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

We have created a 95% confidence interval for µ with the result (10, 15). What decision will we make if we test H₀ : µ = 16 versus H₁ : µ × 16 at ? = 0.025?

A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you were to conduct a test to determine whether the auditor should conclude that there is evidence that the average balance is different from $75, which test would you use?

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 9-7, what will be the p-value if these data were used to perform a two-tailed test?

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 18 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.

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, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance?

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.

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, which test would you use?

For a given level of significance, if the sample size is increased, the probability of committing a Type I error will increase.

A _____ is a numerical quantity computed from the data of a sample and is used in reaching a decision on whether or not to reject the null hypothesis.

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 bank can conclude that the average age is greater than 45 at a significance level of ? = 0.01.

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, the population of interest 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.

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 web site was 10.1% The web site at the company was redesigned in an attempt to increase its conversion rates. Samples of 200 browsers at the redesigned site were 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 company officials can conclude that there is sufficient evidence that the conversion rate at the company's web site has increased using a level of significance of 0.05.

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, what is the power of the test if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance?

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, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance?

$\text { TABLE 9-1 }$ 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, if these data were used to perform a two-tailed test, the p-value would be 0.042.

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.

If a researcher accepts a true null hypothesis, she has made a(n)______ decision.

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 ? = 0.01.

We have created a 95% confidence interval for µ with the result (10, 15). What decision will we make if we test H₀ : µ = 16 versus H₁ : µ ? 16 at ? = 0.05?