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

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 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: -"What conclusions and interpretations can you reach from the results of the hypothesis test?" is not an important question to ask when performing a hypothesis test.

TABLE 9-9 The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years, but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. -Referring to Table 9-9, 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, 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. -Referring to Table 9-4, if the level of significance had been chosen as 0.05, the null hypothesis would be rejected.

TABLE 9-9 The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years, but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. -Referring to Table 9-9, the president can conclude that the mean SAT score of the entering class this year is lower than previous years using a level of significance of 0.10.

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 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: -Suppose, in testing a hypothesis about a mean, 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 9-1 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 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: -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 accepted at a level of significance of 0.01.

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 (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. -Referring to Table 9-7, 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.

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

TABLE 9-5 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45 years, versus an alternative that the mean age is greater than 45 years. They take a sample and calculate a p-value of 0.0202. -Referring to Table 9-5, the null hypothesis would be rejected at a significance level of α = 0.05.

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, state the null hypothesis for this study.

TABLE 9-9 The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years, but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. -Referring to Table 9-9, the highest level of significance at which the null hypothesis cannot 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 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: -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 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: -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 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: -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 0.05.

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?

TABLE 9-9 The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years, but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. -Referring to Table 9-9, the null hypothesis would be rejected if a 10% probability of committing a Type I error is allowed.

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?

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

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, the student 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), the student 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 "actual confidence coefficient"?