Population Mean Hypothesis Test Exam Prep Chapter 9

CHAPTER 9

TRUE/FALSE

1. The probability of making a Type I error is typically denoted by β.

2. The level of significance in hypothesis testing, , measures the maximum probability of making a Type II error, which is the probability of accepting a false null hypothesis.

3. In hypothesis testing, we reject the null hypothesis if the p-value < .

4. The p-value measures the probability that, if the null hypothesis is true, we would randomly produce a sample result at least as unlikely as the sample result that we actually produced.

5. The level of significance in hypothesis testing is the probability of accepting a true null hypothesis.

6. In hypothesis testing, the critical value is a number that establishes the boundary of the reject H₀ region of the test.

7. Type I error occurs when a false null hypothesis is rejected.

8. In hypothesis testing, we reject the null hypothesis if the p-value > .

9. The magnitude of z for a two-tailed hypothesis test of a population mean is greater than it would be in a one-tail hypothesis test given the same level of significance.

10. In the standard forms for the null and alternative hypotheses, the equality part of a hypothesis can appear in either hypothesis.

11. The level of significance,  is the probability of making a Type II error, which is rejecting a true null hypothesis.

12. When the p-value is used for hypothesis testing, the null hypothesis is rejected when the p-value < .

13. The p-value is the probability associated with the test statistic while alpha ( is the probability associated with the critical value.

14. The value of the test statistic for a two-tail hypothesis test for a population mean is the same as for a one-tail hypothesis test.

15. To decide whether to reject a null hypothesis about a population mean, a 95% confidence interval or two-tailed hypothesis test with a significance level of 2.5% could be used.

16. To decide whether to reject a null hypothesis about a population mean, a 95% confidence interval or two-tailed hypothesis test with a significance level of 5% could be used.

Multiple Choice

17. Which of the following is NOT a common step in hypothesis testing:

a. determine the null and alternative hypotheses

b. specify the level of significance 

c. collect the sample data and compute the value of the test statistic

d. use  to determine the critical value and the rejection rule

e. compute the p-value based on the test statistic and compare it to  to

determine whether to reject H₀

18. Which of the following does not need to be known in order to compute the p-value?

a. knowledge of whether the test is one-tailed or two-tailed

b. the value of the test statistic

c. the level of significance

d. a, b, and c are all unnecessary

e. a, b and c are all needed

19. A student believes that the average grade on the final examination in statistics is greater than 85. To make her case, she will take a sample of scores and test a skeptical null hypothesis. The correct set of hypotheses is

a. H₀: μ < 85; H_a: μ > 85

b. H₀: μ > 85; H_a: μ < 85

c. H₀: μ < 85; H_a: μ > 85

d. H₀: μ = 85; H_a: μ ≠ 85

e. none of the above

20. Which of the following best describe(s) the significance level?

a. it is the probability of rejecting a true null hypothesis

b. it is denoted by 

c. it is the probability of making a Type I error

d. it is used to define the boundary of the reject H₀ region

e. all of the above

21. When computing the p-value for a two-tailed hypothesis test, we need to

a. multiply two times the one-tail value

b. divide the upper p-value by two

c. multiply the  value by two

d. subtract the one-tail lower p-value from the one-tail upper p-value

e. none of the above

22. A meteorologist stated that the average temperature during July in Orlando, Florida is exactly 80 degrees. A sample of 32 Julys was taken. If you wanted to challenge the statement, the correct set of hypothesis would be

a. H₀: μ < 80; H_a: μ ≤ 80

b. H₀: μ ≤ 80; H_a: μ > 80

c. H₀: μ = 80; H_a: μ ≠ 80

d. H₀: μ ≠ 80; H_a: μ = 80

e. none of the above

23. Which of the following is NOT a common step in hypothesis testing?

a. state the null and alternative hypotheses

b. use the significance level to establish the reject H₀ region of the test

c. compute the p-value for the sample result

d. apply the decision rule and make your decision

e. none of the above; all are steps

24. In hypothesis testing, the critical value is

a. a number that establishes the boundary of the reject H₀ region

b. based on the significance level 

c. a and b only

d. the probability of a Type I error

e. used to set the p-value

25. Whenever the population standard deviation is unknown, which distribution is used in developing a hypothesis test for a population mean?

a. standard distribution

b. z distribution

c. binomial distribution

d. t distribution

e. none of the above

26. The normal distribution can be used to approximate the t distribution when

a. standard deviations are small

b. sample size is large

c. the t distribution is skewed

d. a normal table is unavailable

e. none of the above

27. In hypothesis testing, if the null hypothesis is rejected when the alternative hypothesis is true

a. a Type I error has been committed

b. a Type II error has been committed

c. either a Type I or Type II error has been committed

d. the correct decision has been made

e. none of the above

28. A p-value is the

a. probability, when the null hypothesis is true, of obtaining a sample result that is at least as unlikely as what is observed

b. value of the test statistic

c. probability of a Type II error

d. probability corresponding to the critical value(s) in a hypothesis test

e. none of the above

29. A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. Any over-filling or under-filling results in the shutdown and readjustment of the machine. To determine whether the machine is properly adjusted, the most appropriate set of hypotheses is

a. H₀: μ < 12; H_a: μ ≤ 12

b. H₀: μ ≤ 12; H_a: μ > 12

c. H₀: μ = 12; H_a: μ ≠ 12

d. H₀: μ ≠ 12; H_a: μ = 12

e. none of the above

30. If a null hypothesis is rejected at a 1% significance level,

a. It may or may not be rejected at a 5% significance level.

b. It will not be rejected at a 5% significance level.

c. It will also be rejected at a 5% significance level.

d. The survey needs to be repeated in order to determine the conclusion at a 5% significance level.

e. none of the above

31. If a null hypothesis is rejected at a 10% significance level,

a. It may or may not be rejected at a 5% significance level.

b. It will not be rejected at a 5% significance level.

c. It will be rejected at a 5% significance level.

d. The test needs to be repeated in order to determine the conclusion at a 5% significance level.

e. none of the above

32. Changing the significance level of a hypothesis test from 5% to 10%,

a. Reduces the chances of Type I error.

b. Increases the chances of Type I error.

c. Reduces the chances of Type II error.

d. Both a and c are true.

e. Both b and c are true.

33. Which of the following is NOT a common step in hypothesis testing?

a. state the null and alternative hypotheses

b. use the significance level to establish the decision rule

c. compute the value of the test statistic

d. apply the decision rule and make your decision

e. none of the above; all are common steps

34. In hypothesis testing, the critical value is

a. a number that establishes the boundary of the reject H₀ region

b. the probability of a Type I error

c. the probability of a Type II error

d. the same as the p-value

e. none of the above

35. When using the confidence interval approach for a two-tailed test, which of the following is true?

a. the rejection region is in the upper tail of the distribution

b. if the confidence interval contains the hypothesized value, then reject H₀

c. if the confidence interval contains the hypothesized value, do not reject the H₀

d. use the F distribution

e. none of the above

36. The average life expectancy of tires produced by the Whitney Tire Company has been 40,000 miles. Management believes that due to a new production process, the life expectancy of its tires has increased. In order to convince others of the validity of management’s belief, the correct set of hypotheses is

a. H₀: μ < 40,000; H_a: μ ≥ 40,000

b. H₀: μ ≤ 40,000; H_a: μ > 40,000

c. H₀: μ > 40,000; H_a: μ ≤ 40,000

d. H₀: μ ≥ 40,000; H_a: μ < 40,000

e. H₀: μ = 40,000; H_a: μ ≠ 40,000

37. Which of the following is NOT a common step in hypothesis testing?

a. determine the p-value and compare it to 

b. state the null and alternative hypothesis

c. use the significance level to establish the decision rule

d. compare the sample size to the population size

e. apply the decision rule and make your decision

38. In hypothesis testing, the critical value is:

a. a number that establishes the boundary of the reject H₀ region

b. based on the significance level 

c. compared to the test statistic to determine if the null hypothesis is rejected

d. all of the above

e. none of the above

39. Type I error in hypothesis testing is

a. accepting a false null hypothesis

b. rejecting a true alternative hypothesis

c. rejecting a true null hypothesis

d. rejecting the sample result when it should have been accepted

e. none of the above

40. Type I error in hypothesis testing is

a. accepting the sample result when it should have been rejected

b. rejecting an alternative hypothesis that is actually false

c. rejecting a null hypothesis when it is actually true

d. rejecting the sample result when it should have been accepted

e. none of the above

41. Type II error in hypothesis testing is

a. accepting a false null hypothesis

b. rejecting a true alternative hypothesis

c. rejecting a true null hypothesis

d. rejecting the sample result when it should have been accepted

e. none of the above

42. Type II error in hypothesis testing is

a. accepting the sample result when it should have been rejected

b. rejecting an alternative hypothesis that is actually true

c. accepting a null hypothesis when it is actually false

d. rejecting the sample result when it should have been accepted

e. none of the above

43. As sample size increases, and so degrees of freedom increase for a t distribution, the shape of the distribution.

a. becomes less skewed

b. is unaffected

c. looks more like a normal distribution

d. becomes bimodal

e. none of the above

44. Using the t distribution for a hypothesis test regarding a population mean is generally appropriate when

a. sample size is no more than 30

b. the sample standard deviation is being used

c. the population distribution is normal

d. a and b

e. a and c

45. In interval estimation, sample results are used to estimate the value of a population parameter. In hypothesis testing,

a. A hypothesis regarding a sample statistic is tested against the actual sample results.

b. A hypothesis regarding a population parameter is tested using sample results.

c. Sample results are used to form a hypothesis regarding a population.

d. The known values of population parameters are used to form a hypothesis about a sample statistic.

e. none of the above

46. The average household size in the US in 2012 was 2.55 persons (Source: US Census Bureau). In order to test for a change in the average size of a household, the correct null and alternative hypothesis would be:

a.

H₀: µ > 2.55 persons (The average household size is greater than 2.5 persons.)

H_a: µ ≤ 2.55 persons (The average household size is no more than 2.55 persons.)

b.

H₀: µ = 2.55 persons (The average household size is 2.55 persons.)

H_a: µ ≠ 2.55 persons (The average household size is not 2.55 persons.)

c.

H₀: µ ≠ 2.55 persons (The average household size is not 2.55 persons.)

H_a: µ = 2.55 persons (The average household size is 2.55 persons.)

d.

H₀: µ ≤ 2.55 persons (The average household size is no more than 2.55 persons.)

H_a: µ > 2.55 persons (The average household size is greater than 2.55 persons.)

e.

H₀: µ < 2.55 persons (The average household size is less than 2.55 persons.)

H_a: µ > 2.55 persons (The average household size is greater than 2.55 persons.)

47. In 2011, adults who used text messaging on their cell phones sent and received an average of 41.5 texts per day (Source: PewInternet.org). You would like to determine whether there is evidence of an increase since 2011 in the average number of texts sent and received. Taking the “status quo” approach to form your null hypothesis, what would be the appropriate pair of hypotheses for your test?

a.

H₀: µ > 41.5 texts (The average number of texts sent/received is greater than 41.5.)

H_a: µ ≤ 41.5 texts (The average number of texts sent/received is no more than 41.5.)

b.

H₀: µ = 41.5 texts (The average number of texts sent/received is 41.5.)

H_a: µ ≠ 41.5 texts (The average number of texts sent/received is not 41.5.)

c.

H₀: µ ≠ 41.5 texts (The average number of texts sent/received is not 41.5.)

H_a: µ = 41.5 texts (The average number of texts sent/received is 41.5 texts.)

d.

H₀: µ ≤ 41.5 texts (The average number of texts sent/received is no more than 41.5.)

H_a: µ > 41.5 texts (The average number of texts sent/received is greater than 41.5.)

e.

H₀: µ < 41.5 texts (The average number of texts sent/received is less than 41.5.)

H_a: µ > 41.5 texts (The average number of texts sent/received is greater than 41.5.)

48. You are to test the following hypotheses:

H₀: µ < 1500

H_a: µ > 1500

A sample of size 16 produces a sample mean of 1540, with a standard deviation of 100.The p-value for this test is .0652. If the significance level set for the test is .05, we should

a. Reject the null hypothesis since  < p-value.

b. Reject the null hypothesis since p-value < 

c. not Reject the null hypothesis since  > p-value

d. not Reject the null hypothesis since p-value > 

e. Accept the null hypothesis since  > p-value

49. You are to test the following hypotheses:

H₀: µ < 900

H_a: µ > 900

A sample of size 25 produces a sample mean of 926, with a standard deviation of 50.The p-value for this test is .0079. If the significance level set for the test is .01, we should

a. Reject the null hypothesis since  < p-value.

b. Reject the null hypothesis since t_stat > t_c

c. not Reject the null hypothesis since  > p-value

d. not Reject the null hypothesis since t_stat > t_c

e. Accept the null hypothesis since p-value < 

50. You are to test the following hypotheses:

H₀: µ > 900

H_a: µ < 900

A sample of size 16 produces a sample mean of 882, with a standard deviation of 40.The p-value for this test is .0460. If the significance level set for the test is .01, we should

a. Reject the null hypothesis since  < p-value.

b. Reject the null hypothesis since p-value < .

c. not Reject the null hypothesis since p-value > .

d. not Reject the null hypothesis since  > p-value

e. Accept the null hypothesis since p-value < 

51. You are to test the following hypotheses:

H₀: µ = 200

H_a: µ ≠ 200

A sample of size 25 produces a sample mean of 193, with a standard deviation of 28.The p-value for this two-tailed test is .2233. If the significance level set for the test is .10, we should

a. Reject the null hypothesis since  < p-value.

b. Reject the null hypothesis since p-value < .

c. not Reject the null hypothesis since  > p-value

d. not Reject the null hypothesis since p-value > .

e. Accept the null hypothesis since p-value < 

52. You are to test the following hypotheses:

H₀: µ = 650

H_a: µ ≠ 650

A sample of size 20 produces a sample mean of 672, with a standard deviation of 42.The p-value for this two-tailed test is .0302. If the significance level set for the test is .05, we should

a. Reject the null hypothesis since  < p-value.

b. Reject the null hypothesis since p-value < .

c. not Reject the null hypothesis since  > p-value

d. not Reject the null hypothesis since p-value > .

e. Accept the null hypothesis since p-value < 

53. A report stated that the average number of overtime hours for IRS workers during the past week was more than 10 hours per worker. You will test that statement with a sample of workers. What null and alternative hypotheses would you recommend here if you want to choose a null hypothesis that is skeptical of the report’s number?

a.

H₀: µ > 10 units (The average overtime is more than 10 hours.)

H_a: µ ≤ 10 units (The average overtime is no more than 10 hours.)

b.

H₀: µ = 10 units (The average overtime is 10 hours.)

H_a: µ ≠ 10 units (The average overtime is not 10 hours.)

c.

H₀: µ ≠ 10 units (The average overtime is not be 10 hours.)

H_a: µ = 10 units (The average overtime is 10 hours.)

d.

H₀: µ ≤ 10 units (The average overtime is no more than 10 hours.)

H_a: µ > 10 units (The average overtime is more than 10 hours.)

e.

H₀: µ < 10 units (The average overtime is less than 10 hours.)

H_a: µ > 10 units (The average overtime is more than 10 hours.)

54. In recent years the average speed along a dangerous stretch of Highway 26 has been reported to be 55.8 mph. You plan to take a sample of cars along that stretch of highway to determine if the average speed recently is different from the reported average. What hypotheses would be appropriate for a hypothesis test here? Use a "status quo" approach to choose your null hypothesis.

a.

H₀: µ = 55.8 mph

H_a: µ ≠ 55.8 mph

b.

H₀: µ ≠ 55.8 mph

H_a: µ = 55.8 mph

c.

H₀: µ > 55.8 mph

H_a: µ ≠ 55.8 mph

d.

H₀: µ < 55.8 mph

H_a: µ = 55.8 mph

e.

H₀: µ > 55.8 mph

H_a: µ = 55.8 mph

55. The fundamental hypothesis test in the American judicial system involves the following hypotheses:

H₀: The accused is innocent.

H_a: The accused is guilty.

Describe what a Type I error and a Type II would involve here.

a. A Type I error would be to judge an innocent man guilty. A Type II error would be to judge a guilty man innocent.

b. A Type I error would be to judge a guilty man guilty. A Type II error would be to judge an innocent man innocent.

c. A Type I error would be to judge a guilty man innocent. A Type II error would be to judge an innocent man guilty.

d. A Type I error would be to judge an innocent man innocent. A Type II error would be to judge a guilty man guilty.

e. A Type I error would be to judge an innocent man guilty. A Type II error would be to judge a guilty man guilty.

56. The competing hypotheses for a hypothesis test are as follows:

H₀: The earth is flat.

H_a: The earth is round.

Describe what a Type I error and a Type II error would be here.

a. A Type I error here would be to believe the earth is the center of the solar system. A Type II error would be to believe that the earth is the only place in the universe where you can get a good hamburger.

b. A Type I error here would be to believe the earth is round when it’s actually flat. A Type II error would be to believe that the earth is flat when it’s actually round.

c. A Type I error here would be to believe the earth is flat when it’s actually round. A Type II error would be to believe that the earth is round when it’s actually flat.

d. A Type I error here would be to believe the earth revolves around the moon. A Type II error would be to believe that the sun revolves around the earth.

e. A Type I error here would be to believe the earth revolves around the sun. A Type II error would be to believe that the sun revolves around the earth.

Problems

57.

The competing hypotheses for a particular hypothesis test are as follows:

H₀: µ ≤ 110

H_a: µ > 110

A random sample of size 64 is taken from the target population. The sample mean is 112.5. Assume the population standard deviation is known to be 8. Using sample results and a significance level of .05, test the null hypothesis and report your conclusion.

a. Since z_stat < z_c, that is, since 1.85 is less than 1.96, we can't reject the null hypothesis.

b. Since z_stat > z_c, that is, since 2.33 is greater than 1.96, we can reject the null hypothesis.

c. Since p-value < .05, that is, since .0062 is less than , we can reject the null hypothesis.

d. Since p-value > .05, that is, since .0612 is greater than , we can’t reject the null hypothesis.

58.

The competing hypotheses for a particular hypothesis test are as follows:

H₀: µ ≤ 250 (The population mean is less than or equal to 250.)

H_a: µ > 250 (The population mean is greater than 250.)

A random sample of size 36 is taken from the target population. The sample mean is 252.5. Assume the population standard deviation is known to be 12.0. Using sample results and a significance level of .01, test the null hypothesis and report your conclusion.

a. Since z_stat < z_c, that is, since 1.15 is inside 2.58, we can’t reject the null hypothesis. There is just not enough sample evidence to challenge the null hypothesis.

b. Since z_stat < z_c, that is, since 1.55 is inside 2.58, we can’t reject the null hypothesis. There is just not enough sample evidence to challenge the null hypothesis.

c. Since z_stat < z_c, that is, since 1.25 is inside 2.33, we can’t reject the null hypothesis. There is just not enough sample evidence to challenge the null hypothesis.

d. Since z_stat < z_c, that is, since 1.05 is inside 2.33, we can’t reject the null hypothesis. There is just not enough sample evidence to challenge the null hypothesis.

59.

The average cost to irrigate an acre of corn in Ames County, Iowa was $180 last year. If you take a random sample of 49 farms in the county and find that the average cost to irrigate an acre of corn this year is $177.10, would this be sufficient sample evidence to reject a null hypothesis that the average cost for the population of farms in the county this year is $180 or more, at the 5% significance level? Assume the standard deviation of irrigation cost for the population of farms is $14.

a. No. Here z_stat = -1.45; z_c = -1.65. Since z_stat is inside z_c,we can’t reject the null hypothesis. There isn’t sufficient sample evidence to support your belief.

b. No. Here z_stat = -1.35; z_c = -1.95. Since z_stat is inside z_c,we can’t reject the null hypothesis. There isn’t sufficient sample evidence to support your belief.

c. No. Here z_stat = -1.45; z_c = -1.65. Since z_stat is outside z_c,we can reject the null hypothesis. There is sufficient sample evidence to support your belief.

d. Yes. Here z_stat = -1.85; z_c = -1.65. Since z_stat is outside z_c,we can reject the null hypothesis. There is sufficient sample evidence to support your belief.

60.

Drake Homes claims that the average time it takes to assemble its manufactured homes is no more than 475.8 worker hours. You take a sample of 225 Drake homes and find the average assembly time is 489.90 worker hours. Set up a hypothesis test to test the null hypothesis µ ≤ 475.80, against the alternative hypothesis µ > 475.80. Use a significance level of 5% and assume the population standard deviation is 75 hours. Can we use the sample result to reject the null hypothesis?

a. No. Since z_stat < z_c, that is, since 1.52 is less than 1.65, we can't reject the null hypothesis. There isn't sufficient sample evidence to support Drake’s claim.

b. Yes. Since z_stat > z_c, that is, since 2.82 is greater than 1.65, we can reject the null hypothesis. There is sufficient sample evidence to reject Drake’s claim.

c. Yes. Since z_stat > z_c, that is, since 2.42 is greater than 1.65, we can reject the null hypothesis. There is sufficient sample evidence to support Drake’s claim.

d. No. Since z_stat > z_c, that is, since 1.22 is less than 1.65, we can't reject the null hypothesis. There isn't sufficient sample evidence to support Drake’s claim.

61.

The competing hypotheses for a particular hypothesis test are as follows:

H₀: µ ≥ 5000 (The population mean is greater than or equal to 5000.)

H_a: µ < 5000 (The population mean is less than 5000.)

Assume the population standard deviation is known to be 280 and that you intend to take a random sample of size 49. With a significance level of .10, if the sample mean is 4924, should you reject the null hypothesis?

a. Yes. Since the sample mean of 4924 is less than the critical value of 4949, we can reject the null hypothesis

b. No. Since the sample mean of 4824 is less than the critical value of 4954.4, we can't reject the null hypothesis

c. Yes. Since the sample mean of 4884 is less than the critical value of 4954.4, we can reject the null hypothesis

d. No. Since the sample mean of 4964 is greater than the critical value of 4454.4, we can’t reject the null hypothesis

62.

Bargain.com claims to have an average of 637 visitors to its site per hour. You take a random sample of 100 hours and find the average number of visitors for the sample is 632 per hour. Can we use this sample result to reject a µ ≥ 637 visitors null hypothesis at the 5% significance level? Assume the population standard deviation is 18 visitors.

a. Yes. Since the sample mean of 632 is less than test's critical value of 634, we can reject the null hypothesis.

b. No. Since the sample mean of 632 is greater than test's critical value of 628, we can't reject the null hypothesis.

c. Yes. Since the sample mean of 632 is less than test's critical value of 639, we can reject the null hypothesis.

d. No. Since the sample mean of 632 is greater than test's critical value of 614, we can't reject the null hypothesis.

63.

It is reported that the average cost for books per student last semester was $224. You take a random sample of 200 students this semester and find the average cost per student in the sample was $207.

Can we use this sample result to make the case that the average cost per student is less this semester than last? Use a significance level of 10% and a population standard deviation of $206. (Show the null hypothesis as H₀: µ ≥ $224 and the alternative as H_a: µ < $224.)

a. Yes. Since the sample mean of $207 is less than the test's critical value of $208.35, we can reject the null hypothesis. There is enough sample evidence to make the case that the average this semester is less than $224.

b. No. Since the sample mean of $207 is more than the test's critical value of $205.35, we can’t reject the null hypothesis. There is not enough sample evidence to make the case that the average this semester is less than $224.

c. Yes. Since the sample mean of $207 is more than the test's critical value of $207.85, we can reject the null hypothesis. There is enough sample evidence to make the case that the average this semester is less than $224.

d. No. Since the sample mean of $207 is more than the test's critical value of $205.35, we can’t reject the null hypothesis. There is not enough sample evidence to make the case that the average this semester is less than $224.

64.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ ≤ 1000 (The population mean is less than or equal to 1000.)

H_a: µ > 1000 (The population mean is greater than 1000.)

Assume the population standard deviation is known to be 80. A random sample of size 64 has a sample mean of 1020. Use the p-value approach, and a significance level of 5%, to decide whether you can reject the null hypothesis.

a. The p-value for this sample result is .0228. Since this p-value is less than the significance level of .05, we can't reject the null hypothesis.

b. The p-value for this sample result is .0428. Since this p-value is less than the significance level of .05, we can't reject the null hypothesis.

c. The p-value for this sample result is .0428. Since this p-value is less than the significance level of .05, we can reject the null hypothesis.

d. The p-value for this sample result is .0228. Since this p-value is less than the significance level of .05, we can reject the null hypothesis.

65.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ ≤ 2500 (The population mean is less than or equal to 2500.)

H_a: µ > 2500 (The population mean is greater than 2500.)

Assume the population standard deviation is known to be 200. A random sample of size 100 has a sample mean of 2540. Use the p-value approach to determine whether you can reject the null hypothesis. (Use a significance level is .01.)

a. Since the p-value of .0428 is greater than the significance level of .01, we can’t reject the null hypothesis.

b. Since the p-value of .0228 is greater than the significance level of .01, we can’t reject the null hypothesis.

c. Since the p-value of .0428 is greater than the significance level of .01, we can reject the null hypothesis.

d. Since the p-value of .0228 is greater than the significance level of .01, we can reject the null hypothesis.

66.

Dalton Investments reports that on average its clients have $1.65 million invested in the stock market. In a random sample of 40 clients, you find that the average investment is $1.52 million. Is this sample result sufficient to challenge Dalton’s report? Use a two-tailed test and a significance level of 5%. Assume the population standard deviation is $.64 million. Compute the p-value for this sample result and use it to make your decision.

a. The p-value here is approximately .20. Since the p-value is greater than .05, we can reject the null hypothesis. There’s enough sample evidence to challenge Dalton’s claim.

b. The p-value here is approximately .20. Since the p-value is greater than .05, we can’t reject the null hypothesis. There’s not enough sample evidence challenge Dalton’s claim.

c. The p-value here is approximately .16. Since the p-value is greater than .05, we can’t reject the null hypothesis. There’s not enough sample evidence to challenge Dalton’s claim.

d. The p-value here is approximately .16. Since the p-value is greater than .05, we can reject the null hypothesis. There’s enough sample evidence to challenge Dalton’s claim.

67.

Suppose you are testing the following hypotheses:

H₀: µ =100 (The population mean is 100.)

H_a: µ ≠ 100 (The population mean is not 100.)

Sample size is 64. The sample mean is 107. The population standard deviation is 36. The significance level is .01. Compute the sample test statistic, z_stat, and use it to decide whether you can reject the null hypothesis.

a. Here z_stat = 1.96. Since z_stat is between -2.58 and +2.58, we can’t reject the null hypothesis.

b. Here z_stat = 2.96. Since z_stat is outside the interval -2.58 to +2.58, we can reject the null hypothesis.

c. Here z_stat = 1.56. Since z_stat is between -2.58 and +2.58, we can’t reject the null hypothesis.

d. Here z_stat = 2.66. Since z_stat is outside the interval -2.58 to +2.58, we can reject the null hypothesis.

68.

Suppose you are testing the following hypotheses:

H₀: µ = 650 (The population mean is equal to 650.)

H_a: µ ≠ 650 (The population mean is not equal to 650.)

Sample size is 100. The sample mean is 635. The population standard deviation is 140. The significance level is .10. Compute the sample test statistic, z_stat and use it to decide whether to reject the null hypothesis.

a. z_stat = -1.87. Since this value is outside the interval -1.65 to +1.65, we can reject the null hypothesis.

b. z_stat = -2.17. Since this value is outside the interval -1.65 and +1.65, we can reject the null hypothesis.

c. z_stat = -1.37. Since this value is between -1.65 and +1.65, we can’t reject the null hypothesis.

d. z_stat = -1.07. Since this value is between -1.65 and +1.65, we can’t reject the null hypothesis.

69.

Suppose you are testing the following hypotheses:

H₀: µ = 60 (The population mean is equal to 60.)

H_a: µ ≠ 60 (The population mean is not equal to 60.)

Sample size is 49. The sample mean is 68. The population standard deviation is 21. The significance level is .05. Compute the p-value for the sample result and use it to decide whether to reject the null hypothesis.

a. The p-value for this two-tailed test is .0019. Since this p-value is less than of the significance level, that is, since .0019 < .05, we can’t reject H₀.

b. The p-value for this two-tailed test is .0038. Since this p-value is less than the significance level, that is, since .0038 < .05, we can reject H₀.

c. The p-value for this two-tailed test is .0076. Since this p-value is less than the significance level, that is, since .0076 < .05, we can reject H₀.

d. The p-value for this two-tailed test is .0008. Since this p-value is less than the significance level, that is, since .0008 < .05, we can reject H₀.

70.

The average fees paid by students at state universities across the country were $2236 per student last year. The average for a random sample of 500 students this year is $2350. Given the following hypotheses,

H₀: µ = 2236 (The average fees are the same as last year.)

H_a: µ ≠ 2236 (The average fees are not the same as last year.)

compute the p-value for the sample result. Assume the population standard deviation is $724. If the significance level is 5%, can the null hypothesis be rejected?

a. The p-value here is .02. Since .02 is less than the significance level, that is, since .02 < .05, we can’t reject H₀.

b. The p-value here is .04. Since .04 is less than the significance level, that is, since .04 < .05, we can reject H₀.

c. The p-value here is .0004. Since .0004 is less than the significance level, that is, since .0004 < .05, we can reject H₀.

d. The p-value here is .0004. Since .0004 is less than the significance level, that is, since .0004 < .05, we can't reject H₀.

71.

The average number of credit hours accumulated by students entering their third year at George Thomas University in past years has been 69. You take a random sample of 36 students entering their third year this year and find the average in the sample is 72.1 credit hours. Is this sufficient sample evidence to make the case that that average number of credit hours per student is different from—that is, it’s either greater or less than—the 69 hour average of past years? Use a significance level of 5% and a population standard deviation of 12.7 hours.

a. The value of the sample test statistic, z_stat, is 1.46. Since this value is between -1.96 and +1.96, we can reject the null hypothesis. There is sufficient sample evidence to argue that the average number of credit hours this year is different from past years.

b. The value of the sample test statistic, z_stat, is 1.46. Since this value is between -1.96 and +1.96, we can’t reject the null hypothesis. There isn’t sufficient sample evidence to argue that the average number of credit hours this year is different from past years.

c. The value of the sample test statistic, z_stat, is 1.08. Since this value is between -1.96 and +1.96, we can’t reject the null hypothesis. There isn’t sufficient sample evidence to argue that the average number of credit hours is different from past years.

d. The value of the sample test statistic, z_stat, is 1.88. Since this value is between -1.96 and +1.96, we can’t reject the null hypothesis. There isn’t sufficient sample evidence to argue that the average number of average number of credit hours is different from the past years.

72.

Suppose you are testing the following hypotheses:

H₀: µ ≤ 1500 (The population mean is less than or equal to 1500.)

H_a: µ > 1500 (The population mean is greater 1500.)

Sample size is 25. The sample mean is 1545 and the sample standard deviation is 75. The significance level is .05. Assume that the values in the population are normally distributed. Compute the sample test statistic, t_stat, and use it to decide whether you should reject the null hypothesis.

a. t_stat = 3.0. Since t_stat is greater than t_c (1.711), we can reject the null hypothesis.

b. t_stat = 2.0. Since t_stat is greater than t_c (1.711), we can't reject the null hypothesis.

c. t_stat = 2.0. Since t_stat is greater than t_c (1.711), we can reject the null hypothesis.

d. t_stat = 3.0. Since t_stat is greater than t_c (1.711), we can't reject the null hypothesis.

73.

Suppose you are testing the following hypotheses:

H₀: µ = 3200 (The population mean is equal to 3200.)

H_a: µ ≠ 3200 (The population mean is not equal to 3200.)

Sample size is 9. The sample mean is 3260 and the sample standard deviation is 120. The significance level is .10. Assume that the values in the population are normally distributed. Compute the sample test statistic, t_stat, and use it to decide whether you should reject the null hypothesis.

a. t_stat = 1.5. Since t_stat is between -1.860 and +1.860, we can’t reject the null hypothesis.

b. t_stat = 1.0. Since t_stat is between -1.860 and +1.860, we can’t reject the null hypothesis.

c. t_stat = 1.0. Since t_stat is between -2.58 and +2.58, we can’t reject the null hypothesis.

d. t_stat = 1.5. Since t_stat is between -1.860 and +1.860, we can reject the null hypothesis.

74.

The school blog reported the sophomores at Yojii University in Japan spent an average of 77.6 hours studying English last month. In a random sample of 25 sophomores at the school, the average study time was 75.9 hours with a sample standard deviation of 9.8 hours. Is the sample evidence sufficient to reject the null hypothesis shown below? Use a significance level of .05. Assume the population distribution is approximately normal.

H₀: µ ≥ 77.6

H_a: µ < 77.6

a. t_stat = -1.34. Since t_stat is greater than (i.e., inside) -1.711, we can't reject the null hypothesis.

b. t_stat = -.87. Since t_stat is greater than (i.e., inside) -1.711, we can’t reject the null hypothesis.

c. t_stat = -2.24. Since t_stat is less than (i.e., outside) -1.711, we can't reject the null hypothesis.

d. t_stat = -2.14. Since t_stat is less than (i.e., outside) -1.711, we can reject the null hypothesis.

75.

It has been reported that tech companies in Silicon Valley hired an average of 18.7 new employees last year. Suppose you take a random sample of 15 of the companies this year and find that the average number of new hires in the sample is 19.5. The sample standard deviation is 3.6. Is the sample evidence sufficient to reject the null hypothesis shown below? Use a significance level of .05. Assume the population distribution is approximately normal.

H₀: µ = 18.7

H_a: µ ≠ 18.7

a. t_stat = .26. Since t_statis between -2.654 and +2.654, we can’t reject the null hypothesis.

b. t_stat = 1.86. Since t_stat is between -2.234 and +2.234, we can’t reject the null hypothesis.

c. t_stat = .86. Since t_stat is between -2.145 and +2.145, we can reject the null hypothesis.

d. t_stat = .86. Since t_stat is between -2.145 and +2.145, we can’t reject the null hypothesis.

76.

Trend Home Products wants its online service agents to respond to posted customer complaints in an average of 28 minutes. You track a random sample of 12 complaints. The average response time for the sample is 30.2 minutes. The sample standard deviation is 14.3 minutes. Is this sample result sufficient to reject a null hypothesis that the average response time for all customer complaints is exactly 28 minutes? Use a 5% significance level. Assume the population distribution is approximately normal.

a. t_stat = .53. Since t_stat is between -2.201 and + 2.201, we can’t reject the null hypothesis null hypothesis.

b. t_stat = 2.73. Since t_stat is outside the interval -2.332 to + 2.332, we can reject the null hypothesis null hypothesis.

c. t_stat = 2.73. Since t_stat is outside the interval -2.201 to + 2.201, we can reject the null hypothesis null hypothesis.

d. t_stat = 1.33. Since t_stat is between -2.201 and + 2.201, we can’t reject the null hypothesis null hypothesis.

77.

It has been reported that, on average, American kids between the ages of 2 and 7 spend up to 600 minutes per week watching unsuitable television programs. In a telephone survey of 50 randomly selected households with a child in that age group, you find that the average time reported for those in the sample is 685 minutes. Is this sufficient evidence to reject the reported average of 600 minutes? Set up a one-tailed hypothesis test, using  < 600 as the null hypothesis and a 5% significance level. Assume the standard deviation of times for the population of kids in this age group is 240 minutes. Compute the sample test statistic, z_stat, and use it to decide whether to reject the null hypothesis.

a. z_stat = 1.61. Since z_stat < z_c, that is, since 1.61 is less than 1.65, we can't reject the null hypothesis.

b. z_stat = 2.81. Since z_stat > z_c, that is, since 2.81 is greater than 1.65, we can't reject the null hypothesis.

c. z_stat = 2.21. Since z_stat > z_c, that is, since 2.21 is greater than 1.65, we can reject the null hypothesis.

d. z_stat = 2.50. Since z_stat > z_c, that is, since 2.50 is greater than 1.65, we can reject the null hypothesis.

78.

SunEnergy claims that its solar batteries will exceed the requirement that the batteries function properly for an average of 2000 hours before the main element needs to be replaced. You test a simple random sample of 40 batteries and find that the average time before the main element needs to be replaced in the sample is 2110.

Should this be considered sufficient sample evidence to make SunEnergy’s case? That is, is this sufficient evidence to reject a “µ is no more than 2000 hours”? Use a significance level of 5%. Assume the population standard deviation is 370 hours.

a. Since the sample mean is less than the critical value for the test--that is, since 2110 is less than 2226--we can't reject the null hypothesis.

b. Since the sample mean is greater than the critical value for the test--that is, since 2110 is greater than 2083--we can reject the null hypothesis.

c. Since the sample mean is greater than the critical value for the test--that is, since 2110 is greater than 2097--we can reject the null hypothesis.

d. Since the sample mean is less than the critical value for the test--that is, since 2110 is greater than 2146--we can't reject the null hypothesis.

79.

BarnabyToys.com claims that the average delivery time for items purchased on its website is 3.65 days. A random sample of 45 recent deliveries has an average delivery time for the sample of 4.03 days.

Compute the p-value for the sample result and use it to test the null hypothesis that average delivery time for the population of BaranabyToys.com items purchased online is no more than 3.65 days. Use a 5% significance level for the test and assume that the population standard deviation is .92 days. Is the sample result sufficient to reject the null hypothesis?

a. Since the p-value of .0328 is less than the significance level of .05, we can reject the null hypothesis.

b. Since the p-value of .0028 is less than the significance level of .05, we can reject the null hypothesis.

c. Since the p-value of .0328 is less than the significance level of .05, we can't reject the null hypothesis.

d. Since the p-value of .0028 is less than the significance level of .05, we can't reject the null hypothesis.

80.

Young and Company claims that its pressurized diving bell will, on average, maintain its integrity to depths of 2500 feet or more. You take a random sample of 50 of the bells. The average maximum depth for bells in your sample is 2455 feet. Set up an appropriate hypothesis test using Young and Company’s claim as the null hypothesis. Assume the population standard deviation is 200 feet.

Is the sample result of 2455 "statistically significant" at the 10% significance level? Explain.

a. Since the p-value of .0559 is less than the significance level of .10, we can't reject the null hypothesis, which means the sample result is not statistically significant at the 10% significance level.

b. Since the p-value of .0559 is less than the significance level of .10, we can reject the null hypothesis, which means the sample result is statistically significant at the 10% significance level.

c. Since the p-value of .0359 is less than the significance level of .10, we can reject the null hypothesis, which means the sample result is statistically significant at the 10% significance level.

d. Since the p-value of .0359 is less than the significance level of .10, we can't reject the null hypothesis, which means the sample result is not statistically significant at the 10% significance level.

81.

Young and Company claims that its pressurized diving bell will, on average, maintain its integrity to depths of 2500 feet or more. You take a random sample of 50 of the bells. The average maximum depth for bells in your sample is 2455 feet. Set up an appropriate hypothesis test using Young and Company’s claim as the null hypothesis. Assume the population standard deviation is 200 feet.

Is the sample result of 2455 "statistically significant" at the 1% significance level? Explain.

a. Since the p-value of .0559 is greater than the significance level of .01, we can't reject the null hypothesis, which means the sample result is not statistically significant at the 1% significance level.

b. Since the p-value of .0359 is greater than the significance level of .01, we can reject the null hypothesis, which means the sample result is statistically significant at the 1% significance level.

c. Since the p-value of .0559 is greater than the significance level of .01, we can reject the null hypothesis, which means the sample result is statistically significant at the 1% significance level.

d. Since the p-value of .0359 is greater than the significance level of .01, we can't reject the null hypothesis, which means the sample result is not statistically significant at the 1% significance level.

82.

Thornton Renos, a national specialist in kitchen renovations, believes that the average profit on its jobs is at least $19,000. The auditor suspects that the average is less and plans to look at a simple random sample of 50 Thornton jobs. She sets up the competing hypotheses:

H₀: µ ≥ 19,000

H_a: µ < 19,000

She intends to use the following decision rule: Reject the null hypothesis if the sample mean is less than $18,625. What is the significance level implied by this rule? Assume a population standard deviation of $1,050.

a. .01

b. .025

c. .015

d. .0057

83.

You will have a sample of 30 units from a large population in order test the competing hypotheses:

H₀: µ ≥ 125

H_a: µ < 125

If the test uses a significance level of .05, and the sample result produces a p-value of .034, can you reject the null hypothesis?

a. Since the p-value of .034 is less than the significance level of .05, we can't reject the null hypothesis.

b. Since the p-value of .034 is greater than 1/2 of the significance level of .05, we can't reject the null hypothesis.

c. Since the p-value of .034 is less than the significance level of .05, we can reject the null hypothesis.

d. Since the p-value of .034 is greater than 1/2 of the significance level of .05, we can reject the null hypothesis.

84.

You will have a sample of 40 units from a large population in order test the competing hypotheses:

H₀: µ ≥ 230

H_a: µ < 230

If the test uses a significance level of .05, and the sample result produces a z_stat of 2.14, can you reject the null hypothesis?

a. Since z_stat here is 2.14, z_stat < z_c, so we can reject the null hypothesis.

b. Since z_stat here is 2.14, z_stat > z_c, so we can't reject the null hypothesis.

c. Since z_stat here is 2.14, z_stat < z_c, so we can't reject the null hypothesis.

d. Since z_stat here is 2.14, z_stat > z_c, so we can reject the null hypothesis.

85.

Bottles of Langdon Falls water should contain, on average, 15 ounces of liquid— no more and no less. You select a sample of 36 bottles that has a mean of 14.915 ounces per bottle. Can you reject a null hypothesis that  = 15 ounces for the population of Langdon Falls bottles? Use a significance level of 5% and assume that the population standard deviation is .15 ounces.

a. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.65 or z_stat > +1.65.

In this case, z_stat = -4.3. Since z_stat is less than -1.65, we can reject the null hypothesis.

b. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.96 or z_stat > +1.96.

In this case, z_stat = -3.4. Since z_stat is less than -1.96, we can reject the null hypothesis.

c. We can use the Decision Rule: Reject the null hypothesis if z_stat > +1.65.

In this case, z_stat = -4.3. Since z_stat is less than -1.65, we can't reject the null hypothesis..

d. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.96 or z_stat > +1.96.

In this case, z_stat = -3.4. Since z_stat is less than -1.96, we can't reject the null hypothesis.

86.

Bottles of Langdon Falls water should contain, on average, 30 ounces of liquid— no more and no less. You select a sample of 36 bottles that has a mean of 30.09 ounces per bottle. Can you reject a null hypothesis that  = 30 ounces for the population of Langdon Farms bottles? Use a significance level of 5% and assume that the population standard deviation is .3 ounces.

a. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.96 or z_stat > +1.96. In this case, z_stat = -1.2. Since z_stat is between -1.65 and +1.65, we can't reject the null hypothesis.

b. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.65 or z_stat > +1.65. In this case, z_stat = 1.8. Since z_stat is between -1.65 and +1.65, we can't reject the null hypothesis.

c. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.96 or z_stat > +1.96. In this case, z_stat = 1.8. Since z_stat is between -1.96 and +1.96, we can't reject the null hypothesis.

d. We can use the Decision Rule: Reject the null hypothesis if z_stat < -1.96 or z_stat > +1.96. In this case, z_stat = -1.2. Since z_stat is between -1.96 and +1.96, we can reject the null hypothesis.

87.

Lake Huron Industries produces vinyl sheets that, on average, are 9 mm in thickness, with a standard deviation of .25 mm. A random sample of 49 units has an average thickness of 8.98 mm. Set up an appropriate two-tailed hypothesis test to test the null hypothesis that µ = 9 and compute the corresponding p-value for this sample result. If the significance level for the test is .02, what conclusion can you draw? Assume the population standard deviation is .25 mm.

a. The proper p-value here is .5754. Since this p-value is greater than .02, we can’t reject the null hypothesis.

b. The proper p-value here is .2877. Since this p-value is greater than .04, we can’t reject the null hypothesis.

c. The proper p-value here is .0046. Since this two-tailed p-value is less than .02, we can’t reject the null hypothesis.

d. The proper p-value here is .1424. Since this p-value is greater than .02, we can reject the null hypothesis.

88.

UPS drivers in the New York City area are expected to make at least 150 deliveries per week. You observe a random sample of 50 drivers and find that the average number of deliveries in the sample was 136. Can a  > 150 null hypothesis be rejected at the 5% significance level? The standard deviation in the sample is 28 deliveries.

a. z_stat for the sample result is -3.54. Since z_stat is less than -1.65, we can reject the null hypothesis.

b. z_stat for the sample result is -2.54. Since z_stat is less than -1.65, we can reject the null hypothesis.

c. z_stat for the sample result is -2.54. Since z_stat is less than -1.65, we can reject the null hypothesis.

d. z_stat for the sample result is -1.54. Since z_stat is greater than -1.65, we can't reject the null hypothesis.

89.

Dallas National Bank’s goal is to have an average customer waiting time of no more than 2 minutes. You take a random sample of 36 customers and find that the average waiting time for the sample was 2.76 minutes, with a sample standard deviation of .85 minutes. Assume that the population of waiting times is normal. Using  < 2 as the null hypothesis and a significance level of 1%, conduct a test to decide whether to reject the null hypothesis.

1. z_stat for the sample result is 5.36. Since z_stat is greater than 2.33, we can't reject the null hypothesis.

b. z_stat for the sample result is 1.36. Since z_stat is less than 2.33, we can reject the null hypothesis.

c. z_stat for the sample result is 1.36. Since z_stat is less than 2.33, we can't reject the null hypothesis.

d. z_stat for the sample result is 5.36. Since z_stat is greater than 2.33, we can reject the null hypothesis.

90.

The small airport in Madras, Oregon reports that, on average, it has no more than 25 arriving flights per week. You select a sample of five weeks with the following results:

Number of arriving flights: 22 21 24 29 34

Use this data to test a  < 25 null hypothesis. Use a significance level of 5%. Assume that the number of arriving flights per week at this airport is normal.

a. t_stat = .41. Since t_stat is less than 1.862, we can’t reject the null hypothesis.

b. t_stat = 1.41. Since t_stat is less than 2.132, we can’t reject the null hypothesis.

c. t_stat = 1.41. Since t_stat is greater than 1.142, we can reject the null hypothesis..

d. t_stat = .41. Since t_stat is less than 2.132, we can’t reject the null hypothesis.

91.

From a recent production run of portable generators, you select a random sample of 10. You find that the average running time for the units in the sample is 24.4 hours, with a sample standard deviation of .6 hours. Use these sample results to test, at the 5% significance level, the null hypothesis that the average running time for the full population—the entire production run— of generators is precisely 25 hours, no more and no less. Report your conclusion and explain.

a. Here tstat = -2.16. Since tstat is greater than (that is, to the right of) -2.262, we can't reject the null hypothesis. There isn't sufficient sample evidence to believe that the average life for the population of filaments isn’t precisely 2500 hours.

b. Here t_stat = -3.16. Since t_stat is less than (that is, to the left of) -2.262, we can reject the null hypothesis. There is sufficient sample evidence to believe that the average life for the population of filaments isn’t precisely 2500 hours.

c. Here t_stat = -3.16. Since t_stat is less than (that is, to the left of) -2.552, we can't reject the null hypothesis. There isn't sufficient sample evidence to believe that the average life for the population of filaments isn’t precisely 2500 hours.

d. Here t_stat = -2.36. Since t_stat is less than (that is, to the left of) -2.552, we can reject the null hypothesis. There is sufficient sample evidence to believe that the average life for the population of filaments isn’t precisely 2500 hours.

92. You are to test the following hypotheses:

H₀: µ ≥ 1200

H_a: µ < 1200

A sample of size 36 produces a sample mean of 1148, with a standard deviation of 160.The p-value for this test is

a. .0512

b. .0256

c. .0334

d. .0668

93. You are to test the following hypotheses:

H₀: µ ≥ 360

H_a: µ < 360

A sample of size 64 produces a sample mean of 348, with a standard deviation of 60.The p-value for this test is

a. .0548

b. .0274

c. .1096

d. .0438

94. You are to test the following hypotheses:

H₀: µ < 5500

H_a: µ > 5500

A sample of size 100 produces a sample mean of 5560, with a standard deviation of 420.The p-value for this test is

a. .9234

b. .1532

c. .0383

d. .0766

95. You are to test the following hypotheses:

H₀: µ < 940

H_a: µ > 940

A sample of size 49 produces a sample mean of 994, with a standard deviation of 130.The p-value for this test is

a. .0036

b. .0018

c. .0009

d. .0360

96. You are to test the following hypotheses:

H₀: µ = 40

H_a: µ ≠ 40

A sample of size 36 produces a sample mean of 42.2, with a standard deviation of 16.The p-value for this test is

a. .0712

b. .9288

c. .4122

d. .0265

97. You are to test the following hypotheses:

H₀: µ = 800

H_a: µ ≠ 800

A sample of size 81 produces a sample mean of 750, with a standard deviation of 216.The p-value for this test is

a. .0186

b. .0518

c. .0372

d. .0461

98.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ ≥ 1000 (The population mean is greater than or equal to 1000.)

H_a: µ < 1000 (The population mean is less than 1000.)

Assume the population standard deviation is known to be 150 and that you intend to take a random sample of size 49. At a significance level of 0.05, if the sample mean is 965, should you reject the null hypothesis?

a. No. Here z_stat = -1.63; z_c = -1.65. Since z_stat is inside z_c,we can’t reject the null hypothesis.

b. No. Here z_stat = -1.63; z_c = -1.95. Since z_stat is inside z_c,we can’t reject the null hypothesis.

c. Yes. Here z_stat = -11.43; z_c = -1.65. Since z_stat is outside z_c,we can reject the null hypothesis. There is sufficient sample evidence to support your belief.

d. Yes. Here z_stat = -11.43; z_c = -1.95. Since z_stat is outside z_c,we can reject the null hypothesis. There is sufficient sample evidence to support your belief.

99.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ ≥ 1000 (The population mean is greater than or equal to 1000.)

H_a: µ < 1000 (The population mean is less than 1000.)

Assume the population standard deviation is known to be 150. A sample of 49 observations yields a sample mean of 965. The significance level is 0.05. What is the p-value for the hypothesis test?

a. 0.4484

b. 0.0516

c. 0.1032

d. 0.9484

100.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ ≥ 1000 (The population mean is greater than or equal to 1000.)

H_a: µ < 1000 (The population mean is less than 1000.)

Assume the population standard deviation is known to be 150. A sample of 49 observations yields a sample mean of 1035. The significance level is 0.05. What is the p-value for the hypothesis test?

a. 0.4484

b. 0.0516

c. 0.1032

d. 0.9484

101.

On average, adults who use text messaging send or receive 41.5 text messages per day (Source: PewInternet.org). A researcher would like to determine whether younger people send or receive more text messages than do adults in general. The researcher defined the hypothesis test as follows:

H₀: µ ≤ 41.5 texts (The average number of texts sent/received is no more than 41.5.)

H_a: µ > 41.5 texts (The average number of texts sent/received is greater than 41.5.)

A survey of 300 young adults ages 18-29 found that on average, those surveyed sent or received 87.7 text messages per day, with a sample standard deviation of 25.6. Based on the results of the survey, at a significance level of 0.05, the researcher should:

a. Reject the null hypothesis because the test statistic of 31.25 is greater than the critical value, z_c = 1.65.

b. Repeat the survey because the test statistic of 31.25 is too large to be considered a valid value for z_stat.

c. Fail to reject the null hypothesis because the test statistic of 31.25 is to the right of the critical value z_c = -1.65.

d. Reject the null hypothesis because the test statistic of 31.25 is greater than the critical value, z_c = 1.96.

102.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ = 500

H_a: µ ≠ 500

A random sample of size 100 yields a sample mean of 488 and a sample standard deviation of 36. At a 5% significance level, the correct decision regarding the null hypothesis is to:

a. Reject the null hypothesis because the test statistic of -3.33 is less than the critical value z_cl = -1.65.

b. Reject the null hypothesis because the test statistic of -3.33 is less than the critical value z_cl = -1.96.

c. Reject the null hypothesis because the test statistic of 3.33 is to the right of the critical value z_cu = 1.65.

d. Reject the null hypothesis because the test statistic of 3.33 is greater than the critical value, z_cu = 1.96.

103.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ = 500

H_a: µ ≠ 500

A random sample of size 100 has a sample mean of 488 and a sample standard deviation of 36. The p-value for this hypothesis test is:

a. 0.4996

b. 0.9992

c. 0.0004

d. 0.0008

104.

At a local coffee shop last year, the average amount spent per customer (APC) was $5.24. In order to determine whether there has been a change in the APC, the coffee shop took a sample of 500 recent transactions. For the 500 transactions, the APC was $5.39 with a sample standard deviation of $1.20. The median transaction was $5.15.

The hypotheses test was constructed as follows:

H₀: µ = 5.24 (The average transaction is still $5.24)

H_a: µ ≠ 5.24 (The average transaction differs from $5.24.)

At a 5% significance level, which of the following is the correct conclusion?

a. Reject the null hypothesis because the test statistic of 2.80 is not between -1.65 and 1.65. There is not sufficient sample evidence of a change in the average amount of the transaction.

b. Reject the null hypothesis because the test statistic of 2.80 is not between -1.96 and 1.96. There is not sufficient evidence of a change in the average transaction amount.

c. Reject the null hypothesis because the test statistic of 2.80 is not between -1.65 and 1.65. There is sufficient sample evidence to argue that the average transaction amount has changed since last year.

d. Reject the null hypothesis because the test statistic of 2.80 is not between -1.96 and 1.96. There is sufficient sample evidence to argue that the average transaction amount has changed since last year.

105.

The competing hypotheses for a hypothesis test are as follows:

H₀: µ ≥ 1000 (The population mean is greater than or equal to 1000.)

H_a: µ < 1000 (The population mean is less than 1000.)

A sample of 15 observations yields a sample mean of 990 and a sample standard deviation of 16. The significance level is 0.01, and we assume that the variable of interest is normally distributed in the population. What is the conclusion of the hypothesis test?

a. Reject the null hypothesis because the t_stat of -2.42 is less than the t_c of -1.761.

b. Reject the null hypothesis because the t_stat of -9.375 is less than the t_c of -2.624.

c. Do not reject the null hypothesis because the t_stat of -2.42 is not less than the t_c of -2.624.

d. Reject the null hypothesis because the t_stat of -2.42 is less than the t_c of -2.33.

106.

At a local coffee shop last year, the average amount spent per customer (APC) was $5.24. In order to determine whether there has been an increase in the APC, the coffee shop took a sample of 20 recent transactions. They assume that transaction amounts are normally distributed in the population. For the 20 transactions, the APC was $5.39 with a sample standard deviation of $0.65.

The hypotheses test was constructed as follows:

H₀: µ ≤ 5.24 (The average transaction is still $5.24 or less.)

H_a: µ > 5.24 (The average transaction is greater than $5.24.)

At a 5% significance level, which of the following is the correct conclusion?

a. Do not reject the null hypothesis because the test statistic of 1.03 is not greater than the critical value, z_c = 1.65. There is not sufficient sample evidence of an increase in the average amount of the transaction.

b. Do not reject the null hypothesis because the test statistic of 1.03 is not greater than the critical value, t_c =1.729. There is not sufficient sample evidence of an increase in the average transaction amount.

c. Reject the null hypothesis because the test statistic of 1.03 is not greater than the critical value z_c = 1.65. There is sufficient sample evidence to argue that the average transaction amount has increased since last year.

d. Do not reject the null hypothesis because the test statistic of 1.03 is not greater than the critical value t_c = 1.725. There is not sufficient sample evidence of an increase in the average transaction amount.