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

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 average 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 normal 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 9-1 Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. 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 Monarch butterflies: n=46; Arithmetic Mean =28.00; Standard Deviation =25.92; Standard Error =3.82; Null Hypothesis: :\mu\leq20.000;\alpha=0.10;df=45;T Test Statistic =2.09; One-Tailed Test Upper Critical Value =1.3006; p-value =0.021; Decision = Reject. -Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.

$\text { TABLE 9-1 }$ Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wantsto know if the mean number of parasites per butterfly is over 20 . She will make her decision usinga test witha level of significance of $0.10$ . The following informationwas extracted from the Microsoft Excel outputfor the sample of 46 Monarchbutterflies: n=46; Arithmetic Mean =28.00; Standard Deviation =25.92; Standard Error =3.82; Null Hypothesis: :\mu\leq20.000;\alpha=0.10;df=45;T Test Statistic =2.09; One-Tailed Test Upper Critical Value =1.3006; p-value =0.021; Decision = Reject. -Referring to Table 9-1, the value of þ is 0.90.

TABLE 9- 1 Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. 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 Monarch butterflies: n=46; Arithmetic Mean =28.00; Standard Deviation =25.92; Standard Error =3.82; Null Hypothesis: :\mu\leq20.000;\alpha=0.10;df=45;T Test Statistic =2.09; One-Tailed Test Upper Critical Value =1.3006; p-value =0.021; Decision = Reject. -Referring to Table 9-1, the parameter the biologist is interested in is

TABLE 9- 1 Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. 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 Monarch butterflies: n=46; Arithmetic Mean =28.00; Standard Deviation =25.92; Standard Error =3.82; Null Hypothesis: :\mu\leq20.000;\alpha=0.10;df=45;T Test Statistic =2.09; One-Tailed Test Upper Critical Value =1.3006; p-value =0.021; Decision = Reject. -Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per Monarch butterfly in Pismo Beach State Park is over 20 using a level of significance of 0.10.

A Type II error is committed when

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, versus an alternative that the mean age is greater than 45. 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.

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 0.02.

TABLE 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes an average 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 average of 257.3 W. -Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 248 W using a 0.05 level of significance?

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 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 9-7 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 lowest level of significance at which the null hypothesis can be rejected is ______.

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, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202. -Referring to Table 9-5, if the same sample was used to test the opposite one-tailed test, what would be that test's p-value?

TABLE 9-7 A major home improvement store conducted its biggest brand recognitioncampaign in the company's history. A series of new television advertisements featuring well-knownentertainersand sports figures were launched. A key metric for the success of television advertisements is the proportionof viewers who"like the ads a lot." A study of 1,189 adults whoviewed the ads reported that 230 indicated that they "like the ads a lot." The percentage of a typical television advertisement receiving the "like theads a lot"score is believed to be $22 \%$ . Company officials wanted to knowif 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 proportionof "like the ads a lot" for the company's ads is less than0.22) 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 _______ .

An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675 and decide not to reject the null hypothesis, what can you conclude?

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 9-1 Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. 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 Monarch butterflies: n=46; Arithmetic Mean =28.00; Standard Deviation =25.92; Standard Error =3.82; Null Hypothesis: :\mu\leq20.000;\alpha=0.10;df=45;T Test Statistic =2.09; One-Tailed Test Upper Critical Value =1.3006; p-value =0.021; Decision = Reject. -Referring to Table 9-4, for a test with a level of significance of 0.10, 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 value of ? is 0.90.

TABLE 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes an average 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 average 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 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes an average 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 average of 257.3 W. -Referring to Table 9-3, what is the parameter of interest?

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-tailed 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. What would be the p-value of this one-tailed test?