Statistical Intervals for a Single Sample

Chapter 8 continues

Chapter 8B

ENM 500 studentsreacting to yet anotherday of this

A Recap – the main result of Monday’s 8a presentation

/2 /2

/2 /2

1/

with a little algebra,

/ / 1

XP t t

S n

P X t S n X t S n

A 2-sided 100(1- )% confidence Interval on µ, population variance unknown

/2, 1 /nx t s n

Yet Another Example

The following data is from a random sample of NBA player weights in pounds.

Statistic weightAverage: 219.68

Maximum: 285Minimum: 165

standard deviation 27.15sample size 82

/2, 1

27.15/ 219.68 1.9897

82

219.68 5.86 (213.82,225.54)

nx t s n

2-sided 95% CI: t.025,81

Today’s Excitement

The Behind-the-Scenes Probability Statement

2

2 21 /2, 1 /2, 12

2

2 22

2 2/2, 1 1 /2, 1

( 1)Pr 1

Then a 100(1 )% confidence interval on is

( 1) ( 1)

Use square roots to get an interval on the standard devia

n n

n n

n s

n s n s

tion.

Confidence Interval on Standard Deviation and Variance of a Normal Distribution

Chi-square Visually

Note how chi-square moves out to the right as d.f.’s increase.

Some Observations on Chi-Square

d.f. Alpha = .95Upper

Alpha = .05Lower

Ratiolower/upper

5 1.15 11.07 9.63

10 3.94 18.31 4.65

20 10.85 31.41 2.89

40 26.51 55.76 2.10

100 77.93 124.34 1.60

More observations

lead to a tighter interval.

It is Time to Compute an Actual Confidence Interval for

The following sample of response times in hours for restoring power outages for Dayton Power and Light Company has been obtained:

3.4 3.6 4 0.4 23 3.1 4.1 1.4 2.5

1.4 2 3.1 1.8 1.63.5 2.5 1.7 5.1 0.74.2 1.5 3 3.9 3

n = 25Mean = 2.66Variance 1.4075Std dev = 1.186381

2

2 22 2

2 2.025,25 1 .975,25 1

2

Then a 100(1 .05)% confidence interval on is

24 1.4075 24 1.4075( 1) ( 1)

39.36 12.4

.8582 2.7242

.9264 1.6505

n s n s

One-Sided Confidence Bounds

Example 8-6

Confidence Interval For a Population Proportion, p – the Preliminaries

2 2

ˆ

( , ), number of successes in n trials

[ ] , [ ] (1 )

[ ]ˆ ˆ, then [ ]

[ ] (1 ) (1 )ˆ[ ]

(1 )p

X Bin n p

E X np V X np p

X E Xlet p E p p

n nV X np p p p

and V pn n n

p p

n

A Large-Sample Confidence Interval For a Population Proportion

ˆ

ˆ

/2 /2

ˆ ˆIf is large, is approximately standard normal

(1 )

ˆPr 1

(1 )

p

p

P P pn Z

p pn

P pz z

p pn

There is a requirement that np > 5 and n(1 – p) > 5 for using normal approximation to the binomial.

Large Sample C.I. For a Population Proportion cont’d

If we approximate the unknown population parameter p by the estimate of p, we obtain the approximate C.I.

n

ppzpp

n

ppzp

)ˆ1(ˆˆ

)ˆ1(ˆˆ 2/2/

Example 8-7

Sample Size Determinations

)25(.2

2/

E

zn

/2 (1 ) /z p p n error

/2

22/2

2/2

2

(1 ) /

(1 )

(1 )

z p p n E

z p pE

n

z p pn

E

2since ( ) (1 )

( )Max ( ) : 1 2 0; or * .5 and *(1 *) .25

f p p p p p

df pf p p p p p

dp

Problem 8-52

36.050

18ˆ p 50n 96.12/z

/2 /2

ˆ ˆ ˆ ˆ(1 ) (1 )ˆ ˆ

0.36(0.64) 0.36(0.64)0.36 1.96 0.36 1.96 0.227 0.493

50 50

p p p pp z p p z

n n

p p

50 suspension helmets subjected to an impact test in which 18 were damaged.

(a) 95% CI:

2213n76.2212)36.01(36.002.0

96.1)1(

22

2/

ppE

zn

(b) Sample size to reduce error to .02?

2401)5.01(5.002.0

96.1)1(

22

2/

ppE

zn

(c) Sample size regardless of true value of p?

Problem 8-53

The Ohio Department of Transportation wishes to survey state residents to determine what proportion of the population would like to increase state highway speed limits to 75 mph. How many residents need to be surveyed to be at least 99% confident that the sample proportion is within 0.05 of the true proportion?

The worst case would be for p = 0.5, thus with E = 0.05 and = 0.01, z/2 = z0.005 = 2.58 we obtain a sample size of:

64.665)5.01(5.005.0

58.2)1(

22

2/

ppE

zn

n ~ 666 (a devilish result)

One-Sided Bounds

n

ppzpp

n

ppzp

)ˆ1(ˆˆ

)ˆ1(ˆˆ 2/2/

Intuitively, this is like setting the lower bound to zero, or the upper bound to one. Then you lump all of the alpha probability onto the other side.

pn

ppzp

)ˆ1(ˆˆ

n

ppzpp

)ˆ1(ˆˆ

Lower bound

Upper bound

Tests of a Proportion - Example

CBS News Poll. Sept. 14-16, 2007. N=706 adults nationwide. MoE ± 4 (for all adults)."Do you approve or disapprove of the way George W. Bush handled the situation with Iraq?"

Approve Disapprove Unsure    

% % %    

ALL adults 25 70 5    

   Republican 58 31 11    

   Democrat 6 92 2    

   Independent 20 76 4  

A 95 percent Confidence Interval on p

/2 /2

.05/2

ˆ ˆ ˆ ˆ(1 ) (1 )ˆ ˆ

1.96

.7(1 .7) .7(1 .7)0.7 1.96 0.7 1.96

706 706

0.7 1.96 .01725 0.7 1.96 .01725

.6662 .7338

p p p pp z p p z

n nz

p

p

p

ˆ/2 1.96 .01725 .0338pMOE z

Margin Of Error Most surveys report margin of error (MoE) in a manner such

as: "the results of this survey are accurate at the 95% confidence level plus or minus 3 percentage points."

That is the error that can result from the process of selecting the sample. It suggests what the upper and lower bounds of the results are.

Sampling Error is the calculated statistical imprecision due to interviewing a random sample instead of the entire population. The margin of error provides an estimate of how much the

results of the sample may differ due to chance when compared to what would have been found if the entire population was interviewed.

ˆ/2 /2

(1 )p

p pz z

n

Sampling ErrorSample

Size 1,000 750 500 250 100

Percentage near 10

2% 2% 3% 4% 6%

Percentage near 20

3 3 4 5 9

Percentage near 30

3 4 4 6 10

Percentage near 40

3 4 5 7 10

Percentage near 50

3 4 5 7 11

Percentage near 60

3 4 5 7 10

Percentage near 70

3 4 4 6 10

Percentage near 80

3 3 4 5 9

Percentage near 90

2 2 3 4 6

Tolerance and Prediction Intervals

A prediction is that a little more tolerance during our weekly interval will be observed.

Prediction Interval for Future Observation

The prediction interval for Xn+1 will always be longer than the confidence interval for .

Where does that come from?

2221

1

1

)/11(/)(

0)(

n.observatio

additional single a be Let .populationnormal

afromn size ofsampleaofmeanthebeLet

nnXXV

XXE

X

X

n

n

n

use s to estimate

1/2, 1 /2, 1

/2,n-1 1 /2, 1

Pr 11 1/

or

x-t 1 1/ 1 1/

nn n

n n

X Xt t

s n

s n X x t s n

An observation … or two

- Here, note that the assumption of normality cannot be trivially granted under the umbrella of large sample size. Any time you are dealing with a single prediction, distribution is critical

- Also, note that the elongation of the prediction interval comes from the use of the t distribution and the extra term under the radical.

/2,n-1 1 /2, 1x-t 1 1/ 1 1/n ns n X x t s n

Problem 8-56 (99% PI)

983.1517.020

11)25.0(861.225.1

20

11)25.0(861.225.1

11

11

1

1

19,005.0119,005.0

n

n

n

x

x

nstxx

nstx

25.025.120 sxn 861.219,005.0 t

The lower bound of the 99% prediction interval is considerably lower than the 99% confidence interval (1.108 )

Problem 8-64

1

1

124,05.0

91.325

11)08.0(711.105.4

11

n

n

n

x

x

xn

stx

The prediction interval bound is a lot lower than the confidence interval bound of 4.023 mm

To obtain a one sided prediction interval, use t,n-1 instead of t/2,n-1

Since we want a 95% one sided prediction interval, t/2,n-1 = t0.05,24 = 1.711, and xbar = 4.05 s = 0.08

n = 25

8-7.2 Tolerance Interval for a Normal Distribution

Pages 733-734

Example 8-10

Tolerance Intervals for a Normal Distribution

- Table XII in the appendix is the key to this.

- Be careful to understand the meaning of this concept.

A tolerance interval for capturing at least % of the

values in a normal distribution with confidence level

100(1- )% is

(x-ks,x ks) where

k is the factor from Table XI. k values are ta

bulated

for 90, 95, 99 percent tolerance and for 90, 95 and 99

percent confidence.

Tolerance Intervals for a Normal Distribution

Even though 1.96 is the z value appropriate for a two-sided 95% confidence interval, you cannot claim that (xbar – 1.96s, xbar + 1.96s) contains 95% of the population. Sampling variation in x and s affect the size of this interval.

If you knew and , you could say that 95% of the population is in ( - 1.96 , + 1.96 ).

Tolerance interval takes account of this uncertainty in parameter estimation to give us intervals that cover a certain percentage of the population with a given degree of confidence.

A Little More Tolerance – our capstone example

To estimate the tire life resulting from a new rubber compound,16 tires are subjected to an end-of-life road test with the following result: 16, 60,139.7 kilometers, 3645.94 kilometersn x s

95% tolerance interval95% confidence: K = 2.903

95% confidence IntervalFor

95% prediction Interval for x17

This Was The Week That Was (TWTWTW)

Next week

We Hypothesize in Chapter 9

Learn about Type I and Type II errors and how likely that you will be making them.