Exam 9: Fundamentals of Hypothesis Testing: One-Sample Tests

Which of the following would be an appropriate null hypothesis?

The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a side curtain air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the side curtain air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30, which test would you use?

TABLE 9-1 Microsoft 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 over 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: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis: H₀ : μ ≤ 20.000; α = 0.10; df = 45; T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject. -Referring to Table 9-1, the manager can conclude that there is sufficient evidence to show that the mean number of defective bulbs per case is over 20 during the morning shift using a level of significance of 0.10.

It is possible to directly compare the results of a confidence interval estimate to the results obtained by testing a null hypothesis if

Which of the following statements is not true about the level of significance in a hypothesis test?

TABLE 9-7 A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new television advertisements featuring well-known 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 at a 0.01 level of significance. -Referring to Table 9-7, the null hypothesis would be rejected.

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 over 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 be 99% confident in her decision, what conclusion can she make?

The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a side curtain air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the side curtain air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 and decided not to reject the null hypothesis, what conclusion could you draw?

TABLE 9-4 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, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision. -Referring to Table 9-4, the value of the test statistic is ________.

TABLE 9-7 A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new television advertisements featuring well-known 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 at a 0.01 level of significance. -Referring to Table 9-7, state the alternative hypothesis for this study.

TABLE 9-4 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, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision. -Referring to Table 9-4, the null hypothesis will be rejected with a level of significance of 0.10.

TABLE 9-6 The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tail test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test. -Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

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:

If an economist wishes to determine whether there is evidence that mean family income in a community equals $50,000

TABLE 9-1 Microsoft 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 over 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: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis: H₀ : μ ≤ 20.000; α = 0.10; df = 45; T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject. -Referring to Table 9-1, the parameter the manager is interested in is:

TABLE 9-1 Microsoft 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 over 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: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis: H₀ : μ ≤ 20.000; α = 0.10; df = 45; T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject. -Referring to Table 9-1, the evidence proves beyond a doubt that the mean number of defective bulbs per case is over 20 during the morning shift.

For a given level of significance (α), if the sample size n is increased, the probability of a Type II error (β)

A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 18 versus the alternative hypothesis that the mean of the population differs from 18, the null hypothesis could be rejected at a level of significance of 0.05.

TABLE 9-8 One of the biggest issues facing e-retailers is the ability to turn browsers into buyers. This is measured by the conversion rate, the percentage of browsers who buy something in their visit to a site. The conversion rate for a company's website was 10.1%. The website at the company was redesigned in an attempt to increase its conversion rates. A samples of 200 browsers at the redesigned site was selected. Suppose that 24 browsers made a purchase. The company officials would like to know if there is evidence of an increase in conversion rate at the 5% level of significance. -Referring to Table 9-8, the value of the probability of committing a type II error β is 0.95.

A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.10.