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

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 lowest level of significance at which the null hypothesis can be rejected is ________.

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, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

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 null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.

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 null hypothesis would be rejected.

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 null hypothesis will be rejected if the test statistics is

Suppose, in testing a hypothesis about a proportion, the p-value is computed to be 0.034. The null hypothesis should be rejected if the chosen level of significance is 0.01.

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 decision should she make?

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 that the test statistic is -2.20. Can we conclude that H₀ should be rejected at the (a) α = 0.10, (b) α = 0.05, and (c) α = 0.01 level of Type I error?

TABLE 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume an mean of 257.3 W. -Referring to Table 9-3, for a test with a level of significance of 0.05, the critical value would be ________.

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 largest level of significance at which the null hypothesis will not be rejected 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 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 with no more than a 1% probability of incorrectly rejecting the true null hypothesis.

For a given sample size n, if the level of significance (α) is decreased, the power of the test

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 0.05, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

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, what critical value should the company officials use to determine the rejection region?

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 null hypothesis would be rejected.

TABLE 9-2 A student claims that he can correctly identify whether a person is a business major or an agriculture major by the way the person dresses. Suppose in actuality that if someone is a business major, he can correctly identify that person as a business major 87% of the time. When a person is an agriculture major, the student will incorrectly identify that person as a business major 16% of the time. Presented with one person and asked to identify the major of this person (who is either a business or an agriculture major), he considers this to be a hypothesis test with the null hypothesis being that the person is a business major and the alternative that the person is an agriculture major. -Referring to Table 9-2, what would be a Type II error?

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, state the alternative hypothesis for this study.

If, as a result of a hypothesis test, we reject the null hypothesis when it is false, then we have committed

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, for a test with a level of significance of 0.10, the critical value would be ________.

Suppose we wish to test H₀: μ ≤ 47 versus H₁: μ > 47. What will result if we conclude that the mean is greater than 47 when its true value is really 52?