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CHAPTER 13

INFERENCE ABOUT COMPARING TWO POPULATIONS

MULTIPLE CHOICE QUESTIONS

In the following multiple choice ques tions , please circle the correct answer .

1. Two indepen d e n t samples are drawn from two normal populations , where the population variances are assum e d to be equal. The sampling distribution of the ratio of the two sample variances is:a. a normal distributionb. Student t distributionc. an F distributiond. a chi squared distributionANSWER: c

2. The expec ted value of the difference of two sample means equals the difference of the corresponding population means:a. only if the populations are normally distributedb. only if the samples are indepen d e n tc. only if the populations are approxima t ely normal and the sample sizes are

larged. the state m e n t is correct under all circumst a nc e sANSWER: d

3. In constructing confidence interval estimat e for the difference betwe en two population proportions, we:a. pool the population proportions when the populations are normally

distributedb. pool the population proportions when the population means are equalc. pool the population proportions when they are equald. never pool the population proportions

ANSWER: d

227

228 Chapter Thirteen

4. In constructing 95% confidence interval estimat e for the difference betwe en the means of two normally distributed populations , where the unknown population variances are assum e d not to be equal, sum m a ry statistics comput e d from two independ e n t samples are as follows:

501 =n 1751 =x 5.181 =s

422 =n 1582 =x 4.322 =s

The upper confidence limit is:a. 19.123b. 28.212c. 24.911d. 5.788ANSWER: b

5. The ratio of two independ e n t chi squared variables divided by their degree s of freedom is:a. normally distributedb. Student t distributedc. chi squared distributedd. F distributedANSWER: d

6. A sample of size 100 selected from one population has 60 success e s , and a sample of size 150 selected from a second population has 95 success e s . The test statistic for testing the equality of the population proportions equal to:a. 0.5319b. 0.7293c. 0.419d. 0.2702ANSWER: a

7. For testing the difference betwe e n two population proportions , the pooled proportion estimat e should be used to comput e the value of the test statistic when the:

a. populations are normally distributedb. sample sizes are smallc. samples are independ e n t ly drawn from the populationsd. null hypothesis state s that the two population proportions are equalANSWER: d

8. The F distribution is the sampling distribution of the ratio of:a. two normal population variancesb. two normal population meansc. two sample variances provided that the samples are independ e n t ly drawn

from two normal populations with equal variancesd. two sample variances provided that the sample sizes are largeANSWER: c

Inferenc e About Comparing Two Populations 229

9. In testing the null hypothesis 0: 210 =− ppH , if 0H is false, the test could lead to:a. a Type I errorb. a Type II errorc. either a Type I or a Type II errord. None of the aboveANSWER: b

10. A sample of size 150 from population 1 has 40 success e s . A sample of size 250 from population 2 has 30 success e s . The value of the test statistic for testing the null hypothesis that the proportion of success e s in population one exceeds the proportion of success e s in population two by 0.05 is:a. 1.645b. 2.327c. 1.960d. 1.977ANSWER: b

11. Two samples of sizes 25 and 35 are independ e n t ly drawn from two normal populations, where the unknown population variances are assum e d to be equal. The number of degree s of freedo m of the equal variances t test statistic is:a. 60b. 59c. 58d. 35ANSWER: c

12. If some natural relationship exists betwe e n each pair of observa tions that provides a logical reason to compare the first observa tion of sample 1 with the first observa tion of sample 2, the second observa tion of sample 1 with the second observa tion of sample 2, and so on, the samples are referred to as:a. matched samplesb. indepen d e n t samplesc. weighted samplesd. random samplesANSWER: a

13. In testing the difference betwe e n the means of two normally distributed populations, the numb er of degre es of freedom associat ed with the unequalvariances t test statistic usually results in a non integer numb er . It is recom m e n d e d that you:a. round up to the neares t integerb. round down to the neares t integerc. change the sample sizes until the numb er of degree s of freedo m becom e s

an integerd. assum e that the population variances are equal, and then use d.f =

221 −+ nn


ANSWER: b

14. The symbol Dx refers to:a. the difference in the means of two depend e n t populationsb. the difference in the means of two indepen d e n t populationsc. the matched pairs differencesd. the mean difference in the pairs of observa tions taken from two depend e n t

samplesANSWER: d

15. The numb er of degree s of freedo m associat ed with the t test, when the data are gathered from a match ed pairs experime n t with 10 pairs, is:a. 10b. 20c. 9d. 18ANSWER: c

16. The quanti ty 2ps is called the pooled variance estimat e of the common

variance of two unknown but equal population variances . It is the weighted averag e of the two sample variances , where the weights repres en t the:a. sample variancesb. sample stand ard deviationsc. sample sizesd. degree s of freedo m for each sampleANSWER: d

17. Two independ e n t samples of sizes 20 and 30 are randomly selected from two normally distributed populations . Assume that the population variances are unknown but equal. In order to test the difference betwe e n the population means , 21 µµ − , the sampling distribution of the sample mean difference,

21 xx − , is:a. normally distributedb. t distributed with 50 degree s of freedo mc. t distributed with 48 degree s of freedo md. F distributed with 19 and 29 degree s of freedo mANSWER: a

18. Two independ e n t samples of sizes 40 and 50 are randomly selected from two populations to test the difference betwee n the population means 21 µµ − . The sampling distribution of the sample mean difference 21 xx − is:a. normally distributedb. approxima t ely normalc. t distributed with 88 degree s of freedo md. chi squared distributed with 90 degree s of freedo mANSWER: b


19. Two independ e n t samples of sizes 25 and 35 are randomly selected from two normal populations with equal variances . In order to test the difference betwe en the population means , the test statistic is:a. a standard normal rando m variableb. approxima t ely stand ard normal random variablec. Student t distributed with 58 degree s of freedo md. Student t distributed with 33 degree s of freedo mANSWER: c

20. When the necess a ry conditions are met, a two tail test is being conducted to test the difference betwee n two population proportions . If the value of the test statistic z is 2.05, then the p value is:a. 0.4798b. 0.0404c. 0.2399d. 0.0202ANSWER: b

21. Which of the following state m e n t s is not correct for an F distribution?a. Variables that are F distributed range from 0 to 00b. Exact shape of the distribution is deter mine d by two numb ers of degree s

of freedo mc. Degrees of freedom for the denomina tor are always smaller than the

degree s of freedo m for the numer a tord. Degrees of freedom for the numera tor can be larger, smaller, or equal to

the degree s of freedo m for the denomina tor .ANSWER: c

22. In testing the difference betwe e n two population means using two indepen d e n t samples , we use the pooled variance in estimating the standard error of the sampling distribution of the sample mean difference 21 xx − if the:a. sample sizes are both largeb. populations are normal with equal variancesc. populations are non normal with unequal variancesd. all of the above are required conditionsANSWER: b

23. In testing the difference betwe e n two population means using two indepen d e n t samples , the sampling distribution of the sample mean difference

21 xx − is normal if the:a. sample sizes are both grea t e r than 30b. populations are normalc. populations are non normal and the sample sizes are larged. all of the above are required conditionsANSWER: b


24. In testing whether the means of two normal populations are equal, summ ary

statistics comput e d for two independ e n t samples are as follows:

251 =n 30.71 =x 05.11 =s

252 =n 80.62 =x 20.12 =s

Assume that the population variances are equal. Then, the stand ard error of the sampling distribution of the sample mean difference 21 xx − equal to:a. 0.1017b. 1.2713c. 0.3189d. 1.1275ANSWER: c

25. A political analyst in Texas surveys a random sample of register ed Democra t s and compare s the results with those obtained from a rando m sample of register ed Republicans. This would be an example of:a. indepen d e n t samplesb. depend e n t samplesc. indepen d e n t samples only if the sample sizes are equald. depend e n t samples only if the sample sizes are equalANSWER: a

26. In testing the difference betwe e n two population means using two indepen d e n t samples , the population stand ard deviations are assum e d to be known and the calculated test statistic equals 2.56. If the test is two tail and 5% level of significance has been specified, the conclusion should be to:a. reject the null hypothe sisb. not to reject the null hypothesisc. choose two other independ e n t samplesd. none of the above answers is correctANSWER: a

27. In testing the difference betwe e n two population means , for which the population variances are unknown and not assum e d to be equal, two indepen d e n t samples of large sizes are drawn from the populations . Which of the following tests is appropriat e?a. z testb. Pooled variances t testc. Unequal variances t testd. Matched pairs t testANSWER: c


28. Which of the following is a required condition for using the normal approxima tion to the binomial in testing the difference betwe en two population proportions?a. 3011 >pn and 3022 >pn

b. 511 >pn and 522 >pn

c. ,511 >pn ,511 >qn ,522 >pn and 522 >qn

d. ,5ˆ11 >pn ,5ˆ11 >qn ,5ˆ 22 >pn and 5ˆ22 >qnANSWER: c

29. A test is being conducted to test the difference betwe e n two population means using data that are gathered from a matche d pairs experimen t . If the paired differences are normal, then the distribution used for testing is the:a. normalb. binomialc. Student td. FANSWER: c

30. The sampling distribution of the ratio of two sample variances 21s / 2

2s is said to be F distributed provided that:a. the samples are independ e n tb. the populations are normal with equal variancesc. the samples are depend e n t and their sizes are larged. the samples are independ e n t ly drawn from two normal populationsANSWER: d

31. Which of the following state m e n t s is correct regarding the percentile points of the F distribution?a. 0.05,10,20 0.95,10,201/F F=

b. 0.05,10,20 0.05,20,101/F F=

c. 0.95,10,20 0.95,20,101/F F=

d. 0.95,10,20 0.05,20,101/F F=ANSWER: d

32. Which of the following state m e n t s is not correct about the estima t e of the ratio of two population variances?a. We can comput e the estima t e manuallyb. Excel can be used to comput e the estimat ec. Minitab can be used to comput e the estima t ed. All the above are correct state m e n t s .ANSWER: c


33. In testing the difference betwe e n the means of two normal population using two indepen d e n t samples , when the population variances are unequ al, the sampling distribution of the resulting statistic is:a. normalb. Student tc. approxima t ely normald. approxima t ely Studen t tANSWER: d

34. Suppose we randomly selected 200 people, and on the basis of their respons e s we assigned them to one of two groups: high risk group and low risk group. We then recorded the blood pressure for the memb ers of each group. Such data are called:a. observa tionalb. experimen t alc. controlledd. qualitativeANSWER: a

35. In constructing confidence interval estimat e for the difference betwe en the means of two normally distributed populations, using two independ e n t samples , we:a. pool the sample variances when the unknown population variances are

equalb. pool the sample variances when the population variances are known and

equalc. pool the sample variances when the population means are equald. never pool the sample variancesANSWER: a

36. When the necess a ry conditions are met, a two tail test is being conducted to test the difference betwee n two population means , but your statistical software provides only a one tail area of .036 as part of its output . The p value for this test will be:a. 0.018b. 0.009c. 0.072d. 0.964ANSWER: c

37. In testing for the equality of two population variances , when the populations are normally distributed, the 10% level of significance has been used. To deter mine the rejection region, it will be necess a ry to refer to the F table corresponding to an upper tail area of:a. 0.90b. 0.05c. 0.20d. 0.10


ANSWER: b

38. When the necess a ry conditions are met, a two tail test is being conducted to test the difference betwee n two population proportions . The two sample proportions are 1ˆ 0.25p = and 2ˆ 0.20p = , and the standard error of the sampling distribution of 21 ˆˆ pp − is 0.04. The calculated value of the test statistic will be:a. z = 0.25b. z = 1.25c. t = 0.25d. t = 0.80ANSWER: b

39. We can design a matched pairs experimen t when the data collected are:a. observa tionalb. experimen t alc. controlledd. All the above are correctANSWER: d

40. In constructing a 90% interval estima t e for the ratio of two population variances , 2

1σ / 22σ , two indepen d e n t samples of sizes 40 and 60 are drawn

from the populations . If the sample variances are 515 and 920, then the lower confidence limit is:a. 0.352b. 0.341c. 0.890d. 0.918ANSWER: a


TRUE/FALSE QUESTIONS

41. When comparing two population variances , we use the ratio 2 21 2/σ σ rather

than the difference 2 21 2σ σ− .

ANSWER: T

42. In comparing two population means when samples are depend e n t , the variable under considera tion is 1 2p̂ p p= − .ANSWER: F

43. Tests in which samples are not independ e n t are referred to as matched pairs.ANSWER: T

44. Independ e n t samples are those for which the selection process for one is not related to the selection process for the other.ANSWER: T

45. We say that two samples are depend e n t when the selection process for one is related to the selection process for the other. ANSWER: T

46. The pooled variances t test requires that the two population variances are not the same. ANSWER: F.

47. We can use either the z test or the t test to determine whether two population variances are equal.ANSWER: F

48. In testing the difference betwe e n two population means using two indepen d e n t samples , we use the pooled variance in estimating the standard error of the sampling distribution of the sample mean difference 21 xx − if the populations are normal with equal variances .ANSWER: T

49. In testing the difference betwe e n two population means using two indepen d e n t samples , the sampling distribution of the sample mean difference 21 xx − is normal if the sample sizes are both grea t e r than 30.ANSWER: F

50. A political analyst in Iowa surveys a random sample of register ed Democra t s and compare s the results with those obtained from a rando m sample of register ed Republicans. This would be an example of two indepen d e n t samples .ANSWER: T


51. In comparing two means when samples are depend e n t , the variable under considera t ion is Dx , where the subscript D refers to the difference .ANSWER: T

52. In testing the difference betwe e n two population means using two indepen d e n t samples , the population stand ard deviations are assu m e d to be known, and the calculated test statistic equals 2.75. If the test is two tail and 5% level of significance has been specified, the conclusion should be not to reject the null hypothe sis.ANSWER: F

53. When the necess a ry conditions are met, a two tail test is being conducted to test the difference betwe e n two population means , but your statistical software provides only a one tail area of 0.046 as part of its output . The p value for this test will be 0.092.ANSWER: T

54. The best estimator of the difference betwe en two population means 1 2µ µ− is

the difference betwee n two sample means 1 2x x− .ANSWER: T

55. In testing for the equality of two population variances , when the populations are normally distributed, the 5% level of significance has been used. To deter mine the rejection region, it will be necess a ry to refer to the F table corresponding to an upper tail area of 0.05.ANSWER: F

56. When the necess a ry conditions are met, a two tail test is being conducted to test the difference betwe e n two population proportions . The two sample proportions are 1 0.40p = and 2 0.35p = , and the stand ard error of the sampling

distribution of 1 2p p− is 0.04. The calculat ed value of the test statistic will be 1.25.ANSWER: T

57. Two samples of sizes 25 and 20 are independ e n t ly drawn from two normal populations, where the unknown population variances are assu m e d to be equal. The number of degree s of freedo m of the equal variances t test statistic is 44.ANSWER: F

58. The numb er of degree s of freedo m associat ed with the t test, when the data are gathered from a match ed pairs experime n t with 10 pairs, is 9.ANSWER: T

59. When the necess a ry conditions are met, a two tail test is being conducted to test the difference betwee n two population proportions . If the value of the test statistic z is 1.75 , then the p value is 0.0802.ANSWER: T


60. The sampling distribution of 1 2x x− is normal if the sampled populations are normal, and approxima t ely normal if the populations are nonnorm al and the sample sizes 1n and 2n are large.ANSWER: T

61. The expecte d value of 1 2x x− is E( 1 2x x− ) = 1 2µ µ− .ANSWER: T

62. The variance of 1 2x x− is V( 1 2x x− ) = 2 21 2

1 2n n

σ σ− .

ANSWER: F

63. The equal variances test statistic of 1 2µ µ− is Student t distributed with 1n + 2n degree s of freedo m, provided that the two populations are normal.ANSWER: F

64. When the population variances are unequ al, we estimat e each population variance with its sample variance. Hence, the unequal variances test statistic of 1 2µ µ− is Student t distributed with 1n + 2n 2 degre es of freedom. ANSWER: F

65. Statisticians have shown that for given sample sizes 1n and 2n , the number of degre es of freedo m associated with the equal variances test statistic and confidence interval estima tor of 1 2µ µ− is always grea t e r than or equal to numb er of degree s of freedo m associat ed with the un equal variances test statistic and confidence interval estima tor.ANSWER: T

66. Both the equal variances and unequal variances test statistic and confidence interval estimator of 1 2µ µ− require that the two populations be normally distributed.ANSWER: T

67. The Wilcoxon rank sum test for indepen d e n t samples is used to replace the equal variances test of 1 2µ µ− when the sample sizes 1n and 2n are small, but equal.ANSWER: F

68. The Wilcoxon rank sum test for indepen d e n t samples is used to replace the equal variances test of 1 2µ µ− when the populations are not normally distributed.ANSWER: T


69. When comparing two population means using data that are gathered from a matched pairs experimen t , the test statistic for Dµ is Student t distributed

with 1Dnν = − degree s of freedo m, provided that the difference s are normally distributed. ANSWER: T

70. The matched pairs experimen t always produce a larger test statistic than the indepen d e n t samples experimen t .ANSWER: F

71. In comparing two population means of interval data , we must decide whether the samples are indepen d e n t (in which case the para m e t e r of interes t is

1 2µ µ− ) or matched pairs (in which case the para m e t e r is Dµ ) in order to select the correct test statistic.ANSWER: T

72. The Wilcoxon signed rank sum test for matche d pairs is used to replace the ttest of Dµ if the differences are very nonnorm al.ANSWER: T

73. The test statistic employed to test 2 20 1 2 1: /H σ σ = is 2 2

1 2/F s s= , which is F

distributed with 1 1 2 21 and 1n nν ν= − = − degree s of freedo m, provided that the two populations are F distributed .ANSWER: F

74. When the necess a ry conditions are met, a two tail test is being conducted at α = 0.05 to test 2 2

0 1 2 1: /H σ σ = . The two sample variances are 2 21 2400 and 800ss = = , and the sample sizes are 1 225 and 25nn = = . The calculated

value of the test statistic will be F = 2.ANSWER: F

75. When the necess a ry conditions are met, a two tail test is being conducted at α = 0.05 to test 2 2

0 1 2 1: /H σ σ = . The two sample variances are 2 21 2500 and 900ss = = , and the sample sizes are 1 221 and 31nn = = . The rejection

region is F > 2.20 or F < 0.4255.ANSWER: T


TEST QUESTIONS

76. In testing the hypothes e s

25: 210 =− µµH

25: 211 >− µµH

two random samples from two normal populations produced the following results:

421 =n 2151 =x 151 =s

322 =n 1802 =x 202 =s

What conclusion can we draw at the 5% significance level?

ANSWER :Rejection region: 0.05,72 1.667t t> ≈Test statistic: t = 2.459Conclusion: Reject the null hypothes e s

QUESTIONS 77 AND 78 ARE BASED ON THE FOLLOWING INFORMATION:

In random samples of 40 from each of two normal populations, the following statistics were obtained:

761 =x 81 =s

722 =x 5.62 =s

77. Test at the 5% significance level to determine whether we can infer that the two population means differ.

ANSWER:0: 210 =− µµH

0: 211 ≠− µµH

Rejection region: | t| > 0.025,88 1.987t ≈Test Statistic: t = 2.454Conclusion: Reject the null hypothesis . Yes

78. Estimate with 95% confidence the difference betwee n the two population means .

ANSWER :4 3.238 (0.762,0.7238± = )



5:0 =DH µ5:1 >DH µ

two random samples from two normal populations produced the following statistics:

20=Dn 9=Dx 5.7=Ds


ANSWER :Rejection region: 0.01,19 2.539t t> =Test statistic: t = 2.385Conclusion: Don’t reject the null hypothesis


In testing the hypothes e s

0: 210 =− ppH

0: 211 <− ppH ,

we found the following statistics:

4001 =n , 1051 =x , 5002 =n , 1402 =x

80. What conclusion can we draw at the 10% significance level?

ANSWER:Rejection region: 0.10 1.28z z< − = −Test statistic: z = 0.587Conclusion: Don’t reject the null hypothesis

81. Estimate with 90% confidence the difference betwee n the two population proportions .

ANSWER:0.0175 ± 0.049 = (0.0665, 0.0315)




0: 210 =− µµH

0: 211 <− µµH ,

the following statistics were obtained:

101 =n 6.581 =x 45.131 =s

102 =n 6.642 =x 15.112 =s

82. Test the following hypothe s e s at the 5% level of significance

ANSWER:Rejection region: 0.05,18 1.734t t< − = −Test statistic: t = 1.086Conclusion: Don’t reject the null hypothesis


ANSWER: 6.0 ±11.6074

QUESTIONS 84 THROUGH 86 ARE BASED ON THE FOLLOWING INFORMATION:


0 1 2: 0.10H p p− =

1 1 2: 0.10H p p− < ,

we found the following statistics.

4001 =n 2081 =x

2502 =n 1152 =x


ANSWER:Rejection region: 0.10 1.28z z< − = −Test statistic: z = 0.9926Conclusion: Don’t reject the null hypothesis


85. What is the p value of the test in Question 84?

ANSWER:p value = 0.1611


ANSWER:0.06 ±0.0662 = (0.0062 , 0.1262)

87. Random samples from two normal populations produced the following statistics:

=1n 10 =21s 32

=2n 15 =22s 22

Is there enough evidence at the 5% significance level to infer that the variance of population 1 is larger than the variance of population 2?

ANSWER:=2

2210 /: σσH 1

>22

211 /: σσH 1

Rejection region: F > 0.05,9,14F = 2.65Test statistics: F = 1.455Conclusion: Don’t reject the null hypothesis. No



0: 210 =− µµH

0: 211 ≠− µµH ,

two rando m samples from two normal populations produced the following statistics:

501 =n 351 =x 281 =s

502 =n 302 =x 102 =s


ANSWER :There is reason to believe that the population variances are unequ al.Rejection region: | t | > 0.025,61 2.0t ≈Test statistic: t = 1.189


Conclusion: Don’t reject the null hypothes e s


ANSWER:5.0 ±8.409 = (3.409, 13.409)

90. Explain how to use the confidence interval in Question 89 for testing the hypothe s e s .

ANSWER:Since the hypothesized value 00 =µ is included in the 95% confidence interval, we fail to reject the null hypothesis at =α 0.05.


The following data were genera t e d from a matched pairs experimen t .

Pair: 1 2 3 4 5 6 7 Sample 1: 8 15 7 9 10 13 11Sample 2: 12 18 8 9 12 11 10

91. Determine whether these data are sufficient to infer at the 10% significance level that the two population means differ.

ANSWER:0:0 =DH µ0:1 ≠DH µ

Rejection region: | t | > 0.05,6t = 1.943Test Statistics: t = 1.225Conclusion: Don’t reject the null hypothesis. No

92. Estimate with 90% confidence the mean difference.

ANSWER: –1.0 ±1.587 = (2.587, 0.587)

93. Briefly describe what the interval estima t e in Question 92 tells you, and explain how to use it to test the hypothes e s .

ANSWER:


We estimat e that the mean difference is betwe en –2.587 and 0.587. Since the hypothe sized value 0 is included in the 90% interval estimat e , we fail to reject the null hypothe sis at =α 0.10.



0 1 2: 0.10H p p− =

1 1 2: 0.10H p p− ≠


1501 =n 721 =x

1752 =n 702 =x


ANSWER:Rejection region: | z | > 0.05 1.96z =Test statistic: z = 0.36Conclusion: Don’t reject the null hypothesis

95. What is the p value of the test?

ANSWER:p value = 0.7188 …(actually 0.7166)

96. Briefly explain how to use the p value to test the hypothes e s .

ANSWER:Since p value = 0.147 > =α 0.05, we fail to reject the null hypothesis


ANSWER:0.08 ± 0.107 = (0.027, 0.117)

98. Explain how to use the confidence interval in Question 97 to test the hypothe s e s .

ANSWER:


We estima t e that the difference betwee n the population proportions lies betwe en –0.028 and 0.118. Since the hypothe sized value 0 is included in the 95% interval estima t e , we fail to reject the null hypothesis at =α 0.05.


Random samples from two normal populations produced the following statistics:

=1n 16 =21s 55

=2n 14 =22s 118

99. Is there enough evidence at the 10% significance level to infer that the two population variances differ?

ANSWER:=2

2210 /: σσH 1

≠22

211 /: σσH 1

Rejection region: F > 0.05,15,13F = 2.53 or F < 0.95,15,13F = 1/ 0.05,13,15F ≈ 0.403Test statistics: F = 0.466Conclusion: Don’t reject the null hypothesis. No

100. Estimate with 90% confidence the ratio of the two population variances .

ANSWER:

LCL = ( /21s

22s ) / 0.05,15,13F = 0.1842

UCL = ( /21s

22s ) . 0.05,13,15F = 1.1566

101. Briefly describe what the interval estimat e in Question 100 tells you.

ANSWER: We estima t e that ( 2

221 /σσ ) lies betwee n 0.1842 and 1.1566.

102. Briefly explain how to use the interval estimat e in Question 100 to test the hypothe s e s .

ANSWER: Since the hypothe sized value 1 is included in the 90% interval estima t e , we fail to reject the null hypothesis at α = 0.10.




0: 210 =− ppH

0: 211 >− ppH ,

we found the following statistics:

2001 =n , 801 =x , 4002 =n , 1402 =x103. What conclusion can we draw at the 5% significance level?

ANSWER:Rejection region: 0.05z z> = 1.645Test statistic: z = 1.199Conclusion: Don’t reject the null hypothesis



105. Briefly explain how to use the p value for testing the hypothe s e s .

ANSWER:Since p value = 0.1151 > =α 0.05, we fail to reject the null hypothe sis.


ANSWER:0.05 ±0.0824 = (0.0324, 0.1324)


0: 210 =− ppH

0: 211 ≠− ppH


601 =n 361 =x

802 =n 522 =x


ANSWER:


Rejection region: | z | > 0.005z = 2.575Test statistic: z = 0.61Conclusion: Don’t reject the null hypothesis



In order to test the hypothe s e s

0: 210 =− µµH

0: 211 ≠− µµH ,

we independ e n t ly draw a rando m sample of 18 observa tions from a normal population with stand ard deviation of 15, and another random sample of 12 from a second normal population with stand ard deviation of 25.

108. If we set the level of significance at 10%, determine the power of the test when 521 =− µµ

ANSWER:Power = 1 β = 0.1631

109. Re do Question 108 if the level of significance is reduced from 10% to 5%.

ANSWER: Power = 1 β = 0.095

110. Describe the effect of reducing the level of significance on the power of the test.

ANSWER:As the level of significance decrea s e s from 10% to 5%, the power of the test decreas e s from 0.1631 to 0.095.


The general manag er of a chain of fast food chicken restaura n t s wants to determine how effective their promotional campaigns are. In these campaigns “20% off” coupons are widely distributed. These coupons are only valid for one week. To examine their effectivenes s , the executive records the daily gross sales (in $1,000s) in one restaura n t during the campaign and during the week after the camp aign ends. The data is shown below. Can they infer at the 5% significance level that sales increas e during the campaign?

Day Sales During Campaign

Sales After Campaign

Sunday 18.1 16.6Monday 10.0 8.8Tuesday 9.1 8.6Wednesd ay 8.4 8.3Thursday 10.8 10.1Friday 13.1 12.3Saturday 20.8 18.9


ANSWER:0:0 =DH µ0:1 >DH µ

Rejection region: t > 1.943Test statistic: t = 4.111Conclusion: Reject the null hypothesis . Yes

112. Estimate with 95% confidence the mean difference.

ANSWER:0.957 ± 0.57 = (0.387, 1.527)

113. Briefly explain what the interval estima t e in Question 112 tells you.

ANSWER:We estima t e that the daily sales during the camp aign increase on averag e betwe en 0.387 and 1.527 thousan d dollars


A survey of 1,500 Canadians reveals that 945 believe that there is too much violence on television. In a survey of 1,500 Americans, 810 believe that there is too much television violence.

114. Can we infer at the 99% significance level that the proportion of Canadians and Americans who believe that there is too much violence on television differ?

ANSWER: 0: 210 =− ppH

0: 211 ≠− ppH

Rejection region: | z | > 0.005z = 2.575 Test statistic: z = 5.0 Conclusion: Reject the null hypothesis . Yes

115. Estimate with 99% confidence the difference in the proportion of Canadians and Americans who believe that there is too much violence on television.

ANSWER: 0.09 ±0.0461 = (0.0439, 0.1361)


116. Briefly explain what the interval estima t e in Question 115 tells you.

ANSWER:We estima t e that the proportion of Canadians who believe that there is too much violence on television is betwee n 4.39% and 13.61% higher than the proportion of Americans who share the same view.


A statistician wants to test for the equality of means in two indepen d e n t samples drawn from normal populations. However, he will not perform the equal variance t test of the difference betwe e n the population means if the condition necess ary for its use is not satisfied. The data follow:

Sample 1: 7 9 6 15 7 10 8 12Sample 2: 2 25 9 15 10 18 5 22 27 3

117. Given the data above, can the statistician conclude at the 5% significance level that the required condition is not satisfied?

ANSWER: =2

2210 /: σσH 1

≠22

211 /: σσH 1

Rejection region: F > 0.025,7,9F = 4.20 or F < 0.975,7,9F = 1/ 0.025,9,7F = 0.207 Test statistics: F = 0.108 Conclusion: Reject the null hypothesis . Yes


ANSWER:

LCL = ( /21s

22s ) / 0.025,7,9F = 0.0257

UCL = ( /21s

22s ) . 0.025,9,7F = 0.5214


ANSWER:We estima t e that ( 2


120. Briefly explain how to use the interval estimat e in Question 118 to test the hypothe s e s .

ANSWER:Since the hypothesized value 1 is not included in the 95% interval estima t e , we reject the null hypothesis at α = 0.05.


121. The owner of a service station wants to determine if owners of new cars (two years old or less) change their cars’ oil more frequen tly than owners of older cars (more than two years old). From his records he takes a random sample of ten new cars and ten older cars and determines the number of times the oil was change d in the last 12 months . The data follow. Do these data allow the service station owner to infer at the 10% significance level that new car owners change their cars’ oil more frequen tly than older car owners?

Frequency of Oil Changes in Past 12 Months

New Car Owners Old Cars Owners6 43 23 13 24 33 26 25 35 24 1


0: 211 >− µµH

Rejection region: t > 0.10,18t = 1.33Test statistic: t = 2.914Conclusion: Reject the null hypothesis . Yes

122. Because of the rising costs of industrial accident s , many chemical, mining, and manufacturing firms have instituted safety courses . Employees are encourag e d to take these courses designed to heighten safety awaren e s s . A company is trying to decide which one of two courses to institute . To help make a decision eight employees take course 1 and another eight take course 2. Each employee takes a test , which is graded out of a possible 25. The safety test results are shown below. Assume that the scores are normally distributed. Do these data provide sufficient evidence at the 5% level of significance to infer that the marks from course 1 are lower than those of course 2?

Course 1 14 21 17 14 17 19 20 16Course 2 20 18 22 15 23 21 19 15


0: 211 <− µµH

Rejection region: t < 0.05,14t = 1.761Test statistic: t = 1.336Conclusion: Don’t reject the null hypothesis. No



Ten functionally illiterat e adults were given an experime n t al one week crash course in reading. Each of the ten adults was given a reading test prior to the course and another test after the course. The results are shown below.

Adult 1 2 3 4 5 6 7 8 9 10Score after course 48 42 43 34 50 30 43 38 41 38Score before course 31 34 18 30 44 28 34 33 27 32

123. Is there enough evidence to infer at the 5% significance level that the reading scores have improved?

ANSWER:=DH µ:0 0

>DH µ:1 0


124. Estimate the mean improve m e n t with 95% confidence.

ANSWER:8.6 ±5.072 = (3.528, 13.672)

125. Briefly describe what the interval estima t e in Question 124 tells you.

ANSWER:We estima t e that the scores after taking the course improve on the averag e betwe en 3.528 and 13.672 points.

126. A politician regularly polls her constituency to gauge her level of support among voters. This month, 652 out of 1158 voters support her. Five months ago, 412 out of 982 voters support ed her. With a 5% significance level, can she infer that support has increase d by at least 10 percent a g e points?

ANSWER:10.: 210 =− ppH

10.: 211 >− ppH

Rejection region: z > =05.z 1.645 Test statistic: z = 2.0 Conclusion: Reject the null hypothesis . Yes



An investor is considering two types of investm e n t . She is quite satisfied that the expecte d return on investm e n t 1 is higher than the expec ted return on investm e n t 2. However, she is quite concerne d that the risk associat ed with invest m e n t 1 is higher than that of investm e n t 2. To help make her decision, she randomly selects seven monthly returns on investm e n t 1 and ten monthly returns on investm e n t 2. She finds that the sample variances of invest m e n t s 1 and 2 are 225 and 118, respec tively.

127. Can she infer at the 5% significance level that the population variance of investm e n t 1 exceed s that of invest m e n t 2?

ANSWER:=2

2210 /: σσH 1

>22

211 /: σσH 1

Rejection region: F > 0.05,6,9F = 3.37Test statistics: F = 1.907Conclusion: Don’t reject the null hypothesis. No

128. Estimate with 95% confidence the ratio of the two population variances , and briefly describe what the interval estima t e tells you.

ANSWER:

LCL = ( /21s

22s ) / 0.025,6,9F = 0.4414

UCL = ( /21s

22s ) . 0.025,9,6F = 10.525

We estima t e that ( 22



A political poll immediat ely prior to a congressional election reveals that 145 out of 250 male voters and 105 out of 200 female voters intend to vote for the Democra t candida t e .

129. Can we infer at the 5% significance level that the proportion of male and female voters who intend to vote for the Democra t candida t e differ?.

ANSWER:0: 210 =− ppH

0: 211 ≠− ppH

Rejection region: | z | > =025.z 1.96Test statistic: z = 1.17Conclusion: Don’t reject the null hypothesis. No




131. Estimate with 95% confidence the difference in the proportion of male and female voters who intend to vote for the Democra t candida t e .

ANSWER:0.055 ±0.0924 = (0.0374, 0.1474)

132. Explain how to use the interval estima t e in Question 131 to test the hypothe s e s .

ANSWER:Since the hypothesized value 0 is included in the 95% confidence interval, we fail to reject the null hypothesis at =α 0.05.


Thirty five employee s who complet e d two years of college were asked to take a basic mathe m a t ics test. The mean and stand ard deviation of their scores were 75.1 and 12.8, respec tively. In a random sample of 50 employee s who only complet e d high school, the mean and stand ard deviation of the test scores were 72.1 and 14.6, respec tively.

133. Can we infer at the 10% significance level that a difference exists betwe en the two groups?


0: 211 ≠− µµH

Rejection region: | t | > 0.05,83t ≈ 1.664Test statistic: t = 0.98Conclusion: Don’t reject the null hypothesis. No

134. Estimate with 90% confidence the difference in mean scores betwe e n the two groups of employe es .

ANSWER:3.0 ±5.094 = (2.094, 8.094)

135. Explain how to use the interval estima t e in Question 134 to test the hypothe s e s .

ANSWER:


Since the hypothesized value 0 is included in the 90% confidence interval, we fail to reject the null hypothesis at =α 0.10.


Do govern m e n t employee s take longer coffee breaks than private sector workers? That is a question that interes t e d a manag e m e n t consultant . To examine the issue, he took a rando m sample of ten governm e n t employee s and another random sample of ten private sector workers and measur e d the amount of time (in minutes) they spent in coffee breaks during the day. The results are listed below.

Govern m e n t Employees

Private Sector Workers

23 2518 1934 1831 2228 2833 2525 2127 2132 2021 16

136. Do these data provide sufficient evidence at the 5% significance level to support the consultan t’s claim?


0: 211 >− µµH


137. Estimate with 95% confidence the difference in coffee breaks mean time betwe en the two groups and explain what the interval estimat e tells you.

ANSWER:5.7 ±4.309 = (1.371, 10.029)We estimat e that govern m e n t employee s on averag e take betwe e n 1.371 and 10.029 minutes longer for coffee breaks than private sector workers do.



A quality control inspector keeps a tally shee t of the numb er of accept able and unaccep t a ble products that come off two different production lines. The complet ed shee t is shown below.

ProductsProduction line

Acceptable Unaccept a ble

1 152 482 136 54

138. Can the inspector infer at the 5% significance level that production line 1 is doing a better job than production line 2?


0: 211 >− ppH

Rejection region: z > 0.05z = 1.645Test statistic: z = 1.782Conclusion: Reject the null hypothesis . Yes

139. What is the p value of the test? Explain how to use it for testing the hypothe s e s .

ANSWER:p value = 0.0375Since p value = 0.0375 < α = 0.05, we reject the null hypothesis .

140. Estimate with 95% confidence the difference in population proportions .

ANSWER:0.08 ±0.0877 = (0.0077, 0.1677)


A politician has commissioned a survey of blue collar and white collar employee s in her constituency. The survey reveals that 286 out of 542 blue collar workers intend to vote for her in the next election whereas 428 out of 955 white collar workers intend to vote for her.

141. Can she infer at the 5% level of significance that the level of support differs betwe en the two groups of workers?


0: 211 ≠− ppH

Rejection region: | z | > 0.025z = 1.96


Test statistic: z = 2.96Conclusion: Reject the null hypothesis . Yes

142. What is the p value of the test? Explain how to use it to test the hypothes e s .

ANSWER:p value = 0.0015. Since p value = 0.0015 < =α 0.05, we reject the null hypothesis .


ANSWER:0.0795 ±0.0525 = (0.027, 0.132)


ANSWER:We estima t e that the proportion of blue collar workers intend to vote for the politician in the next election is betwe en 2.7% and 13.2% higher than the white collar workers who intend to vote for her.


An industrial statistician wanted to deter mine if efforts to promote safety have been successful. By checking the records of 250 employees , he found that 30 of them suffered either minor or major injuries that year. A random sample of 400 employee s last year revealed that 80 suffered some form of injury.

145. Can the statistician infer at the 5% significance level that efforts to promote safety have been successful?


0: 211 <− ppH

Rejection region: z < =05.z 1.645Test statistic: z = 2.65Conclusion: Reject the null hypothesis . Yes

146. What is the p value of the test? Explain how to use it for testing the hypothe s e s .

ANSWER:p value = 0.004Since p value = 0.004 < =α 0.05, we reject the null hypothe sis.


ANSWER: 0.08 ±0.0562 = (0.1362, 0.0238)


148. Automobile insuranc e appraisers examine cars that have been involved in accident al collisions to asses s the cost of repairs. An insurance executive is concerne d that different appraisers produce significantly different asses s m e n t s . In an experime n t 10 cars that have recently been involved in accident s were shown to two appraisers . Each asses s e d the estimat e d repair costs. These results are shown below. Can the executive conclude at the 5% significance level that the appraisers differ in their asses s m e n t s?

Car Appraiser 1 Appraiser 21 1650 14002 360 3803 640 6004 1010 9205 890 9306 750 6507 440 4108 1210 10809 520 480

10 690 770


Rejection region: | t| > 0.025,9t = 2.262Test statistic: t = 1.802Conclusion: Don’t reject the null hypothesis. No


A marke ting consultant was in the process of studying the perceptions of married couples concerning their weekly food expenditures . He believed that the husband’s perception would be higher than the wife’s. To judge his belief, he takes a random sample of ten married couples and asks each spouse to estima t e the family food expenditure (in dollars) during the previous week. The data are shown below.

Couple Husband Wife1 380 2702 280 3003 215 1854 350 3205 210 1806 410 3907 250 2508 360 3209 180 170

10 400 330


149. Can the consultant conclude at the 5% significance level that the husban d’s estima t e is higher than the wife’s estimat e?

ANSWER:0:0 =DH µ0:0 >DH µ

Rejection region: t > 0.05,9t = 1.833Test statistics: t = 2.776Conclusion: Reject the null hypothesis . Yes.

150. Estimate with 95% confidence the population mean difference .

ANSWER:32 ±26.076 = (5.924, 58.076)


ANSWER:We estimat e that the husban d’s perception of their weekly food expenditures would be on average betwe e n $5.924 and $78.076 higher than the wife’s.


Do out ofstate motorists violate the speed limit more frequently than instate motorists? This vital ques tion was address e d by the highway patrol in a large east ern state . A random sample of the speeds of 2,500 randomly selected cars was categorized according to whether the car was register ed in the state or in some other state and whether or not the car was violating the speed limit. The data follow.

In State Cars Out of State CarsSpeeding 521 328Not speeding 1141 510

152. Do these data provide enough evidence to support the highway patrol’s claim at the 5% significance level?


0: 211 <− ppH

Rejection region: z < 0.05z = 1.645 Test statistic: z = 3.88 Conclusion: Reject the null hypothesis . Yes



ANSWER:0.0779 ± 0.0399 = (0.1178, 0.0380)


ANSWER:We estima t e that the proportion of motorists that violate the speed limit is betwe en 3.8% and 11.78% less for cars that were registered in the state than for those register ed in some other state .


In a rando m sample of 20 patients who visited the emerg ency room of hospital 1, a researche r found that the variance of the waiting time (in minutes) was 128.0. In a random sample of 15 patients in the emerge ncy room of hospital 2, the research e r found the variance to be 178.8.

155. Can we infer at the 5% level of significance that the population variances differ?

ANSWER:=2

2210 /: σσH 1

≠22

211 /: σσH 1

Rejection region: F > 0.025,19,14F ≈ 2.84 or F < 0.975,19,14 0.025,14,19F F= ≈ 0.382Test statistics: F = 0.716Conclusion: Don’t reject the null hypothesis. No


ANSWER:

LCL = ( /21s

22s ) / 0.025,19,14F ≈ 0.252

UCL = ( /21s

22s ) . 0.025,14,19F ≈ 1.874


ANSWER:We estima t e that ( 2




The marke ting manag er of a pharm a c e u tical company believes that more girls than boys use its acne medicine. In a recent survey, 2500 teenag e r s are asked whether or not they use that particular product . The respons e s , categorized by sex, are sum m arized below.

Sex Use acne Don’t use acne

Female 540 810Male 391 759

158. Do these data provide enough evidence at the 10% significance level to support the manag er’s claim?


0: 211 >− ppH

Rejection region: z > 0.10z = 1.28 Test statistic: z = 3.09 Conclusion: Reject the null hypothesis . Yes

159. Estimate with 90% confidence the difference in the proportion of male and female users of the acne medicine.

ANSWER:0.06 ±0.0318 = (0.0282, 0.0918)

160. Describe what the interval estimat e in Question 159 tells you.

ANSWER:We estima t e that the proportion of girls who use the acne medicine is betwe en 2.82% and 9.18% more than the proportion of boys who use acne.


The presiden t of a breakfas t cereal manufac turer believes that families where both spouses work are much more likely to be consu m e rs of his product . To prove his point, he commissions a survey of 300 families where both spouse s work and 300 families with only one working spouse. Each family is asked whether the company’s cereal is eaten for breakfas t . The results are shown below.

Two spouse s working

One spouse working

Eat cereal 114 87


Don’t eat cereal

186 213

161. Do these data provide enough evidence at the 1% significance level to infer that the proportion of families with two work ing spouse s who eat the cereal is at least 5% larger than the proportion of families with one working spouse who eat the cereal?

ANSWER:0 1 2: 0.05H p p− =

1 1 2: 0.05H p p− > Rejection region: z > 0.01z = 2.33 Test statistic: z = 1.04 Conclusion: Don’t reject the null hypothesis. No

162. What is the p value of the test? Briefly explain how to use it for testing the hypothe s e s .

ANSWER:p value = 0.1492Since p value = 0.1492 > α = 0.01, we fail to reject the null hypothe sis.


ANSWER:0.09 ± 0.098 = (0.008, 0.188)

164. A manag e m e n t consultant wants to compare the income of gradua t e s of MBA progra m s with gradua t e s of BBA progra m s . In a random sample of 20 incomes five years after getting an MBA degree , the consultan t found the mean salary and the stand ard deviation to be $45,300 and $9,600, respec tively. A random sample of 25 incomes five years after getting a BBA degree yielded a mean salary of $43,600 with a standard deviation of $12,300.a. Can we infer at the 5% level of significance that the population means

differ?b. Estimate with 95% confidence the difference in mean salaries betwe e n

MBA and BBA gradua t e s .

ANSWERS:a. 0: 210 =− µµH

0: 211 ≠− µµH

Rejection region: | t| > 0.025,43t ≈ 2.014 Test statistic: t = 0.507 Conclusion: Don’t reject the null hypothesis. Nob. 1.7 ±6.76 = (5.06, 8.46)


165. A psychologist has performe d the following experimen t . For each of 10 sets of identical twins who were born 30 years ago, he recorded their annual incomes according to which twin was born first. The results (in $1,000s) are shown below. Can he infer at 5% significance level that there is a difference in income betwe en the twins?

Twin Set First Born Second Born1 32 442 36 433 21 284 30 395 49 516 27 257 39 328 38 429 56 64

10 44 44


Rejection region: | t | > 0.025,9t = 2.262Test statistics: t = 2.191Conclusion: Don’t reject the null hypothesis. No

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