08 ch ken black solution

Chapter 8: Statistical Inference: Estimation for Single Populations 1

Chapter 8 Statistical 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 to estimate population means when σ is unknown. Emphasize that this applies only when the population is normally distributed. 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. Here 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, exercise extreme caution is using this technique. Some statisticians omit this technique from consideration. 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 for 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 Estimate Chi-square Distribution Robust Degrees of Freedom(df) Sample-Size Estimation Error of Estimation t Distribution Interval 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

x + zn

σ = 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

x + zn

σ = 119.6 + 2.33

75

89.2 = 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

x + zn

σ = 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

x ± z1−

−N

nN

n

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 = 23 95% C.I. z.025 = 1.96

x ± zn

σ = 211 ± 1.96

36

2 = 211 ± 7.51 = 203.49 < µ < 218.51

8.3 n = 81 x = 47 σ = 5.89 90% C.I. z.05=1.645

x ± zn

σ = 47 ± 1.645

81

89.5 = 47 ± 1.08 = 45.92 < µ < 48.08

8.4 n = 70 σ2 = 49 x = 90.4 x = 90.4 Point Estimate 94% C.I. z.03 = 1.88

x + zn

σ = 90.4 ± 1.88

70

49 = 90.4 ± 1.57 = 88.83 < µ < 91.97

8.5 n = 39 N = 200 x = 66 σ = 11 96% C.I. z.02 = 2.05

x ± z1−

−N

nN

n

σ = 66 ± 2.05

1200

39200

9

11

−−

=

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


8.6 n = 120 x = 18.72 σ = 0.8735 99% C.I. z.005 = 2.575 x = 18.72 Point Estimate

x + zn

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 years 95% C.I. z.025 = 1.96 x = 5.3 years Point Estimate

x ± z1−

−N

nN

n

σ = 5.3 ± 1.96

11500

1871500

187

28.1

−−

=

5.3 ± .17 = 5.13 < µ < 5.47 8.8 n = 24 x = 5.656 σ = 3.229 90% C.I. z.05 = 1.645

x ± zn

σ = 5.625 ± 1.645

24

23.3 = 5.625 ± 1.085 = 4.540 < µ < 6.710

8.9 n = 36 x = 3.306 σ = 1.17 98% C.I. z.01 = 2.33

x ± zn

σ = 3.306 ± 2.33

36

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


8.10 n = 36 x = 2.139 σ = .113 x = 2.139 Point Estimate 90% C.I. z.05 = 1.645

x ± zn

σ = 2.139 ± 1.645

36

)113(. = 2.139 ± .03 = 2.109 < µ < 2.169

8.11 µ = 27.4 95% confidence interval n = 45 x = 24.533 σ = 5.124 z = + 1.96

Confidence interval: x + zn

σ = 24.533 + 1.96

45

124.5 =

24.533 + 1.497 = 23.036 < µµµµ < 26.030 8.12 The point estimate is 0.5765. n = 41 The assumed standard deviation is 0.1394 99% level of confidence: z = + 1.96 Confidence interval: 0.5336 < µ < 0.6193 Error of the estimate: 0.6193 - 0.5765 = 0.0428


8.13 n = 13 x = 45.62 s = 5.694 df = 13 – 1 = 12 95% Confidence Interval 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


n

stx ± = 128.4 ± 2.423

27

6.20 = 128.4 ± 9.61 = 118.79 < µ < 138.01

x = 128.4 Point Estimate


8.16 n = 15 x = 2.364 s2 = 0.81 df = 15 – 1 = 14 90% Confidence interval α/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

x = 16.088 Point Estimate 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 < µµµµ < 1,341.78


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 < µµµµ < 2.45346

Point Estimate = 2.36116 Error = 0.0923


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 < µµµµ < 162.37 The point estimate is 152.16 8.23 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


8.24 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


8.25 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 ± .106 = .364 < p < .576

99% C.I. z.005 = 2.575

n

qpzp

ˆˆˆ

⋅± = .47 ± 2.57585

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

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

8.26 n = 1003 p̂ = .245 99% CI z.005 = 2.575

n

qpzp

ˆˆˆ

⋅± = .245 + 2.5751003

)755)(.245(. = .245 + .035 = .21 < p < .28


8.27 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

8.28 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

8.29 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 ± .02 =

.246 < p < .286


8.30 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

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

n

qpzp

ˆˆˆ

⋅± = .63 + 1.96672

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

8.32 a) n = 12 x = 28.4 s2 = 44.9 99% C.I. df = 12 – 1 = 11 χ2

.995,11 = 2.60321 χ2.005,11 = 26.7569

7569.26

)9.44)(112( − < σ2 <

60321.2

)9.44)(112( −

18.46 < σ2 < 189.73

b) n = 7 x = 4.37 s = 1.24 s2 = 1.5376 95% C.I. df = 12 – 1 = 11

χ2

.975,6 = 1.237347 χ2.025,6 = 14.4494

4494.14

)5376.1)(17( − < σ2 <

237347.1

)5376.1)(17( −

0.64 < σ2 < 7.46

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


χ2.95,19 = 10.117 χ2

.05,19 = 30.1435

1435.30

)1024)(120( − < σ2 <

117.10

)1024)(120( −

645.45 < σ2 < 1923.10 d) n = 17 s2 = 18.56 80% C.I. df = 17 – 1 = 16 χ2

.90,16 = 9.31223 χ2.10,16 = 23.5418

5418.23

)56.18)(117( − < σ2 <

31223.9

)56.18)(117( −

12.61 < σσσσ2 < 31.89 8.33 n = 16 s2 = 37.1833 98% C.I. df = 16-1 = 15 χ2

.99,15 = 5.22935 χ2.01,15 = 30.5779

5779.30

)1833.37)(116( − < σ2 <

22935.5

)1833.37)(116( −

18.24 < σσσσ2 < 106.66 8.34 n = 12 s = 4.3 s2 = 18.49

98% C.I. df = 20 – 1 = 19 χ2

.99,19 = 7.63273 χ2.01,19 = 36.1980

1980.36

)49.18)(120( − < σ2 <

63273.7

)49.18)(120( −

9.71 < σ2 < 46.03 Point Estimate = s2 = 18.49 8.35 n = 152 s2 = 3.067 99% C.I. df = 15 – 1 = 14 χ2

.995,14 = 4.07468 χ2.005,14 = 31.3193


3193.31

)067.3)(115( − < σ2 <

07468.24

)067.3)(115( −

1.37 < σ2 < 10.54 8.36 n = 14 s2 = 26,798,241.76 95% C.I. df = 14 – 1 = 13 Point Estimate = s2 = 26,798,241.76 χ2

.975,13 = 5.00874 χ2.025,13 = 24.7356

7356.24

)76.241,798,26)(114( − < σ2 <

00874.5

)76.241,798,26)(114( −

14,084,038.51 < σ2 < 69,553,848.45 8.37 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 88% Confidence z.06 = 1.555

n = 2

22

2

22

3

)5.14()555.1(=E

z σ = 56.5

Sample 57 8.38 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 8.39 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 8.40 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 8.41 E = $100 Range = $2,500 - $600 = $1,900 σ ≈ 1/4 Range = (.25)($1,900) = $475 90% Confidence z.05 = 1.645

n = 2

22

2

22

100

)475()645.1(=E

z σ = 61.05

Sample 62


8.42 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 8.43 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 8.44 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


8.45 x = 45.6 σ = 7.75 n = 35 80% confidence z.10 = 1.28

35

75.728.16.45 ±=±

nzx

σ = 45.6 + 1.677

43.923 < µµµµ < 47.277 94% confidence z.03 = 1.88

35

75.788.16.45 ±=±

nzx

σ = 45.6 + 2.463

43.137 < µµµµ < 48.063 98% confidence z.01 = 2.33

35

75.733.26.45 ±=±

nzx

σ = 45.6 + 3.052

42.548 < µµµµ < 48.652

8.46 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 < µµµµ < 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 < µµµµ < 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 < µµµµ < 12.48 8.47 a) n = 715 x = 329

715

329ˆ =p = .46

95% confidence z.025 = 1.96

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

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

χ2

.95,9 = 3.32511 χ2.05,9 = 16.919

919.16

)7667.54)(110( − < σ2 <

32511.3

)7667.54)(110( −

29.133 < σ2 < 148.236 95% confidence, α/2 = .025 1 - α/2 = .975 χ2

.975,9 = 2.70039 χ2.025,9 = 19.0228

0228.19

)7667.54)(110( − < σ2 <

70039.2

)7667.54)(110( −

4.258 < σ2 < 182.529


8.49 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 = 1/4(range) = (.25)(68) = 17 90% confidence z.05 = 1.645

2

22

2

22

2

)17()645.1(=E

z σ = 195.5

Sample 196 c) E = .04 p = .50 q = .50 98% confidence z.01 = 2.33

2

2

2

2

)04(.

)50)(.50(.)33.2(=⋅E

qpz = 848.3

Sample 849 d) E = .03 p = .70 q = .30 95% confidence z.025 = 1.96

2

2

2

2

)03(.

)30)(.70(.)96.1(4=⋅E

qpz = 896.4

Sample 897


8.50 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 8.51 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 8.52 n = 17 s2 = 4.941 99% C.I. df = 17 – 1 = 16 χ2

.995,16 = 5.14224 χ2.005,16 = 34.2672

2672.34

)941.4)(117( − < σ2 <

14224.5

)941.4)(117( −

2.307 < σ2 < 15.374

8.53 n = 45 x = 213 σ = 48 98% Confidence z.01 = 2.33

45

4833.2213±=±

nzx

σ = 213 ± 16.67

196.33 < µ < 229.67


8.54 n = 39 x = 37.256 σ = 3.891 90% confidence z.05 = 1.645

39

891.3645.1256.37 ±=±

nzx

σ = 37.256 ± 1.025

36.231 < µ < 38.281 8.55 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 8.56 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


8.57 n = 41 s = 21 x = 128 98% C.I. df = 41 – 1 = 40 t.01,40 = 2.423 Point Estimate = $128

25

21423.2128±=±

n

stx = 128 + 7.947

120.053 < µµµµ < 135.947 Interval Width = 135.947 – 120.053 = 15.894

8.58 n = 60 x = 6.717 σ = 3.06 N =300 98% Confidence z.01 = 2.33

1300

60300

60

06.333.2717.6

1 −−±=

−−±

N

nN

nzx

σ =

6.717 ± 0.825 = 5.892 < µ < 7.542 8.59 E = $20 Range = $600 - $30 = $570 1/4 Range = (.25)($570) = $142.50 95% Confidence z.025 = 1.96

n = 2

22

2

22

20

)50.142()96.1(=E

z σ = 195.02

Sample 196


8.60 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 8.61 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

8.62 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 < µµµµ < 53.29 χ2

.975,11 = 3.81575 χ2.025,11 = 21.92

92.21

)228)(112( − < σ2 <

81575.3

)228)(112( −

114.42 < σ2 < 657.28


8.63 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 We are 95% confident that µ does not equal 4.50. 8.64 n = 77 x = 2.48 σ = 12 95% Confidence z.025 = 1.96

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.

8.65 n = 560 p̂ =.33 99% Confidence z.005= 2.575

560

)67)(.33(.575.233.

ˆˆˆ ±=⋅±

n

qpzp = .33 ± (2.575) = .33 ± .05

.28 < p < .38


8.66 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 8.67 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 ± (2.479) = 2.10 ± 0.41

1.69 < µ < 2.51 8.68 n = 23 df = 23 – 1 = 22 s = .0631455 90% C.I. χ2

.95,22 = 12.338 χ2.05,22 = 33.9244

9244.33

)0631455)(.123( 2− < σ2 <

338.12

)0631455)(.123( 2−

.0026 < σ2 < .0071 8.69 n = 39 x = 1.294 σ = 0.205 99% Confidence z.005 = 2.575

39

205.0575.2294.1 ±=±

nzx

σ = 1.294 ± (2.575) = 1.294 ± .085

1.209 < µ < 1.379


8.70 The sample mean fill for the 58 cans is 11.9788 oz. with a standard deviation of .0556 oz. The 99% confidence interval for the population fill is 11.9607 oz. to 11.9970 oz. which does not include 12 oz. We are 99% confident that the population mean is not 12 oz. indicating an underfill from the machine. 8.71 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. 8.72 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 24.02 years and 31.24 years. 8.73 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 = .648. A 95% confidence interval to estimate the population proportion shows that we are 95% confident that the actual value lies between .613 and .681. The error of this interval is + .034.

08 ch ken black solution

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