Fundamentals of Hypothesis Testing Sample Assignment
Chapter 11 â€“ Fundamentals of Hypothesis Testing: OneSample Tests
1) Which of the following would be an appropriate null hypothesis?
A) The mean of a population is equal to 55.
B) The mean of a sample is equal to 55.
C) The mean of a population is greater than 55.
D) Only A and C are appropriate.
Answer: A
2) Which of the following would be an appropriate null hypothesis?
A) The population proportion is less than 0.65.
B) The sample proportion is less than 0.65.
C) The population proportion is not less than 0.65.
D) The sample proportion is no less than 0.65.
Answer: C
3) Which of the following would be an appropriate alternative hypothesis?
A) The mean of a population is equal to 55.
B) The mean of a sample is equal to 55.
C) The mean of a population is greater than 55.
D) The mean of a sample is greater than 55.
Answer: C
4) Which of the following would be an appropriate alternative hypothesis?
A) The population proportion is less than 0.65.
B) The sample proportion is less than 0.65.
C) The population proportion is not less than 0.65.
D) The sample proportion is not less than 0.65.
Answer: A
5) A Type II error is committed when
A) you reject a null hypothesis that is true.
B) you don't reject a null hypothesis that is true.
C) you reject a null hypothesis that is false.
D) you don't reject a null hypothesis that is false.
Answer: D
6) A Type I error is committed when
A) you reject a null hypothesis that is true.
B) you don't reject a null hypothesis that is true.
C) you reject a null hypothesis that is false.
D) you don't reject a null hypothesis that is false.
Answer: A
7) The power of a test is measured by its capability of
A) rejecting a null hypothesis that is true.
B) not rejecting a null hypothesis that is true.
C) rejecting a null hypothesis that is false.
D) not rejecting a null hypothesis that is false.
Answer: C
8) If an economist wishes to determine whether there is evidence that mean family income in a community exceeds $50,000,
A) either a onetail or twotail test could be used with equivalent results.
B) a onetail test should be utilized.
C) a twotail test should be utilized.
D) none of the above
Answer: B
9) If an economist wishes to determine whether there is evidence that mean family income in a community equals $50,000,
A) either a onetail or twotail test could be used with equivalent results.
B) a onetail test should be utilized.
C) a twotail test should be utilized.
D) none of the above
Answer: C
10) If the pvalue is less than Î± in a twotail test,
A) the null hypothesis should not be rejected.
B) the null hypothesis should be rejected.
C) a onetail test should be used.
D) no conclusion should be reached.
Answer: B
11) If a test of hypothesis has a Type I error probability (Î±) of 0.01, it means that
A) if the null hypothesis is true, you don't reject it 1% of the time.
B) if the null hypothesis is true, you reject it 1% of the time.
C) if the null hypothesis is false, you don't reject it 1% of the time.
D) if the null hypothesis is false, you reject it 1% of the time.
Answer: B
12) If the Type I error (Î±) for a given test is to be decreased, then for a fixed sample size n
A) the Type II error (Î²) will also decrease.
B) the Type II error (Î²) will increase.
C) the power of the test will increase.
D) a onetail test must be utilized.
Answer: B
13) For a given sample size n, if the level of significance (Î±) is decreased, the power of the test
A) will increase.
B) will decrease.
C) will remain the same.
D) cannot be determined.
Answer: B
14) For a given level of significance (Î±), if the sample size n is increased, the probability of a Type II error (Î²)
A) will decrease.
B) will increase.
C) will remain the same.
D) cannot be determined.
Answer: A
15) If you know that the probability of committing a Type II error ( Î²) is 5%, you can tell that the power of the test is
A) 2.5%.
B) 95%.
C) 97.5%.
D) unknown.
Answer: B
16) If a researcher rejects a true null hypothesis, she has made a ________ error.
Answer: Type I
17) If a researcher does not reject a true null hypothesis, she has made a ________ decision.
Answer: correct
18) If a researcher rejects a false null hypothesis, she has made a ________ decision.
Answer: correct
19) If a researcher does not reject a false null hypothesis, she has made a ________ error.
Answer: Type II
20) The power of a statistical test is
A) the probability of not rejecting H0 when it is false.
B) the probability of rejecting H0 when it is true.
C) the probability of not rejecting H0 when it is true.
D) the probability of rejecting H0 when it is false.
Answer: D
21) The symbol for the power of a statistical test is
A) Î±.
B) 1  Î±.
C) Î².
D) 1  Î².
Answer: D
22) The symbol for the probability of committing a Type I error of a statistical test is
A) Î±.
B) 1  Î±.
C) Î².
D) 1  Î².
Answer: A
23) The symbol for the level of significance of a statistical test is
A) Î±.
B) 1  Î±.
C) Î².
D) 1  Î².
Answer: A
24) The symbol for the probability of committing a Type II error of a statistical test is
A) Î±.
B) 1  Î±.
C) Î².
D) 1  Î².
Answer: C
25) The symbol for the confidence coefficient of a statistical test is
A) Î±.
B) 1  Î±.
C) Î².
D) 1  Î².
Answer: B
26) Suppose we wish to test H0: Î¼ â‰¤ 47 versus H 1: Î¼ > 47. What will result if we conclude that the mean is greater than 47 when its true value is really 52?
A) We have made a Type I error.
B) We have made a Type II error.
C) We have made a correct decision.
D) None of the above are correct.
Answer: C
27) How many tissues should the Kimberly Clark Corporation package of KleenexÂ® contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: = 52, S = 22. Give the null and alternative hypotheses to determine if the number of tissues used during a cold is less than 60.
A) H0: Î¼ â‰¤ 60 and H1: Î¼ > 60.
B) H0: Î¼ â‰¥ 60 and H1: Î¼ < 60.
C) H0: â‰¥ 60 and H1: < 60.
D) H0: = 52 and H1: â‰ 52.
Answer: B
28) How many tissues should the Kimberly Clark Corporation package of KleenexÂ® contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: = 52, S = 22. Using the sample information provided, calculate the value of the test statistic.
A) t = (52  60) / 22
B) t = (52  60) / (22/100)
C) t = (52  60) / (22/1002)
D) t = (52  60) / (22/10)
Answer: D
29) How many tissues should the Kimberly Clark Corporation package of KleenexÂ® contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: = 52, S = 22. Suppose the alternative you wanted to test was H1: Î¼ < 60. State the correct rejection region for Î± = 0.05.
A) Reject H0 if t > 1.6604.
B) Reject H0 if t < 1.6604.
C) Reject H0 if t > 1.9842 or Z < 1.9842.
D) Reject H0 if t < 1.9842.
Answer: B
30) How many tissues should the Kimberly Clark Corporation package of KleenexÂ® contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: = 52, S = 22. Suppose the test statistic does fall in the rejection region at Î± = 0.05. Which of the following decisions is correct?
A) At Î± = 0.05, you do not reject H0.
B) At Î± = 0.05, you reject H0.
C) At Î± = 0.10 you reject H0.
D) At Î± = 0.10, you do not reject H0.
Answer: B
31) How many tissues should the Kimberly Clark Corporation package of KleenexÂ® contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: = 52, S = 22. Suppose the test statistic does fall in the rejection region at Î± = 0.05. Which of the following conclusions is correct?
A) At Î± = 0.05, there is not sufficient evidence to conclude that the mean number of tissues used during a cold is 60 tissues.
B) At Î± = 0.05, there is sufficient evidence to conclude that the mean number of tissues used during a cold is 60 tissues.
C) At Î± = 0.05, there is insufficient evidence to conclude that the mean number of tissues used during a cold is not 60 tissues.
D) At Î± = 0.10, there is sufficient evidence to conclude that the mean number of tissues used during a cold is not 60 tissues.
Answer: D
32) You have created a 95% confidence interval for Î¼ with the result 10 â‰¤ Î¼ â‰¤ 15. What decision will you make if you test H0: Î¼ =16 versus H1: Î¼ â‰ 16 at Î± = 0.05?
A) Reject H0 in favor of H1.
B) Do not reject H0 in favor of H1.
C) Fail to reject H0 in favor of H1.
D) We cannot tell what our decision will be from the information given.
Answer: A
33) Which of the following statements is NOT true about the level of significance in a hypothesis test?
A) The larger the level of significance, the more likely you are to reject the null hypothesis.
B) The level of significance is the maximum risk we are willing to accept in making a Type I error.
C) The significance level is also called the Î± level.
D) The significance level is another name for Type II error.
Answer: D
34) The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. The appropriate hypotheses to test are:
A) H0: Î¼ â‰¥ 30 versus H1: Î¼ < 30.
B) H0: Î¼ â‰¤ 30 versus H1: Î¼ > 30.
C) H0: â‰¥ 30 versus H1: < 30.
D) H0: â‰¤ 30 versus H1: > 30.
Answer: B
35) The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. If she wants to have a level of significance at 0.01, what rejection region should she use?
A) Reject H0 if t < 2.3263.
B) Reject H0 if t < 2.5758.
C) Reject H0 if t > 2.3414.
D) Reject H0 if t > 2.5758.
Answer: C
36) The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to have a level of significance at 0.01, what decision should she make?
A) Reject H0.
B) Reject H1.
C) Do not reject H0.
D) We cannot tell what her decision should be from the information given.
Answer: C
37) The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to have a level of significance at 0.01 what conclusion can she make?
A) There is not sufficient evidence that the mean age of her customers is greater than 30.
B) There is sufficient evidence that the mean age of her customers is greater than 30.
C) There is not sufficient evidence that the mean age of her customers is not greater than 30.
D) There is sufficient evidence that the mean age of her customers is not greater than 30.
Answer: A
38) The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. What is the pvalue associated with the test statistic?
A) 0.3577
B) 0.1423
C) 0.0780
D) 0.02
Answer: C
39) An entrepreneur is considering the purchase of a coinoperated laundry. The current owner claims that over the past five years, the mean daily revenue was $675 with a population standard deviation of $75. A sample of 30 days reveals a daily mean revenue of $625. If you were to test the null hypothesis that the daily mean revenue was $675, which test would you use?
A) Z test of a population mean
B) Z test of a population proportion
C) t test of population mean
D) t test of a population proportion
Answer: A
40) An entrepreneur is considering the purchase of a coinoperated laundry. The current owner claims that over the past five years, the mean daily revenue was $675 with a population standard deviation of $75. A sample of 30 days reveals a daily mean revenue of $625. If you were to test the null hypothesis that the daily mean revenue was $675 and decide not to reject the null hypothesis, what can you conclude?
A) There is not enough evidence to conclude that the daily mean revenue was $675.
B) There is not enough evidence to conclude that the daily mean revenue was not $675.
C) There is enough evidence to conclude that the daily mean revenue was $675.
D) There is enough evidence to conclude that the daily mean revenue was not $675.
Answer: B
41) A manager of the credit department for an oil company would like to determine whether the mean monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the mean owed is $83.40 with a sample standard deviation of $23.65. If you were to conduct a test to determine whether the auditor should conclude that there is evidence that the mean balance is different from $75, which test would you use?
A) Z test of a population mean
B) Z test of a population proportion
C) t test of population mean
D) t test of a population proportion
Answer: C
42) A manager of the credit department for an oil company would like to determine whether the mean monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the mean owed is $83.40 with a sample standard deviation of $23.65. If you wanted to test whether the mean balance is different from $75 and decided to reject the null hypothesis, what conclusion could you reach?
A) There is not evidence that the mean balance is $75.
B) There is not evidence that the mean balance is not $75.
C) There is evidence that the mean balance is $75.
D) There is evidence that the mean balance is not $75.
Answer: D
43) A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to ________.
A) 4.12
B) 2.33
C) 1.86
D) 0.07
Answer: B
44) A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose you reject the null hypothesis. What conclusion can you reach?
A) There is not sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90.
B) There is sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90.
C) There is not sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90.
D) There is sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90.
Answer: D
45) A major DVD rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with DVD players. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have DVD players. State the test of hypothesis that is of interest to the rental chain.
A) H0: Ï€ â‰¤ 0.32 versus H1: Ï€ > 0.32
B) H0: Ï€ â‰¤ 0.25 versus H1: Ï€ > 0.25
C) H0: Ï€ â‰¤ 5,000 versus H1: Î¼ > 5,000
D) H0: Î¼ â‰¤ 5,000 versus H1: Î¼ > 5,000
Answer: B
46) A major DVD rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with DVD players. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have DVD players. The value of the test statistic in this problem is approximately equal to ________.
A) 2.80
B) 2.60
C) 1.94
D) 1.30
Answer: A
47) A major DVD rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with DVD players. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have DVD players. The pvalue associated with the test statistic in this problem is approximately equal to ________.
A) 0.0100
B) 0.0051
C) 0.0026
D) 0.0013
Answer: C
48) A major DVD rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with DVD players. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have DVD players. The decision on the hypothesis test using a 5% level of significance is:
A) to reject H0 in favor of H1.
B) to accept H0 in favor of H1.
C) to fail to reject H0 in favor of H1.
D) We cannot tell what the decision should be from the information given.
Answer: A
49) A major DVD rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with DVD players. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have DVD players. The rental chain's conclusion from the hypothesis test using a 5% level of significance is:
A) to open a new store.
B) not to open a new store.
C) to delay opening a new store until additional evidence is collected.
D) We cannot tell what the decision should be from the information given.
Answer: A
TABLE 1
MicrosoftÂ® Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases:
50) Referring to Table 1, the parameter the manager is interested in is:
A) the mean number of defective light bulbs per case produced at the plant.
B) the mean number of defective light bulbs per case among the 46 cases.
C) the mean number of defective light bulbs per case produced during the morning shift.
D) the proportion of cases with defective light bulbs produced at the plant.
Answer: C
51) Referring to Table 1, state the alternative hypothesis for this study.
Answer: H1: Î¼ > 20
52) Referring to Table 1, what critical value should the manager use to determine the rejection region?
A) 1.6794
B) 1.3011
C) 1.3006
D) 0.6800
Answer: C
53) True or False: Referring to Table 1, the null hypothesis would be rejected.
Answer: TRUE
54) True or False: Referring to Table 1, the null hypothesis would be rejected if a 5% probability of committing a Type I error is allowed.
Answer: TRUE
55) True or False: Referring to Table 1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.
Answer: FALSE
56) True or False: Referring to Table 1, the manager can conclude that there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift using a level of significance of 0.10.
Answer: TRUE
TABLE 2
A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an mean of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normally distributed with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, it will continue to produce the older drug. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision.
57) Referring to Table 2, the appropriate hypotheses are:
A) H0: Î¼ = 7.4 versus H1: Î¼ â‰ 7.4.
B) H0: Î¼ â‰¤ 7.4 versus H1: Î¼ > 7.4.
C) H0: Î¼ â‰¥ 7.4 versus H1: Î¼ < 7.4.
D) H0: Î¼ > 7.4 versus H1: Î¼ â‰¤ 7.4.
Answer: C
58) Referring to Table 2, for a test with a level of significance of 0.10, the critical value would be ________.
Answer: 1.28
59) Referring to Table 2, the value of the test statistic is ________.
Answer: 1.50
60) Referring to Table 2, the pvalue of the test is ________.
Answer: 0.0668
61) True or False: Referring to Table 2, the null hypothesis will be rejected with a level of significance of 0.10.
Answer: TRUE
TABLE 3
A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new television advertisements featuring wellknown entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who "like the ads a lot". A study of 1,189 adults who viewed the ads reported that 230 indicated that they "like the ads a lot." The percentage of a typical television advertisement receiving the "like the ads a lot" score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e., if there is evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22) at a 0.01 level of significance.
62) Referring to Table 3, the parameter the company officials is interested in is
A) the mean number of viewers who "like the ads a lot".
B) the total number of viewers who "like the ads a lot".
C) the mean number of company officials who "like the ads a lot".
D) the proportion of viewers who "like the ads a lot".
Answer: D
63) Referring to Table 3, state the null hypothesis for this study.
Answer: H0: Ï€ â‰¥ 0.22
64) Referring to Table 3, state the alternative hypothesis for this study.
Answer: H1: Ï€ < 0.22
65) Referring to Table 3, what critical value should the company officials use to determine the rejection region?
Answer: âˆ’2.3263
66) Referring to Table 3, the null hypothesis will be rejected if the test statistic is
A) greater than 2.3263.
B) less than 2.3263.
C) greater than 2.3263.
D) less than 2.3263.
Answer: D
67) True or False: Referring to Table 3, the null hypothesis would be rejected.
Answer: FALSE
68) True or False: Referring to Table 3, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.01.
Answer: FALSE
69) True or False: Referring to Table 3, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.05.
Answer: TRUE
70) Referring to Table 3, what will be the pvalue if these data were used to perform a twotail test?
Answer: 0.027
71) A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. Suppose that a storm is in progress with a severe storm class rating. Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? Is the Pvalue area on the left, right, or on both sides of the mean?

72) Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 7.6 seconds. Suppose that you want to set up a statistical test to challenge the claim of 7.6 seconds. What would you use for the null hypothesis?

73) Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 8.6 seconds. Suppose that you want to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer than 8.6 seconds. What would you use for the alternative hypothesis?

74) Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 8.4 seconds. Suppose that you want to test the claim that the average time to accelerate from 0 to 60 miles per hour is less than 8.4 seconds. What would you use for the alternative hypothesis?

75) Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with Ïƒ = 2.0%. A random sample of 19 Australian bank stocks has a mean x = 6.33%. For the entire Australian stock market, the mean dividend yield is Î¼ = 7.5%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 7.5%? Use Î± = 0.01. What is the level of significance?

76) Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with Ïƒ = 2.0%. A random sample of 23 Australian bank stocks has a sample mean of x = 7.67%. For the entire Australian stock market, the mean dividend yield is Î¼ = 7.6%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 7.6%? Use Î± = 0.05. What is the value of the test statistic?

77) Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with . A random sample of Australian bank stocks has a sample mean of . For the entire Australian stock market, the mean dividend yield is . Do these data indicate that the dividend yield of all Australian bank stocks is higher than %? Use . Find (or estimate) the Pvalue.

78) Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with Ïƒ = 2.6%. A random sample of 17 Australian bank stocks has a sample mean of x = 8.76%. For the entire Australian stock market, the mean dividend yield is Î¼ = 6.5%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 6.5%? Use Î± = 0.01. Are the data statistically significant at the given level of significance? Based on your answers, will you reject or fail to reject the null hypothesis?

79) A professional employee in a large corporation receives an average of Î¼ = 38.3 emails per day. Most of these emails are from other employees in the company. Because of the large number of emails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an email. One month after the new priority list was put into place, a random sample of 33 employees showed that they were receiving an average of x = 28.7 emails per day. The computer server through which the emails are routed showed that Ïƒ = 17.7 Has the new policy had any effect? Use a 10% level of significance to test the claim that there has been a change (either way) in the average number of emails received per day per employee. What is the level of significance?

80) A professional employee in a large corporation receives an average of Î¼ = 38.8 emails per day. Most of these emails are from other employees in the company. Because of the large number of emails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an email. One month after the new priority list was put into place, a random sample of 39 employees showed that they were receiving an average of x = 31 emails per day. The computer server through which the emails are routed showed that Ïƒ = 15.4. Has the new policy had any effect? Use a 5% level of significance to test the claim that there has been a change (either way) in the average number of emails received per day per employee. What are the null and alternate hypotheses?

81) A professional employee in a large corporation receives an average of Î¼ = 42.7 emails per day. Most of these emails are from other employees in the company. Because of the large number of emails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an email. One month after the new priority list was put into place, a random sample of 38 employees showed that they were receiving an average of x = 35.3 emails per day. The computer server through which the emails are routed showed that Ïƒ = 19.6. Has the new policy had any effect? Use a 5% level of significance to test the claim that there has been a change (either way) in the average number of emails received per day per employee. What is the value of the test statistic?

82) A professional employee in a large corporation receives an average of Î¼ = 40.3 emails per day. Most of these emails are from other employees in the company. Because of the large number of emails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an email. One month after the new priority list was put into place, a random sample of 32 employees showed that they were receiving an average of x = 33.3 emails per day. The computer server through which the emails are routed showed that Ïƒ = 15.4. Has the new policy had any effect? Use a 1% level of significance to test the claim that there has been a change (either way) in the average number of emails received per day per employee. Find (or estimate) the Pvalue.

83) A professional employee in a large corporation receives an average of Î¼ = 39.8 emails per day. Most of these emails are from other employees in the company. Because of the large number of emails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an email. One month after the new priority list was put into place, a random sample of 38 employees showed that they were receiving an average of x = 33.1 emails per day. The computer server through which the emails are routed showed that Ïƒ = 16.2. Has the new policy had any effect? Use a 10% level of significance to test the claim that there has been a change (either way) in the average number of emails received per day per employee. Are the data statistically significant at level Î±? Based on your answers, will you reject or fail to reject the null hypothesis?

84) Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n =351 numerical entries from the file and r = 108 of the entries had a first nonzero digit of 1. Let represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that is less than 0.301 by using Î± = 0.05. What is the level of significance?

85) Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 466 numerical entries from the file and r = 113 of the entries had a first nonzero digit of 1. Let represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that is less than 0.301 by using Î± = 0.05. What is the value of the test statistic?

86) Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 250 numerical entries from the file and r = 60 of the entries had a first nonzero digit of 1. Let represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that is less than 0.301 by using Î± = 0.01. What does the area of the sampling distribution corresponding to your Pvalue look like?

87) Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 247 numerical entries from the file and r = 60 of the entries had a first nonzero digit of 1. Let represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that is less than 0.301 by using Î± = 0.1. Are the data statistically significant at the significance level? Based on your answers, will you reject or fail to reject the null hypothesis?

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