76197264 ken black qa 5th chapter 8 solution

Chapter 8: Statistical Inference: Estimation for Single Populations 1

Chapter 8Statistical Inference: Estimation for Single

Populations

LEARNING OBJECTIVES

The overall learning objective of Chapter 8 is to help you understand estimating

parameters of single populations, thereby enabling you to:

1. Know the difference between point and interval estimation.2. Estimate a population mean from a sample mean when σ

is known.3. Estimate a population mean from a sample mean when σ

is unknown.4. Estimate a population proportion from a sample proportion.5. Estimate the population variance from a sample variance.6. Estimate the minimum sample size necessary to achieve

given statistical goals.

CHAPTER TEACHING STRATEGY

Chapter 8 is the student's introduction to interval estimation and estimation of sample size. In this chapter, the concept of point estimate is discussed along with the notion that as each sample changes in all likelihood so will the point estimate. From this, the student can see that an interval estimate may be more usable as a one-time proposition than the point estimate. The confidence interval formulas for large sample means and proportions can be presented as mere algebraic manipulations of formulas developed in chapter 7 from the Central Limit Theorem.

It is very important that students begin to understand the difference between mean and proportions. Means can be generated by averaging some sort of measurable item such as age, sales, volume, test score, etc. Proportions are


computed by counting the number of items containing a characteristic of interest out of the total number of items. Examples might be proportion of people carrying a VISA card, proportion of items that are defective, proportion of market purchasing brand A. In addition, students can begin to see that sometimes single samples are taken and analyzed; but that other times, two samples are taken in order to compare two brands, two techniques, two conditions, male/female, etc.

In an effort to understand the impact of variables on confidence intervals, it may be useful to ask the students what would happen to a confidence interval if the sample size is varied or the confidence is increased or decreased. Such consideration helps the student see in a different light the items that make up a confidence interval. The student can see that increasing the sample size reduces the width of the confidence interval, all other things being constant, or that it increases confidence if other things are held constant. Business students probably understand that increasing sample size costs more and thus there are trade-offs in the research set-up.

In addition, it is probably worthwhile to have some discussion with students regarding the meaning of confidence, say 95%. The idea is presented in the chapter that if 100 samples are randomly taken from a population and 95% confidence intervals are computed on each sample, that 95%(100) or 95 intervals should contain the parameter of estimation and approximately 5 will not. In most cases, only one confidence interval is computed, not 100, so the 95% confidence puts the odds in the researcher's favor. It should be pointed out, however, that the confidence interval computed may not contain the parameter of interest.

This chapter introduces the student to the t distribution for estimating population means when σ is unknown. Emphasize that this applies only when the population is normally distributed because it is an assumption underlying the t test that the population is normally distributed, albeit that this assumption is robust. The student will observe that the t formula is essentially the same as the z formula and that it is the table that is different. When the population is normally distributed and σ is known, the z formula can be used even for small samples.

A formula is given in chapter 8 for estimating the population variance; and it is here that the student is introduced to the chi-square distribution. An assumption underlying the use of this technique is that the population is normally

distributed. The use of the chi-square statistic to estimate the population variance is extremely sensitive to violations of this assumption. For this reason, extreme

caution should be exercised in using this technique. Because of this, some statisticians omit this technique from consideration presentation and usage.

Lastly, this chapter contains a section on the estimation of sample size. One of the more common questions asked of statisticians is: "How large of a sample size should I take?" In this section, it should be emphasized that sample size estimation gives the researcher a "ball park" figure as to how many to sample. The “error of estimation “ is a measure of the sampling error. It is also equal to the + error of the interval shown earlier in the chapter.


CHAPTER OUTLINE

8.1 Estimating the Population Mean Using the z Statistic (σ known).

Finite Correction Factor

Estimating the Population Mean Using the z Statistic when the

Sample Size is Small

Using the Computer to Construct z Confidence Intervals for the

Mean

8.2 Estimating the Population Mean Using the t Statistic (σ unknown).

The t Distribution

Robustness

Characteristics of the t Distribution.

Reading the t Distribution Table

Confidence Intervals to Estimate the Population Mean Using the t

Statistic

Using the Computer to Construct t Confidence Intervals for the

Mean

8.3 Estimating the Population Proportion

Using the Computer to Construct Confidence Intervals of the

Population Proportion

8.4 Estimating the Population Variance


8.5 Estimating Sample Size

Sample Size When Estimating µ

Determining Sample Size When Estimating p

KEY WORDS

Bounds Point EstimateChi-square Distribution RobustDegrees of Freedom(df) Sample-Size EstimationError of Estimation t DistributionInterval Estimate t Value

SOLUTIONS TO PROBLEMS IN CHAPTER 8

8.1 a) x = 25 σ = 3.5 n = 60 95% Confidence z.025 = 1.96

n

zxσ±

= 25 + 1.96

60

5.3 = 25 + 0.89 = 24.11 < µ < 25.89

b) x = 119.6 σ = 23.89 n = 75 98% Confidence z.01 = 2.33

n

zxσ± = 119.6 + 2.33 75

89.23 = 119.6 ± 6.43 = 113.17 < µ <

126.03 c) x = 3.419 σ = 0.974 n = 32

90% C.I. z.05 = 1.645

n

zxσ± = 3.419 + 1.645

32

974.0 = 3.419 ± .283 = 3.136 < µ < 3.702

d) x = 56.7 σ = 12.1 N = 500 n = 47


80% C.I. z.10 = 1.28

1−−±

N

nN

nzx

σ= 56.7 + 1.28

1500

47500

47

1.12

−−

=

56.7 ± 2.15 = 54.55 < µ < 58.85


8.2 n = 36 x = 211 σ = 2395% C.I. z.025 = 1.96

nzx

σ± = 211 ± 1.9636

23 = 211 ± 7.51 = 203.49 < µ < 218.51

8.3 n = 81 x = 47 σ = 5.8990% C.I. z.05=1.645

n

zxσ± = 47 ± 1.645

81

89.5 = 47 ± 1.08 = 45.92 < µ < 48.08

8.4 n = 70 σ 2 = 49 x = 90.4 x = 90.4 Point Estimate

94% C.I. z.03 = 1.88

nzx

σ± = 90.4 ± 1.8870

49 = 90.4 ± 1.57 = 88.83 < µ < 91.97

8.5 n = 39 N = 200 x = 66 σ = 1196% C.I. z.02 = 2.05

1−−±

N

nN

nzx

σ = 66 ± 2.05

1200

39200

39

11

−−

=

66 ± 3.25 = 62.75 < µ < 69.25 x = 66 Point Estimate

8.6 n = 120 x = 18.72 σ = 0.873599% C.I. z.005 = 2.575

x = 18.72 Point Estimate

nzx

σ± = 18.72 ± 2.575

120

8735.0 = 8.72 ± .21 = 18.51 < µ < 18.93

8.7 N = 1500 n = 187 x = 5.3 years σ = 1.28 years95% C.I. z.025 = 1.96


x = 5.3 years Point Estimate

1−

−±N

nN

nzx

σ= 5.3 ± 1.96

11500

1871500

187

28.1

−−

=

5.3 ± .17 = 5.13 < µ < 5.47

8.8 n = 24 x = 5.625 σ = 3.2390% C.I. z.05 = 1.645

nzx

σ± = 5.625 ± 1.64524

23.3 = 5.625 ± 1.085 = 4.540 < µ < 6.710

8.9 n = 36 x = 3.306 σ = 1.1798% C.I. z.01 = 2.33

n

zxσ± = 3.306 ± 2.33

36

17.1 = 3.306 ± .454 = 2.852 < µ < 3.760

8.10 n = 36 x = 2.139 σ = .113


90% C.I. z.05 = 1.645

nzx

σ± = 2.139 ± 1.645

36

)113(. = 2.139 ± .031 = 2.108 < µ < 2.170


8.11 95% confidence interval n = 45

x = 24.533 σ = 5.124 z = + 1.96

nzx

σ± = 24.533 + 1.9645

124.5 =

24.533 + 1.497 = 23.036 < µ < 26.030

8.12 The point estimate is 0.5765. n = 41

The assumed standard deviation is 0.14

95% level of confidence: z = + 1.96

Confidence interval: 0.533647 < µ < 0.619353

Error of the estimate: 0.619353 - 0.5765 = 0.042853

8.13 n = 13 x = 45.62 s = 5.694 df = 13 – 1 = 12

95% Confidence Interval and α /2=.025

t.025,12 = 2.179

n

stx ± = 45.62 ± 2.179

13

694.5 = 45.62 ± 3.44 = 42.18 < µ < 49.06

8.14 n = 12 x = 319.17 s = 9.104 df = 12 - 1 = 11

90% confidence interval

α /2 = .05 t.05,11 = 1.796

n

stx ± = 319.17 ± (1.796)

12

104.9 = 319.17 ± 4.72 = 314.45 < µ <

323.89


8.15 n = 41 x = 128.4 s = 20.6 df = 41 – 1 = 40

98% Confidence Intervalα /2 = .01

t.01,40 = 2.423

n

stx ± = 128.4 ± 2.423

41

6.20 = 128.4 ± 7.80 = 120.6 < µ < 136.2


8.16 n = 15 x = 2.364 s2 = 0.81 df = 15 – 1 = 14

90% Confidence intervalα /2 = .05

t.05,14 = 1.761

n

stx ± = 2.364 ± 1.761

15

81.0 = 2.364 ± .409 = 1.955 < µ < 2.773

8.17 n = 25 x = 16.088 s = .817 df = 25 – 1 = 24

99% Confidence Interval

α /2 = .005

t.005,24 = 2.797

n

stx ± = 16.088 ± 2.797

25

)817(. = 16.088 ± .457 = 15.631 < µ < 16.545



8.18 n = 22 x = 1,192 s = 279 df = n - 1 = 21

98% CI and α /2 = .01 t.01,21 = 2.518

n

stx ± = 1,192 + 2.518

22

279 = 1,192 + 149.78 = 1,042.22 < µ < 1,341.78

The figure given by Runzheimer International falls within the confidence interval. Therefore, there is no reason to reject the Runzheimer figure as different from what we are getting based on this sample.

8.19 n = 20 df = 19 95% CI t.025,19 = 2.093

x = 2.36116 s = 0.19721

2.36116 + 2.09320

1972.0 = 2.36116 + 0.0923 = 2.26886 < µ < 2.45346

Point Estimate = 2.36116

Error = 0.0923

8.20 n = 28 x = 5.335 s = 2.016 df = 28 – 1 = 27

90% Confidence Interval α /2 = .05

t.05,27 = 1.703

n

stx ± = 5.335 ± 1.703

28

016.2 = 5.335 + .649 = 4.686 < µ < 5.984

8.21 n = 10 x = 49.8 s = 18.22 df = 10 – 1 = 9

95% Confidence α /2 = .025 t.025,9 = 2.262

n

stx ± = 49.8 ± 2.262

10

22.18 = 49.8 + 13.03 = 36.77 < µ < 62.83

8.22 n = 14 98% confidence α /2 = .01 df = 13


t.01,13 = 2.650

from data: x = 152.16 s = 14.42

confidence interval:n

stx ± = 152.16 + 2.65

14

42.14 =

152.16 + 10.21 = 141.95 < µ < 162.37

The point estimate is 152.16

8.23 n = 17 df = 17 – 1 = 16 99% confidence α /2 = .005

t.005,16 = 2.921

from data: x = 8.06 s = 5.07

confidence interval:n

stx ± = 8.06 + 2.921

17

07.5 =

8.06 + 3.59 = 4.47 < µ < 11.65

8.24 The point estimate is x which is 25.4134 hours. The sample size is 26 skiffs. The confidence level is 98%. The confidence interval is:

x ts

nx t

s

n± ≤ ≤ ±µ = 22.8124 < µ < 28.0145

The error of the confidence interval is 2.6.011.


8.25 a) n = 44 p̂ =.51 99% C.I. z.005 = 2.575

n

qpzp

ˆˆˆ

⋅± = .51 ± 2.57544

)49)(.51(. = .51 ± .194 = .316 < p< .704

b) n = 300 p̂ = .82 95% C.I. z.025 = 1.96

n

qpzp

ˆˆˆ

⋅± = .82 ± 1.96300

)18)(.82(. = .82 ± .043 = .777 < p < .863

c) n = 1150 p̂ = .48 90% C.I. z.05 = 1.645

n

qpzp

ˆˆˆ

⋅± = .48 ± 1.6451150

)52)(.48(. = .48 ± .024 = .456 < p < .504

d) n = 95 p̂ = .32 88% C.I. z.06 = 1.555

n

qpzp

ˆˆˆ

⋅± = .32 ± 1.55595

)68)(.32(. = .32 ± .074 = .246 < p < .394

8.26 a) n = 116 x = 57 99% C.I. z.005 = 2.575

p̂ = 116

57=n

x = .49

n

qpzp

ˆˆˆ

⋅± = .49 ± 2.575116

)51)(.49(. = .49 ± .12 = .37 < p < .61

b) n = 800 x = 479 97% C.I. z.015 = 2.17

p̂ = 800

479=n

x = .60

n

qpzp

ˆˆˆ

⋅± = .60 ± 2.17800

)40)(.60(. = .60 ± .038 = .562 < p < .638

c) n = 240 x = 106 85% C.I. z.075 = 1.44


p̂ = 240

106=n

x = .44

n

qpzp

ˆˆˆ

⋅± = .44 ± 1.44240

)56)(.44(. = .44 ± .046 = .394 < p < .486

d) n = 60 x = 21 90% C.I. z.05 = 1.645

p̂ = 60

21=n

x = .35

n

qpzp

ˆˆˆ

⋅± = .35 ± 1.64560

)65)(.35(. = .35 ± .10 = .25 < p < .45

8.27 n = 85 x = 40 90% C.I. z.05 = 1.645

p̂ = 85

40=n

x = .47

n

qpzp

ˆˆˆ

⋅± = .47 ± 1.64585

)53)(.47(. = .47 ± .09 = .38 < p < .56

95% C.I. z.025 = 1.96

n

qpzp

ˆˆˆ

⋅± = .47 ± 1.9685

)53)(.47(. = .47 ± .11 = .36 < p < .58

99% C.I. z.005 = 2.575

n

qpzp

ˆˆˆ

⋅± = .47 ± 2.57585

)53)(.47(. = .47 ± .14 = .33 < p < .61

All other things being constant, as the confidence increased, the width of the interval increased.

8.28 n = 1003 p̂ = .255 99% CI z.005 = 2.575


n

qpzp

ˆˆˆ

⋅± = .255 + 2.5751003

)745)(.255(. = .255 + .035 = .220 < p < .290

n = 10,000 p̂ = .255 99% CI z.005 = 2.575

n

qpzp

ˆˆˆ

⋅± = .255 + 2.575000,10

)745)(.255(. = .255 + .011 = .244 < p < .266

The confidence interval constructed using n = 1003 is wider than the confidence interval constructed using n = 10,000. One might conclude that, all other things being constant, increasing the sample size reduces the width of the confidence interval.

8.29 n = 560 p̂ = .47 95% CI z.025 = 1.96

n

qpzp

ˆˆˆ

⋅± = .47 + 1.96560

)53)(.47(. = .47 + .0413 = .4287 < p < .5113

n = 560 p̂ = .28 90% CI z.05 = 1.645

n

qpzp

ˆˆˆ

⋅± = .28 + 1.645560

)72)(.28(. = .28 + .0312 = .2488 < p < .3112

8.30 n = 1250 x = 997 98% C.I. z.01 = 2.33

p̂ = 1250

997=n

x = .80

n

qpzp

ˆˆˆ

⋅± = .80 ± 2.331250

)20)(.80(. = .80 ± .026 = .774 < p < .826


8.31 n = 3481 x = 927

p̂ = 3481

927=n

x = .266

a) p̂ = .266 Point Estimate

b) 99% C.I. z.005 = 2.575

n

qpzp

ˆˆˆ

⋅± = .266 + 2.5753481

)734)(.266(. = .266 ± .019 =

.247 < p < .285

8.32 n = 89 x = 48 85% C.I. z.075 = 1.44

p̂ = 89

48=n

x = .54

n

qpzp

ˆˆˆ

⋅± = .54 ± 1.4489

)46)(.54(. = .54 ± .076 = .464 < p < .616

8.33 p̂ = .63 n = 672 95% Confidence z.025 = + 1.96

n

qpzp

ˆˆˆ

⋅± = .63 + 1.96672

)37)(.63(. = .63 + .0365 = .5935 < p < .6665

8.34 n = 275 x = 121 98% confidence z.01 = 2.33

p̂ = 275

121=n

x = .44

n

qpzp

ˆˆˆ

⋅± = .44 ± 2.33275

)56)(.44(. = .44 ± .07 = .37 < p < .51


8.35 a) n = 12 x = 28.4 s2 = 44.9 99% C.I. df = 12 – 1 = 11

χ 2.995,11 = 2.60320 χ 2

.005,11 = 26.7569

7569.26

)9.44)(112( − < σ 2 <

60320.2

)9.44)(112( −

18.46 < σ 2 < 189.73

b) n = 7 x = 4.37 s = 1.24 s2 = 1.5376 95% C.I. df = 7 – 1 = 6

χ 2.975,6 = 1.23734 χ 2

.025,6 = 14.4494

4494.14

)5376.1)(17( − < σ 2 <

23734.1

)5376.1)(17( −

0.64 < σ 2 < 7.46

c) n = 20 x = 105 s = 32 s2 = 1024 90% C.I. df = 20 – 1 = 19

χ 2.95,19 = 10.11701 χ 2

.05,19 = 30.1435

1435.30

)1024)(120( − < σ 2 <

11701.10

)1024)(120( −

645.45 < σ 2 < 1923.10

d) n = 17 s2 = 18.56 80% C.I. df = 17 – 1 = 16

χ 2.90,16 = 9.31224 χ 2

.10,16 = 23.5418

5418.23

)56.18)(117( − < σ 2 <

31224.9

)56.18)(117( −

12.61 < σ 2 < 31.89


8.36 n = 16 s2 = 37.1833 98% C.I. df = 16-1 = 15

χ 2.99,15 = 5.22936 χ 2

.01,15 = 30.5780

5780.30

)1833.37)(116( − < σ 2 <

22936.5

)1833.37)(116( −

18.24 < σ 2 < 106.66

8.37 n = 20 s = 4.3 s2 = 18.49 98% C.I. df = 20 – 1 = 19

χ 2.99,19 = 7.63270 χ 2

.01,19 = 36.1908

1908.36

)49.18)(120( − < σ 2 <

63270.7

)49.18)(120( −

9.71 < σ 2 < 46.03

Point Estimate = s2 = 18.49

8.38 n = 15 s2 = 3.067 99% C.I. df = 15 – 1 = 14

χ 2.995,14 = 4.07466 χ 2

.005,14 = 31.3194

3194.31

)067.3)(115( − < σ 2 <

07466.4

)067.3)(115( −

1.37 < σ 2 < 10.54

8.39 n = 14 s2 = 26,798,241.76 95% C.I. df = 14 – 1 = 13

Point Estimate = s2 = 26,798,241.76

χ 2.975,13 = 5.00874 χ 2

.025,13 = 24.7356

7356.24

)76.241,798,26)(114( − < σ 2 <

00874.5

)76.241,798,26)(114( −

14,084,038.51 < σ 2 < 69,553,848.45


8.40 a) σ = 36 E = 5 95% Confidence z.025 = 1.96

n = 2

22

2

22

5

)36()96.1(=E

z σ = 199.15

Sample 200

b) σ = 4.13 E = 1 99% Confidence z.005 = 2.575

n = 2

22

2

22

1

)13.4()575.2(=E

z σ = 113.1

Sample 114

c) E = 10 Range = 500 - 80 = 420

1/4 Range = (.25)(420) = 105

90% Confidence z.05 = 1.645

n = 2

22

2

22

10

)105()645.1(=E

z σ = 298.3

Sample 299

d) E = 3 Range = 108 - 50 = 58

1/4 Range = (.25)(58) = 14.5


n = 2

22

2

22

3

)5.14()555.1(=E

z σ = 56.5

Sample 57


8.41 a) E = .02 p = .40 96% Confidence z.02 = 2.05

n = 2

2

2

2

)02(.

)60)(.40(.)05.2(=⋅E

qpz = 2521.5

Sample 2522 b) E = .04 p = .50 95% Confidence z.025 = 1.96

n = 2

2

2

2

)04(.

)50)(.50(.)96.1(=⋅E

qpz = 600.25

Sample 601

c) E = .05 p = .55 90% Confidence z.05 = 1.645

n = 2

2

2

2

)05(.

)45)(.55(.)645.1(=⋅E

qpz = 267.9

Sample 268 d) E =.01 p = .50 99% Confidence z.005 = 2.575

n = 2

2

2

2

)01(.

)50)(.50(.)575.2(=⋅E

qpz = 16,576.6

Sample 16,577

8.42 E = $200 σ = $1,000 99% Confidence z.005 = 2.575

n = 2

22

2

22

200

)1000()575.2(=E

z σ = 165.77

Sample 166


8.43 E = $2 σ = $12.50 90% Confidence z.05 = 1.645

n = 2

22

2

22

2

)50.12()645.1(=E

z σ = 105.7

Sample 106

8.44 E = $100 Range = $2,500 - $600 = $1,900

σ ≈ 1/4 Range = (.25)($1,900) = $475


n = 2

22

2

22

100

)475()645.1(=E

z σ = 61.05

Sample 62

8.45 p = .20 q = .80 E = .02

90% Confidence, z.05 = 1.645

n = 2

2

2

2

)02(.

)80)(.20(.)645.1(=⋅E

qpz = 1082.41

Sample 1083

8.46 p = .50 q = .50 E = .05

95% Confidence, z.025 = 1.96

n = 2

2

2

2

)05(.

)50)(.50(.)96.1(=⋅E

qpz = 384.16

Sample 385

8.47 E = .10 p = .50 q = .50


95% Confidence, z.025 = 1.96

n = 2

2

2

2

)10(.

)50)(.50(.)96.1(=⋅E

qpz = 96.04

Sample 97

8.48 x = 45.6 σ = 7.75 n = 35

80% confidence z.10 = 1.28

35

75.728.16.45 ±=±

nzx

σ = 45.6 + 1.68

43.92 < µ < 47.28


35

75.788.16.45 ±=±

nzx

σ = 45.6 + 2.46

43.14 < µ < 48.06


35

75.733.26.45 ±=±

nzx

σ = 45.6 + 3.05

42.55 < µ < 48.65


8.49 x = 12.03 (point estimate) s = .4373 n = 10 df = 9

For 90% confidence: α /2 = .05 t.05,9= 1.833

10

)4373(.833.103.12 ±=±

n

stx = 12.03 + .25

11.78 < µ < 12.28

For 95% confidence: α /2 = .025 t.025,9 = 2.262

10

)4373(.262.203.12 ±=±

n

stx = 12.03 + .31

11.72 < µ < 12.34

For 99% confidence: α /2 = .005 t.005,9 = 3.25

10

)4373(.25.303.12 ±=±

n

stx = 12.03 + .45

11.58 < µ < 12.48

8.50 a) n = 715 x = 329 95% confidence z.025 = 1.96

715

329ˆ =p = .46

715

)54)(.46(.96.146.

ˆˆˆ ±=⋅±

n

qpzp = .46 + .0365

.4235 < p < .4965

b) n = 284 p̂ = .71 90% confidence z.05 = 1.645

284

)29)(.71(.645.171.

ˆˆˆ ±=⋅±

n

qpzp = .71 + .0443

.6657 < p < .7543

c) n = 1250 p̂ = .48 95% confidence z.025 = 1.96


1250

)52)(.48(.96.148.

ˆˆˆ ±=⋅±

n

qpzp = .48 + .0277

.4523 < p < .5077

d) n = 457 x = 270 98% confidence z.01 = 2.33

457

270ˆ =p = .591

457

)409)(.591(.33.2591.

ˆˆˆ ±=⋅±

n

qpzp = .591 + .0536

.5374 < p < .6446

8.51 n = 10 s = 7.40045 s2 = 54.7667 df = 10 – 1 = 9

90% confidence, α /2 = .05 1 - α /2 = .95

χ 2.95,9 = 3.32512 χ 2

.05,9 = 16.9190

9190.16

)7667.54)(110( − < σ 2 <

32512.3

)7667.54)(110( −

29.133 < σ 2 < 148.235

95% confidence, α /2 = .025 1 - α /2 = .975

χ 2.975,9 = 2.70039 χ 2

.025,9 = 19.0228

0228.19

)7667.54)(110( − < σ 2 <

70039.2

)7667.54)(110( −

25.911 < σ 2 < 182.529


8.52 a) σ = 44 E = 3 95% confidence z.025 = 1.96

n = 2

22

2

22

3

)44()96.1(=E

z σ = 826.4

Sample 827

b) E = 2 Range = 88 - 20 = 68

use σ = 1/4(range) = (.25)(68) = 17


2

22

2

22

2

)17()645.1(=E

z σ = 195.5

Sample 196

c) E = .04 p = .50 q = .50


2

2

2

2

)04(.

)50)(.50(.)33.2(=⋅E

qpz = 848.3

Sample 849

d) E = .03 p = .70 q = .30


2

2

2

2

)03(.

)30)(.70(.)96.1(=⋅E

qpz = 896.4

Sample 897


8.53 n = 17 x = 10.765 s = 2.223 df = 17 - 1 = 16

99% confidence α /2 = .005 t.005,16 = 2.921

17

223.2921.2765.10 ±=±

n

stx = 10.765 + 1.575

9.19 < µ < 12.34

8.54 p = .40 E=.03 90% Confidence z.05 = 1.645

n = 2

2

2

2

)03(.

)60)(.40(.)645.1(=⋅E

qpz = 721.61

Sample 722

8.55 n = 17 s2 = 4.941 99% C.I. df = 17 – 1 = 16

χ 2.995,16 = 5.14216 χ 2

.005,16 = 34.2671

2671.34

)941.4)(117( − < σ 2 <

14216.5

)941.4)(117( −

2.307 < σ 2 < 15.374

8.56 n = 45 x = 213 σ = 48


45

4833.2213 ±=±

nzx

σ = 213 ± 16.67

196.33 < µ < 229.67


8.57 n = 39 x = 37.256 σ = 3.891


39

891.3645.1256.37 ±=±

nzx

σ = 37.256 ± 1.025

36.231 < µ < 38.281

8.58 σ = 6 E = 1 98% Confidence z.98 = 2.33

n = 2

22

2

22

1

)6()33.2(=E

z σ = 195.44

Sample 196

8.59 n = 1,255 x = 714 95% Confidence z.025 = 1.96

1255

714ˆ =p = .569

255,1

)431)(.569(.96.1569.

ˆˆˆ ±=⋅±

n

qpzp = .569 ± .027

.542 < p < .596


8.60 n = 41 s = 21 128=x 98% C.I. df = 41 – 1 = 40

t.01,40 = 2.423

Point Estimate = $128

41

21423.2128 ±=±

n

stx = 128 + 7.947

120.053 < µ < 135.947

Interval Width = 135.947 – 120.053 = 15.894

8.61 n = 60 x = 6.717 σ = 3.06 N = 300


1300

60300

60

06.333.2717.6

1 −−±=

−−±

N

nN

n

szx =

6.717 ± 0.825

5.892 < µ < 7.542

8.62 E = $20 Range = $600 - $30 = $570

1/4 Range = (.25)($570) = $142.50


n = 2

22

2

22

20

)50.142()96.1(=E

z σ = 195.02

Sample 196


8.63 n = 245 x = 189 90% Confidence z.05= 1.645

245

189ˆ ==

n

xp = .77

245

)23)(.77(.645.177.

ˆˆˆ ±=⋅±

n

qpzp = .77 ± .044

.726 < p < .814

8.64 n = 90 x = 30 95% Confidence z.025 = 1.96

90

30ˆ ==

n

xp

= .33

90

)67)(.33(.96.133.

ˆˆˆ ±=⋅±

n

qpzp = .33 ± .097

.233 < p < .427

8.65 n = 12 x = 43.7 s2 = 228 df = 12 – 1 = 11 95% C.I.

t.025,11 = 2.201

12

228201.27.43 ±=±

n

stx = 43.7 + 9.59

34.11 < µ < 53.29

χ 2.99,11 = 3.05350 χ 2

.01,11 = 24.7250

7250.24

)228)(112( − < σ 2 <

05350.3

)228)(112( −

101.44 < σ 2 < 821.35


8.66 n = 27 x = 4.82 s = 0.37 df = 26

95% CI: t.025,26 = 2.056

27

37.0056.282.4 ±=±

n

stx = 4.82 + .1464

4.6736 < µ < 4.9664

Since 4.50 is not in the interval, we are 95% confident that µ does not equal 4.50.

8.67 n = 77 x = 2.48 σ = 12


77

1296.148.2 ±=±

nzx

σ = 2.48 ± 2.68

-0.20 < µ < 5.16

The point estimate is 2.48

The interval is inconclusive. It says that we are 95% confident that the average arrival time is somewhere between .20 of a minute (12 seconds) early and 5.16 minutes late. Since zero is in the interval, there is a possibility that, on average, the flights are on time.

8.68 n = 560 p̂ =.33

99% Confidence z.005= 2.575

560

)67)(.33(.575.233.

ˆˆˆ ±=⋅±

n

qpzp = .33 ± .05

.28 < p < .38


8.69 p = .50 E = .05 98% Confidence z.01 = 2.33

2

2

2

2

)05(.

)50)(.50(.)33.2(=⋅E

qpz = 542.89

Sample 543

8.70 n = 27 x = 2.10 s = 0.86 df = 27 - 1 = 26

98% confidence α /2 = .01 t.01,26 = 2.479

27

86.0479.210.2 ±=±

n

stx = 2.10 ± 0.41

1.69 < µ < 2.51

8.71 n = 23 df = 23 – 1 = 22 s = .0631455 90% C.I.

χ 2.95,22 = 12.33801 χ 2

.05,22 = 33.9245

9245.33

)0631455)(.123( 2− < σ 2 <

33801.12

)0631455)(.123( 2−

.0026 < σ 2 < .0071

8.72 n = 39 x = 1.294 σ = 0.205 99% Confidence z.005 = 2.575

39

205.0575.2294.1 ±=±

nzx

σ = 1.294 ± .085

1.209 < µ < 1.379


8.73 The sample mean fill for the 58 cans is 11.9788 oz. with a standard deviation of

.0536 oz. The 99% confidence interval for the population fill is 11.9607 oz. to 11.9969 oz. which does not include 12 oz. We are 99% confident that the population mean is not 12 oz., indicating that the machine may be under filling the cans.

8.74 The point estimate for the average length of burn of the new bulb is 2198.217 hours. Eighty-four bulbs were included in this study. A 90% confidence interval can be constructed from the information given. The error of the confidence interval is + 27.76691. Combining this with the point estimate yields the 90% confidence interval of 2198.217 + 27.76691 = 2170.450 < µ < 2225.984.

8.75 The point estimate for the average age of a first time buyer is 27.63 years. The sample of 21 buyers produces a standard deviation of 6.54 years. We are 98% confident that the actual population mean age of a first-time home buyer is between 24.0222 years and 31.2378 years.

8.76 A poll of 781 American workers was taken. Of these, 506 drive their cars to work. Thus, the point estimate for the population proportion is 506/781 = .647887. A 95% confidence interval to estimate the population proportion shows that we are 95% confident that the actual value lies between .61324 and .681413. The error of this interval is + .0340865.

76197264 ken black qa 5th chapter 8 solution

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