⎠

⎞⎜⎝

⎛ −<<−−=+=<<=−

c) ARL = 1/ p where 1 - p = P(-3<Z<3)= 0.9973. Thus, ARL = 1/0.0027 = 370.4.

If k = 2, 1 - p = P(-2<Z<2) = 0.9545 and ARL = 1/ p = 22.0.

The 2-sigma limits result in a false alarm for every 22 points on the average. This is a high number of false

alarms for the routine use of a control chart.

d) From part (b), ARL = 1/ p, where 1 - p = P Z( )− − < < − =3 5 3 5 0 7764. . ARL = 4.47 assuming 3-sigma control limits.

16-86 Determine n such that

16-61



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−−

−

−<<

−

−−

−

−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=−

−−

+<

−

−<

−

−−

−=

=<<=

)1(

)1(

)1()1(

)1(

)1(

|)1(

)1(

)1()1(

)1(

)|ˆ(5.0

cccc

c

cccc

c

c

cc

c

cc

c

cc

c

c

p p

p pk

n

p p

p p Z

p p

p pk

n

p p

p pP

p p

n

p p

pn

p pk p

n

p p

pP

n

p p

pn

p pk p

P

p pUCLP LCLP

Use the fact that if pc > p then the probability is approximately equal to the probability that Z is less than the

upper limit.

⎟

⎟⎟⎟

⎠

⎞

⎜

⎜⎜⎜

⎝

⎛

−

−−

−

−<≅

)1(

)1(

)1(cccc

c

p p

p pk

n

p p

p p Z P

Then, the probability above approximately equals 0.5 if

)1(

)1(

)1( cccc

c

p p

p pk

n

p p

p p

−

−=

−

−

Solving for n,2

2

)(

)1(

c p p

p pk n

−−

=

16-87 The LCL is

n

p pk p

)1( −−

n

p pk p

)1( −− = 0 or

p

pk n

)1(2 −=

16-88 The P is desired. Now, using the normal approximation:LCL P UCL p( | .< < = 0 08)

90.0)29.1(

100

)08.01(08.0

08.0115.0

100

)08.01(08.0

08.0ˆ

)08.0|115.0ˆ()08.0|115.0ˆ0(

115.0100

)95.0(05.0305.0

)1(3

0015.0100

)95.0(05.0305.0

)1(3

=<=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−<

−

−=

=<==<<

=+=−

+=

→−=−=−

−=

Z PP

P

pPP pPP

n

p p pUCL

n

p p p LCL

16-62



Therefore, the probability of detecting shift on the first sample following the shift is 1 − 0.90 = 0.10.

The probability of detecting a shift by at least the third sample following the shift can be determined from the

geometric distribution to be 0.10 + 0.90(0.10) + 0.902(0.10) = 0.27

16-89 The process should be centered at the middle of the specifications; that is, at 100.

For an x chart:

84.0)15(

2/5

1055.107

2/5

105

2/5

1055.92)105|(

5.1072/)5(3100/

5.922/)5(3100/

100

5.1072/)5(3100/

0

0

0

0

=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−==<<

=+=+=

=−=−=

===+=−=

Z P

X PUCL X LCLP

nk UCL

nk LCL

CL

nk UCL

μ

σ μ

σ μ

μ

σ μ

The requested probability is then 1 − 0.84 = 0.16. The ARL = 1/0.16 = 6.25. With μ = 105, the

specifications at 100 ± 15 and σ = 5, the probability of a defective item is

0228.0)2()4(

5105115

5105

510585

5105)115()85(

=>+−<=

⎟ ⎠ ⎞⎜

⎝ ⎛ −>−+⎟

⎠ ⎞⎜

⎝ ⎛ −<−=>+<

Z P Z P

X P X P X P X P

Therefore, the average number of observations until a defective occurs, follows from the geometric

distribution to be 1/0.0228 = 43.86. However, the x chart only requires 6.25 samples of 4 observations each

= 6.25(4) = 25 observations, on average, to detect the shift.

16-90 Let X denote the number of defectives in a sample of n. Then X has a binomial distribution with E( X ) = np

and V( X ) = np(1 - p). Therefore, the estimates of the mean and standard deviation of X are np and

)1( p pn − , respectively. Using these estimates results in the given control limits.

b) Data from example 16-4

20100

55

45

35

25

Sample Number

S a m p l e C o u n t

NP Chart for defectiv

NP=40.00

3.0SL=54.70

-3.0SL=25.30

c) Note that

n

p p pP

n

p p p

p pn pnPn p pn pn

)1(3ˆ)1(

3

)1(3ˆ)1(3

−+<<

−−

−+<<−−

16-63



Therefore, the np control chart always provides results equivalent to the p chart.

16-91 a)

The center line 8C =

The UCL=16.49

The LCL=0

Sample


2018161412108642

18

16

14

12

10

8

6

4

2

0

_ C=8

UCL=16.49

LCL=0

C Chart of Number of defects

b)Yes.

16-92 Because −3 < Z < 3 if and only if

n

p p pP

n

p p por

n

p p

pPi

)1(3ˆ)1(

33)1(

ˆ3

−+<<

−−<

−

−<−

a point is in control on this chart if and only if the point is in control on the original p chart.

16-93 For unequal sample sizes, the p control chart can be used with the value of n equal to the size of each sample.

That is,

i

i

n

p p

pP Z

)1(

ˆ

−

−= where ni is the size of the ith sample.

109876543210

3

2

1

0

-1

-2

-3

Observation Number

z i

Standardized Chart for P

X=0.000

3.0SL=3.0003

-3.0SL=-3.000-3

16-64



16-65



CHAPTER 16

Note to the Instructor: Many of the control charts and control chart summaries were created using

Statgraphics. Minitab can be used to provide similar results.

Section 16-5

16-1 a) 286.3435

120022335

7805 ==== r x

0)286.34(0

286.34

51.72)286.34(115.2

22.203)286.34(577.0223

223

78.242)286.34(577.0223

3

4

2

2

===

=

===

=−=−=

=

=+=+=

r D LCL

CL

r DUCL

chart R

r ACL LCL

CL

r ACLUCL

chart x

b)

74.14326.2

286.34ˆ

223ˆ

2

===

==

d

r

x

σ

μ

16-2 a) 1456.025

64.3344.0

25

60.8510.14

25

75.362====== sr x

0)344.0(0

344.0728.0)344.0(115.2

312.14)344.0(577.0510.14

510.14

708.14)344.0(577.0510.14

3

4

2

2

===

====

=−=−=

=

=+=+=

r D LCL

CLr DUCL

chart R

r ACL LCL

CL

r ACLUCL

chart x

16-1



b) c4 = 0.94

00129.094.0194.0

1456.031456.013

1456.0

3041.094.0194.0

1456.031456.013

417.142594.0

1456.03510.143

510.14

603.142594.0

1456.03510.143

224

4

22

4

4

4

4

→−=−⎟ ⎠ ⎞⎜⎝ ⎛ −=−−=

=

=−⎟ ⎠

⎞⎜⎝

⎛ +=−+=

=−=−=

=

=+=+=

cc

ss LCL

CL

cc

ssUCL

chart S

nc

sCL LCL

CL

nc

sCLUCL

chart x

16-3 a) 58.1320

6.271223

20

4460==== s x

061.31512.09213.0

58.13358.1313

58.13

77.301512.09213.0

58.13358.1313

89.200

209213.0

58.1332233

223

11.24549213.0

58.1332233

2

4

4

2

4

4

4

4

→−=⎟ ⎠

⎞⎜⎝

⎛ −=−−=

=

=⎟ ⎠

⎞⎜⎝

⎛ +=−+=

=⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=−=

=

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=+=

cc

ss LCL

CL

cc

ssUCL

chart S

nc

sCL LCL

CL

nc

sCLUCL

chart x

b) Process mean and standard deviation

74.149213.0

58.13

ˆ223ˆ4 ===== c

s

x σ μ

16-2



16-4 a) 5476.3)534.2(4.1534.24.10.20 2

2

===== r d d

r x

0)5476.3(0

5476.3

11.7)53476.3(004.2

29.18)5476.3(483.00.20

0.20

71.21)5476.3(483.00.20

3

4

2

2

===

=

===

=−=−=

=

=+=+=

r D LCL

CL

r DUCL

chart R

r ACL LCL

CL

r ACLUCL

chart x

043.00946.0)5.1(3427.113

427.1

811.20946.0)5.1(3427.113

427.1so9515.0

16.186/)5.1(30.203

0.20

84.216/)5.1(30.203

5.1/where

2

4

4

2

4

4

4

4

4

4

=−=−−=

=

=+=−+=

==

=−=−=

=

=+=+=

=

cc

sCL LCL

CL

c

c

sCLUCL

scchart S

nc

sCL LCL

CL

nc

sCLUCL

cschart x

16-5 a)X-bar and Range - Initial Study

Problem 16-1

X-bar | Range

----- | -----

UCL: + 3.0 sigma = 37.5789 | UCL: + 3.0 sigma = 11.9461

Centerline = 34.32 | Centerline = 5.65

LCL: - 3.0 sigma = 31.0611 | LCL: - 3.0 sigma = 0

|

out of limits = 1 | out of limits = 0

--------------------------------------------------------------------------------

Chart: Both Normalize: No

20 subgroups, size 5 0 subgroups excluded

Estimated

process mean = 34.32

process sigma = 2.42906

mean Range = 5.65

16-3



0 4 8 12 16 2031

33

35

37

39

34.32

37.5789

31.0611

0 4 8 12 16 2002468

1012

5.65

11.9461

0

Charting xbar

X-bar

Problem 16-5

subgroup

Range

b)

X-bar | Range

----- | -----




|


--------------------------------------------------------------------------------


Estimated



mean Range = 5.73684

0 4 8 12 16 2030

32

34

36

38

40

34.0947

37.4038

30.7857

0 4 8 12 16 200

3

6

9

12

15

5.73684

12.1297

0

X-bar

Problem 16-5

subgroup

Range

16-4




Charting Problem 16-8

X-bar | Range

----- | -----




|


Estimated



mean Range = 1.1328

0 5 10 15 20 25

5.15.55.96.36.77.17.5

6.2836

7.44253

5.12467

0 5 10 15 20 250

0.51

1.52

2.53

1.1328

2.9153

0

X-bar

Problem 16-6

subgroup

Range

There are no points beyond the control limits. The process appears to be in control.

b) No points fell beyond the control limits. The limits do not need to be revised.

16-7 a) X-bar and Range - Initial Study


X-bar | Range

----- | -----




|

Test Results: X-bar One point more than 3.00 sigmas from center line.

Test Failed at points: 4 6 7 10 12 15 16 20

Test Results for R Chart: One point more than 3.00 sigmas from center line.

16-5



Test Failed at points: 19

2010Subgroup 0

20

15

10 S a m p l e

M e a n

1

1

1

1

1

1

11

Mean=15.09

UCL=17.40

LCL=12.79

.

b) Removed points 4, 6, 7, 10, 12, 15, 16, 19, and 20 and revised the control limits. The control limits are not

as wide after being revised: X-bar UCL=17.96, CL=15.78, LCL=13.62 and R UCL = 5.453, R-bar = 2.118,

LCL=0.

c) X-bar and StDev - Initial Study


X-bar | StDev

----- | -----


Centerline = 15.09 | Centerline = 1.188LCL: - 3.0 sigma = 12.77 | LCL: - 3.0 sigma = 0

|

Test Results: X-bar One point more than 3.00 sigmas from center line.

Test Failed at points: 4 6 7 10 12 15 16 20

Test Results for S Chart:One point more than 3.00 sigmas from center line.

Test Failed at points: 19

.

6

5

4

3

2

1

0

1

S a m p l e R a n g e

R=2.25

UCL=5.792

LCL=0

Xbar/R Chart for x1-x3

105Subgroup 0

18

17

16

15

14

13

S a m p l e M e a n

Mean=15.78

UCL=17.95

LCL=13.62

6

5

4

3

2

1

0 S a m p l e R a n g e

R=2.118

UCL=5.453

LCL=0

Revised Control Limits

Xbar/R Chart for x1-x3

16-6



0Subgroup 10 20

10

15

20

S a m p

l e M e a n

1

1

1

1

1

1

11

Mean=15.09

UCL=17.42

LCL=12.77

1

Removed points 4, 6, 7, 10, 12, 15, 16, 19, and 20 and revised the control limits. The control limits

are not as wide after being revised: X-bar UCL = 17.95, CL =1 5.78, LCL=13.62 and S UCL =

2.848, S-bar=1.109, LCL=0.

0

1

2

3

S a m p l e S t D e v

S=1.188

UCL=3.051

LCL=0

Xbar/S Chart for x1-x3

105Subgroup 0

18

17

16

15

14

13

S a m p l e M e a n

Mean=15.78

UCL=17.95

LCL=13.62

3

2

1

0 S a m p l e S t D e v

S=1.109

UCL=2.848

LCL=0

Revised Control Limits

Xbar/S Chart for x1-x3

16-7



16-8

a) The average range is used to estimate the standard deviation. Sample 2, 9, and 17 are out-of-control

Sample

S a m

p l e

M

e a n

2018161412108642

4.98

4.97

4.96

4.95

4.94

_ _ X=4.96065

UC L=4.97331

LCL=4.94799

Sample

S a

m

p l e

R a n g e

2018161412108642

0.060

0.045

0.030

0.015

0.000

_ R=0.02195

UC L=0.04641

LCL=0

1

1

1

Xbar-R Chart of x1, ..., x5

b)

Sample

S a

m

p l e

M

e a n

161412108642

4.970

4.965

4.960

4.955

4.950

_ _ X=4.96124

UC L=4.97284

LCL=4.94963

Sample

S a m

p l e

R a n g e

161412108642

0.04

0.03

0.02

0.01

0.00

_ R=0.02012

UC L=0.04254

LCL=0

Xbar-R Chart of x1_ 1, ..., x5_1

16-8



c) For X -S chart

Sample

S a m

p l e

M e a n

2018161412108642

4.98

4.97

4.96

4.95

4.94

_ _ X=4.96065

UC L=4.97333

LCL=4.94797

Sample

S a m

p l e

S t D e v

2018161412108642

0.020

0.015

0.010

0.005

0.000

_ S=0.00888

UC L=0.01855

LCL=0

1

1

1

Xbar-S Chart of x 1, ..., x5

Sample

S a m

p l e

M

e a n

161412108642

4.970

4.965

4.960

4.955

4.950

_ _ X=4.96124

UC L=4.97305

LCL=4.94942

Sample

S a m

p l e

S t D e v

161412108642

0.020

0.015

0.010

0.005

0.000

_ S=0.00828

UC L=0.01730

LCL=0

1

Xbar-S Chart of x1_ 1, ..., x5_ 1

16-9 a)

The control limits for the following chart were obtained from an estimate of σ obtained from the pooled standard deviation from each subgroup of size 5.

16-9



Sample

S a m

p l e

M

e a n

24222018161412108642

0.0640

0.0635

0.0630

0.0625

0.0620

_ _ X=0.062938

UC L=0.063505

LCL=0.062370

Sample

S a m

p l e

R a n g e

24222018161412108642

0.0024

0.0018

0.0012

0.0006

0.0000

_ R=0.000984

UC L=0.002080

LCL=0

1

11

1

1

1

1

Xbar-R Chart of x1 , ..., x5

b) The test failed at points 1, 6, 14, 15, 17, 21 and 22. After eliminating these points, the test failed

at points 4, 6, 9 and 15 (new sample numbers based on remaining samples). After eliminating

these points, the test failed at point 4. After eliminating this point, the points are in control:

Sample

S a m

p l e M e a n

13121110987654321

0.0634

0.0632

0.0630

0.0628

0.0626

_ _ X=0.0629877

UCL=0.0633116

LCL=0.0626638

Sample

S a m

p l e R a n g e

13121110987654321

0.0012

0.0009

0.0006

0.0003

0.0000

_ R=0.000562

UCL=0.001187

LCL=0

Xbar-R Chart of C6

c) The test failed at points 1, 6, 14, 15, 17, 21, 22.

16-10



Sample

S a m p l e M e a n

252321191715131197531

0.0640

0.0635

0.0630

0.0625

0.0620

_ _ X=0.062938

UCL=0.063461

LCL=0.062414

Sample

S a m p l e S t D e v

252321191715131197531

0.00100

0.00075

0.00050

0.00025

0.00000

_ S=0.000367

UCL=0.000766

LCL=0

1

11

1

1

1

1

Xbar-S Chart of C6

Removal of these points left points 4, 6, 9 and 15 out of control (newly labeled).

Removal of these points left point 4 out of control. After the removal of this point, the

remaining points were in control.

Sample

S a m

p l e M e a n

13121110987654321

0.0634

0.0632

0.0630

0.0628

0.0626

_ _ X=0.0629877

UCL=0.0633197

LCL=0.0626556

Sample

S a m

p l e S t D e v

13121110987654321

0.00048

0.00036

0.00024

0.00012

0.00000

_ S=0.0002326

UCL=0.0004860

LCL=0

Xbar-S Chart of C6

16-10 It violates the rule of two out of three consecutive points plot beyond a 2-sigma limit.

16-11 The sample standard deviation is 2.956. It is larger than the estimate obtained from the revised X and R

chart, because the mean of the process has been shifted in the original data.

16-12 The sample standard deviation is 0.0127. It is larger than the estimate 0.010301 obtained from the revised

X and R chart, because the mean of the process has been shifted in the original data.

Section 16-6

16-11



Section 16-6

16-11



16-13 a) Individuals and MR(2) - Initial Study

------------------------------------------------------------------------------


Ind.x | MR(2)

----- | -----




|


------------------------------------------------------------------------------



Estimated



mean MR = 2.94737


0Subgroup 10 20

45

50

55

60


Mean=53.05

UCL=60.89

LCL=45.21

0

5

10

M o v i n g R a n g e

R=2.947

UCL=9.630

LCL=0

I and MR Chart for hardness

b) Estimated process mean and standard deviation

613.2128.1

94737.2ˆ05.53ˆ2

=====d

mr x σ μ

16-14 a) The process appears to be in statistical control. There are no points beyond the control limits.

16-12



302010Subgroup 0

20

19

18

17

16

1514

13

12


Mean=15.99

UCL=19.15

LCL=12.83

4

3

2

1

0 M o v i n g R a n g e

R=1.190

UCL=3.887

LCL=0

I and MR Chart for wafer


05.1128.1

190.1ˆ99.15ˆ

2

=====d

mr x σ μ

16-15

a) Ind.x and MR(2) - Initial Study------------------------------------------------------------------------------

Charting diameter

Ind.x | MR(2)

----- | -----



|


------------------------------------------------------------------------------



Estimated



mean MR(2) = 0.19125

16-13



0Subgroup 5 10 15 20 25

9.5

10.0

10.5


Mean=10.03

UCL=10.54

LCL=9.519

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


R=0.1912

UCL=0.6249

LCL=0

I and MR Chart for Dia



16955.0128.1

19125.0ˆ0272.10ˆ

2

=====d

mr x σ μ

16-16 a) Ind.x and MR(2) - Initial Study

--------------------------------------------------------------------------------Charting Problem 16-16

Ind.x | MR(2)

----- | -----




|


--------------------------------------------------------------------------------



Estimated



mean MR(2) = 19.3684

16-14



0Subgroup 10 20

450

500

550


Mean=500.6

UCL=552.1

LCL=449.1

0

10

20

30

40

50

60

70


R=19.37

UCL=63.28

LCL=0

I and MR Chart for viscosity


17.17128.1

3684.19ˆ6.500ˆ

2

=====d

mr x σ μ

16-17

a)

Observat ion

I n d i v i d u a

l V a l u e

24222018161412108642

140

120

100

80

_ X=100.78

UC L=130.50

LCL=71.06

Observat ion

M

o v i n g

R a n g e

24222018161412108642

40

30

20

10

0

__ MR=11.18

UC L=36.51

LCL=0

1

I -MR Chart of SI ZE

Remove the out-of-control observation:

16-15



Observa t ion


24222018161412108642

120

100

80

_ X=99.48

UC L=127.08

LCL=71.88

Observa t ion

M

o v i n g

R a n g e

24222018161412108642

30

20

10

0

__ MR=10.38

UC L=33.91

LCL=0

I -MR Chart of SI ZE

b) The estimated mean is 99.4792 and estimated standard deviation is 9.20059.



Section 16-7

16-18. a)

The natural tolerance limits are100 18± .

401.1111

6 36

USL LSLPCR

σ

−= = =

Since the mean of the process is centered at the nominal dimension,

1.1111k

PCR PCR= =Since the process natural tolerance limits lie inside the specifications, very few defective units will be

produced.

The fraction defective is 2 ( 20 /6) 0.0858%Φ − =

b)

The natural tolerance limits are106 18± .

401.1111

6 36

USL LSLPCR

σ

−= = =

Since the mean of the process is not centered at the nominal dimension,

14 26min , 0.7778

18 18k

PCR⎡ ⎤= =⎢ ⎥⎣ ⎦

The small PCRK indicates that the process is likely to produce units outside the specification limits.

The fraction defective is

( ) ( )( ) ( ) 14 / 6 26 / 6 0.0098+0=0.0098P X LSL P X USL P Z P Z < + > = < − + > =

16-19 a)12

1.56 6

USL LSLPCR

σ σ

−= = = so 1.3333σ =

b) (20+32)/2=26 When the process is centered at the nominal dimension, the fraction defective is minimized

for any σ .

16-16



16-20 a) The natural tolerance limits are 20 6± .

201.6667

6 12

USL LSLPCR

σ

−= = =

Since the mean of the process is centered at the nominal dimension,

1.6667k

PCR PCR= =

Since the process natural tolerance limits lie inside the specifications, very few defective units will be produced.

The fraction defective is 2 ( 10 / 2) 0Φ − =

b) The natural tolerance limits is 23 6± .

201.6667

6 12

USL LSLPCR

σ

−= = =

Since the mean of the process is not centered at the nominal dimension,

13 7min , 1.1667

6 6k

PCR⎡ ⎤= =⎢ ⎥⎣ ⎦

The fraction defective is

00023.000023.00)2/7()2/13()()( =+=>+−<=>+< Z P Z PUSL X P LSL X P

c) The measure of actual capability decreases and the fraction defective increases when the

process mean is shifted from the center of the specification limits.

16-21 a) If the process uses 66.7% of the specification band, then 6σ = 0.667(USL-LSL). Because the process is

centered

LSU USL

USL LSLUSL

−=−=

−=−=−=

μ μ σ

μ μ μ σ

5.4

)(667.0)(677.0)(667.03

5.13

5.4

,3

5.4

min =⎥⎦

⎤

⎢⎣

⎡

== σ

σ

σ

σ

K PCRPC

Because PCR and PCR k exceed unity, the natural tolerance limits are inside the specification limits and few

defective units should be produced.

b) Assuming a normal distribution with 6σ = 0.667(USL − LSL) and a centered process, then

3σ = 0.667(USL − μ). Consequently, USL − μ = 4.5σ and μ − LSL = 4.5σ

011

)5.4(1)5.4(5.4

)(

=−=

<−=>=⎟ ⎠

⎞⎜⎝

⎛ >=> Z P Z P Z PUSL X Pσ

σ

By symmetry, the fraction defective is 2[P(X > USL)] = 0.

16-22 a) If the process uses 85% of the spec band then 6σ = 0.85(USL − LSL) and

18.185.0

1

)(85.0==

−−

= LSLUSL

LSLUSLPCR

Then 3σ = 0.85(USL − μ) = 0.85(μ − LSL)

Therefore,

18.13

53.3,

3

53.3min =⎥⎦

⎤⎢⎣

⎡=σ

σ

σ

σ

k PCR

16-17





b) Assuming a normal distribution with 6σ = 0.85(USL − LSL) and a centered process, then

3σ = 0.85(USL − μ). Consequently, USL − μ = 3.5σ and μ − LSL = 3.5σ

000233.0999767.01

)5.3(1)5.3(5.3

)(

=−=

<−=>=⎟ ⎠

⎞⎜⎝

⎛ >=> Z P Z P Z PUSL X P

σ

σ

By symmetry, the fraction defective is 2[P(X > USL)] = 0.00046

16-23 Assume a normal distribution with = 223 and =μ σ =326.2

286.3414.74

( ) ( )

00604.099396.01)51.2(1

)51.2(74.14

223260

ˆ

ˆ

)(

00175.0

92.274.14

223180

ˆ

ˆ

=−=<−=>=⎟ ⎠

⎞

⎜⎝

⎛ −

>=⎟ ⎠

⎞

⎜⎝

⎛ −

>=>

=

−<=⎟ ⎠

⎞⎜⎝

⎛ −<=⎟

⎠

⎞⎜⎝

⎛ −<=<

Z P

Z P Z PUSL

Z PUSL X P

Z P Z P LSL

Z P LSL X P

σ

μ

σ

μ

Therefore, the proportion nonconforming is given by

P(X<LSL) + P(X>USL) = 0.00175 + 0.00604 = 0.00779

[ ]837.0

972.0,837.0min

)74.14(3

180223,)74.14(3

223260min

ˆ3,

ˆ3min

905.0)74.14(6

180260

)ˆ(6

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦

⎤⎢⎣

⎡ −−=

=−

=−

=

σ σ

σ

LSL x xUSLPCR

LSLUSLPCR

K

The process capability is marginal.

16-24. a) Assume a normal distribution with = 14.510 andμ .

..σ = = =

r

d2

0 344

23260148

( ) ( )

00028.0

45.3148.0

51.1400.14

ˆ

ˆ

=

−<=⎟ ⎠

⎞⎜⎝

⎛ −<=⎟

⎠

⎞⎜⎝

⎛ −<=< Z P Z P

LSL Z P LSL X P

σ

μ

00047.099953.01)31.3(1

)31.3(148.0

51.1400.15

ˆ

ˆ)(

=−=<−=

>=⎟ ⎠

⎞⎜⎝

⎛ −>=⎟

⎠

⎞⎜⎝

⎛ −>=>

Z P

Z P Z PUSL

Z PUSL X Pσ

μ


P(X<LSL) + P(X>USL) =0.00028 + 0.00047 = 0.00075

16-18



b)

[ ]104.1

15.1,104.1min

)148.0(3

00.1451.14,

)148.0(3

51.1400.15min

ˆ3,

ˆ3min

13.1)148.0(6

00.1400.15

)ˆ(6

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦

⎤⎢⎣

⎡ −−=

=−

=−

=

σ σ

σ

LSL x xUSLPCR

LSLUSLPCR

K



Because PCR K ≅ PCR the process appears to be centered.

16-25 a) Assume a normal distribution with = 223 andμ 74.14

9213.0

58.13ˆ

4

===c

sσ

( ) ( )

00016.0

60.374.14

223170

ˆ

ˆ

=

−<=⎟ ⎠

⎞⎜⎝

⎛ −<=⎟

⎠

⎞⎜⎝

⎛ −<=< Z P Z P

LSL Z P LSL X P

σ

μ

00074.099926.01)18.3(1

)18.3(75.14

223270

ˆ

ˆ)(

=−=<−=

>=⎟ ⎠

⎞⎜⎝

⎛ −>=⎟

⎠

⎞⎜⎝

⎛ −>=>

Z P

Z P Z PUSL

Z PUSL X Pσ

μ

Probability of producing a part outside the specification limits is 0.00016 + 0.00074 = 0.0009

[ ]06.1

19.1,06.1min

)75.14(3

170223,

)75.14(3

223270min

ˆ3,

ˆ3min

13.1)74.14(6

170270

)ˆ(6

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦

⎤⎢⎣

⎡ −−=

=−

=−

=

σ σ

σ

LSL x xUSLPCR

LSLUSL

PCR

K


defective units should be produced. The estimated proportion nonconforming is given by P(X < LSL) + P(X> USL) = 0.00016 + 0.00074 = 0.0009

16-19



16-26 Assuming a normal distribution with = 20.0 and = 1.4μ σ

[ ]19.1

19.1,19.1min

)4.1(3

1520,

)4.1(3

2025min

ˆ3

ˆ,

ˆ3

ˆmin

19.1)4.1(6

1525

)ˆ(6

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥

⎦

⎤⎢

⎣

⎡ −−=

=−

=−

=

σ

μ

σ

μ

σ

LSLUSLPCR

LSLUSLPCR

K

The process is capable.

16-27 a) 446.2326.2

737.5ˆ

2

===d

r σ 0002466.0ˆ =σ

[ ]217.1

217.1,489.1min

)0002466.0(3

5025.05034.0,

)0002466.0(3

5034.05045.0min

ˆ3,

ˆ3min

35.1)0002466.0(6

5025.05045.0)ˆ(6

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦

⎤⎢⎣

⎡ −−=

=−=−=

σ σ

σ

LSL x xUSLPCR

LSLUSLPCR

K

Because PCR and PCR k exceed unity, the natural tolerance limits are inside the specification limits and fewdefective units should be produced.

Because PCR K ≠PCR the process is slightly off center.

b) Assume a normal distribution with = 0.5034 and = 0.0002466μ σ

( ) ( )

011

)46.4(1)46.4(ˆ

ˆ)(

00013.065.3ˆ

ˆ

=−=

<−=>=⎟ ⎠

⎞⎜⎝

⎛ −>=>

=−<=⎟ ⎠

⎞⎜⎝

⎛ −<=<

Z P Z PUSL

Z PUSL X P

Z P LSL

Z P LSL X P

σ

μ

σ

μ


P(X<LSL) + P(X>USL) = 0.00013 + 0 = 0.00013

16-20



16-28 Assuming a normal distribution with = 6.284 and =μ σ693.1

1328.1= 0.669

[ ]357.0

640.0,357.0min

)669.0(3

5284.6,

)669.0(3

284.67min

ˆ3,

ˆ3min

50.0)669.0(6

57

)ˆ(6

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦⎤⎢

⎣⎡ −−=

=−

=−

=

σ σ

σ

LSL x xUSLPCR

LSLUSLPCR

K

The process capability is poor.

16-29 329.1693.1

25.2ˆ

2

===d

r σ

09.15= x

5120.00.05120.0

)96.5()053.0(

693.1

09.155

693.1

09.1515

)5()15(

=+=

−<+−>=

⎟ ⎠

⎞⎜⎝

⎛ −<+⎟

⎠

⎞⎜⎝

⎛ −>=

<+>

Z P Z P

Z P Z P

X P X P

985.0)693.1(6

515 =−=PCR

Because the estimated PCR is less than unity, the process capability is not good.

16-30 Assuming a normal distribution with = 500.6 and = 17.17μ σ

[ ]474.0

50.0,474.0min

)17.17(34756.500,

)17.17(36.500525min

ˆ3,

ˆ3min

49.0)17.17(6

475525

)ˆ(6

=

=

⎥⎦⎤⎢

⎣⎡ −−=

⎥⎦

⎤⎢⎣

⎡ −−=

=−

=−

=

σ σ

σ

LSL x xUSLPCR

LSLUSLPCR

K

Because the process capability ratios are less than unity, the process capability appears to be poor.

16-21



Section 16-8

16-31 a) This process is out of control statistically

Sample

P r o p o r t i o n

2018161412108642

0.14

0.12

0.10

0.08

0.06

0.04

0.02

_ P=0.0645

UCL=0.0835

LCL=0.0455

1

1111

111

1

1

1

1

1

1

1

1

P Char t of C1

b)

Sample

P r o p o r t i o n

2018161412108642

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

_ P=0.0645

UCL=0.1382

LCL=0

11

P Chart of C1

The process is still out of control, but not as many points fall outside of the control limits. The

control limits are wider for smaller values of n.Test Failed at points: 5 and 7.

16-22



The following charts eliminates points 5 and 7.

Sample

P r o p o r t i o n

18161412108642

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

_ P=0.0561

UCL=0.1252

LCL=0

P Chart of C2

c) The larger sample size leads to smaller standard deviation of proportions and thus narrower

control limits.

16-32 a)

0 1 0 2 0 3 0

0 .0

0 .1

0 .2

S a m p l e N u m b e r

P r o p o r t i o n

P C har t f o r x

P = 0 . 0 9 9 3 3

U C L = 0 . 1 8 9 1

L C L = 0 . 0 0 9 6 0 1

b) The process appears to be in statistical control.

16-33P Chart - Initial Study


P Chart

-----

UCL: + 3.0 sigma = 0.198553

Centerline = 0.150571

LCL: - 3.0 sigma = 0.10259

out of limits = 12

Estimated

mean P = 0.150571

16-23



sigma = 0.0159937

0 4 8 12 1 6 2 0 2 4

0

0.1

0.2

0.3

0.4

0.150571

0.198553

0.10259

subgroup

P

Problem 16-28

The samples with out-of-control points are 1, 2, 3, 6, 7, 8, 11, 12, 13, 15, 17, and 20. The control limits need

to be revised.

P Chart - Revised Limits


P Chart

-----

UCL: + 3.0 sigma = 0.206184

Centerline = 0.157333

LCL: - 3.0 sigma = 0.108482

out of limits = 0

Estimated

mean P = 0.157333

sigma = 0.0162837

0 4 8 12 16 20 24

0

0.1

0.2

0.3

0.4

0.157333

0.206184

0.108482

subgroup

P

Problem 16-28Revised Limits

There are no further points out of control for the revised limits.

16-34 The process does not appear to be in control.

35302520151050

0.02

0.01

0.00

Sample Number


U Chart for defects per 1000 ft

1

1

1

1

U=0.008143

UCL=0.01670

LCL=0

16-24



16-35 a)

2520151050

5

4

3

2

1

0

Sample Number


U Chart for defects

1

1

U=1.942

3.0SL=3.811

-3.0SL=0.07217

Samples 5 and 24 are points beyond the control limits. The limits need to be revised.

b)

0 10 20

0

1

2

3

4

Sample Number


U Chart for defects_

U=1.709

UCL=3.463

LCL=0

The control limits are calculated without the out-of-control points. There are no further points out of control

for the revised limits.

16-36 a)

16-25



Sample

S a m p l e C o u n t P e r U

n i t

1009080706050403020101

45

40

35

30

25

20

15

10

5

_ U=19.51

UCL=32.77

LCL=6.261

1

1

11

1

1

U Chart of Number of Eart hquakes

b) No. The process is out-of-control.



Section 16-9

16-37 a) (112-100)/3=4

b)

( )88 96 112 96

(88 112) 2 44 4

( 4) ( 2) 0.9772

X

X P X P P Z

P Z P Z

μ

σ

⎛ ⎞− − −< < = < < = − < <⎜ ⎟

⎝ ⎠= < − < − =

The probability of detecting is 1-0.9772=0.0228.

c) 1/0.0228=43.8596 ARL to detect the shift is about 44.

16-38 a) UCLn =+

σ μ 3

4)100106(3

2

1064

3100

=−=

=+

σ

σ

b) 96,22

4ˆˆ ==== μ

σ σ

n x

( )

8413.01587.01)1()5(

512

961062

9694)10694(

=−=−<−<=

<<−=⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ −<−<−=<<

Z P Z P

Z P X P X P xσ

μ

The probability that this shift will be detected on the next sample is p = 1−0.8413 = 0.1587.

c) 301.61587.0

11===

p ARL

16-26



16-39 a) x = 7401. σx= 0.0045 μ = 74.01

7823.0)11(7823.0)]22.5(1[)78.0(

)22.5()78.0()78.022.5(

0045.0

01.740135.74

ˆ0045.0

01.749865.73

)0135.749865.73(

=−−=<−−<=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−=

<<

Z P Z P

Z P Z P Z P

X P

X P

xσ

μ


b) 6.42177.0

11===

p ARL

16-40 a) 148.0326.2

344.0ˆ

2

===d

Rσ σx

= 066.05

148.0ˆ==

n

σ , μ = 14.6

94950.0)0(94950.0

)36.4()64.1()64.136.4(

066.0

6.14708.14

066.0

6.14312.14)708.14312.14(

=−=

−<−<=<<−=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

x

σ

μ


b) 8.190505.0

11===

p ARL

16-41 a) 74.14ˆ2

==d

Rσ σx

= .

.σ

n= =

1474

56592 , μ = 210

8485.01515.01

)03.1()97.4()97.403.1(

592.6

21078.242

592.6

21022.203)78.24222.203(

=−=

−<−<=<<−=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

x

σ

μ

The probability that this shift will be detected on the next sample is p = 1−.8485= 0.1515.

b) 6.61515.0

11===

p ARL

16-42 a) 4.1ˆ2

==d

Rσ σx

= 5715.06

4.1ˆ==

n

σ , μ = 17

0119.09881.01

)26.2()24.8()24.826.2(

5715.0

1771.21

5715.0

1729.18)71.2129.18(

=−=

<−<=<<=⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

xσ

μ


b) 012.19881.0

11===

p ARL

16-27



16-43 a) σx= 103.1

5

4664.2ˆ==

n

σ , μ = 36

8980.008980.0 )73.4()27.1()27.173.4(

103.1

36404.37

ˆ103.1

3678.30)404.3778.30(

=−=−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

xσ

μ


b) 8.9102.0

11===

p ARL

16-44 a) 329.1693.1

25.2ˆ

2

===d

Rσ σx

= 767.03

329.1ˆ==

n

σ , μ = 13

6064.03936.01

)27.0()74.5()74.527.0(

767.0

134.17

767.0

1379.12)4.1779.12(

=−=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

xσ

μ

The probability that this shift will be detected on the next sample is p = 1−06064 = 0.3936.

b) 54.23936.0

11===

p ARL

16-45 a) 000397.0

326.2

000924.0ˆ

2

===d

Rσ σx

= 000178.0

5

000397.0ˆ==

n

σ , μ = 0.0625

7123.02877.01

)56.0()62.5()62.556.0(

000178.0

0625.00635.0

000178.0

0625.00624.0)0635.00624.0(

=−=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

xσ

μ

The probability that this shift will be detected on the next sample is p = 1 − 0.7123 = 0.2877.

b) 48.32877.0

11===

p ARL

16-46 a) 669.0ˆ2

==d

Rσ σx

= 386.0

3

669.0ˆ==

n

σ , μ = 5.5

83397.016603.01

)97.0()03.5()03.597.0(

386.0

5.5443.7

0386

5.5125.5)5.5|443.7125.5(

=−=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−==<<

Z P Z P Z P

X P X P

xσ

μ μ


16-28



b) 02.616603.0

11===

p ARL



Section 16-10

16-47

a) Yes, this process is in-control. CUSUM chart with h = 4 and k = 0.5 is shown.

Sample

C u m u l a t i v e S u m

151413121110987654321

4

3

2

1

0

-1

-2

-3

-4

0

UCL=3.875

LCL=-3.875

CUSUM Chart of v iscosit y

b) Yes, this process has shifted out-of-control. For the CUSUM estimated from all the data

observation 20 exceeds the upper limit.

16-29



Sample


2018161412108642

7.5

5.0

2.5

0.0

-2.5

-5.0

0

UCL=3.88

LCL=-3.88

CUSUM Char t of v iscosi ty

For a CUSUM with standard deviation estimated from the moving range of the first 15 observation, the

moving range is 1.026 and the standard deviation estimate is 0.9096. If this standard deviation is used with a

target of 14.1, the following CUSUM is obtained.

Sample

C u m u l a t i v e

S u m

191715131197531

7.5

5.0

2.5

0.0

-2.5

-5.0

0

UCL=3.64

LCL=-3.64

CUSUM Char t of C1

16-48 a) CUSUM Control chart with k = 0.5 and h = 4

16-30



3

2

1

0

-1

-2

-3

3.2

-3.2

20100

Subgroup Number


Upper CUSUM

Lower CUSUM

CUSU M C hart for Purity

The CUSUM control chart for purity does not indicate an out-of-control situation. The SH values do not plot

beyond the values of − H and H .

b) CUSUM Control chart with k = 0.5 and h = 4

3

2

1

0

-1

-2

-3

3.2

-3.2

2520151050

Subgroup Number


Upper CUSUM

Lower CUSUM

CUSUM Chart for New Purity Data

The process appears to be moving out of statistical control.

16-49a) σ ˆ = 0.1736

b) CUSUM Control chart with k = 0.5 and h = 4

16-31



-0.5

0.0

0.5

-6.8E-01

0.678191

0 5 10 15 20 25

Subgroup Number


Upper CUSUM

Lower CUSUM

CUSUM Chart for Diameter

The process appears to be out of control at the specified target level.

16-50 a) CUSUM Control chart with k = 0.5 and h = 4

-30

-20

-10

0

10

20

30

-32

32

0 10 20

Subgroup Number


Upper CUSUM

Lower CUSUM

CUSUM Chart for Concentration

The process appears to be in statistical control.

b) With the target = 100 a shift to 104 is a shift of 104 – 100 = 4 = 0.5σ. From Table 16-9 with h = 4 and a

shift of 0.5, ARL = 26.6

16-51 a) A shift to 51 is a shift of μ μ

σ

−=

−0 51 50

2= 0.5 standard deviations. From Table 16-9, ARL = 38.0

b) If n = 4, the shift to 51 is a shift of

μ μ

σ

−

=

−0 51 50

2 4 / / n = 1 standard deviation. From Table 16-9, ARL =

10.4

16-32



16-52 a) The process appears to be in control.

Sample

E W M A

2018161412108642

91.0

90.5

90.0

89.5

89.0

_ _ X=90

UCL=90.800

LCL=89.200

EWMA Char t o f C1

b) The process appears to be in control .

Sample

E W M A

2018161412108642

91.5

91.0

90.5

90.0

89.5

89.0

88.5

_ _ X=90

UCL=91.386

LCL=88.614

EWMA Char t o f C1

c) For part (a), there is no evidence that the process has shifted out of control.

16-33



Sample

E W M A

24222018161412108642

91.0

90.5

90.0

89.5

89.0

_ _ X=90

UCL=90.800

LCL=89.200

EWMA Char t of C1

For part b), there is no evidence that the process has shifted out of control.

Sample

E W M

A

24222018161412108642

91.5

91.0

90.5

90.0

89.5

89.0

88.5

_ _

X=90

UCL=91.386

LCL=88.614

EWMA Char t of C1

16-53 a) The estimated standard deviation is 0.169548.

b) The process appears to be in control.

16-34



Sample

E W M A

24222018161412108642

10.2

10.1

10.0

9.9

9.8

_ _ X=10

UCL=10.1695

LCL=9.8305

EWMA Char t of C1

c) The process appears to be out of control at the observation 13.

Sample

E W M A

24222018161412108642

10.3

10.2

10.1

10.0

9.9

9.8

9.7

_ _ X=10

UCL=10.2937

LCL=9.7063

EWMA Char t of C1


16-35



Sample

E W M A

2018161412108642

110

105

100

95

90

_ _ X=100

UCL=108.00

LCL=92.00

EWMA Chart of C1


Sample

E W M A

2018161412108642

115

110

105

100

95

90

85

_ _ X=100

UCL=113.86

LCL=86.14

EWMA Char t of C1

c) Since the shift is 0.5σ , smaller λ is preferred to detect the shift fast. so the chart in part (a)

is preferred.

16-55 a) The shift of the mean is1σ . So we prefer 0.1λ = and L=2.81 since this setting has the

smaller ARL 10.3.

b) The shift of the mean is 2 X

σ . So we prefer 0.5λ = and L=3.07 since this setting has the

smaller ARL 3.63

c) The shift of the mean is 3 X

σ . Solving2

2 / 3n

⎛ ⎞=⎜ ⎟

⎝ ⎠for n gives us the required sample size

of 9.

16-36



16-56 a) With a target = 100 and a shift to 102 results in a shift of 4

100102 −= 0.5 standard deviations.

From Table 16-9, ARL = 38. The hours of production are 2(38) = 76.

b) The ARL = 38. However, the time to obtain 38 samples is now 0.5(38) = 19.

c) From Table 16-9, the ARL when there is no shift is 465. Consequently, the time between false alarms is0.5(465) = 232.5 hours. Under the old interval, false alarms occurred every 930 hours.

d) If the process shifts to 102, the shift is =−

=−

4/4

100102

/

0

nσ

μ μ 1 standard deviation. From Table 16-9, the

ARL for this shift is 10.4. Therefore, the time to detect the shift is 2(10.4) = 20.8 hours. Although this time

is slightly longer than the result in part (b), the time between false alarms is 2(465) = 930 hours, which is

better than the result in part (c).



Supplementary Exercises


--------------------------------------------------------------------------------

X-bar | Range

---- | -----


Centerline = 64 | Centerline = 0.01764


|


--------------------------------------------------------------------------------


Estimated

process mean = 64


mean Range = 0.01764

0Subgroup 5 10 15 20 25

63.98

63.99

64.00

64.01

64.02

S a m p l e M e a n

Mean=64.00

UCL=64.02

LCL=63.98

0.00

0.01

0.02

0.03

0.04

0.05

S a m p l e R a n g e

R=0.01764

UCL=0.04541

LCL=0

Xbar/R Chart for C1-C3

The process is in control.

b) μ = =x 64 0104.0693.1

01764.0ˆ

2

===d

Rσ

c) 641.0)0104.0(6

98.6302.64

ˆ6 =

−

=

−

= σ

LSLUSL

PCR

16-37



The process does not meet the minimum capability level of PCR ≥ 1.33.

d)

[ ]

641.0

641.0,641.0min

)0104.0(398.6364,

)0104.0(36402.64min

ˆ3,

ˆ3min

=

=

⎥⎦⎤⎢

⎣⎡ −−=

⎥⎦

⎤⎢⎣

⎡ −−=

σ σ

LSL x xUSLPCRk

e) In order to make this process a “six-sigma process”, the variance σ2 would have to be decreased such that

PCR k = 2.0. The value of the variance is found by solving PCR k =x LSL−

=3

2 0σ

. for σ:

0033.0

6

98.630.64

98.630.646

0.23

98.6364

=

−=

−=

=−

σ

σ

σ

σ

Therefore, the process variance would have to be decreased to σ2 = (0.0033)2 = 0.000011.

f) σx= 0.0104

8295.00020.08315.0

)88.2()96.0()96.088.2(

0104.0

01.6402.64

0104.0

01.6498.63)02.6498.63(

=−=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

xσ

μ

The probability that this shift will be detected on the next sample is p = 1−0.8295 = 0.1705

87.51705.0

11===

p ARL

16-38



16-58 a)

0Subgroup 5 10 15 20 25

63.98

63.99

64.00

64.01

64.02

S a m p l e M e a n

Mean=64.00

UCL=64.02

LCL=63.98

0.00

0.01

0.02

S a

m p l e S t D e v

S=0.009274

UCL=0.02382

LCL=0

Xbar/S Chart for C1-C3

b) μ = =x 64 0104.08862.0

009274.0ˆ

4

===c

sσ

c) Same as 16-57 641.0)0104.0(6

98.6302.64

ˆ6=

−=

−=

σ

LSLUSLPCR

The process does not meet the minimum capability level of PCR ≥ 1.33.

d) Same as 16-57

[ ]

641.0

641.0,641.0min

)0104.0(3

98.6364,

)0104.0(3

6402.64min

ˆ3,

ˆ3min

=

=

⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦⎤⎢

⎣⎡ −−=

σ σ

LSL x xUSLPCRk

e) Same as 16-57 e). In order to make this process a “six-sigma process”, the variance σ2would have to be

decreased such that PCR k = 2.0. The value of the variance is found by solving PCR k =x LSL−

=3

2 0σ

. for σ:

0033.0

6

98.630.64

98.630.646

0.23

98.6364

=

−=

−=

=−

σ

σ

σ

σ

Therefore, the process variance would have to be decreased to σ2 = (0.0033)2 = 0.000011.

16-39



f) Same as 16-57 σx= 0.0104

8295.00020.08315.0

)88.2()96.0()96.088.2(

0104.0

01.6402.64

0104.0

01.6498.63)02.6498.63(

=−=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−<

−=<<

Z P Z P Z P

X P X P

xσ

μ

The probability that this shift will be detected on the next sample is p = 1−0.8295 = 0.1705

87.51705.0

11===

p ARL

16-59 a)

0 10 20

0.0

0.1

0.2

Sample Number

P r o p o r t i o n

P Chart for def

P=0.11

UCL=0.2039

LCL=0.01613

There are no points beyond the control limits. The process is in control.

b)

0 10 20

0.04

0.09

0.14

0.19

Sample Number

P r o p o r t i o n

P Chart for def2

1

P=0.11

UCL=0.1764

LCL=0.04363

There is one point beyond the upper control limit. The process is out of control. The revised limits are:

16-40



0 10 20

0.04

0.09

0.14

0.19

Sample Number

P r o p o r t i o n

P Chart for def2

P=0.1063

UCL=0.1717

LCL=0.04093

There are no further points beyond the control limits.

c) A larger sample size with the same percentage of defective items will result in more narrow control limits.The control limits corresponding to the larger sample size are more sensitive to process shifts.

16-60 a)

0 5 10 15 20 25

0

1

2

Sample Number


U Chart for Defects

1

1

U=0.528

UCL=1.503

LCL=0

Points14 and 23 are beyond the control limits. The process is out of control.

16-41



b) After removing points 14 and 23, the limits are narrowed.

0 10 20

0.0

0.5

1.0

1.5

Sample Number


U Chart for Defects

U=0.4261

UCL=1.302

LCL=0

c) The control limits are narrower for a sample size of 10

2520151050

1.0

0.5

0.0

Sample Number


U C hart for defects n=10

1

1

U=0.264

UCL=0.7514

LCL=0

20100

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Sample Number


U Chart for defects n=10

U=0.2130

UCL=0.6509

LCL=0

16-42



16-61 a) Using I-MR chart.

Observa t ion


2018161412108642

60.3275

60.3270

60.3265

60.3260

60.3255

_ X=60.32641

UC L=60.327362

LCL=60.325458

Observa t ion


2018161412108642

0.00100

0.00075

0.00050

0.00025

0.00000

__ MR=0.000358

UC L=0.001169

LCL=0

I -MR Chart of C1

b) The chart is identical to the chart in part (a) except for the scale of the individuals chart.

Observa t ion


2018161412108642

0.0015

0.0010

0.0005

0.0000

-0.0005

_ X=0.00041

UC L=0.001362

LCL=-0.000542

Observa t ion


2018161412108642

0.00100

0.00075

0.00050

0.00025

0.00000

__ MR=0.000358

UC L=0.001169

LCL=0

I -MR Chart of C1

c)The estimated mean is 60.3264. The estimated standard deviation is 0.0003173.

0.0021.0505

6 6(0.0003173)

USL LSLPCR

σ

−= = =

0.0009 0.0011min , 0.9455

3 3k PCR

σ σ

⎡ ⎤= =⎢ ⎥⎣ ⎦

16-43



16-62 a)

Observat ion


30272421181512963

7000

6000

5000

4000

_

X=5832

UC L=6669

LCL=4995

Observat ion


30272421181512963

1000

750

500

250

0

__ MR=315

UC L=1028

LCL=0

11

1

1

111

1

I -MR Chart of enery

b) The data does not appear to be generated from an in-control process. The average tends to drift

to larger values and then drop back off over the last 5 values.

16-63 a)Trial control limits :

S chart

UCL= 170.2482

CL = 86.4208

LCL = 2.59342

X bar chart

UCL= 670.0045

CL = 558.766

LCL = 447.5275

5 10 15 20 25

4 0 0

6 0 0

8 0 0

Index

x b a r

5 10 15 20 25

0

5 0

1 5 0

Index

s

16-44



b) An estimate of σ is given by 8259.909515.0/4208.86/ 4 ==cS

PCR=500/(6*90.8259)=0.9175 and PCR k = ⎥⎦

⎤⎢⎣

⎡ −−)8259.90(3

33077.558,

)8259.90(3

77.558830min =0.8396

Based on the capability ratios above (both <1), the process is operating off-center and will result

in a large number of non-conforming units.

c) To determine the new variance, solve

for σ. Since PCR 2=k PCR k =σ 3

33077.558 −in this exercise, we find σ=38.128 or σ2=1453.77.

d) The probability that X falls within the control limits is

( )

9706.0)89.111.4(

6

8259.90

6000045.670

6

8259.90

6005275.4470045.6705275.447

=<<−

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

<<−

=<<

Z P

Z P X P

Thus, p=0.0294 and ARL=1/p=34.01. The probability that the shift will be detected in the next

sample is 0.0294.

For part b) through d), if we remove the out-of-control samples in part a) and recalculate the control limits, we

will get the following control limits for S chart and Xbar chart.

S chart UCL= 158.9313

CL = 80.67615

LCL = 2.421028

X bar chart

UCL= 655.7918CL = 551.9477

LCL = 448.1035

b)σ ˆ =84.78839

min ,ˆ ˆ3 3

830 551.95 551.95 330min ,

3(84.79) 3(84.79)

0.8725

K

USL x x LSLPCR

σ σ

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦

⎡ ⎤− −= ⎢ ⎥

⎣ ⎦=

c)

min ,ˆ ˆ3 3

551.95 330

ˆ3( )

2

ˆSo 36.9917

K

USL x x LSLPCR

σ σ

σ

σ

− −⎡ ⎤= ⎢ ⎥⎣ ⎦

−=

=

=

16-45



d) In-control distribution X ~ N(551.9477, )234.6

Out-of-control distribution X ~ N(600, )234.6

P[448.1 ≤ X ≤ 655.8, when μ =600]

= ]6.346008.655

6.346001.448[ −≤≤− Z P

= ]6124.139.4[ ≤≤− Z P

=0.9463

Out-of-control ARL= =− 9463.01

118.6 ≈ 19.

16-64 ⎯ X Control Chart with 2-sigma limits

n LCL

nUCL

CL

σ μ

σ μ

μ

2,2 −=+=

=

02275.097725.01)2(1)2(2/

2

02275.097725.01)2(1)2(2/

2

=−=<−=−<=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −<

−=⎟

⎠

⎞⎜⎝

⎛ −<

=−=<−=>=⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ >−=⎟

⎠ ⎞⎜

⎝ ⎛ +>

Z P Z Pn

X P

n X P

and

Z P Z Pn

X Pn

X P

σ

μ σ μ

σ

μ σ μ

The answer is 0.02275 + 0.02275 = 0.0455. The answer for 3-sigma control limits is 0.0027. The 3-sigma

control limits result in many fewer false alarms.

16-65

a) The following control chart use the average range from 25 subgroups of size 3 to estimate the process

standard deviation. Minitab uses a pooled estimate of variance as the default method for an EWMA control

chart so that the range method was selected from the options. Points are clearly out of control.

Sample

E W M A

24222018161412108642

64.7

64.6

64.5

64.4

64.3

64.2

64.1

64.0

63.9

_ _ X=64.1334

UCL=64.2759

LCL=63.9910

EWMA Char t of C3, ..., C5

16-46



b) The following control chart use the average range from 25 subgroups of size 3 to estimate the process

standard deviation. There is a big shift in the mean at sample 10 and the process is out of control at

this point.

Sample

E W M A

24222018161412108642

65.75

65.50

65.25

65.00

64.75

64.50

64.25

64.00

_ _ X=64.133

UCL=64.380

LCL=63.887

EWMA Chart of C3, . .., C5

16-66

a) The data appears to be generated from an out-of-control process.

Sample

E W M A

30272421181512963

6600

6400

6200

6000

5800

5600

5400

5200

5000

_ _

X=5832

UCL=6111

LCL=5553

EWMA Chart of C1

b) The data appears to be generated from an out-of-control process.

16-47



Sample

E W M A

30272421181512963

7000

6500

6000

5500

5000

4500

_ _ X=5832

UCL=6315

LCL=5349

EWMA Chart of C1


Sample

E W M A

2018161412108642

60.3268

60.3267

60.3266

60.3265

60.3264

60.3263

60.3262

60.3261

_ _ X=60.32641

UCL=60.3267273

LCL=60.3260927

EWMA Chart of C1


Sample

E W M A

2018161412108642

60.32700

60.32675

60.32650

60.32625

60.32600

_ _ X=60.32641

UCL=60.326960

LCL=60.325860

EWMA Chart of C1

16-48



16-68

Sample


30272421181512963

5.0

2.5

0.0

-2.5

-5.0

0

UCL=5.27

LCL=-5.27

CUSUM Chart of C1

Process standard deviation is estimated using the average moving range of size 2 with MR/d 2, where d 2 =

1.128. The estimate is 1.05. Recommendation for k and h are 0.5 and 4 or 5, respectively for n =1. For this

chart h = 5 was used.

16-69 The process is not in control.

K = k σ = 1, so that k = 0.5

H = hσ = 10, so that h = 5

40

30

20

10

0

-10

10

-10

20100

Subgroup Number


Upper CUSUM

Lower CUSUM

CUSUM Chart for hardness

16-49



16-70

-80

0

80

-85.8529

85.8529

0 10 20

Subgroup Number


Upper CUSUM

Lower CUSUM

CUSUM Chart for Viscosity

Process standard deviation is estimated using the average moving range of size 2 with MR/d 2, where d 2 =

1.128 for a moving range of 2. The estimate is 17.17. Recommendation for k and h are 0.5 and 4 or 5,

respectively, for n = 1.

16-71 a)

Sample


24222018161412108642

0.020

0.015

0.010

0.005

0.000

-0.005

-0.010

0

UCL=0.01152

LCL=-0.01152

CUSUM Char t of x

σ is estimated using the moving range: 0.0026/1.128=0.0023. H and K were computed using

k=0.5 and h=5. The process is not in control.

b) EWMA gives similar results.

16-50



Sample

E W M A

24222018161412108642

0.404

0.403

0.402

0.401

0.400

0.399

_ _ X=0.401184

UCL=0.403489

LCL=0.398879

EWMA Chart of x

16-72 a) Let p denote the probability that a point plots outside of the control limits when the mean has shifted from

μ0 to μ = μ0 + 1.5σ. Then,

5.0]0[5.0

)6()0()06(

3/

5.1

/3

/

5.1

33

)( 00

=−=

−<−<=<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +<

−<−

−=

⎟ ⎠

⎞

⎜⎝

⎛

+<<−=<<

Z P Z P Z P

nn

X

nP

n X nPUCL X LCLP

σ

σ

σ

μ

σ

σ

σ

μ

σ

μ

Therefore, the probability the shift is undetected for three consecutive samples is (1 − p)3 =

(0.5)3 = 0.125.

b) If 2-sigma control limits were used, then

15866.0]0[84134.01

)5()1()15(

2/

5.1

/2

/

5.1

22

)(1 00

=−−=

−<−−<=−<<−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +<

−<−

−=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

+<<−=<<=−

Z P Z P Z P

nn

X

nP

n X nPUCL X LCLP p

σ

σ

σ

μ

σ

σ

σ

μ

σ

μ

Therefore, the probability the shift is undetected for three consecutive samples is (1 − p)3 = (0.15866)3 =

0.004.

c) The 2-sigma limits are narrower than the 3-sigma limits. Because the 2-sigma limits have a smaller

probability of a shift being undetected, the 2-sigma limits would be better than the 3-sigma limits for a mean

shift of 1.5σ. However, the 2-sigma limits would result in more signals when the process has not shifted

(false alarms).

16-51



16-73. ARL = 1/ p where p is the probability a point falls outside the control limits.

a) μ μ and n = 1σ= +0

02275.0

]0[97725.01

1)4()2(

)3()3(

/

3

/

3

)()(

0000

=

+−=

=−<+>=

−−<+−>=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −−−

<+⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −−+

>=

<+>=

nwhen Z P Z P

n Z Pn Z P

n

n

Z Pn

n

Z P

LCL X PUCL X P p

σ

σ μ σ

μ

σ

σ μ σ

μ

Therefore, ARL = 1/ p = 1/0.02275 = 43.9.

b) μ μ σ= +0 2

15866.0

]0[84134.01

1)5()1(

)23()23(

/

23

/

23

)()(

0000

=

+−=

=−<+>=

−−<+−>=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −−−<+

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −−+>=

<+>

nwhen Z P Z P

n Z Pn Z P

n

n Z P

n

n Z P

LCL X PUCL X P

σ

σ μ σ

μ

σ

σ μ σ

μ

Therefore, ARL = 1/ p = 1/0.15866 = 6.30.

c) μ μ σ= +0 3

50.0

]0[50.01

1)6()0(

)33()33(

/

33

/

33

)()(

0000

=+−=

=−<+>=

−−<+−>=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −−−<+

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −−+>=

<+>

nwhen Z P Z P

n Z Pn Z P

n

n Z P

n

n Z P

LCL X PUCL X P

σ

σ μ σ

μ

σ

σ μ σ

μ

Therefore, ARL = 1/ p = 1/0.50 = 2.00.

d) The ARL is decreasing as the magnitude of the shift increases from σ to 2σ to 3σ. The ARL decrease as

the magnitude of the shift increases since a larger shift is more likely to be detected earlier than a smaller

shift.

16-52



16-74 a) Because ARL = 370, on the average we expect there to be one false alarm every 370 hours. Each 30-day

month contains 30 × 24 = 720 hours of operation. Consequently, we expect 720/370 = 1.9 false alarms each month

0027.0)00135.0(2)3()3()ˆ3()ˆ3( ==−<+>=−<++> zP zP X X P X X P σ σ

ARL=1/p=1/0.0027=370.37

b) With 2-sigma limits the probability of a point plotting out of control is determined as follows, when

μ μ σ= +0

P X UCL P X LCL

PX

PX

P Z P Z

P Z P Z

( ) ( )

( ) ( )

( ) [ ( )]

. .

.

> + <

=− −

>+ − −⎛

⎝ ⎜⎞ ⎠⎟

+− −

<− − −⎛

⎝ ⎜⎞ ⎠⎟

= > + < −

= − < + − <

= − + −

=

μ σ

σ

μ σ μ σ

σ

μ σ

σ

μ σ μ σ

σ0 0 0 0 0 02 2

1 3

1 1 1 3

1 084134 1 0 99865

0160

Therefore, ARL=1/ p = 1/0.160 = 6.25. The 2-sigma limits reduce the ARL for detecting a shift in the

mean of magnitude σ. However, the next part of this solution shows that the number of false alarms increases

with 2-sigma limits.

c) 2σ limits

0455.0)02275.0(2)2()2()ˆ2()ˆ2( ==−<+>=−<++> zP zP X X P X X P σ σ

AR L= 1/ p= 1/0.0455 = 21.98. This ARL is not satisfactory. There would be too many false alarms. We

would expect 32.76 false alarms per month.


Charting xbar

X-bar | Range

----- | -----





Estimated

process mean = 139.49process sigma = 0.505159

mean Range = 1.175

0 4 8 12 16 20

13 7

13 8

13 9

14 0

14 1

139.49

140.168

138.812

0 4 8 12 16 20

0

0.5

1

1.5

2

2.5

1.175

2.48437

0

X-bar

Problem 16-51

subgroup

Range

There are points beyond the control limits. The process is out of control. The points are 8, 10, 11,

12, 13, 14, 15, 16, and 19.

16-53



b) Revised control limits are given in the table below:

X-bar and Range - Initial Study

Charting Xbar

X-bar | Range

----- | -----




1050

140.5

140.0

139.5

139.0

Subgroup Number

x b a r

Xbar Chart

Mean=139.7

UCL=140.4

LCL=139.0

1050

3 .0

1 .5

0

S u b g r o u p N u m b e r

r

R C hart

M ean=1.227

U C L = 2 . 5 9 6

L C L = 0

There are no further points beyond the control limits.

The process standard deviation estimate is given by 5276.0326.2

227273.1ˆ

2

===d

Rσ

c) 26.1)528.0(6

138142

ˆ6=

−=

−=

σ

LSLUSLPCR

[ ] 08.108.1,45.1min

)528.0(3

138709.139,

)528.0(3

709.139142min

ˆ3,

ˆ3min

==

⎥⎦

⎤⎢⎣

⎡ −−=⎥

⎦

⎤⎢⎣

⎡ −−=

σ σ

LSL x xUSLPCRk

Because the process capability ratios are less than unity, the process capability appears to be poor. PCR is

slightly larger than PCR k indicating that the process is somewhat off center.

16-54



d) In order to make this process a “six-sigma process”, the variance σ2would have to be decreased such that

PCR k = 2.0. The value of the variance is found by solving PCR k = 0.23

=−

σ

LSL xfor σ:

2848.0

6

138709.139

138709.1396

0.23

138709.139

=

−=

−=

=−

σ

σ

σ

σ

Therefore, the process variance would have to be decreased to σ2= (0.2848)

2= 0.081.

e) σx= 0.528

8197.0

093418.0913085.0

)32.1()36.1(

)35.132.1(528.0

7.139417.140

528.0

7.139001.139

)7.139|417.140001.139(

=

−=

−<−<=

<<−=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −<

−<

−=

=<<=

Z P Z P

Z P

X P

X P p

xσ

μ

μ

The probability that this shift will be detected on the next sample is 1 − p = 1−0.8197 = 0.1803.

55.51803.0

1

1

1==

−=

p ARL

16-76

a) The probability of having no signal is

( 3 3) 0.9973P X − < < = P(No signal in 3 samples)=(0.9973)

3=0.9919

P(No signal in 6 samples)=(0.9973)6= 0.9839

P(No signal in 10 samples)=(0.9973)10= 0.9733

16-77 PCR = 2 but μ σ = +USL 3

P X USL P Z P Z( )( )

( ) .< = <− −⎛

⎝ ⎜

⎞ ⎠⎟ = < − =

μ σ μ

σ3

3 0 00135

16-78 a) The P(LCL < < UCL), when p = 0.08, is needed.P

115.0100

)05.01(05.0305.0

)1(3

0015.0

100

)05.01(05.0305.0

)1(3

=−

+=−

+=

→−=−

−=−

−=

n

p p pUCL

n

p p p LCL

Therefore, when p = 0.08

16-55



100

)92.0(08.0

08.0115.0

100

)92.0(08.0

08.0ˆ)115.0ˆ()115.0ˆ0(

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −

≤−

=≤=≤≤P

PPPPP

solution manual stats

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