1st Edition Test Bank Ch.10 Hypothesis Tests Proportions

CHAPTER 10

TRUE/FALSE

1. A matched sample design often leads to a smaller sampling error than the independent sample design because variation between sampled items is reduced or eliminated as a source of sampling error.

2. The sampling distribution of the difference between two sample proportions is approximately normal whenever n ≥ 30.

3. One of the differences between conducting a two-tailed hypothesis test for the difference between two population means versus a one-tailed test is that the value of the test statistic becomes larger.

4. For a hypothesis test involving the difference between two proportions, the pooled sample proportion is a weighted average of the two individual sample proportions.

5. The sampling distribution of the sample proportion is the probability distribution of all possible values of the sample proportion when a sample size n is taken from a particular population.

6. In a matched sample design, one uses the average for each pair of data values when building a confidence interval.

7. The positive critical value for a two-tailed hypothesis test of the difference between two population means is larger than for a one-tailed hypothesis test given the same level of significance.

8. The level of significance in a hypothesis test for a population proportion is the probability of accepting a false null hypothesis.

9. In hypothesis test for the difference between two population means, the critical value is a number that establishes the boundary of the reject H₀ region.

10. You are using a sample of size 100 to conduct a hypothesis test in which you want to determine whether a certain population proportion  has increased since last year. If the null hypothesis is π < .33 and the test statistic turns out to be 2.13, you should conclude, at the .05 significance level, that the population proportion has, in fact, increased.

11. You are using a sample of size 120 to conduct a hypothesis test in which you want to determine whether a certain population proportion  has changed since last year. If the null hypothesis is π = .25 and the test statistic turns out to be 1.83, you should conclude, at the .05 significance level, that the population proportion has, in fact, changed.

12. You are using a sample of size 200 to conduct a hypothesis test in which you want to determine whether a certain population proportion  has decreased since last year. If the null hypothesis is π > .5 and the test statistic turns out to be -1.74, you should conclude, at the .05 significance level, that the population proportion has, in fact, decreased.

13. You are using a sample of size 150 to conduct a hypothesis test in which you want to determine whether a certain population proportion  has decreased since last year. If the null hypothesis is π > .5 and the p-value for the test turns out to be .0324, you should conclude, at the .05 significance level, that the population proportion has, in fact, decreased.

14. You are using a sample of size 225 to conduct a hypothesis test in which you want to determine whether a certain population proportion  has changed since last year. If the null hypothesis is π = .25 and the p-value for the this two-tailed test turns out to be .0263, you should conclude, at the .05 significance level, that the population proportion has, in fact, changed.

15. You are using independent samples of size 12 each to test whether two population means are equal, with ₁ = ₂ as the null hypothesis. If the p-value for the test turns out to be .0113, you should conclude, at the .01 significance level, that the population means are not the equal.

16. You are using independent samples to test whether two population means are equal, with ₁ = ₂ as the null hypothesis. If the p-value for the test leads you to reject the null hypothesis at the 5% significance level, it will also lead you to reject the null hypothesis at the 1% significance level.

17. You are using independent samples of size 10 to test whether two population means are equal, with ₁ = ₂ as the null hypothesis. If the populations are normal and have equal standard deviations, it would be appropriate to use a pooled sample standard deviation in your test.

18. You are using independent samples of size 14 to test whether two population means are equal, with ₁ = ₂ as the null hypothesis. If the populations are normal and have unequal variances, it would be appropriate to use a pooled sample standard deviation in your test.

19. You are using independent samples of size 100 to test whether two population proportions are equal, with ₁ = ₂ as the null hypothesis. In this test, it would be appropriate to use the pooled sample proportion in an estimate of the standard error for your test.

20. You are using independent samples of size 150 to test whether two population proportions are equal, with ₁ = ₂ as the null hypothesis. If the sample proportion difference is statistically significant at the 5% significance level, it will also be statistically significant at the 10% significance level.

Multiple Choice

21.

In which of the following situations would a matched sample experiment be the appropriate way to test for a difference between two variables of interest?

a. To determine whether the mean income of all public employees in California differs from the mean income of all public employees in Oregon.

b. To determine whether the proportion of unemployed workers is higher in California than in Oregon.

c. To determine whether the number of cars sold by employees at a car dealership tends to be higher on Fridays than on Thursdays.

d. To determine whether average Friday sales at your car dealership is greater than average Friday sales at all dealerships nationwide.

22. For large sample cases, in a two-tailed test of the difference between two population means (significance level = 5%), the decision rule is:

a. Reject the null hypothesis if the test statistic, z_stat, is either less than −1.96 or greater than +1.96.

b. Reject the null hypothesis if the test statistic, z_stat, is either greater than −1.96 or less than +1.96.

c. Reject the null hypothesis if the p-value is greater than .05.

d. Reject the null hypothesis if the test statistic, t_stat, is either less than −1.022 or greater than +1.022.

23. In large sample cases, using a significance level of 5% in a one-tailed test in which the null hypothesis is ₁ < ₂, the decision rule is:

a. Reject the null hypothesis if ₁ - ₂is greater than 0.

b. Reject the null hypothesis if the test statistic, z_stat, is either greater than −1.65 or less than +1.65.

c. Reject the null hypothesis if the p-value for the sample mean difference is less than .05.

d. Reject the null hypothesis if the difference in sample proportions is greater than 0.

24. In a one-tailed hypothesis test in which the null hypothesis is ₁ < ₂, suppose sample results lead you to reject the null hypothesis at the 1% significance level. Which of the following statements would be accurate?

a. ₁ - ₂must be greater than 0.

b. the test statistic, z_stat, is between −2.33 and +2.33.

c. the test statistic, z_stat, is less −2.33.

d. the p-value for the sample proportion difference is less than .01.

25. In a two-tailed hypothesis test in which the null hypothesis is ₁ = ₂, suppose sample results lead you to reject the null hypothesis at the 5% significance level. Which of the following statements must be true?

a. ₁ - ₂must be greater than 0.

b. you would also reject the null hypothesis at the 1% significance level.

c. the sample mean difference is statistically significant at the 5% level and above.

d. the p-value for the sample mean difference could be.075.

26. In a one-tailed hypothesis test in which the null hypothesis is ₁ > ₂, suppose sample results lead you to reject the null hypothesis at the 1% significance level. Which of the following statements must be true?

a. ₁ - ₂must be greater than 0.

b. you would also reject the null hypothesis at the 5% significance level.

c. the sample mean difference is not statistically significant at the 5% significance level.

d. the p-value for the sample mean difference could be.025.

27. In a one-tailed hypothesis test in which the null hypothesis is ₁ > ₂, sample sizes were 12 for sample 1 and 16 for sample 2. Which of the following statements is true?

a. If the two populations are normal, with equal standard deviations, we can pool sample standard deviations to estimate the common population standard deviation.

b. We could use the t distribution with df = 28 to conduct the test.

c. The sample mean difference can only be approximated.

d. The p-value for the sample mean difference would necessarily be less than .

28. For a one-tailed hypothesis test in which the null hypothesis is  >  and sample size is 150, which of the following statements is true?

a. Pooling sample standard deviations would be appropriate.

b. We could use the normal distribution to conduct the test.

c. The sample proportion can only be approximated.

d. cannot be less than .7.

29. For a hypothesis test in which the null hypothesis is ₁ - ₂ = , which of the following statements must be true?

a. To reject the null hypothesis, ₁ cannot be less than ₂.

b. Failing to reject the null hypothesis will mean that ₁ and ₂ are equal.

c. The sample proportions can only be approximated.

d. The null hypothesis can only be rejected if the sample proportions are not equal.

30. For a hypothesis test in which the null hypothesis is ₁ - ₂ > 0, which of the following statements must be true?

a. To reject the null hypothesis, ₁ cannot be less than ₂.

b. Failing to reject the null hypothesis will mean that ₁ and ₂ are equal.

c. The sample means will be different.

d. The null hypothesis can only be rejected if the sample means are not equal.

31. If we are interested in testing a null hypothesis that the mean of Population 1 is smaller than the mean of Population 2, the

a. null hypothesis should state μ₁ - μ₂ < 0

b. null hypothesis should state μ₁ - μ₂ > 0

c. alternative hypothesis should state μ₁ - μ₂ ≥ 0

d. alternative hypothesis should state μ₁ - μ₂ < 0

e. none of the above

32. You are testing the difference between two population means. The sample sizes are small. Under the assumption that the standard deviations of the two populations are equal and using a pooled estimator for the standard deviation, which of the following statements is true?

a. the sample size must be increased in order to conduct a proper test

b. the standard normal distribution can be used

c. the t distribution with n₁ + n₂ degrees of freedom is used

d. the t distribution with n₁ + n₂ – 2 degrees of freedom is used

e. none of the above

33. You are testing the difference between two population means. The pooled standard deviation is only appropriate if the two populations

a. are normally distributed

b. have equal variances (standard deviations)

c. have equal means (averages)

d. are skewed

e. have equal sample sizes

34. You are testing the difference between two population means. Which of the following best describes the pooled sample standard deviation under the assumption that the two population standard deviations are equal for small samples?

a. it is the square root of the weighted average of the sample variances, weighted by their respective degrees of freedom

b. it is a simple average of the sample standard deviations

c. it cannot be calculated for small samples

d. it is the square of the differences of the sample variances

e. none of the above

35. Which of the following statements about the sampling distribution of the sample proportion is TRUE?

a. it is approximately normal, so long as n > 5 and n(1 -  > 5

b. it is centered on the population proportion 

c. it has standard deviation equal to the square root of (1-)/n.

d. b and c only

e. all of the above

36. Independent samples are obtained from two normal populations with unknown but equal standard deviations (variances) in order to construct a hypothesis test for the difference between the population means. If the first sample contains 20 items and the second sample contains 25 items, the correct form to use for the sampling distribution is the

a. normal distribution

b. t distribution with 43 degrees of freedom

c. t distribution with 45 degrees of freedom

d. F distribution with 20 and 25 degrees of freedom

e. none of the above

37. Assume we are interested in determining whether the proportion of voters planning to vote for candidate A (define this proportion as _A) is less than the proportion of voters planning to vote for candidate B (define this proportion as _B) using the contrary position as the null hypothesis. The correct set of hypotheses for testing here is

a. H₀: _A - _B < 0; H_a: _A - _B ≥ 0

b. H₀: _A - _B ≥ 0; H_a: _A - _B < 0

c. H₀: _A - _B = 0; H_a: _A - _B ≠ 0

d. H₀: _A - _B ≤ 0; H_a: _A - _B > 0

e. none of the above

38. In a hypothesis test in which the null hypothesis states that the difference between two population proportions is 0, that is, when ₁ = ₂, which of the following occurs?

a. a common pooled value is calculated as a weighted average of the individual sample proportions

b. a one-tailed hypothesis test is performed

c the t distribution is used

d. the sampling distribution is positively skewed

e. all of the above

39. If we are interested in making the case that the proportion in Population 1 is larger than the proportion in Population 2, and we plan to use the contrary position as the null hypothesis, then the

a. null hypothesis should state ₁-₂< 0.

b. null hypothesis should state ₁-₂> 0.

c. alternative hypothesis should state ₁-₂≥ 0.

d. alternative hypothesis should state ₁-₂ < 0.

e. none of the above.

40. Independent samples are obtained from two normal populations with unknown but equal variances in order to construct a hypothesis test for the difference between the population means. If the first sample contains 16 items and the second sample contains 26 items, the correct form to use for the sampling distribution is the

a. normal distribution

b. t distribution with 15 degrees of freedom

c. t distribution with 25 degrees of freedom

d. t distribution with 40 degrees of freedom

e. none of the above

PROBLEMS

41.

You want to set up a test for the following hypotheses:

H₀: π ≥ .45

H_a: π < .45

You will have a random sample of 200 population members to make your decision. You intend to compute the value of the sample test statistic, z_stat. Using a significance level of .05, the decision rule here should be:

Reject the null hypothesis if z_stat < _______.

a. 1.65

b. -1.65

c. -1.96

d. 2.33

e. 1.96

42.

You want to set up a test for the following hypotheses:

H₀: π ≥ .45

H_a: π < .45

You have a random sample of 200 population members to make your decision. Suppose the sample proportion turns out to be .395. Compute the value of the sample test statistic, z_stat.

a. 2.16

b. -1.23

c. 1.99

d. -1.56

e. -1.24

43.

You want to set up a test for the following hypotheses:

H₀: π ≥ .45

H_a: π < .45

You have a random sample of 200 population members to make your decision. Suppose the sample proportion turns out to be .395. Compute z_stat for this sample result. If the significance level for your test is 5%, what is your conclusion?

a. Reject the null hypothesis since z_stat < z_c.

b. Don't reject the null hypothesis since z_stat < z_c.

c. Reject the null hypothesis since z_stat > z_c.

d. Don't reject the null hypothesis since z_stat > z_c.

44.

You want to set up a test for the following hypotheses:

H₀: π ≥ .45

H_a: π < .45

You have a random sample of 200 population members to make your decision. Suppose the sample proportion turns out to be .385. Compute the p-value for the sample result.

a. .0590

b. .1253

c. .0323

d. .2168

e. .0912

45.

You want to set up a test for the following hypotheses:

H₀: π ≥ .45

H_a: π < .45

You have a random sample of 200 population members to make your decision. Suppose the sample proportion turns out to be .405. Compute the p-value for this sample result. If the significance level for your test is 5%, what is your conclusion?

a. Reject the null hypothesis since p-value < significance level.

b. Reject the null hypothesis since p-value > significance level.

c. Don't reject the null hypothesis since p-value < significance level.

d. Don't reject the null hypothesis since p-value > significance level.

46.

You want to set up a test for the following hypotheses:

H₀: π = .75

H_a: π ≠ .75

You will have a random sample of 225 population members to make your decision. Using a significance level of .01, your decision rule could be: Reject the null hypothesis if the sample proportion is less than .xxx or greater than ________. (Determine the proper value to fill in the blank.)

a. .786

b. .697

c. .824

d. .812

e. .806

47.

You want to set up a test for the following hypotheses:

H₀: π ≤ .32

H_a: π > .32

You will have a random sample of 100 population members to make your decision. Suppose the sample proportion turns out to be .39. Compute the sample statistic, z_stat, and report it here.

a. 1.09

b. .86

c. 2.12

d. 1.39

e. 1.50

48.

You want to set up a test for the following hypotheses:

H₀: π ≤ .32 (The population proportion is .32 or less.)

H_a: π > .32 (The population proportion is greater than .32.)

You will have a random sample of 100 population members to make your decision. Suppose the sample proportion turns out to be .39. Compute the sample test statistic, z_stat, and report your decision. Use a significance level of 5%.

a. Reject the null hypothesis since z_stat < z_c.

b. Don't reject the null hypothesis since z_stat > z_c.

c. Don't reject the null hypothesis since z_stat < z_c.

d. Reject the null hypothesis since z_stat > z_c.

49.

You want to set up a test for the following hypotheses:

H₀: π ≤ .32 (The population proportion is .32 or less.)

H_a: π > .32 (The population proportion is greater than .32.)

You will have a random sample of 100 population members to make your decision. Suppose the sample proportion turns out to be .39. Compute the p-value for this sample result and report it here.

a. .0667

b. .0213

c. .0726

d. .0520

e. .1344

50.

You want to set up a test for the following hypotheses:

H₀: π ≤ .32 (The population proportion is .32 or less.)

H_a: π > .32 (The population proportion is greater than .32.)

You will have a random sample of 100 population members to make your decision. Suppose the sample proportion turns out to be .41. Compute the p-value for this sample result. If the significance level for the test is .05, report your conclusion.

a. Don't reject the null hypothesis since p-value < significance level.

b. Reject the null hypothesis since p-value > significance level.

c. Reject the null hypothesis since p-value < significance level.

d. Don't reject the null hypothesis since p-value > significance level.

51.

Over the past few years, 44% of Hobart Sporting Goods’ orders have come from states west of the Rocky Mountains. The company suspects that this number is changing and has set up a two-tailed hypothesis test to assess its suspicion. In a sample of 200 randomly selected orders this year, 96 were from customers west of the Rocky Mountains. Compute the p-value for the sample result in this two-tailed test, and report it here.

a. .1136

b. .2544

c. .0906

d. .1272

e. .1632

52.

Over the past few years, 44% of Hobart Sporting Goods’ orders have come from states west of the Rocky Mountains. The company suspects that this number is changing and has set up a two-tailed hypothesis test to assess its suspicion. In a sample of 200 randomly selected orders this year, 101 were from customers west of the Rocky Mountains. Compute p-value for the sample result, and report your conclusion. Use a significance level of 5%.

a. Reject the null hypothesis since p-value is less than .05.

b. Reject the null hypothesis since p-value is greater than .10.

c. Don't reject the null hypothesis since p-value is less than .10.

d. Don't reject the null hypothesis since p-value is greater than .05.

53.

Dexter Industries states that least 95% of all customer orders are shipped within 3 days of the order being received by the company. In a random sample of 600 orders shipped this year by the company, 555 met the 3 day requirement. You have set up a hypothesis test to determine whether this sample evidence is sufficient to reject, at the 5% significance level, a null hypothesis that the proportion of all orders shipped by the company this year meets Dexter’s standard. (Use H₀:  > .95) Compute the p-value for the sample result in this one-tailed test.

a. .0213

b. .0025

c. .0096

d. .0145

e. .0656

54.

Dexter Industries states that least 95% of all customer orders are shipped within 3 days of the order being received by the company. In a random sample of 500 orders shipped this year by the company, 465 met the 3 day requirement. You have set up a hypothesis test to determine whether this sample evidence is sufficient to reject, at the 5% significance level, a null hypothesis that the proportion of all orders shipped by the company this year meets Dexter’s standard. (Use H₀:  > .95) Compute the p-value for the sample result in this one-tailed test and use it to reach the proper conclusion.

a. Conclusion: Reject the null hypothesis since p-value < .05.

b. Conclusion: Don't reject the null hypothesis since p-value > .05.

c. Conclusion: Don't reject the null hypothesis since p-value < .05.

d. Conclusion: Reject the null hypothesis since p-value > .05.

55.

Organizers of a large conference on workplace environment would like an equal representation of male and female participants. You are given a random sample of 120 invitation responses. Your job is to use this sample to test a null hypothesis that exactly 50% of the total invitation responses are from males.

Suppose a particular sample of 120 contains 74 from males. For your test, compute the appropriate p-value for the sample result in this two-tailed test.

a. .0436

b. .0100

c. .0242

d. .0550

e. .0618

56.

Organizers of a large conference on workplace environment would like an equal representation of male and female participants. You are given a random sample of 140 invitation responses. Your job is to use this sample to test a null hypothesis that exactly 50% of the total invitation responses are from males.

Suppose a particular sample of 140 contains 62 from males. For your test, compute the appropriate p-value for the sample result in this two-tailed test and use it to reach the proper conclusion. Use a significance level of .10.

a. Don't reject the null hypothesis since the p-value is greater than .10.

b. Reject the null hypothesis since the p-value is greater than .10.

c. Reject the null hypothesis since the p-value is less than .10.

d. Don't reject the null hypothesis since the p-value is less than .20.

57.

A random sample of 500 current college freshmen is selected to test a null hypothesis that the proportion of military veterans in the freshman class is no higher than .13 (13%). Suppose in the sample of 500 freshmen, 76 were veterans. Compute z_stat for this sample result.

a. 1.23

b. 1.04

c. .91

d. 1.88

e. 1.46

58.

a. Conclusion: Reject the null hypothesis since z_stat is greater than 1.28.

b. Conclusion: Don't Reject the null hypothesis since z_stat is less than 1.65.

c. Conclusion: Don't Reject the null hypothesis since z_stat is less than 2.58.

d. Conclusion: Reject the null hypothesis since z_stat is less than 1.28.

59.

You have selected two random samples of size 50, one from Population 1 and one from Population 2. You intend to use the samples to test the following hypotheses regarding the difference in means for the two populations:

H₀: µ1 = µ2

H_a: µ1 ≠ µ2

Suppose the mean for the Population 1 sample is 122 and mean for the Population 2 sample is 115. Compute the proper p-value for the sample result in this two tailed test. Assume the standard deviation for Population 1 is 14 and for Population 2 it is 18.

a. .112

b. .015

c. .007

d. .030

e. .107

60.

H₀: µ1 = µ2

H_a: µ1 ≠ µ2

Suppose the mean for the Population 1 sample is 123 and mean for the Population 2 sample is 118. Compute the proper p-value for the sample result in this two-tailed test and use it to reach a proper conclusion. Assume the standard deviation for Population 1 is 14 and for Population 2 it is 16. Use a significance level of 10%.

a. Conclusion: Since the p-value is less than .10, we can reject the null hypothesis. That is, we can conclude that there is a difference in the population means.

b. Conclusion: Since the p-value is less than .05, we can reject the null hypothesis. That is, we can't conclude that there is a difference in the population means.

c. Conclusion: Since the p-value is less than .10, we can’t reject the null hypothesis. That is, we cannot conclude that there is a difference in the population means.

d. Conclusion: Since the p-value is less than .05, we can't reject the null hypothesis. That is, we cannot conclude that there is a difference in the population means.

61.

H₀: µ1 - µ2 = 0

H_a: µ1 - µ2 ≠ 0

Suppose the mean for the Population 1 sample is 123 and mean for the Population 2 sample is 118. Assume the standard deviation for Population 1 is 14 and for Population 2 it is 16. Compute the standard error (standard deviation) of the appropriate sampling distribution for the test.

a. 3.01

b. 2.46

c. 3.59

d. 4.13

62.

H₀: µ1 - µ2 ≤ 0 (The mean of Population 1 is no greater than the mean of Population 2.)

H_a: µ1 - µ2 > 0 (The mean of Population 1 is greater than the mean of Population 2.)

Suppose the mean for the Population 1 sample is 85 and the mean for the Population 2 sample is 82. Assume the standard deviation for Population 1 is 9 and for Population 2 it is 11. Compute the appropriate p-value for the sample result

a. .045

b. .002

c. .054

d. .034

e. .017

63.

H₀: µ1 - µ2 ≤ 0 (The mean of Population 1 is no greater than the mean of Population 2.)

H_a: µ1 - µ2 > 0 (The mean of Population 1 is greater than the mean of Population 2.)

Suppose the mean for the Population 1 sample is 85 and the mean for the Population 2 sample is 82. Assume the standard deviation for Population 1 is 9 and for Population 2 it's 11. Use a significance level of .01. Compute the appropriate p-value for this sample result and use it to reach the proper conclusion.

a. Conclusion: Since the p-value is less than .01, we can’t reject the null hypothesis. That is, we can’t conclude that the mean of Population 1 is greater than the mean of Population 2.

b. Conclusion: Since the p-value is greater than .01, we can reject the null hypothesis. That is, we can conclude that the mean of Population 1 is greater than the mean of Population 2.

c. Conclusion: Since the p-value is less than .01, we can reject the null hypothesis. That is, we can conclude that the mean of Population 1 is greater than the mean of Population 2.

d. Conclusion: Since the p-value is greater than .01, we can’t reject the null hypothesis. That is, we can’t conclude that the mean of Population 1 is greater than the mean of Population 2.

64.

TriCal Power conducted a study of average sustained wind velocity for two possible sites for its wind farm. For a sample of 125 days in North Canyon, average sustained velocity was 16.2 mph. For a sample of 160 days at Table Mesa, average sustained velocity was 14.3 mph. Your job is to determine whether these sample results are sufficient to reject a “no difference in (population) average sustained velocity” null hypothesis. Assume the population standard deviations are 4.9 mph for the North Canyon site and 5.1 mph for the Table Mesa site. Compute z_stat, the sample test statistic and report it here.

a. 1.01

b. 3.19

c. 2.25

d. 3.71

e. 2.58

65.

TriCal Power conducted a study of average sustained wind velocity for two possible sites for its wind farm. For a sample of 125 days in North Canyon, average sustained velocity was 16.2 mph. For a sample of 160 days at Table Mesa, average sustained velocity was 15.3 mph. Your job is to determine whether these sample results are sufficient to reject a “no difference in (population) average sustained velocity” null hypothesis. Assume the population standard deviations are 4.9 mph for the North Canyon site and 5.1 mph for the Table Mesa site. Compute z_stat, the sample test statistic and use it to reach the proper conclusion. Use a significance level of 5%.

a. Since the test statistic puts the sample result inside the two critical values of -1.96 and +1.96, we can't reject the null hypothesis. There’s not enough sample evidence to show that there is a difference in the averages for the two populations. The difference in sample means is not statistically significant at the 5% significance level.

b. Since the test statistic puts the sample result outside the upper critical value of 1.96, we can reject the null hypothesis. There’s enough sample evidence to show that there is a difference in the averages for the two populations. The difference in sample means is statistically significant at the 5% significance level.

c. Since the test statistic puts the sample result outside the upper critical value of 1.96, we can't reject the null hypothesis. There’s not enough sample evidence to show that there is a difference in the averages for the two populations. The difference in sample means is not statistically significant at the 5% significance level.

d. Since the test statistic puts the sample result inside the two critical values of -1.96 and +1.96, we can reject the null hypothesis. There’s enough sample evidence to show that there is a difference in the averages for the two populations. The difference in sample means is statistically significant at the 5% significance level.

66.

You will have a random sample of 10 population members from each of two normal populations having equal standard deviations (variances) and you want to set up a test using the following hypotheses:

H₀: µ1 = µ2

H_a: µ1 ≠ µ2

Suppose the sample mean for the sample selected from Population 1 is 17.5 (with a standard deviation of 4.2) and the sample mean for the sample selected from Population 2 is 15 (with a standard deviation of 4.6). Compute the appropriate test statistic, t_stat, for the test and report it here.

a. 2.14

b. 3.81

c. 4.25

d. 4.62

e. 1.27

67.

You will have a random sample of 10 population members from each of two normal populations with equal standard deviations and you want to set up a test using the following hypotheses:

H₀: µ1 - µ2 = 0

H_a: µ1 - µ2 ≠ 0

Suppose the sample mean for the sample selected from Population 1 is 18.8 (with a standard deviation of 4.2) and the sample mean for the sample selected from Population 2 is 14.6 (with a standard deviation of 4.6). Compute the appropriate test statistic, t_stat, for the test and use it to reach the proper conclusion. Use a significance level of 5%.

a. Since t_stat is outside the critical values of ± 2.101, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. That is, the sample mean difference is statistically significant at the 5% significance level.

b. Since t_stat is not outside the critical values of ± 2.101, we can't reject the null hypothesis. There isn't enough sample evidence to convince us that the population means are different. That is, the sample mean difference is not statistically significant at the 5% significance level.

c. Since t_stat is outside the critical values of ± 1.601, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. That is, the sample mean difference is statistically significant at the 5% significance level.

d. Since t_stat is not outside the critical values of ± 3.101, we can't reject the null hypothesis. There isn't enough sample evidence to convince us that the population means are different. That is, the sample mean difference is not statistically significant at the 5% significance level.

68.

You will have a random sample of 10 Population 1 members and a sample of 13 Population 2 members, and you want to set up a test using the following hypotheses:

H₀: µ1 - µ2 ≤ 0 (The mean of Population 1 is no greater than the mean of Population 2.)

H_a: µ1 - µ2 > 0 (The mean of Population 1 is greater than the mean of Population 2.)

Assume that the two population distributions are normal and that the population standard deviations are equal.

Suppose the sample mean for the sample taken from Population 1 is 95 and the sample mean for the sample taken from Population 2 is 89. The respective sample standard deviations are 5.2 and 5.6. Compute the test statistic, t_stat for this sample result and report it here.

a. 3.25

b. 2.63

c. 2.03

d. 3.77

e. 3.99

69.

You will have a random sample of 10 Population 1 members and a sample of 13 Population 2 members, and you want to set up a test using the following hypotheses:

H₀: µ1 - µ2 ≤ 0 (The mean of Population 1 is no greater than the mean of Population 2.)

H_a: µ1 - µ2 > 0 (The mean of Population 1 is greater than the mean of Population 2.)

Assume that the two population distributions are normal and that the population standard deviations are equal.

Suppose the sample mean for the sample taken from Population 1 is 98 and the sample mean for the sample taken from Population 2 is 90. The respective sample standard deviations are 6.2 and 5.6. Compute the test statistic, t_stat for this sample result and use this test statistic to reach the proper conclusion here. Use a significance level of .01.

a. Since t_stat is less than the critical value of 2.518, we can't reject the null hypothesis. There isn't enough sample evidence to convince us that the mean of population 1 is greater than the mean of population 2.

b. Since t_stat is less than the critical value of 2.118, we can reject the null hypothesis. There is enough sample evidence to convince us that the mean of population 1 is greater than the mean of population 2.

c. Since t_stat is above the critical value of 2.518, we can reject the null hypothesis. There is enough sample evidence to convince us that the mean of population 1 is greater than the mean of population 2.

d. Since t_stat is above the critical value of 2.118, we can't reject the null hypothesis. There isn't enough sample evidence to convince us that the mean of population 1 is greater than the mean of population 2.

70.

You are conducting a study of unemployment patterns during the most recent economic downturn. You want to determine if the average time that a person is unemployed is different for those with a college degree and those without. You interview a random sample of 10 people with a college degree who had been unemployed but have recently found a job, and a random sample of 15 people without degrees who had been unemployed but have recently found a job. Workers in the ‘degree’ sample were unemployed for an average of 5.2 months, with a sample standard deviation of 1.8 months, while workers in the ‘no degree’ sample were unemployed for an average of of 4.1 months, with a sample standard deviation of 1.3 months. Assume that the population distributions are normal and have equal standard deviations. Compute the sample statistic, t_stat, for your test and report it here.

a. 1.63

b. 1.99

c. 1.78

d. 1.34

e. 2.81

71.

You are conducting a study of unemployment patterns during the most recent economic downturn. You want to determine if the average time that a person is unemployed is different for those with a college degree and those without. You interview a random sample of 10 people with a college degree who had been unemployed but have recently found a job, and a random sample of 15 people without degrees who had been unemployed but have recently found a job. Workers in the ‘degree’ sample were unemployed for an average of 5.8 months, with a sample standard deviation of 1.8 months, while workers in the ‘no degree’ sample were unemployed for an average of 3.9 months, with a sample standard deviation of 1.3 months. Assume that the population distributions are normal and have equal standard deviations. Compute the sample statistic, t_stat, for your test and use it to reach the proper conclusion.

a. Conclusion: Since t_stat is outside the critical values of ± 2.807, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. That is, the sample mean difference is statistically significant at the 1% significance level.

b. Conclusion: Since t_stat is inside the critical values of ± 2.807, we can’t reject the null hypothesis. There isn’t enough sample evidence to convince us that the population means are different. That is, the sample mean difference isn’t statistically significant at the 1% significance level.

c. Conclusion: Since t_stat is outside the critical values of ± 1.271, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. That is, the sample mean difference is statistically significant at the 1% significance level.

d. Conclusion: Since t_stat is inside the critical values of ± 1.271, we can't reject the null hypothesis. There isn't enough sample evidence to convince us that the population means are different. That is, the sample mean difference is not statistically significant at the 1% significance level.

72.

You will have a random sample of 100 population members from each of two populations and you want to set up a test for the following hypotheses:

H₀: π1 = π2 (The two population proportions are equal.)

H_a: π1 ≠ π2 (The two population proportions are not equal.)

Suppose the sample proportion for the Population 1 sample is .44 and the sample proportion for the Population 2 sample is .36. Calculate the sample statistic, z_stat,that would be appropriate for your test and report it here.

a. .82

b. 1.49

c. 3.41

d. 1.15

e. 1.65

73.

You will have a random sample of 100 population members from each of two populations and you want to set up a test for the following hypotheses:

H₀: π1 = π2 (The two population proportions are equal.)

H_a: π1 ≠ π2 (The two population proportions are not equal.)

Suppose the sample proportion for the Population 1 sample is .43 and the sample proportion for the Population 2 sample is .37. Calculate the sample statistic, z_stat, that would be appropriate for your test and use it to reach the proper conclusion. Use a significance level of 10%.

a. Conclusion: Since the z_stat value is inside the critical values of ±1.96, we can’t reject the null hypothesis. There isn’t enough sample evidence to show that the population means are different.

b. Conclusion: Since the z_stat value is inside the critical values of ±1.65, we can’t reject the null hypothesis. There isn’t enough sample evidence to show that the population means are different.

c. Conclusion: Since the z_stat value is inside the critical values of ±1.96, we can’t reject the null hypothesis. There isn’t enough sample evidence to show that the population means are different.

d. Conclusion: Since the z_stat value is outside the critical values of ±1.65, we can reject the null hypothesis. There is enough sample evidence to show that the population means are different.

74.

You will have a random sample of 500 population members from each of two populations and you want to set up a test for the following hypotheses:

H₀: π1 - π2 ≤ 0 (The Population 1 proportion is no greater than the Population 2 proportion.)

H_a: π1 - π2 > 0 (The Population 1 proportion is greater than the Population 2 proportion.)

Suppose the sample proportion for the Population 1 sample is .31 and the sample proportion for the Population 2 sample is .25. Calculate the p-value that would be appropriate for your test.

a. .162

b. .004

c. .068

d. .034

e. .017

75.

You will have a random sample of 200 population members from each of two populations and you want to set up a test for the following hypotheses:

H₀: π1 - π2 ≤ 0 (The Population 1 proportion is no greater than the Population 2 proportion.)

H_a: π1 - π2 > 0 (The Population 1 proportion is greater than the Population 2 proportion.)

Suppose the sample proportion for the Population 1 sample is .31 and the sample proportion for the Population 2 sample is .25. Calculate the sample p-value that would be appropriate for your test and use it to make your decision. Use a significance level of 5%.

a. Since p-value > .05, we can reject the null hypothesis. There is enough sample evidence to show that the population 1 proportion is greater than the population 2 proportion.

b. Since p-value > .05, we can’t reject the null hypothesis. There is not enough sample evidence to show that the population 1 proportion is greater than the population 2 proportion.

c. Since p-value < .05, we can reject the null hypothesis. There is enough sample evidence to show that the population 1 proportion is greater than the population 2 proportion.

d. Since p-value < .05, we can’t reject the null hypothesis. There is not enough sample evidence to show that the population 1 proportion is greater than the population 2 proportion.

76.

In a survey involving a national random sample of 600 graduating high school seniors (250 from suburban high schools and 350 from city high schools), 73% of the suburban seniors in the survey and 66% of the city seniors said they planned to attend college next year. Can these survey results be used to reject a null hypothesis that the proportion of suburban seniors who plan to attend college is (exactly) the same as the proportion of city seniors who plan to attend college? Construct a hypothesis test using a significance level of 5%. Report your conclusion.

a. Since the .034 p-value < .05, we can reject the null hypothesis. There is enough sample evidence to show that the proportions in the two populations are different.

b. Since the .068 p-value > .05, we can’t reject the null hypothesis. There isn’t enough sample evidence to show that the proportions in the two populations are different.

c. Since the .136 p-value > .05, we can’t reject the null hypothesis. There isn’t enough sample evidence to show that the proportions in the two populations are different.

d. Since the .034 p-value < .10, we can reject the null hypothesis. There is enough sample evidence to show that the proportions in the two populations are different.

77.

In a survey involving a national random sample of 600 graduating high school seniors (300 from suburban high schools and 300 from city high schools), 73% of the suburban seniors in the survey and 66% of the city seniors said they planned to attend college next year. Can these survey results be used to make the case that the proportion of suburban seniors who plan to attend college is greater than the proportion of city seniors who plan to attend college? Construct a hypothesis test using a significance level of 5%. Report your conclusion.

a. The proper p-value in this one-tailed test is .031. Since the p-value is greater than .025, we can’t reject the 1 < 2 null hypothesis. There’s not enough sample evidence to conclude that the proportion of suburban seniors is greater than the proportion of city seniors.

b. The proper p-value in this one-tailed test is .031. Since the p-value is less than .05, we can reject the 1 < 2 null hypothesis. There’s enough sample evidence to conclude that the proportion of suburban seniors is greater than the proportion of city seniors.

c. The proper p-value in this one-tailed test is .062. Since the p-value is greater than .05, we can’t reject the 1 < 2 null hypothesis. There’s not enough sample evidence to conclude that the proportion of suburban seniors is greater than the proportion of city seniors.

d. The proper p-value in this one-tailed test is .015. Since the p-value is less than .05, we can reject the 1 < 2 null hypothesis. There’s enough sample evidence to conclude that the proportion of suburban seniors is greater than the proportion of city seniors.

78.

In a random sample of 100 stock fund managers and 200 real estate fund managers, 65 of the stock fund managers predicted a sharp upturn in the economy over the next 12 months while 116 of the real estate fund managers held this view.

Use this sample data, and a significance level of 5%, to test the null hypothesis that the proportion of all stock fund managers who hold this view is the same as the proportion of all real estate fund managers. Report your conclusion.

a. Since the z_stat of 1.17 puts the sample result inside side the critical z values of ±2.58, we can’t reject the null hypothesis. There is not enough sample evidence to show that the two population proportions are not the same.

b. Since the z_stat of 1.17 puts the sample result inside the critical z values of ±1.96, we can’t reject the null hypothesis. There is not enough sample evidence to show that the two population proportions are not the same.

c. Since the z_stat of 2.89 puts the sample result outside the critical z values of ±1.96, we can't reject the null hypothesis. There isn't enough sample evidence to show that the two population proportions are not the same.

d. Since the z_stat of 2.89 puts the sample result inside the critical z values of ±2.58, we can reject the null hypothesis. There is enough sample evidence to show that the two population proportions are not the same.

79.

Cromwell, Inc. wants to determine whether two different assembly procedures for its product will produce different assembly times. Company analysts use a matched sample experiment to test a null hypothesis that the average assembly times for the two procedures are the same. The sample consists of five pairs of assemblers, matched according to skill levels and prior assembly experience. The table below shows the results. Assuming that the necessary population conditions are satisfied, construct a hypothesis to test the null hypothesis that there is no difference in average assembly times for the populations represented. Report the value of the sample test statistic, t_stat.

Assembly time (hrs)

assembler pairs	1	2	3	4	5
Procedure 1	5.6	7.6	8.4	8.2	6.0
Procedure 2	5.8	6.0	7.2	7.5	5.7

a. 1.05

b. 2.63

c. 3.07

d. .82

e. 2.26

80.

Cromwell, Inc. wants to determine whether two different assembly procedures for its product will produce different assembly times. Company analysts use a matched sample experiment to test a null hypothesis that the average assembly times for the two procedures are the same. The sample consists of five pairs of assemblers, matched according to skill levels and prior assembly experience. The table below shows the results. Assuming that the necessary population conditions are satisfied, compute the value of the sample test statistic, t_stat and use it to reach the proper conclusion here. Use a significance level of .05.

Assembly time (hrs)

assembler pairs	1	2	3	4	5
Procedure 1	5.2	7.6	8.9	8.2	6.4
Procedure 2	5.4	6.0	7.2	6.3	5.6

a. Since the t_stat of 2.989 is outside the interval -2.776 to +2.776, we can reject the “no difference” null hypothesis. Sample evidence is strong enough for us to conclude that the two populations have different means.

b. Since the t_stat of 1.947 is in the interval -4.604 to +4.604, we can’t reject the “no difference” null hypothesis. Sample evidence is strong enough for us to conclude that the two populations represented here would have different means.

c. Since the t_stat of -3.362 is outside the interval -2.776 to +2.776, we can reject the “no difference” null hypothesis. Sample evidence is strong enough for us to conclude that the two populations have different means.

d. Since the t_stat of 2.081 is between -2.306 and +2.306, we can’t reject the “no difference” null hypothesis. Sample evidence is not strong enough for us to conclude that the two promotions produce different average sales changes.

81.

Quality inspectors perform periodic inspections of the units being produced at Tolkien Industries. As a matter of policy, if the process is producing no more than 2% defective units, it is considered "in control" and should be allowed to continue unadjusted. If, however, the process is producing at a rate above 2%, it will be considered "out of control" and shut down for re-adjustment. A standard hypothesis test has been set up to enable the inspectors to decide whether to shut down the process, using the "in control" position as the null hypothesis. In this test, a Type I error would be concluding that the process

a. needs adjusting when, in fact, it doesn’t.

b. does not need adjusting when, in fact, it does.

is ‘in control.’
is in control when the defectives rate is actually less than 2%.

82.

a. needs adjusting when, in fact, it doesn’t.

b. does not need adjusting when, in fact, it does.

c. is ‘out of control.’

d. is in control when the defectives rate is actually less than 2%.

83.

You are conducting a survey to assess hiring expectations for companies in the region. You take a sample of 40 manufacturing companies and 60 service companies. On average, the companies in the manufacturing sample reported that they expected to add 12.7 jobs over the next 6 months, with a standard deviation of 4.8 jobs. On average, companies in the service sector sample expected to add 10.8 jobs, with a standard deviation of 3.6 jobs. Set up a hypothesis test to test a “no difference in population means” null hypothesis. Compute the proper p-value for the sample result in this two-tailed test.

a. .0441

b. .0328

c. .0267

d. .0164

e. .0882

84.

You are conducting a survey to assess hiring expectations for companies in the region. You take a sample of 50 manufacturing companies and 60 service companies. On average, the companies in the manufacturing sample reported that they expected to add 13.9 jobs over the next 6 months, with a standard deviation of 4.8 jobs. On average, companies in the service sector sample expected to add 11.7 jobs, with a standard deviation of 4.6 jobs. Set up a hypothesis test to test a “no difference in population means” null hypothesis. Compute the proper p-value for the sample result in this two-tailed test. Use a significance level of 1%.

a. Since the p-value of .007 is less than .01, we can reject the null hypothesis. There is enough sample evidence to convince us that the population averages are different. The sample mean difference is statistically significant at the 1% significance level.

b. Since the p-value of .001 is less than .01, we can reject the null hypothesis. There is enough sample evidence to convince us that the population averages are different. The sample mean difference is statistically significant at the 1% significance level.

c. Since the p-value of .031 is greater than .01, we can’t reject the null hypothesis. There is not enough sample evidence to convince us that the population averages are different. The sample mean difference is not statistically significant at the 1% significance level.

d. Since the p-value of .015 is greater than .01, we can’t reject the null hypothesis. There is not enough sample evidence to convince us that the population averages are different. The sample mean difference is not statistically significant at the 1% significance level.

85.

In a recent study, Helix Inc. reported that a sample of 100 randomly selected customers gave the company and average rating of 85.4 out of 100 for customer service, with a sample standard deviation of 19.2. The average rating given to its primary competitor, Horizon International, by an independent sample of 100 Horizon customers was 79.3, with a sample standard deviation of 23.2.

You are to set up a hypothesis test to decide whether this is sufficient evidence to reject a null hypothesis which states that Helix’s average rating is no greater than Horizon’s average rating for the two populations represented. Use a 5% significance level.

Compute the test statistic, z_stat, for the sample result and report your conclusion.

a. z_stat = 2.02. Since z_stat is above the critical value of 1.65, we can reject the null hypothesis. The sample mean difference is statistically significant at the 5% significance level.

b. z_stat = 1.52. Since z_stat is less than the critical value of 1.65, we can’t reject the null hypothesis. The sample mean difference is not statistically significant at the 5% significance level.

c. z_stat = 2.42. Since z_stat is above the critical value of 1.65, we can reject the null hypothesis. The sample mean difference is not statistically significant at the 5% significance level.

d. z_stat = 2.9. Since z_stat is above the critical value of 1.65, we can't reject the null hypothesis. The sample mean difference is not statistically significant at the 5% significance level.

86.

Two different brands of golf balls are being tested by ProGolf Associates to determine if one travels farther than the other. Ten randomly selected Hyperdive balls are selected for the test along with and another 10 randomly selected Score 60 balls. The balls in each sample are struck with equal force by a robotic driver. The average distance for the Hyperdrive balls in the sample is 252 yards, with a standard deviation of 4.9 yards. The Score 60 balls in the sample have an average distance of 246 yards, with a standard deviation of 5.4 yards.

a. 3.86

b. 2.11

c. 2.60

d. 2.28

e. 1.69

87.

Two different brands of golf balls are being tested by Pro Golf Associates to determine if one travels farther than the other. Ten randomly selected Hyperdrive balls are selected for the test along with and another 10 randomly selected Score 60 balls. The balls in each sample are struck with equal force by a robotic driver. The average distance for the Hyperdrive balls in the sample is 252 yards, with a standard deviation of 4.9 yards. The Score 60 balls in the sample have an average distance of 246 yards, with a standard deviation of 5.4 yards.

Set up a hypothesis test to test a null hypothesis that there is no difference in the average distance for the two brands of balls. Assume that the population distributions are normal and that the population standard deviations are equal. Compute the value of the t_stat for the sample result in this two-tailed test and report your conclusion. Use a 5% significance level. (You can assume that the population distributions are normal and that the population standard deviations are equal.)

a. Since the t_stat of 1.79 is inside the critical t values of -2.101 and +2.101, we can’t reject the null hypothesis. There is not enough sample evidence to convince us that the population means are different.

b. Since the t_stat of 2.60 is outside the critical t values of -2.101 and +2.101, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different.

c. Since the t_stat of .97 is between the critical t values of -1.661 and +1.661, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different.

d. Since the t_stat of 1.52 is between -2.101 and +2.101, we can’t reject the null hypothesis. There isn’t enough sample evidence to convince us that the population means are different.

88.

A sample of 6 Ford drivers were asked to test drive a new Toyota and to rate the car after completing the test drive. Ratings for the sample are given below:

Driver	#1	#2	#3	#4	#5	#6
Rating	52.2	48.0	61.0	55.2	39.0	56.0

A sample of 5 Chevrolet drivers was similarly selected to test drive the new Toyota, with the following rating results:

Driver	#1	#2	#3	#4	#5
Rating	60.0	58.0	41.0	35.0	43.0

You will use these results to test a null hypothesis there would be no difference in average ratings for the two populations represented. Compute the value of the sample test statistic, t_stat, that would be appropriate here and report it here. Assume the populations are normal and the population standard deviations are equal.

a. 1.9

b. .8

c. .03

d. 2.4

e. 1.3

89.

A sample of 6 Ford drivers were asked to test drive a new Toyota and to rate the car after completing the test drive. Ratings for the sample are given below:

Driver	#1	#2	#3	#4	#5	#6
Rating	52.2	48.0	61.0	55.2	39.0	56.0

A sample of 5 Chevrolet drivers was similarly selected to test drive the new Toyota, with the following rating results:

Driver	#1	#2	#3	#4	#5
Rating	60.0	58.0	41.0	35.0	43.0

You will use these results to test a null hypothesis there would be no difference in average ratings for the two populations represented. Assume the populations are normal and the population standard deviations are equal. Compute the value of the sample test statistic, t_stat, that would be appropriate and use it to reach a conclusion. Use a significance level of 5%.

a. Since t_stat is outside the interval -2.862 to +2.862, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. The sample mean difference is statistically significant at the 5% significance level.

b. Since t_stat is between -2.262 and +2.262, we can’t reject the null hypothesis. There is not enough sample evidence to convince us that the population means are different. The sample mean difference is not statistically significant at the 5% significance level.

c. Since t_stat is between -1.883 and +1.883, we can’t reject the null hypothesis. There is not enough sample evidence to convince us that the population means are different. The sample mean difference is not statistically significant at the 5% significance level.

d. Since t_stat is outside the interval -2.262 to +2.262, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. The sample mean difference is statistically significant at the 5% significance level.

90.

Coach Parker is testing two new offenses for his Tyler University basketball team. In a sample of 10 games in which his team used offense A, the team had an average of 20 turnovers (turnover = loss of possession of the ball) per game, with a standard deviation of six turnovers. In a sample of 10 other games in which the team used offense B, the team had an average of 17 turnovers, with a standard deviation of four turnovers. You are to set up hypothesis test to test a “no difference in average turnovers for the two populations represented” null hypothesis using a significance level of 5%. Assume the population distributions are normal, with equal standard deviations. Compute the value of the sample test statistic, t_stat, for your test and report it.

a. .88

b. 1.63

c. 1.89

d. 1.32

e. 2.16

91.

Coach Parker is testing two new offenses for his Tyler University basketball team. In a sample of 10 games in which his team used offense A, the team had an average of 20 turnovers (turnover = loss of possession of the ball) per game, with a standard deviation of six turnovers. In a sample of 10 other games in which the team used offense B, the team had an average of 14 turnovers, with a standard deviation of four turnovers. You are to set up hypothesis test to test a “no difference in average turnovers for the two populations represented” null hypothesis using a significance level of 5%. Assume the population distributions are normal, with equal standard deviations. Is the difference in sample means here statistically significant at the 5% significance level? Explain.

a. Since the t_stat of 2.631 is outside the interval -2.101 to +2.101, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. The sample mean difference is statistically significant at the 5% significance level.

b. Since the t_stat of 1.94 is inside the interval of -2.101 to +2.101, we can’t reject the null hypothesis. There is not enough sample evidence to convince us that the population means are different. The sample mean difference is not statistically significant at the 5% significance level.

c. Since the t_stat of 1.724 is outside the critical t values of -1.501 and +1.501, we can reject the null hypothesis. There is enough sample evidence to convince us that the population means are different. The sample mean difference is statistically significant at the 5% significance level.

d. Since the t_stat of 1.091 is between -1.501 and +1.501, we can’t reject the null hypothesis. There isn’t enough sample evidence to convince us that the population means are different. The sample mean difference is not statistically significant at the 5% significance level.

92.

A recent study of 300 college-educated voters and 400 non-college-educated voters reported that 84% of the college-educated sample and 77% of the non-college-educated sample supported a complete overhaul of the federal income tax system. You set up a hypothesis to test a “no difference in population proportions” null hypothesis. Compute the p-value for the sample result in this two-tailed test and report it here.

a. .0334

b. .0532

c. .0109

d. .0432

e. .0219

93.

A recent study of 200 college-educated voters and 400 non-college-educated voters reported that 83% of the college-educated sample and 74% of the non-college-educated sample supported a complete overhaul of the federal income tax system. You set up a hypothesis to test a “no difference in population proportions” null hypothesis. Compute the p-value for the sample result in this two-tailed test and use it to reach a decision. Use a significance level of .05.

a. The p-value here is .0436. Since this p-value is less than .05, we can reject the null hypothesis. There’s enough sample evidence to conclude that the proportion of college-educated voters who would support the overhaul is not the same as the proportion of non-college-educated voters who would support the overhaul.

b. The p-value here is .0135. Since this p-value is less than .05, we can reject the null hypothesis. There’s enough sample evidence to conclude that the proportion of college-educated voters who would support the overhaul is not the same as the proportion of non-college-educated voters who would support the overhaul.

c. The p-value here is .0622. Since this p-value is greater than .05, we can’t reject the null hypothesis. There is not enough sample evidence to conclude that the proportion of college-educated voters who would support the overhaul is not the same as the proportion of non-college-educated voters who would support the overhaul.

d. The p-value here is .0714. Since this p-value is greater than .05, we can’t reject the null hypothesis. There’s not enough sample evidence to conclude that the proportion of college-educated voters who would support the overhaul is not the same as the proportion of non-college-educated voters who would support the overhaul.

94.

In a simple random sample of 100 managers and a simple random sample of 150 rank-and-file production workers, 62% of the management sample and 54% of the production-worker sample report that they are “completely satisfied” with the company’s current health plan. Using a "no difference in population proportions" null hypothesis, you are to set up an appropriate hypothesis test. Compute the test statistic, z_stat, for your test and report it here.

a. 1.96

b. 2.21

c. 1.58

d. 3.67

e. 1.25

95.

In a simple random sample of 100 managers and a simple random sample of 150 rank-and-file production workers, 64% of the management sample and 54% of the production-worker sample report that they are “completely satisfied” with the company’s current health plan. Using a "no difference in population proportions" null hypothesis, you are to set up an appropriate hypothesis test. Compute the p-value for the sample result in this two-tailed test and use it to reach the appropriate conclusion. Use a significance level of 5%.

a. The proper p-value is .1166. Since the p-value is greater than .05, we can’t reject the null hypothesis. There is not enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is not statistically significant at the 5% significance level.

b. The proper p-value is .0203. Since the p-value is less than .05, we can reject the null hypothesis. There is enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is statistically significant at the 5% significance level.

c. The proper p-value is .0583. Since the p-value is greater than .05, we can’t reject the null hypothesis. There is not enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is not statistically significant at the 5% significance level.

d. The proper p-value is .0406. Since the p-value is greater than .05, we can’t reject the null hypothesis. There isn’t enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is not statistically significant at the 5% significance level.

96.

ProductsDirect.com has tracked a sample of 150 USPS deliveries of customer orders and 180 FedEx deliveries. In the samples, 141 of the USPS deliveries were on time, while 162 of the FedEx deliveries were on time. You are to set up a hypothesis test to test a null hypothesis that the USPS on-time rate is no better than FedEx’s. Compute the proper p-value for the sample result in this one-tailed test and report it here.

a. .0312

b. .0144

c. .0934

d. .0821

e. .1005

97.

ProductsDirect.com has tracked a sample of 150 USPS deliveries of customer orders and 180 FedEx deliveries. In the samples, 144 of the USPS deliveries were on time, while 162 of the FedEx deliveries were on time. You are to set up a hypothesis test to test a null hypothesis that the USPS on-time rate is no better than FedEx’s. Compute the proper p-value for the sample result in this one-tailed test and use it to reach the proper conclusion. Use a 10% significance level.

a. Since the p-value of .116 is greater than .10, we can’t reject the null hypothesis. There’s not enough sample evidence to conclude that the proportion of USPS on-time deliveries of ProductDirect.com’s orders is greater than the proportion of FedEx’s.

b. Since the p-value of .062 is less than .10, we can reject the null hypothesis. There’s enough sample evidence to conclude that the proportion of USPS on-time deliveries of ProductDirect.com’s orders is greater than the proportion of FedEx’s.

c. Since the p-value of .018 is less than .10, we can reject the null hypothesis. There’s enough sample evidence to conclude that the proportion of USPS on-time deliveries of ProductDirect.com’s orders is greater than the proportion of FedEx’s.

d. Since the p-value of .046 is less than .05, we can reject the null hypothesis. There’s enough sample evidence to conclude that the proportion of USPS on-time deliveries of ProductDirect.com’s orders is greater than the proportion of FedEx’s.

98.

A recent poll of 400 randomly-selected residents of Minnesota and 500 randomly-selected residents of Florida showed the following results: 63% of the Minnesota sample and 57% of the Florida sample expressed approval of their state’s handling of environmental matters. You are to set up a hypothesis test to test whether this is sufficient sample evidence to reject a “no difference in population proportions” null hypothesis for the two populations represented. Compute the value of the test statistic, z_stat, for the sample result.

a. 4.53

b. 1.82

c. 2.79

d. .96

e. 5.94

99.

A recent poll of 200 randomly-selected residents of Minnesota and 400 randomly-selected residents of Florida showed the following results: 63% of the Minnesota sample and 52% of the Florida sample expressed approval of their state’s handling of environmental matters. You are to set up a hypothesis test to test whether this is sufficient sample evidence to reject a “no difference in population proportions” null hypothesis for the two populations represented. Compute the value of the test statistic, z_stat, for the sample result and use it to reach the proper conclusion. Use a significance level of .05.

a. Since the value of z_stat(2.56) puts the sample result outside ±1.96, we can reject the null hypothesis. There is enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is statistically significant at the 5% significance level.

b. Since the value of z_stat (3.21) puts the sample result outside ±1.96, we can't reject the null hypothesis. There isn't enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is not statistically significant at the 5% significance level.

c. Since the value of z_stat (1.27) puts the sample result inside ±1.96, we can't reject the null hypothesis. There isn't enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is not statistically significant at the 5% significance level.

d. Since the value of z_stat (1.75) puts the sample result inside ±1.96, we can reject the null hypothesis. There is enough sample evidence to show that the proportions in the two populations are different. The difference in sample proportions is statistically significant at the 5% significance level.

100.

Five college baseball players were randomly selected to test two new bat designs. Each was asked to swing each of the bats. The order in which the bats were swung to was random. The maximum bat speed was recorded for each player using each for each bat. Results are shown in the table below.

Bat speed (mph)

Player	1	2	3	4	5
Bat 1	65.4	72.1	59.7	61.2	68.8
Bat 2	60.3	65.8	61.5	59.0	64.2

Assuming that all necessary conditions are satisfied, you are to construct a matched samples hypothesis test to test a null hypothesis that there is no difference in the average maximum bat speed for the populations represented. Compute the value of the test statistic, t_stat, and report it here.

a. 5.76

b. 1.05

c. 3.18

d. 2.29

e. 4.63

101.

Bat speed (mph)

Player	1	2	3	4	5
Bat 1	65.4	72.1	59.7	61.2	68.8
Bat 2	60.3	65.8	61.5	59.0	64.2

Assuming that all necessary conditions are satisfied, you are to construct a matched samples hypothesis test to test a null hypothesis that there is no difference in the average number of items checked for the populations represented. Compute the value of the test statistic, t_stat, and use it to reach the proper conclusion. Use a 5% significance level.

a. Since t_stat is outside ±2.306, we can reject the “no difference” null hypothesis. The difference in sample average bats speed is statistically significant at the 5% significance level.

b. Since t_stat is inside ±2.306, we can't reject the “no difference” null hypothesis. The difference in sample average bat speed is not statistically significant at the 5% significance level.

c. Since t_stat is inside ±2.776, we can't reject the “no difference” null hypothesis. The difference in sample average bat speed is not statistically significant at the 5% significance level.

d. Since t_stat is outside ±2.776, we can reject the “no difference” null hypothesis. The difference in sample average bat speed is statistically significant at the 5% significance level.

102.

Suppose you test the following hypotheses at a significance level of 0.05.

H₀: π ≤ .32

H_a: π > .32

If based on a random sample of 200 population members, the null hypothesis is rejected, this means

a. There is only a 5% chance that the population proportion is less than 0.32.

b. There is a 95% probability that the population proportion is greater than 0.32.

c. There is strong evidence that the sample proportion is greater than 0.32.

d. The sample provides strong evidence that the population proportion is greater than 0.32.

103.

Suppose you test the following hypotheses at a significance level of 0.05.

H₀: π ≤ .32

H_a: π > .32

A random sample of 200 population members provides a sample proportion of 0.34.

Based on this result, the correct conclusion is:

a. Reject the null hypothesis since z_stat < z_c.

b. Do not reject the null hypothesis since z_stat < z_c.

c. Reject the null hypothesis since z_stat > z_c.

d. Do not reject the null hypothesis since z_stat > z_c.

104.

In the United States, about 13 percent of adults are military veterans. A researcher conducts a survey to determine whether the share of veterans differs across regions of the US. A survey of 2,000 people from the Southeastern US finds the proportion of veterans in the sample is 14.6% (Source: Gallup.com). The researcher frames the hypotheses for the test as follows:

H₀: π = 0.13 (the proportion of veterans in the Southeast equals the US proportion of 0.13)

H_a: π ≠ 0.13 (the proportion of veterans in the Southeast differs from the proportion in the US overall).

The correct p-value for the hypothesis test is:

a. 0.4843

b. 0.0157

c. 0.0334

d. The p-value cannot be determined because the significance level is not known.

105.

In a random sample of 100 National Football League (NFL) players, the players’ average weight was 248 pounds with a sample standard deviation of 32 pounds. In a random sample of 80 National Basketball Association (NBA) players, the players’ average weight was 221 pounds with a sample standard deviation of 27 pounds.

A researcher uses these sample results to conduct a hypothesis test to determine whether the average weights of NFL and NBA players differ. The hypotheses tested are the following:

H₀: µ1 - µ2 = 0 (The population means are equal.)

H_a: µ1 - µ2 ≠ 0 (The population means are not equal.)

At a significance level of 10%, the null hypothesis will be rejected if

a. The z_stat value is inside the critical values of ±1.96.

b. The z_stat value is inside the critical values of ±1.65

c. The z_stat value is outside the critical values of ±1.96.

d. The z_stat value is outside the critical values of ±1.65

106.

A researcher uses these sample results to conduct a hypothesis test to determine whether the average weights of NFL and NBA players differ. The NFL players are labeled group 1; the NBA players are labeled group 2. The hypotheses tested are the following:

H₀: µ1 - µ2 = 0 (The population means are equal.)

H_a: µ1 - µ2 ≠ 0 (The population means are not equal.)

Compute the standard error of the appropriate sampling distribution for the test.

a. 4.40

b. 19.35

c. 0.81

d. 6.13

107.

H₀: µ1 - µ2 = 0 (The population means are equal.)

H_a: µ1 - µ2 ≠ 0 (The population means are not equal.)

The test statistic for the hypothesis test is:

a. z_stat = 1.40

b. z_stat = 6.13

c. z_stat = 33.30

d. The z_stat value is too large to be valid for the test.

108.

Greg would like to test whether he runs faster in the afternoon than in the morning, on average. Based on a sample of 10 morning runs and 12 afternoon runs he calculates the average and standard deviation of his pace in the morning and in the afternoon. He measures his pace in seconds per mile.

He assumes that running times are normally distributed in each population and that the population standard deviations are equal. Calling morning runs “population 1” and afternoon runs “population 2,” he sets up the hypothesis test as follows:

H₀: µ1 - µ2 ≤ 0

H_a: µ1 - µ2 > 0

Suppose Greg is able to reject the null hypothesis at a 1% significance level. Which of the following statements best summarizes the results of the hypothesis test?

a. There is little evidence that on average, his pace is faster in the afternoon than in the morning.

b. There is strong evidence that on average, Greg’s pace is slower in the morning than in the afternoon.

c. There is strong evidence that on average, Greg’s pace is slower in the afternoon than in the morning.

d. There is strong evidence of a difference between Greg’s average morning pace and his average afternoon pace.

109.

Greg would like to test whether he runs faster in the afternoon than in the morning, on average. In a sample of 10 morning runs, his average pace was 7 minutes 50 seconds per mile (490 seconds/mile), with a sample standard deviation of 35 seconds per mile. In a sample of 12 afternoon runs, his average pace was 7 minutes, 30 seconds per mile (450 seconds/mile), with a sample standard deviation of 32 seconds per mile.

H₀: µ1 - µ2 ≤ 0

H_a: µ1 - µ2 > 0

Using a significance level of 5%, Greg should reject the null hypothesis if:

a. The sample statistic t_stat is less than the critical value of -1.725.

b. The sample statistic t_stat is outside the critical values of ± 2.086.

c. The sample statistic t_stat is greater than the critical value of 1.717.

d. The sample statistic t_stat is greater than the critical value of 1.725.

110.

Greg would like to test whether he runs faster in the afternoon than in the morning, on average. In a sample of 10 morning runs, his average pace was 7 minutes 50 seconds per mile (470 seconds/mile), with a sample standard deviation of 45 seconds per mile. In a sample of 12 afternoon runs, his average pace was 7 minutes, 30 seconds per mile (450 seconds/mile), with a sample standard deviation of 32 seconds per mile.

H₀: µ1 - µ2 ≤ 0 (his average morning pace is equal or less than his afternoon pace).

H_a: µ1 - µ2 > 0 (his average morning pace is slower than his afternoon pace).

The pooled standard deviation to be used in calculating the test statistic is:

a. 32

b. 33.4

c. 38.5

d. 1474.5

111.

Greg would like to test whether he runs faster on in the afternoon than in the morning, on average. In a sample of 10 morning runs, his average pace was 7 minutes 50 seconds per mile (470 seconds/mile), with a sample standard deviation of 45 seconds per mile. In a sample of 12 afternoon runs, his average pace was 7 minutes, 30 seconds per mile (450 seconds/mile), with a sample standard deviation of 32 seconds per mile.

H₀: µ1 - µ2 ≤ 0

H_a: µ1 - µ2 > 0

The test statistic for the hypothesis test is:

a. 1.96

b. 2.36

c. 1.22

d. 16.44

112.

H₀: µ1 - µ2 ≤ 0

H_a: µ1 - µ2 > 0

At a significance level of 5%, the null hypothesis will be rejected if:

a. The test statistic is greater than 1.65.

b. The test statistic is greater than 1.717.

c. The test statistic is greater than 1.725.

d. The test statistic is greater than 2.086.

113.

H₀: µ1 - µ2 ≤ 0

H_a: µ1 - µ2 > 0

At a significance level of 5%, the correct decision for this hypothesis test is:

a. Reject the null hypothesis because the p-value of 0.012 is less than the significance level of 0.05.

b. Do not reject the null hypothesis because the p-value of 0.12 is greater than the significance level of 0.05.

c. Reject the null hypothesis because the p-value of 0.007 is less than the significance level of 0.05.

d. Do not reject the null hypothesis because the p-value of 0.012 is less than the significance level of 0.05.

114.

A 2011 survey of US adults (population 1) found that in a random sample of 1,000 adults, 800 had home internet access. A corresponding survey in Canada found that of 1,000 Canadian adults (population 2), 870 had home internet access (Gallup World, 2013).

You use this information to test for a difference in the proportion of adults in the two countries that have home internet access. The hypotheses tested are as follows:

H₀: π1 = π2 (The two population proportions are equal.)

H_a: π1 ≠ π2 (The two population proportions are not equal.)

The test statistic for this hypothesis test is:

a. -4.24

b. 0.017

c. 4.24

d. The test statistic cannot be calculated because the sample standard deviations are not given.

115.

You use this information to test for a difference in the proportion of adults in the two countries that have home internet access. The hypotheses tested are as follows:

H₀: π1 = π2 (The two population proportions are equal.)

H_a: π1 ≠ π2 (The two population proportions are not equal.)

At a 5% significance level, the correct decision for this hypothesis test is:

a. Do not reject the null hypothesis because the test statistic of 0.017 is inside the critical values of ± 1.96.

b. Reject the null hypothesis because the test statistic of 4.24 is outside the critical values of ± 1.65.

c. Reject the null hypothesis because the test statistic of -4.24 is outside the critical values of ± 1.96.

d. Do not the null hypothesis because the test statistic of 4.24 is outside the critical values of ± 1.65.

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Chapter 10 Hypothesis Tests Proportions

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Ned Freed

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