Math 150 Test 3 Review

1.- According to WHO, Europeans are less religious than Americans. In a sample of 1000Europeans, 300 said that they believe in a higher power. In a sample of 900 Americans,50%, said that they believe God. Is there enough evidence too show that there is a di↵erencebetween Americans and Europeans when it comes to religious beliefs? Use ↵ = 0.05

HH

ni = 1000,

4=300,

F. = o . }← 3¥00

F= 3,00000++494=0.39hz= 900 , 1/2=450, 152=0.5

* ( 900 )

0.3 - 0.5

51 ) Ho :P ,=Pz ( no difference ) sz ) Z * =-

Ha '. Pit Pz ( Difference ) (0.39110-61)/5150+901=-8.8

s } ) IRIR : 20.025=1-9654 ) Reject Ho

⇒ There is a difference

There Is enough evidence to show that

there Is a difference betweenH@Dqggamen.ca, and European ,.

Z*= - 8.8 Table0.9998

p.UA/ueMethod_si ) sz ) Same µ0%253 ) P=2P( z > 8.8 ) calculator → or -

K 3.59

= 2 hovmulcdf ( 8.8 ,1000 )

= 1.39 E - 18

= 0.00000000000000000139

54 ) Pt

< X=o -05 ⇒ Reject( 0.0004 )

:able

2.- A Special Forces Military Recruiter claims that the height of his recruits have a standarddeviation of 5inches. A sample of 24 of its recruits find that the sample standard deviation is5.5inches. At ↵ = 0.01, is there enough evidence to support the his claim (Do both methods,this is a two tailed test)?

§¥_#

n= 24,

5=5.5 , X=o . 01

51 ) Ho : 0=5 sz ) X2*= ( n . 1) 52

=z= ( 23 )( 5. 5)

2

Ha : 0=15

52=27.8353 ) IRR : = 28

RRxxiiiixiieiiiiiii

.ie?iiieoss4) Fail to reject Ho ( Accept Ho )

×2*=zg⇒ 0=5

⇒ There is enough evidence to support the recruiter 's

claim that the standard deviation is 5 inches .

R.vaweme.tk# Table0.9

s } ) D= 2P(X2 > X '* ) -70.1

IY.in?IInooo.m?/~/I.= 0.43,

14.848 32.007

iX2*=28

54 ) 117=0.43 D &= 0.01,

D= 2 ( 0.2 )⇒ Accept Ho

1=0.4C Fail to reject )

:(

3.- Apple claims that the battery life of the iPhone is better than that of the Google phone.A sample is taken from each company. The Apple sample (n = 20) had a mean of 12hoursand standard deviation of 3hour, while the Google sample (n = 24) had a mean of 10hourswith a standard deviation of 4hours.

Is there a di↵erence in the battery life of the two companies? Use ↵ = 0.05 (Both Meth-

ods)

§±

15=12 S,

=3 ni=2o

152=10 52=4 n2= 24

51 ) Ho :µ=µz ( no difference In battery life )

Ha : µitµz

so

t*x¥÷÷f=E⇐÷" "

s } ) RR :

to.ozs.la:2093#uM54 ) Fail to reject Ho

o

⇒ ( no difference )

⇒ There Is not enough evidence to support t*- 1.89

Apple 's claim.

P . value Method xvan '

-0¥- Hit 0.025

s } ) 117=2 Pct > 1.89 )tOR #l!ggT '

1 2.093= 2 tcdf ( 1. 89,1000119 )

,1.729

= 0.07 1

s } ) 117=210.03 )1

54 ) 117=0.07 > X= 0.05,

= 0.06

54 ) p -0.06 > X=O . 05

⇒ Accept

4.- A new volley-ball shoe manufacturer claims that its shoes increase the vertical jump (ininches) of any athlete. The following tables shows that result of five athletes with regularshoes and the new shoes. Is there enough evidence to suggest that there is a di↵erence? Use↵ = 0.05.

Table 1: Vertical Jumps in Inchesx1 33 28 27 33x2 34 27 30 35

§9@

n=4 gc@±x= d- OI

33 34 1 1

I = I= 1.25 SI = 15 - (5) 2/4

28 27 - 1 1 4 -

27 30 3 9 3

33 35 2- 4-= 2.92

5 15 ⇒sd =27=1-751) Ho : Md = 0 ( no difference between before latter )

Ha : Md to

sz ) t*=rgaI= ftp.t#=l.5537 112112 : to -025,3=3.182 .PE#E.

- 3.182

54 ) Fail to reject Ho t*=t5

( Accept Ho ) ⇒ no difference

There is Not enough evidence that the new shoes . make you

jump higher .

P.ba#thls3)lP=2P(t> 1. 5)

=2 tcdf ( 1.5, 1000,3 )

= 0.23

54 ) IP = 0.23 > X= 0.05

⇒ Accept to.

5.- Consider the following data.

x y x2 y2 xy

2 153 155 106 89 3

(a) Find r

(b) Find y

(c) Predict y when x is 4.

(d) Test the hypothesis ⇢ = 0 vs ⇢ 6= 0, with ↵ = 0.05

§ 10.1/10.2

Sxx = 155 - @5) 2/5 ( t )

= 30

SYy= 623 - ( 51 ) 2/5 ( t )

= 102.8

Sxy = 200 - ( 253151 ) 15

25 51 155 623 200= - 55

r.se#a=iEon=o.aa

FITEII .I¥l¥IIga,

5=+83×+4

Y"

= - 1.83 (4) + 19.4

= 12

Sl ) Ho : f=o ( no correlation ) 53 ) IRK : to -025,3=3.182

Ha : ft0n

-2s.it#.o.aaftfoarf=flfp.= - 12.15

54 ) Reject Ho ( f=o )t*= - 12.15

⇒ j=o ⇒ correlation

÷-

value : 54 ) 117=0.001 < X=o . os

531 117=2 Pct > 12.15 )= Ztcdf C 12.15 ,

1000,3 )⇒ Reject to

= 0.001

6.- An instructor wishes to see if the way people obtain information is independent of the ed-ucational background. A survey of high school and college graduates yielded this information.at ↵ = 0.05, test the claim that the way people obtain information is independent of theireducational background.

- T.V Newspaper Other

High School 150 90 50College 40 40 30

§ 10.3

d. f= ( 2 - 1) (3-1)= 2

C 138 ) ( 94 ) ( 58 )

C 52 ) ( 36 ) ( 22 )

Sl ) Ho : 12 and C Var are IndHa '

.K ' ' l ' " cc De p

54 X2*= (138=35805+1945405+158*02 +5*245+1361.412+422021

" " "÷ :};I=" "

FETE "

541 Reject Ho

⇒ R and C variables are deP .

×2= g. y

Fvalues} ) P=P(×2 > ×2* '

'

'

; #y|= PCXZ > 8.4 )

= X2cdf( 8.4,

1000,2 )

= 0.015

54 ) 117=0.015 ← X= 0.05 ⇒ Reject

7.- Test the following �1 = �2 vs �1 6= �2.

Where,

S1 = 16, S2 = 12, n1 = 8, n2 = 6

§ 9.4

d. f. N= 7 d. f. D= 5

Sl ) Ho : 9=02sz ) F * =

II= 1.78

Ha : at 02 122

s } ) Foozs, 7,5=6.85 µg ,

54 ) Fail to reject Ho-

F*=1 . 8

⇒ a=E

IP - Value Method- $4 ) 113=0.54 > X= 0.05

537 113=21 > ( f > 1.78 )⇒ Accept Ho

= 2 Fcdf ( 1.78 ,1000 , 7. 5) -

= 0.54

8.- A researcher wishes to try three di↵erent techniques to lower blood pressure. The subjectsare randomly assigned to three groups. After four weeks, the reduction in each person’s bloodpressure is recorded. At ↵ = 0.05, test the claim that there is no di↵erence among the means.The data are in the following table.

Medication Exercise Diet

10 6 58 8 6

6 3 712 5 87 5 4

1<=3

N= 15

5,32=14.46,

SI = 3.87

Sl ) Ho :µ=µz=µzHa '

.At least one µ is different

site.IE?FIIII=..t.#E"

sy ) Accept Ho( Fail to reject Ho )f*=z . 7

⇒ no difference between

treatments to lower blood pressure .

÷awemethod53 ) D= 2P( F > 3.7 )

= 2 Fcdf ( 3. 7,1000 ,2 ,

12 )

I 0.11

⇒ Accept HoSu ) 117=0.11 > X = 0.05-

9.- True/False

(a) For any random sample P (�2 < �2L) = P (�2 > �2

R)

(b) For any random sample F⇤ � 1

(c) For any random sample �2L = �2

R

(d) If Ha : p < p0 then P = P (t < t⇤)

(e) For any random sample if b < 0 then r < 0

(f) For any two random samples p1 > p2

True r< 0

True m<°

SH

False m =

#Z Z

falseS#[r =

fastFalse

b= g- MI

False