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

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 auditor should conclude that there is evidence that the mean balance is different from $75, which test would you use?

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 null hypothesis for this study.

If a test of hypothesis has a Type I error probability (α) of 0.01, we mean

"Is the intended sample size large enough to achieve the desired power of the test for the level of significance chosen?" should be among the questions asked when performing a hypothesis test.

Suppose, in testing a hypothesis about a proportion, the Z test statistic is computed to be 2.04. The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tail test is used.

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 appropriate hypotheses are:

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 is the power of the test?

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, if the level of significance had been chosen as 0.05, the null hypothesis would be rejected.

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 will be rejected if the test statistics 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, the parameter the company officials is interested in 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 less than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

The smaller is the p-value, the stronger is the evidence against the null hypothesis.

The statement of the null hypothesis always contains an equality.

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 draw?

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

Which of the following would be an appropriate null hypothesis?

A Type I error is committed when

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 parameter the company officials 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, what critical value should the manager 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, state the alternative hypothesis for this study.