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```					CHAPTER 14

STATISTICAL INFERENCE: REVIEW
OF CHAPTERS 12 AND 13

SECTION 1

MULTIPLE CHOICE QUESTIONS

In the following multiple-choice questions, please circle the correct answer.

1.     When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population proportions. The two sample proportions are
p1  0.20 and p2  0.15 , and the standard error of the sampling distribution of
ˆ             ˆ
p1  p2 is 0.25. The calculated value of the test statistic will be:
ˆ     ˆ
a. z = 2.0
b. z = 1.15
c. t = 2.0
d. t = 1.2

2.     A random sample of size 15 taken from a normally distributed population resulted in a
sample variance of 25. The upper limit of a 99% confidence interval for the population
variance would be:
a. 12.868
b. 92.032
c. 85.896
d. 75.100

489
490   Chapter Fourteen

3.    From a sample of 500 items, 30 were found to be defective. The point estimate of the
population proportion defective will be:
a. 0.06
b. 30.0
c. 16.667
d. None of the above

4.    A random sample of 20 observations taken from a normally distributed population
revealed a sample mean of 65 and a sample variance of 16. The lower limit of a 90%
confidence interval for the population mean would equal:
a. 66.546
b. 63.454
c. 63.812
d. 66.188

5.    In testing the hypotheses H 0 : p  0.50 vs. H1 : p  0.50 , at the 10% significance level, if
the sample proportion is 0.56, and the standard error of the sample proportion is 0.025,
then the appropriate conclusion would be:
a. to reject H 0
b. not to reject H 0
c. to reject H 1
d. to reject both H 0 and H 1

6.    Two independent samples of sizes 35 and 40 are randomly selected from two normally
distributed populations. Assume that the population variances are unknown but equal. In
order to test the difference between the population means, 1   2 , the sampling
distribution of the sample mean difference, x1  x2 , is:
a. normally distributed
b. t-distributed with 75 degrees of freedom
c. t-distributed with 73 degrees of freedom
d. F-distributed with 34 and 39 degrees of freedom

7.    A sample of size 400 had 60 successes. The upper limit of a 90% confidence interval for
the population proportion is:
a. 0.1206
b. 0.1794
c. 0.1271
d. 0.1729
Statistical Inference: Review of Chapters 12 and 13                      491

8.    In testing the null hypothesis: H 0 : p1  p2  0 , if H 0 is true, the test could lead to:
a. a Type I error
b. a Type II error
c. either a Type I or a Type II error
d. neither a Type I nor a Type II error

9.    In a hypothesis test for the population variance, the hypotheses are H 0 :  2  25 vs.
H1 :  2  25 . If the sample size is 15 and the test is being carried out at the 5% level of
significance, the null hypothesis will be rejected if:
a.  2  6.5706
b.  2  24.9958
c.  2  7.2609
d.  2  23.6848

10.   Two independent samples of sizes 50 and 50 are randomly selected from two populations
to test the difference between the population means 1   2 . The sampling distribution of
the sample mean difference x1  x2 is:
a. normally distributed
b. approximately normal
c. t-distributed with 98 degrees of freedom
d. chi-squared distributed with 99 degrees of freedom

11.   Two independent samples of sizes 20 and 25 are randomly selected from two normal
populations with equal variances. In order to test the difference between the population
means, the test statistic is:
a. a standard normal random variable
b. approximately standard normal random variable
c. Student t distributed with 45 degrees of freedom
d. Student t distributed with 43 degrees of freedom

12.   The number of degrees of freedom associated with the t-test, when the data are gathered
from a matched pairs experiment with 15 pairs, is:
a. 30
b. 15
c. 28
d. 14
492   Chapter Fourteen

13.   Based on sample data, the 95% confidence interval limits for the population mean are
LCL = 124.6 and UCL = 148.2. If the 5% level of significance were used in testing the
hypotheses: H 0 :   140 vs. H1 :   140 , the null hypothesis:
a. would be rejected
b. would not be rejected
c. would have to be revised
d. None of the above

14.   After calculating the sample size needed to estimate a population proportion to within
0.05, you have been told that the maximum allowable error must be reduced to just 0.025.
If the original calculation led to a sample size of 1000, the sample size will now have to
be:
a. 2000
b. 4000
c. 1000
d. 8000

15.   In a hypothesis test for the population variance, the hypotheses are H 0 :  2  175 vs.
H 1 :  2  175 . If the sample size is 25 and the test is being carried out at the 5% level of
significance, the rejection region will be:
a.  2 < 15.6587 or  2 > 33.1963
b.  2 <12.4011 or  2 >39.3641
c.  2 < 16.4734 or  2 >34.3816
d.  2 < 13.1197 or  2 <37.6525

16.   In testing for the equality of two population variances, when the populations are normally
distributed, the 5% level of significance has been used. To determine the rejection region,
it will be necessary to refer to the F table corresponding to an upper-tail area of:
a. 0.950
b. 0.050
c. 0.025
d. 0.100

17.   A random sample of 30 observations is selected from a normally distributed population.
The sample variance is 12. In the 90% confidence interval for the population variance,
the upper limit will be:
a. 15.176
b. 8.177
c. 19.652
d. 16.941
Statistical Inference: Review of Chapters 12 and 13                  493

18.   In testing the hypotheses: H 0 :   76.5 vs. H1 :   76.5 , suppose that we rejected the
null hypothesis at  = .05. Then for which of the following  values do we also reject
the null hypothesis?
a. 0.025
b. 0.010
c. 0.100
d. All  values that are smaller than .05

19.   For a sample of size 25 observations taken from a normally distributed population with
standard deviation of 6, a 95% confidence interval estimate for the population mean
would require the use of:
a. t = 2.064
b. t = 1.711
c. z = 1.40
d. z = 1.96

20.   When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population proportions, but your statistical software provides only
a one-tail area of .058 as part of its output. The p-value for this test will be:
a. 0.058
b. 0.116
c. 0.029
d. 0.942

21.   In constructing a 95% interval estimate for the ratio of two population variances,  12 /  2 ,
2

two independent samples of sizes 30 and 40 are drawn from the populations. If the
sample variances are 425 and 675, then the upper confidence limit is about:
a. 1.2215
b. 0.3132
c. 1.2656
d. 0.3246

22.   A sample of size 200 from population 1 has 50 successes. A sample of size 200 from
population 2 has 40 successes. The value of the test statistic for testing the null hypothesis
that the proportion of successes in population 1 exceeds the proportion of successes in
population 2 by 0.025 is:
a. 1.96
b. 1.25
c. 0.5998
d. 1.20
494   Chapter Fourteen

23.   In constructing 90% confidence interval estimate for the difference between the means of
two normally distributed populations, where the unknown population variances are
assumed not to be equal, summary statistics computed from two independent samples are
as follows: n1  40 , x1  95 , s1  12.5 , n2  30 , x2  75 , and s2  35.5 . The lower
confidence limit is:
a. 30.086
b. 8.542
c. 0.914
d. 31.458

24.   Which of the following is a required condition for using the normal approximation to the
binomial in constructing interval estimate for the difference between two population
proportions?
a. n1 p1  5, n1 (1  p1 )  5, n2 p2  5, and n2 (1  p2 )  5
ˆ               ˆ            ˆ                   ˆ
b. n1 p1  30 and n2 p2  30
c. n1 p1  5 and n2 p2  5
d. n1 p1  5, n1 (1  p1 )  5, n2 p2  5, and n2 (1  p2 )  5

25.   Suppose that a one-tail t-test is being applied to find out if the population mean is at least
75. The level of significance is .05 and 20 observations were sampled. The rejection
region is:
a. t > 1.729
b. t < 2.086
c. t > 2.093
d. t < 1.725

26.   In testing the hypotheses H 0 :   150 vs. H1 :   150 , the sample mean is found to be
125. The null hypothesis:
a. should be rejected
b. should not be rejected
c. should be rejected only if n > 30
d. None of the above

27.   In testing the difference between two population means, for which the population
variances are unknown and assumed to be equal, two independent samples are drawn
from the populations. Which of the following tests is appropriate?
a. z-test
b. Equal-variances t-test
c. F-test
d. Matched pairs t-test
Statistical Inference: Review of Chapters 12 and 13        495

28.   When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population proportions. The two sample proportions are
p1  0.25 and p2  0.20 , and the sample sizes are n1  160 and n2  200. Then the pooled
ˆ             ˆ
estimate of the population proportion is
a. 0.250
b. 0.225
c. 0.222
d. 0.200

29.   When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population proportions. If the value of the test statistic is 1.86,
then the p-value is:
a. 0.4686
b. 0.0314
c. 0.0628
d. 0.0942

30.   Two samples of sizes 32 and 38 are independently drawn from two normal populations,
where the unknown population variances are assumed to be equal. The number of degrees
of freedom of the equal-variances t test statistic is:
a. 70
b. 68
c. 66
d. 64

31.   Which of the following statements is correct regarding the percentile points of the F-
distribution?
a. F0.10,10,15  1/ F0.90,15,10
b. F0.10,10,15  1/ F0.90,10,15
c. F0.90,10,15  1/ F0.10,10,15
d. F0.90,10,15  1/ F0.90,15,10

32.   A sample of size 125 selected from one population has 55 successes, and a sample of size
140 selected from a second population has 70 successes. The test statistic for testing the
equality of the population proportions equal to:
a. -0.060
b. -0.977
c. -0.940
d. -0.472
496   Chapter Fourteen

33.   In testing whether the means of two normal populations are equal, summary statistics
computed for two independent samples are as follows: n1  20 , x1  10.8 , s1  0.90 ,
n2  18 , x2  9.6 , and s2  1.10 . Assume that the population variances are unequal.
Then, the standard error of the sampling distribution of the sample mean difference
x1  x2 equals to:
a. 0.3247
b. 0.3282
c. 0.1054
d. 0.1125

34.   Assuming that all necessary conditions are met, what needs to be changed in the formula
( x1  x2 )  z / 2 s 2 ( n1  n12 ) so that we can use it to construct a confidence interval
p
1

estimate for the difference of two population means when the population variances are
assumed to be equal?
a. The x1  x2 should be replaced by 1   2
b. The z  / 2 should be replaced by z
c. The z  / 2 should be replaced by t / 2
d. The s 2 should be replaced by s12  s 2
p
2

35.   In testing the difference between two population means using two independent samples,
the population standard deviations are assumed to be known and the calculated test
statistic equals 1.05. If the test is upper-tail and 10% level of significance has been
specified, the conclusion should be:
a. reject the null hypothesis
b. do not to reject the null hypothesis
c. choose two other independent samples
d. None of the above
Statistical Inference: Review of Chapters 12 and 13                  497

TRUE / FALSE QUESTIONS

36.   When testing the equality of two population variances, the null hypothesis would be
H0 : 12   2  0 .
2

37.   We use the F-test to determine whether two population variances are equal.

38.   When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population means, but your statistical software provides only a
one-tail area of 0.0409 as part of its output. The p-value for this test will be 0.0818.

39.   A two-tail test of the population proportion produces a test statistic z = -1.57. The p-value
of the test is equal to 0.0582.

40.   When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population proportions. The two sample proportions are
p1  0.32 and p2  0.38 , and the standard error of the sampling distribution of p1  p2 is
0.046. The calculated value of the test statistic will be 1.3043.

41.   In a one-tail test, the p-value is found to be equal to 0.0624. If the test had been two-tail,
the p-value would have been 0.0312.

42.   When the necessary conditions are met, a two-tail test is being conducted at  = 0.10 to
test H0 : 12 /  2  1 . The two sample variances are s12  736 and s2  1024 , and the sample
2                                                   2

sizes are n1  16 and n2  25 . The rejection region is F > 2.11 or F < 0.4367.

43.   If a sample has 20 observations and a 95% confidence estimate for  is needed, the
appropriate t-score is 2.093

44.   When the necessary conditions are met, a two-tail test is being conducted to test the
difference between two population proportions. If the value of the test statistic z is 0.97,
then the p-value is 0.332.

45.   Both the equal-variances and unequal variances t-test statistic of 1  2 require that the
two populations be Student t distributed.
498   Chapter Fourteen

46.   If a sample has 12 observations and a 90% confidence estimate for  is needed, the
appropriate t-score is 1.363.

47.   The equal-variances test statistic of 1  2 is Student t distributed with n1 + n2 -2 degrees
of freedom, provided that the two populations are normal.

48.   If a sample has 300 observations and a 96.6% confidence estimate for p is needed, the
appropriate z-score is 2.12.

49.   If a null hypothesis about the population proportion p is rejected at the 0.05 level of
significance, it must be rejected at the 0.10 level.

50.   When the necessary conditions are met, a two-tail test is being conducted at  = 0.025 to
test H0 : 12 /  2  1 . The two sample variances are s12  375 and s2  625 , and the sample
2                                                    2

sizes are n1  36 and n2  36 . The calculated value of the test statistic will be F = 0.60.

51.   The pooled-variance estimator, s 2 , requires that the two population variances be equal.
p

52.   If we reject a null hypothesis at the 0.05 level of significance, then we must also reject it
at the 0.10 level.

53.   If a sample of size 36 is selected, the value of A for the probability P(-A  t  A) = 0.99
is 2.724.

54.   If a sample of size 400 is selected, the value of A for the probability P(-A  t  A) = 0.95
is 1.96.

55.   The upper limit of the 89.9% confidence interval for the population proportion p, given
ˆ
that n = 80; and p = 0.40 is 0.4898.

56.   The lower limit of the 87.4% confidence interval for the population proportion p, given
ˆ
that n = 250; and p = 0.15 is 0.1492.
Statistical Inference: Review of Chapters 12 and 13             499

57.   The number of degrees of freedom associated with the t test, when the data are gathered
from a matched pairs experiment with 9 pairs, is 16.

58.   If a sample has 25 observations and a 99% confidence estimate for  is needed, the
appropriate t-score is 2.797.

59.   If a sample of size 25 is selected, the value of A for the probability P(t  A) = 0.05 is
1.708.

60.   Two samples of size 30 each are independently drawn from two normal populations,
where the unknown population variances are assumed to be equal. The number of degrees
of freedom of the equal-variances t-test statistic is 59.
500      Chapter Fourteen

STATISTICAL CONCEPTS & APPLIED QUESTIONS

FOR QUESTIONS 61 THROUGH 69, USE THE FOLLOWING NARRATIVE:
Narrative: Fitness Training
There are different approaches to fitness training. To judge which one of two approaches is
better, 200 twenty-five year old men are randomly selected to participate in an experiment. For
four weeks, 100 men are trained by approach 1 while the other 100 men are trained by approach
2. The percentage improvement in fitness was measured for each man and the statistics shown
below were computed. The percentage figures are known to be normally distributed.
Approach 1            Approach 2
x1  27.3             x2  33.6
s12  47.614           s 2  28.09
2

61.     {Fitness Training Narrative} Determine whether these data are sufficient to infer at the
5% significance level that the two population variances differ.

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F > F0.025,99,99  1.486 or F < F0.975,99,99  1/ F0.025,99,99  0.673
Test statistics: F = 1.6949
Conclusion: Reject the null hypothesis. Yes

62.     {Fitness Training Narrative} Estimate with 95% confidence the ratio of the variances of
the percentage improvement in fitness, and briefly describe what the interval estimate
tells you.

LCL = ( s12 / s2 ) / F0.025,99,99  1.1406
2

UCL = ( s12 / s2 ).F0.025,99,99  2.5186
2

We estimate that  12 /  2 lies between 1.1406 and 2.5186.
2

63.     {Fitness Training Narrative} Do these results allow us to conclude at the 5% significance
level that approach 2 is superior?

H 0 : 1   2 vs. H1 : 1   2
Rejection region: t < - t0.05,186  -1.653
Test statistic: t = -7.241
Conclusion: Reject the null hypothesis. Yes, these results allow us to conclude at the 5%
significance level that approach 2 is superior
Statistical Inference: Review of Chapters 12 and 13              501

64.   {Fitness Training Narrative} Estimate with 95% confidence the difference in the mean
percentage improvement between approaches 1 and 2, and briefly describe what this
interval estimate tells you.

-6.3  1.716. Thus, LCL = -8.016, and UCL = -4.584. We estimate that the mean
percentage improvement with approach 1 is between 4.584 and 8.016 less than that with
approach 2.

65.   {Fitness Training Narrative} Estimate with 95% confidence the mean percentage
improvement with approach 2.

33.6  1.052. Thus, LCL = 32.548, and UCL = 34.652

66.   {Fitness Training Narrative} Do these results allow us to conclude at the 5% significance
level that the mean percentage improvement with approach 1 is at least 25%?

H 0 : 1  25 vs. H1 : 1  25
Rejection region: t > t0.05,99  1.66
Test statistic: t = 3.33
Conclusion: Reject the null hypothesis. Yes, these results allow us to conclude at the 5%
significance level that the mean percentage improvement with approach 1 is at least 25%

67.   {Fitness Training Narrative} Scientists are interested in determining which of the two
approaches more consistently improves fitness. Do these results allow us to conclude at
the 10% significance level that approach 2 results in a more consistent improvement in
fitness than approach 1?

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F > F0.10,99,99  1.295
Test statistics: F = 1.6949
Conclusion: Don’t reject the null hypothesis. These results don’t allow us to conclude at
the 10% significance level that approach 2 results in a more consistent improvement in
fitness than approach 1
502      Chapter Fourteen

68.     {Fitness Training Narrative} Do these results allow us to conclude at the 5% significance
level that the variance of the percentage improvement with approach 2 is less than 40?

H 0 :  2  40 vs. H 1 :  2 < 40
2                  2

Rejection region:  2   0.95,99  77.9295
2

Test statistics:  2  69.5228
Conclusion: Reject the null hypothesis. Yes, these results allow us to conclude at the 10%
significance level that approach 2 results in a more consistent improvement in fitness than
approach 1

69.     {Fitness Training Narrative} Estimate with 95% confidence the variance of the
percentage improvement with approach 1.

LCL = (n1  1) s12 /  0.025,99  36.3797
2

UCL = (n1  1) s12 /  0.975,99  63.5040
2

FOR QUESTIONS 70 THROUGH 85, USE THE FOLLOWING NARRATIVE:
Narrative: VCR Tapes
Videocassette recorder (VCR) tapes are designed so that users can repeatedly record new
material over old material. However, after a number of re-recordings the tape begins to
deteriorate. A VCR tape manufacturer is experimenting with a new technology, which hopefully
will produce longer-lasting tapes. Thirty of the old-style tapes and 30 utilizing the new
technology were used in an experiment. The tapes were used to record and rerecord programs
until they began to deteriorate. The number of re-recordings is assumed to be normally
distributed. It is generally accepted that the number of re-recordings should exceed 55. Any that
do not are considered to be unacceptable. The number of re-recordings were observed and shown
in the accompanying table.

Old-Style                    New Technology
Tapes                           Tapes
60      61       48               68   70     58
51      46       66               74   72     69
66      63       61               77   73     49
73      55       71               59   66     61
71      49       76               52   58     59
47      56       55               66   51     49
60      62       64               62   59     57
52      51       63               51   56     66
64      68       52               50   55     76
47      55       58               63   68     78
Statistical Inference: Review of Chapters 12 and 13           503

70.   {VCR Tapes Narrative} Determine whether these data are sufficient to infer at the 10%
significance level that the two population variances differ.

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F > F0.05,29,29  1.8608 or F < F0.95,29,29  1/ F0.05,29,29  0.5374
Test statistics: F = 0.9005
Conclusion: Don’t reject the null hypothesis. These data are not sufficient to infer at the
10% significance level that the two population variances differ.

71.   {VCR Tapes Narrative} Estimate with 90% confidence the ratio of the variances of the
number of re-recordings of the two types of tape, and briefly describe what the interval
estimate tells you.

LCL = ( s12 / s2 ) / F0.05,29,29  0.484
2

UCL = ( s12 / s2 ).F0.05,29,29  1.676
2

We estimate that the ratio  12 /  2 is between 0.484 and 1.676.
2

72.   {VCR Tapes Narrative} Can we conclude at the 10% significance level that the new tapes
last longer than the old tapes?

H 0 : 1   2 vs. H1 : 1   2
Rejection region: t <- t0.10,58  -1.296
Test statistic: t = -1.528
Conclusion: Reject the null hypothesis. Yes, we can conclude at the 10% significance
level that the new tapes last longer than the old tapes

73.   {VCR Tapes Narrative} Estimate with 90% confidence the difference in the mean
number of re-recordings between the old and new tapes.

-3.4  3.717. Thus, LCL = -7.117, and UCL = 0.317.

74.   {VCR Tapes Narrative} Can we conclude at the 10% significance level that the mean
number of re-recordings of the new tapes is at least 55?

H 0 :  2  55 vs. H1 :  2  55
Rejection region: t > t0.10,29  1.311, Test statistic: t = 4.586
Conclusion: Reject the null hypothesis. Yes, we conclude at the 10% significance level
that the mean number of re-recordings of the new tapes is at least 55
504   Chapter Fourteen

75.   {VCR Tapes Narrative} Estimate with 90% confidence the mean number of re-recordings
of the new tapes.

62.4  2.742. Thus, LCL = 59.658, and UCL = 65.142

76.   {VCR Tapes Narrative} Do the data allow us to infer at the 10% significance level that
the new technology tapes are superior to the old-style tapes in terms of the number of
unacceptable tapes?

H 0 : p1  p2  0 vs. H1 : p1  p2  0
Rejection region: |z| > z0.10  1.28
Test statistics: z = 1.2209
Conclusion: Don’t reject the null hypothesis. The data allow us to infer at the 10%
significance level that the new technology tapes are superior to the old-style tapes in
terms of the number of unacceptable tapes

77.   {VCR Tapes Narrative} Find the p-value of the test in the previous question, and explain
how to use it to test the hypotheses.

p-value = 0.1112. Since p-value = 0.1112 >  = 0.10, don’t reject the null hypothesis.

78.   {VCR Tapes Narrative} Estimate with 90% confidence the difference in the proportions
of unacceptable tapes between the old and new tapes.

0.1333  0.1774. Thus, LCL = -0.0441, and UCL = 0.3107

79.   {VCR Tapes Narrative} Can we infer at the 10% significance level that the number of re-
recordings of the old tapes is more consistent than the re-recordings of the new tapes?

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F < F0.90,29,29  1/ F0.10,29,29  0.617
Test statistics: F = 0.9005
Conclusion: Don’t reject the null hypothesis. No, we can’t infer at the 10% significance
level that the number of re-recordings of the old tapes is more consistent than the re-
recordings of the new tapes
Statistical Inference: Review of Chapters 12 and 13              505

80.   {VCR Tapes Narrative} Do the data allow us to infer at the 10% significance level that
the proportion of unacceptable old tapes exceeds 20%?

H 0 : p1  0.20 , H1 : p1  0.20
Rejection region: z > z0.10  1.28
Test statistics: z = 1.37
Conclusion: Reject the null hypothesis. Yes, the data allow us to infer at the 10% level of
significance that the proportion of unacceptable old tapes exceeds 20%.

81.   {VCR Tapes Narrative}Find the p-value of the test in the previous question.

p-value = 0.0853

82.   {VCR Tapes Narrative} Estimate with 90% confidence the proportion of unacceptable
old tapes, and briefly describe what this interval estimate tells you.

0.30  0.138. Thus, LCL = 0.162, and UCL = 0.438
We estimate that the proportion of unacceptable old tapes is between 16.2% and 43.8%.

83.   {VCR Tapes Narrative} Do the data allow us to infer at the 10% significance level that
the proportion of unacceptable new tapes is less than 20%?

H 0 : p2  0.20 vs. H1 : p2  0.20
Rejection region: z < - z0.10  -1.28
Test statistics: z = -0.489
Conclusion: Don’t reject the null hypothesis. No, the data don’t allow us to infer at the
10% significance level that the proportion of unacceptable new tapes is less than 20%

84.   {VCR Tapes Narrative} Can we infer at the 10% significance level that the variance of
the number of re-recordings of the new tape is less than 100?

H 0 :  2  100 vs. H 1 :  2  100
2                   2

Rejection region:  2   0.90,29  19.7687
2

Test statistics:  2  22.652
Conclusion: Don’t reject the null hypothesis. No, we can’t infer at the 10% significance
level that the variance of the number of re-recordings of the new tape is less than 100
506     Chapter Fourteen

85.    {VCR Tapes Narrative} Estimate with 90% confidence the population variance of the
number of re-recordings of the new tape.

LCL = (n2  1) s2 /  0.05,29  53.227
2     2

UCL = (n2  1) s2 /  0.95,29  127.917
2     2

FOR QUESTIONS 86 THROUGH 100, USE THE FOLLOWING NARRATIVE:
Narrative: Car Door Hinges
Many parts of cars are mechanically tested to be certain that they do not fail prematurely. In an
experiment to determine which one of two types of metal alloy produces superior door hinges, 40
of each type were tested until they failed. Car manufacturers consider any hinge that does not
survive 1 million openings and closings to be a failure. The number of openings and closings as
observed and recorded in the accompanying table (to the closest .1 million). A statistician has
determined that the number of openings and closings is normally distributed.

Number of Openings and Closings
Alloy 1                            Alloy 2
1.5     1.5      0.9     1.3        1.4     0.9      1.3              0 .8
1.8     1.6      1.3     1.5        1.3     1.3      0.9                1.4
1.6     1.2      1.2     1.8        0.7     1.2      1.1                0.9
1.3     0.9      1.5     1.6        1.2     0.8      1.2                1.1
1.2     1.3      1.4     1.4        0.8     0.7      1.1                1.4
1.1     1.5      1.1     1.5        1.1     1.4      0.8                0.8
1.3     0.8      0.8     1.1        1.3     1.1      1.5                0.9
1.1     1.6      1.6     1.3        1.4     1.2      1.3                1.6
0.9     1.4      1.7     0.9        0.6     0.9      1.8                1.4
1.1     1.3      1.9     1.3        1.5     0.8      1.6                1.3

86.    {Car Door Hinges Narrative} Determine whether these data are sufficient to infer at the
5% significance level that the two population variances differ.

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F > F0.025,39,39  1.89 or F < F0.975,39,39  1/ F0.025,39,39  0.529
Test statistics: F = 0.9166
Conclusion: Don’t reject the null hypothesis. No, these data are not sufficient to infer at
the 5% significance level that the two population variances differ.

87.    {Car Door Hinges Narrative} Estimate with 95% confidence the ratio of the population
variances of number of openings and closings between the two types of alloy.

LCL = ( s12 / s2 ) / F0.025,39,39  0.4850, and UCL = ( s12 / s2 ).F0.025,39,39  1.7324
2                                               2
Statistical Inference: Review of Chapters 12 and 13             507

88.   {Car Door Hinges Narrative} Can we conclude at the 5% significance level that hinges
made with alloy1 last longer than hinges made with alloy 2?

H 0 : 1   2 vs. H1 : 1   2
Rejection region: t > t0.05,88  1.662
Test statistic: t = 2.841
Conclusion: Reject the null hypothesis. Yes, we can conclude at the 5% significance level
that hinges made with alloy1 last longer than hinges made with alloy 2

89.   {Car Door Hinges Narrative}Estimate with 95% confidence the difference in the mean
number of openings and closings between the two types of alloy.

0.183  0.128. Thus, LCL = 0.055, and UCL = 0.310

90.   {Car Door Hinges Narrative} Can we conclude at the 5% significance level that the mean
number of door openings and closings with hinges made from alloy 1 is at least 1.25
million?

H 0 : 1  1.25 , H1 : 1  1.25
Rejection region: t > t0.05,39  1.684
Test statistic: t = 1.744
Conclusion: Reject the null hypothesis. Yes, we can conclude at the 5% significance level
that the mean number of door openings and closings with hinges made from alloy 1 is at
least 1.25 million

91.   {Car Door Hinges Narrative} Estimate with 95% confidence the mean number of
openings and closings with hinges made from alloy 1.

1.3275  0.0898. Thus, LCL = 1.2377, and UCL = 1.4173

92.   {Car Door Hinges Narrative} Critics of the automobile industry accuse car makers of
producing products that are designed to last for a certain amount of time and then fail.
Are critics more likely to draw that conclusion on the basis of hinges made from alloy 1
than from hinges made with alloy 2? Test with  = .05.

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F > F0.05,39,39  1.69
Test statistics: F = 0.9166
Conclusion: Don’t reject the null hypothesis. No
508    Chapter Fourteen

93.   {Car Door Hinges Narrative} Can we conclude at the 5% significance level that the
variance of the number of openings and closings with the hinges made from alloy 1 is less
than 0.2 million2?

H 0 :  12  0.20 vs. H 1 :  12 < 0.20
Rejection region:  2   0.95,39  26.509
2

Test statistics:  2  15.397
Conclusion: Reject the null hypothesis. Yes

94.   {Car Door Hinges Narrative} Estimate with 95% confidence the variance of the number
of openings and closings with the hinges made from alloy 2.

LCL = (n2  1) s2 /  0.025,39  0.0566
2     2

UCL = (n2  1) s2 /  0.975,39  0.1375
2     2

95.   {Car Door Hinges Narrative} Can we conclude at the 5% significance level that alloy 2
produces a greater number of failures than alloy 1?

H 0 : p1  p2  0 vs. H1 : p1  p2  0
Rejection region: z < - z0.05  -1.645
Test statistics: z = -2.066
Conclusion: Reject the null hypothesis. Yes, we can conclude at the 5% significance level
that alloy 2 produces a greater number of failures than alloy 1.

96.   {Car Door Hinges Narrative} Find the p-value of the test in the previous question, and
explain how to use it to test the hypotheses.

p-value = 0.0192.
Since p-value = 0.0192 <  = 0.05, we reject the null hypothesis.

97.   {Car Door Hinges Narrative}Estimate with 95% confidence the difference in the
proportion of failures between the two types of alloy.

-0.2  0.185. Thus, LCL = -0.385, and UCL = -0.015
Statistical Inference: Review of Chapters 12 and 13       509

98.       {Car Door Hinges Narrative}Can we conclude at the 5% significance level that the
proportion of failures with alloy 1 is less than 25%?

H 0 : p1  0.25 vs. H1 : p1  0.25
Rejection region: z < - z0.05  -1.645
Test statistics: z = -1.46
Conclusion: Don’t reject the null hypothesis. We can’t conclude at the 5% significance
level that the proportion of failures with alloy 1 is less than 25%.

99.       {Car Door Hinges Narrative}Find the p-value of the test in the previous question, and
briefly explain how to use it to test the hypotheses.

p-value = 0.0721. Since p-value >  , we fail to reject the null hypothesis.

100.      {Car Door Hinges Narrative} Estimate with 95% confidence the proportion of failures
among the hinges made with alloy 2.

0.35  0.1478. Thus, LCL = 0.2022, and UCL = 0.4978

FOR QUESTIONS 101 THROUGH 110, USE THE FOLLOWING NARRATIVE:
The irradiation of food to destroy bacteria is a growing phenomenon. In order to determine which
one of two methods of irradiation is best, a scientist took a random sample of 100 one- pound
packages of minced meat and subjected 50 of them to irradiation method 1 and the remaining 50
to irradiation method 2. The bacteria count was measured and the following statistics were
computed.

Method 1         Method 2
x1  86          x2  98
s12  324     s 2  841
2

The scientist noted that the data are normally distributed.
510     Chapter Fourteen

101.   {Food Irradiation Narrative} Determine whether these data are sufficient to infer at the
5% significance level that the two population variances differ.

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F > F0.025,49,49  1.762 or F < F0.975,49,49  1/ F0.025,49,49  0.568
Test statistics: F = 0.3853
Conclusion: Reject the null hypothesis. Yes, these data are sufficient to infer at the 5%
significance level that the two population variances differ.

102.   {Food Irradiation Narrative} Estimate with 95% confidence the ratio of the variances of
the number of re-recordings of the two types of tape, and briefly describe what the
interval estimate tells you.

LCL = ( s12 / s2 ) / F0.025,49,49  0.2186
2

UCL = ( s12 / s2 ).F0.025,49,49  0.6788
2

We estimate that the ratio  12 /  2 is between 0.2186 and 0.6788.
2

103.   {Food Irradiation Narrative} Do these results allow us to infer at the 5% significance
level that there is a difference in bacteria count between methods 1 and 2?

H 0 : 1   2 vs. H1 : 1   2
Rejection region: |t| > t0.025,82  1.989
Test statistic: t = -2.486
Conclusion: Reject the null hypothesis. Yes, these results allow us to infer at the 5%
significance level that there is a difference in bacteria count between methods 1 and 2

104.   {Food Irradiation Narrative} Do these results allow us to infer at the 5% significance
level that the mean bacteria count with method 1 is less than 95?

H 0 : 1  95 vs. H1 : 1  95
Rejection region: t < - t0.05,49  -1.677
Test statistic: t = -3.536
Conclusion: Reject the null hypothesis. Yes, these results allow us to infer at the 5%
significance level that the mean bacteria count with method 1 is less than 95
Statistical Inference: Review of Chapters 12 and 13           511

105.   {Food Irradiation Narrative} Estimate with 95% confidence the difference in the mean
bacteria count between method 1 and method 2.

-12  9.604. Thus, LCL = -21.604, and UCL = -2.396

106.   {Food Irradiation Narrative} Estimate with 95% confidence the mean bacteria count with
method 2.

98.0  8.242. Thus, LCL = 89.758, and UCL = 106.242

107.   {Food Irradiation Narrative} An important factor in determining which method to choose
is consistency. That is, all other things being equal we’d prefer to have a method that
leaves all irradiated food with approximately the same bacteria count. Can we infer at the
5% significance level that method 1 is superior to method 2 in this respect?

H 0 :  12 /  2  1 vs. H 1 :  12 /  2  1
2                        2

Rejection region: F < F0.95,49,49  1/ F0.05,49,49  0.622
Test statistics: F = 0.3853
Conclusion: Reject the null hypothesis. Yes, we can infer at the 5% significance level that
method 1 is superior to method 2 in this respect

108.   {Food Irradiation Narrative} Estimate with 95% confidence the ratio of the two
variances.

LCL = ( s12 / s2 ) / F0.025,49,49  0.2186
2

UCL = ( s12 / s2 ).F0.025,49,49  0.6789
2

109.   {Food Irradiation Narrative} Can we conclude at the 5% significance level that the
variance of the bacteria count with method 2 is less than 1,500?

H 0 :  2  1500 vs. H 1 :  2  1500
2                    2

Rejection region:  2   0.95,49  33.93
2

Test statistics:  2  27.473
Conclusion: Reject the null hypothesis. Yes, we can conclude at the 5% significance level
that the variance of the bacteria count with method 2 is less than 1,500
512     Chapter Fourteen

110.   {Food Irradiation Narrative} Estimate with 95% confidence the variance of the bacteria
count with method 1, and briefly describe what this interval estimate tells you.

LCL = (n1  1) s12 /  0.025,49  226.082
2

UCL = (n1  1) s12 /  0.975,49  503.123
2

We estimate that the population variance of the bacteria count with method 1 lies between
226.082 and 503.123

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