qmb3200 ans chp 8
TRANSCRIPT
Chapter 8Statistical Inference: Estimation for Single
Populations
LEARNING OBJECTIVES
The overall learning objective of Chapter 8 is to help you understandestimating parameters of single populations, thereby enabling you to:
1. Estimate the population mean with a known population standarddeviation with the z statistic, correcting for a finite population ifnecessary.
2. Estimate the population mean with an unknown population standarddeviation using the t statistic and properties of the t distribution.
3. Estimate a population proportion using the z statistic.4. Use the chi-square distribution to estimate the population variance given
the sample variance.5. Determine the sample size needed in order to estimate the population
mean and population proportion.
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 means 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 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 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 ( 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 ( 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 = 3.5 n = 60 95% Confidence z.025 = 1.96
x±z
σ
√n
= 25 + 1.96
3 .5
√60 = 25 + 0.89 = 24.11 < µ < 25.89
b) x = 119.6 = 23.89 n = 75 98% Confidence z.01 = 2.33
x±z
σ
√n = 119.6 + 2.33
23 .89
√75 = 119.6 ± 6.43 = 113.17 < µ < 126.03
c) x = 3.419 = 0.974 n = 3290% C.I. z.05 = 1.645
x±z
σ
√n = 3.419 + 1.645
0 .974
√32 = 3.419 ± .283 = 3.136 < µ < 3.702
d) x = 56.7 = 12.1 N = 500 n = 4780% C.I. z.10 = 1.28
x±zσ√n √ N−n
N−1 = 56.7 + 1.28
12 .1√47 √500−47
500−1 =
56.7 ± 2.15 = 54.55 < µ < 58.85
8.2 n = 36 x = 211 = 2395% C.I. z.025 = 1.96
x±zσ
√n = 211 ± 1.96
23
√36 = 211 ± 7.51 = 203.49 < µ < 218.51
8.3 n = 81 x = 47 = 5.8990% C.I. z.05=1.645
x±z
σ
√n = 47 ± 1.645
5. 89
√81 = 47 ± 1.08 = 45.92 < µ < 48.08
8.4 n = 70 2 = 49 x = 90.4
x = 90.4 Point Estimate
94% C.I. z.03 = 1.88
x±zσ
√n = 90.4 ± 1.88
√49√70 = 90.4 ± 1.57 = 88.83 < µ < 91.97
8.5 n = 39 N = 200 x = 66 = 1196% C.I. z.02 = 2.05
x±zσ√n √ N−n
N−1 = 66 ± 2.05
11√39 √200−39
200−1 =
66 ± 3.25 = 62.75 < µ < 69.25
x = 66 Point Estimate
8.6 n = 120 x = 18.72 = 0.873599% C.I. z.005 = 2.575
x = 18.72 Point Estimate
x±zσ
√n
= 18.72 ± 2.575
0 .8735
√120 = 8.72 ± .21 = 18.51 < µ < 18.93
8.7 N = 1500 n = 187 x = 5.3 years = 1.28 years95% C.I. z.025 = 1.96
x = 5.3 years Point Estimate
x±z
σ√n √ N−n
N−1 = 5.3 ± 1.96
1 .28√187 √1500−187
1500−1 =
5.3 ± .17 = 5.13 < µ < 5.47
8.8 n = 24 x = 5.625 = 3.2390% C.I. z.05 = 1.645
x±zσ
√n = 5.625 ± 1.645
3. 23
√24 = 5.625 ± 1.085 = 4.540 < µ < 6.710
8.9 n = 36 x = 3.306 = 1.1798% C.I. z.01 = 2.33
x±z
σ
√n = 3.306 ± 2.33
1.17
√36 = 3.306 ± .454 = 2.852 < µ < 3.760
8.10 n = 36 x = 2.139 = .113
x = 2.139 Point Estimate
90% C.I. z.05 = 1.645
x±zσ
√n
= 2.139 ± 1.645
( . 113)√36 = 2.139 ± .031 = 2.108 < µ < 2.170
8.11 95% confidence interval n = 45
x = 24.533 = 5.124 z = + 1.96
x±zσ
√n = 24.533 + 1.96
5. 124
√45 =
24.533 + 1.497 = 23.036 < < 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 /2=.025
t.025,12 = 2.179
x±ts
√n = 45.62 ± 2.179
5. 694
√13 = 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
/2 = .05 t.05,11 = 1.796
x±ts
√n = 319.17 ± (1.796)
9 .104
√12 = 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 Interval/2 = .01
t.01,40 = 2.423
x±ts
√n = 128.4 ± 2.423
20 .6
√41 = 128.4 ± 7.80 = 120.6 < µ < 136.2
x = 128.4 Point Estimate
8.16 n = 15 x = 2.364 s2 = 0.81 df = 15 – 1 = 14
90% Confidence interval/2 = .05
t.05,14 = 1.761
x±ts
√n = 2.364 ± 1.761
√0.81√15 = 2.364 ± .409 = 1.955 < µ < 2.773
8.17 n = 25 x = 16.088 s = .817 df = 25 – 1 = 24
99% Confidence Interval
/2 = .005
t.005,24 = 2.797
x±ts
√n = 16.088 ± 2.797
( . 817 )√25 = 16.088 ± .457 = 15.631 < µ < 16.545
x = 16.088 Point Estimate
8.18 n = 51 x = 1,192 s = 279 df = n - 1 = 50
98% CI and /2 = .01 t.01,50 = 2.403
x±ts
√n = 1,192 + 2.403 = 1,192 + 93.88 = 1,098.12 < < 1,285.88
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 + 2.093
0 .1972
√20 = 2.36116 + 0.0923 = 2.26886 < < 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 /2 = .05
t.05,27 = 1.703
x±ts
√n = 5.335 ± 1.703
2.016
√28 = 5.335 + .649 = 4.686 < µ < 5.984
8.21 n = 10 x = 49.8 s = 18.22 df = 10 – 1 = 9
95% Confidence /2 = .025 t.025,9 = 2.262
x±ts
√n = 49.8 ± 2.262
18 .22
√10 = 49.8 + 13.03 = 36.77 < µ < 62.83
279
√51
8.22 n = 14 98% confidence /2 = .01 df = 13
t.01,13 = 2.650
from data: x = 152.16 s = 14.42
confidence interval:x±t
s
√n = 152.16 + 2.65
14 .42
√14 =
152.16 + 10.21 = 141.95 < < 162.37
The point estimate is 152.16
8.23 n = 41 df = 41 – 1 = 40 99% confidence /2 = .005
t.005,40 = 2.704
from data: x = 11.098 s = 8.446
confidence interval:x±t
s
√n = 11.098 + 2.704 =
11.098 + 3.567 = 7.531 < < 14.665
8.24 The point estimate is which is 25.4134 hours. The sample size is 26 skiffs. The confidence level is 98%. The confidence interval is:
= 22.8124 < < 28.0145The error of the confidence interval is 26.011.
8 .446
√41
8.25 a) n = 44 p̂ =.51 90% C.I. z.05 = 1.645
p̂±z √ p̂⋅q̂
n = .51 ± 1.645√ ( .51 )( . 49)44 = .51 ± .124 = .386 < p< .634
b) n = 300 p̂ = .82 95% C.I. z.025 = 1.96
p̂±z √ p̂⋅q̂n = .82 ± 1.96√ ( .82 )( .18 )
300 = .82 ± .043 = .777 < p < .863
c) n = 1150 p̂ = .48 90% C.I. z.05 = 1.645
p̂±z √ p̂⋅q̂n = .48 ± 1.645√ ( . 48)( . 52)
1150 = .48 ± .024 = .456 < p < .504
d) n = 95 p̂ = .32 88% C.I. z.06 = 1.555
p̂±z √ p̂⋅q̂n = .32 ± 1.555√ ( .32 )( .68 )
95 = .32 ± .074 = .246 < p < .394
8.26 a) n = 116 x = 57 99% C.I. z.005 = 2.575
p̂ =
xn=57
116 = .49
p̂±z √ p̂⋅q̂n = .49 ± 2.575√ ( . 49)( . 51)
116 = .49 ± .12 = .37 < p < .61
b) n = 800 x = 479 97% C.I. z.015 = 2.17
p̂ =
xn=479
800 = .60
p̂±z √ p̂⋅q̂n = .60 ± 2.17√ ( .60 )( . 40 )
800 = .60 ± .038 = .562 < p < .638 c) n = 240 x = 106 85% C.I. z.075 = 1.44
p̂ =
xn=106
240 = .44
p̂±z √ p̂⋅q̂n = .44 ± 1.44√ ( . 44 )( .56 )
240 = .44 ± .046 = .394 < p < .486
d) n = 60 x = 21 90% C.I. z.05 = 1.645
p̂ =
xn=21
60 = .35
p̂±z √ p̂⋅q̂n = .35 ± 1.645√ ( .35 )( . 65)
60 = .35 ± .10 = .25 < p < .45
8.27 n = 85 x = 40 90% C.I. z.05 = 1.645
p̂ =
xn=40
85 = .47
p̂±z √ p̂⋅q̂
n = .47 ± 1.645√ ( . 47 )(. 53 )85 = .47 ± .09 = .38 < p < .56
95% C.I. z.025 = 1.96
p̂±z √ p̂⋅q̂
n = .47 ± 1.96√ ( . 47 )(. 53 )85 = .47 ± .11 = .36 < p < .58
99% C.I. z.005 = 2.575
p̂±z √ p̂⋅q̂
n = .47 ± 2.575√ ( . 47 )(. 53 )85 = .47 ± .14 = .33 < p < .61
All other things being constant, as the confidence increased, the width of the interval increased.
8.28 a) n = 1003 p̂ = .255 99% CI z.005 = 2.575
p̂±z √ p̂⋅q̂
n = .255 + 2.575√ ( .255 )( .745 )1003 = .255 + .035 = .220 < p < .290
b) n = 10,000 p̂ = .255 99% CI z.005 = 2.575
p̂±z √ p̂⋅q̂
n = .255 + 2.575√ ( .255 )( .745 )10 , 000 = .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
p̂±z √ p̂⋅q̂n = .47 + 1.96√ ( . 47 )(. 53 )
560 = .47 + .0413 = .4287 < p < .5113
n = 560 p̂ = .28 90% CI z.05 = 1.645
p̂±z √ p̂⋅q̂n = .28 + 1.645√ ( .28 )( .72 )
560 = .28 + .0312 = .2488 < p < .3112
8.30 n = 1250 x = 997 98% C.I. z.01 = 2.33
p̂ =
xn=997
1250 = .80
p̂±z √ p̂⋅q̂n = .80 ± 2.33√ ( .80 )( . 20)
1250 = .80 ± .026 = .774 < p < .826
8.31 n = 3481 x = 927
p̂ =
xn=927
3481 = .266
a) p̂ = .266 Point Estimate
b) 99% C.I. z.005 = 2.575
p̂±z √ p̂⋅q̂n = .266 + 2.575√ ( .266 )( . 734 )
3481 = .266 ± .019 =
.247 < p < .285
8.32 n = 89 x = 48 85% C.I. z.075 = 1.44
p̂ =
xn=48
89 = .54
p̂±z √ p̂⋅q̂n = .54 ± 1.44√ ( .54 )( . 46 )
89 = .54 ± .076 = .464 < p < .616
8.33 p̂ = .63 n = 672 95% Confidence z.025 = + 1.96
p̂±z √ p̂⋅q̂n = .63 + 1.96√ ( .63 )( . 37)
672 = .63 + .0365 = .5935 < p < .6665
8.34 n = 275 x = 121 98% confidence z.01 = 2.33
p̂ =
xn=121
275 = .44
p̂±z √ p̂⋅q̂n = .44 ± 2.33√ ( . 44 )( .56 )
275 = .44 ± .07 = .37 < p < .51
8.35 a) n = 12 x = 28.4 s2 = 44.9 99% C.I. df = 12 – 1 = 11
2.995,11 = 2.60320 2
.005,11 = 26.7569
(12−1 )(44 . 9 )26 . 7569 < 2 <
(12−1 )(44 . 9 )2 .60320
18.46 < 2 < 189.73
b) n = 7 x = 4.37 s = 1.24 s2 = 1.5376 95% C.I. df = 7 – 1 = 6
2.975,6 = 1.23734 2
.025,6 = 14.4494
(7−1)(1 .5376 )14 . 4494 < 2 <
(7−1)(1 .5376 )1.23734
0.64 < 2 < 7.46
c) n = 20 x = 105 s = 32 s2 = 1024 90% C.I. df = 20 – 1 = 19
2.95,19 = 10.11701 2
.05,19 = 30.1435
(20−1)(1024 )30 .1435 < 2 <
(20−1)(1024 )10 .11701
645.45 < 2 < 1923.10
d) n = 17 s2 = 18.56 80% C.I. df = 17 – 1 = 16
2.90,16 = 9.31224 2
.10,16 = 23.5418
(17−1)(18 .56 )23 .5418 < 2 <
(17−1)(18 .56 )9 .31224
12.61 < 2 < 31.89
8.36 n = 16 s2 = 37.1833 98% C.I. df = 16-1 = 15
2.99,15 = 5.22936 2
.01,15 = 30.5780
(16−1)(37 . 1833)30 .5780 < 2 <
(16−1)(37 . 1833)5 .22936
18.24 < 2 < 106.66
8.37 n = 20 s = 4.3 s2 = 18.49 98% C.I. df = 20 – 1 = 19
2.99,19 = 7.63270 2
.01,19 = 36.1908
(20−1)(18 .49)36 . 1908 < 2 <
(20−1)(18 . 49)7 .63270
9.71 < 2 < 46.03
Point Estimate = s2 = 18.49
8.38 n = 15 s2 = 3.067 99% C.I. df = 15 – 1 = 14
2.995,14 = 4.07466 2
.005,14 = 31.3194
(15−1 )(3 .067 )31 .3194 < 2 <
(15−1 )(3 .067 )4 .07466
1.37 < 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
2.975,13 = 5.00874 2
.025,13 = 24.7356
(14−1)(26 ,798 ,241.76 )24 . 7356 < 2 <
(14−1)(26 , 798 ,241.76 )5.00874
14,084,038.51 < 2 < 69,553,848.45
8.40 a) = 36 E = 5 95% Confidence z.025 = 1.96
n =
z2 σ2
E2=
(1 . 96)2 (36 )2
52 = 199.15
Sample 200
b) = 4.13 E = 1 99% Confidence z.005 = 2.575
n =
z2 σ2
E2=
(2 . 575)2 (4 . 13)2
12 = 113.1
Sample 114
c) E = 10 Range = 500 - 80 = 420
1/4 Range = (.25)(420) = 105
90% Confidence z.05 = 1.645
n =
z2 σ2
E2=
(1 . 645)2 (105 )2
102 = 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 =
z2 σ2
E2=
(1 .555 )2(14 .5 )2
32 = 56.5
Sample 57
8.41 a) E = .02 p = .40 96% Confidence z.02 = 2.05
n =
z2 p⋅qE2
=(2 . 05)2 ( . 40)( . 60 )
( .02 )2 = 2521.5
Sample 2522 b) E = .04 p = .50 95% Confidence z.025 = 1.96
n =
z2 p⋅qE2
=(1 . 96)2 ( .50 )( .50 )
( . 04 )2 = 600.25
Sample 601
c) E = .05 p = .55 90% Confidence z.05 = 1.645
n =
z2 p⋅qE2
=(1 . 645)2 (. 55 )( . 45)
( . 05 )2 = 267.9
Sample 268 d) E =.01 p = .50 99% Confidence z.005 = 2.575
n =
z2 p⋅qE2
=(2 . 575)2 (. 50 )( .50 )
( . 01)2 = 16,576.6
Sample 16,577
8.42 E = $200 = $1,000 99% Confidence z.005 = 2.575
n =
z2 σ2
E2=
(2 . 575)2 (1000 )2
2002 = 165.77
Sample 166
8.43 E = $2 = $12.50 90% Confidence z.05 = 1.645
n =
z2 σ2
E2=
(1 . 645)2 (12 . 50)2
22 = 105.7
Sample 106
8.44 E = $100 Range = $2,500 - $600 = $1,900
1/4 Range = (.25)($1,900) = $475
90% Confidence z.05 = 1.645
n =
z2 σ2
E2=
(1 . 645)2 (475 )2
1002 = 61.05
Sample 62
8.45 p = .20 q = .80 E = .02
90% Confidence, z.05 = 1.645
n =
z2 p⋅qE2
=(1 . 645)2 (. 20 )( .80 )
( . 02 )2 = 1082.41
Sample 1083
8.46 p = .50 q = .50 E = .05
95% Confidence, z.025 = 1.96
n =
z2 p⋅qE2
=(1 . 96)2 ( .50 )( .50 )
( . 05 )2 = 384.16
Sample 385
8.47 E = .10 p = .50 q = .50
95% Confidence, z.025 = 1.96
n =
z2 p⋅qE2
=(1 . 96)2 ( .50 )( .50 )
( . 10 )2 = 96.04
Sample 97
8.48 x = 45.6 = 7.75 n = 35
80% confidence z.10 = 1.28
x±zσ
√n=45 .6±1 .28
7 .75
√35 = 45.6 + 1.68
43.92 < < 47.28
94% confidence z.03 = 1.88
x±zσ
√n=45 .6±1 .88
7 .75
√35 = 45.6 + 2.46
43.14 < < 48.06
98% confidence z.01 = 2.33
x±zσ
√n=45 .6±2 .33
7 .75
√35 = 45.6 + 3.05
42.55 < < 48.65
8.49 x = 12.03 (point estimate) s = .4373 n = 10 df = 9
For 90% confidence: /2 = .05 t.05,9= 1.833
x±ts
√n=12 .03±1 .833
( . 4373 )√10 = 12.03 + .25
11.78 < < 12.28
For 95% confidence: /2 = .025 t.025,9 = 2.262
x±ts
√n=12 .03±2 .262
( . 4373 )√10 = 12.03 + .31
11.72 < < 12.34
For 99% confidence: /2 = .005 t.005,9 = 3.25
x±ts
√n=12 .03±3 .25
( . 4373 )√10 = 12.03 + .45
11.58 < < 12.48
8.50 a) n = 715 x = 329 95% confidence z.025 = 1.96
p̂=329715 = .46
p̂±z √ p̂⋅q̂n
=. 46±1 .96 √( . 46 )( .54 )715 = .46 + .0365
.4235 < p < .4965
b) n = 284 p̂ = .71 90% confidence z.05 = 1.645
p̂±z √ p̂⋅q̂n
=. 71±1 .645√ ( .71 )( .29 )284 = .71 + .0443
.6657 < p < .7543
c) n = 1250 p̂ = .48 95% confidence z.025 = 1.96
p̂±z √ p̂⋅q̂n
=. 48±1. 96√ ( . 48 )(. 52)1250 = .48 + .0277
.4523 < p < .5077
d) n = 457 x = 270 98% confidence z.01 = 2.33
p̂=270457 = .591
p̂±z √ p̂⋅q̂n
=. 591±2 .33√ ( .591 )( . 409)457 = .591 + .0536
.5374 < p < .6446
8.51 n = 10 s = 7.40045 s2 = 54.7667 df = 10 – 1 = 9
90% confidence, /2 = .05 1 - /2 = .95
2.95,9 = 3.32512 2
.05,9 = 16.9190
(10−1 )(54 . 7667)16 . 9190 < 2 <
(10−1 )(54 . 7667)3 . 32512
29.133 < 2 < 148.235
95% confidence, /2 = .025 1 - /2 = .975
2.975,9 = 2.70039 2
.025,9 = 19.0228
(10−1 )(54 .7667)19 .0228 < 2 <
(10−1 )(54 .7667)2 .70039
25.911 < 2 < 182.529
8.52 a) = 44 E = 3 95% confidence z.025 = 1.96
n =
z2 σ2
E2=
(1 . 96)2 (44 )2
32 = 826.4
Sample 827
b) E = 2 Range = 88 - 20 = 68
use = 1/4(range) = (.25)(68) = 17
90% confidence z.05 = 1.645
z2 σ2
E2=
(1 . 645)2 (17 )2
22 = 195.5
Sample 196
c) E = .04 p = .50 q = .50
98% confidence z.01 = 2.33
z2 p⋅qE2
=(2 . 33)2 (. 50 )( .50 )
( . 04 )2 = 848.3
Sample 849
d) E = .03 p = .70 q = .30
95% confidence z.025 = 1.96
z2 p⋅qE2
=(1 . 96)2 ( .70 )( .30 )
( . 03 )2 = 896.4
Sample 897
8.53 n = 17 x = 10.765 s = 2.223 df = 17 - 1 = 16
99% confidence /2 = .005 t.005,16 = 2.921
x±ts
√n=10 .765±2 .921
2. 223
√17 = 10.765 + 1.575
9.19 < µ < 12.34
8.54 p = .40 E=.03 90% Confidence z.05 = 1.645
n =
z2 p⋅qE2
=(1 . 645)2 (. 40)( . 60)
(. 03 )2 = 721.61
Sample 722
8.55 n = 17 s2 = 4.941 99% C.I. df = 17 – 1 = 16
2.995,16 = 5.14216 2
.005,16 = 34.2671
(17−1)(4 .941 )34 . 2671 < 2 <
(17−1)(4 .941 )5 .14216
2.307 < 2 < 15.374
8.56 n = 45 x = 213 = 48
98% Confidence z.01 = 2.33
x±zσ
√n=213±2 .33
48
√45 = 213 ± 16.67 196.33 < µ < 229.67
8.57 n = 39 x = 37.256 = 3.891
90% confidence z.05 = 1.645
x±zσ
√n=37 .256±1. 645
3 .891
√39 = 37.256 ± 1.025 36.231 < µ < 38.281
8.58 = 6 E = 1 98% Confidence z.98 = 2.33
n =
z2 σ2
E2=
(2 . 33)2 (6)2
12 = 195.44
Sample 196
8.59 n = 1,255 x = 714 95% Confidence z.025 = 1.96
p̂=7141255 = .569
p̂±z √ p̂⋅q̂n
=. 569±1 . 96√( . 569 )(. 431)1,255 = .569 ± .027
.542 < p < .596
8.60 n = 41 s = 21 x=128 98% C.I. df = 41 – 1 = 40
t.01,40 = 2.423
Point Estimate = $128
x±ts
√n=128±2. 423
21
√41 = 128 + 7.947
120.053 < < 135.947
Interval Width = 135.947 – 120.053 = 15.894
8.61 n = 60 x = 6.717 = 3.06 N = 300
98% Confidence z.01 = 2.33
x±zs
√n √ N−nN−1
=6 .717±2 .333 .06√60 √300−60
300−1 =
6.717 ± 0.825
5.892 < µ < 7.542
8.62 E = $20 Range = $600 - $30 = $570
1/4 Range = (.25)($570) = $142.50
95% Confidence z.025 = 1.96
n =
z2 σ2
E2=
(1 . 96)2 (142. 50 )2
202 = 195.02
Sample 196
8.63 n = 245 x = 189 90% Confidence z.05= 1.645
p̂= xn=189
245 = .77
p̂±z √ p̂⋅q̂n
=. 77±1 . 645√( . 77 )( .23 )245 = .77 ± .044
.726 < p < .814
8.64 n = 90 x = 30 95% Confidence z.025 = 1.96
p̂=xn=30 ¿
90 ¿¿
¿ = .33
p̂±z √ p̂⋅q̂n
=. 33±1 . 96√ (. 33 )( .67 )90 = .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
x±ts
√n=43.7±2.201 √228
√12 = 43.7 + 9.59
34.11 < < 53.29
2.99,11 = 3.05350 2
.01,11 = 24.7250
(12−1 )(228)24 . 7250 < 2 <
(12−1 )(228)3 .05350
101.44 < 2 < 821.35
8.66 n = 27 x = 4.82 s = 0.37 df = 26
95% CI: t.025,26 = 2.056
x±ts
√n=4 .82±2 .056
0 .37
√27 = 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 = 12
95% Confidence z.025 = 1.96
x±zσ
√n=2 .48±1.96
12
√77 = 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
p̂±z √ p̂⋅q̂n
=. 33±2 .575√( . 33)( . 67 )560 = .33 ± .05
.28 < p < .38
8.69 p = .50 E = .05 98% Confidence z.01 = 2.33
z2 p⋅qE2
=(2 . 33)2 (. 50 )( .50 )
( . 05 )2 = 542.89
Sample 543
8.70 n = 27 x = 2.10 s = 0.86 df = 27 - 1 = 26
98% confidence /2 = .01 t.01,26 = 2.479
x±ts
√n=2 .10±2 .479
0 .86
√27 = 2.10 ± 0.41 1.69 < µ < 2.51
8.71 n = 23 df = 23 – 1 = 22 s = .0631455 90% C.I.
2.95,22 = 12.33801 2
.05,22 = 33.9245
(23−1)( . 0631455)2
33 . 9245 < 2 <
(23−1)( . 0631455)2
12 .33801
.0026 < 2 < .0071
8.72 n = 39 x = 1.294 = 0.205 99% Confidence z.005 = 2.575
x±zσ
√n=1 . 294±2.575
0 . 205
√39 = 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.