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Chapter 9: Problem 130

A statistician performs the test \(H_{0}: \mu=15\) versus \(H_{1}: \mu \neq 15\) and finds the \(p\) -value to be \(.4546\). a. The statistician performing the test does not tell you the value of the sample mean and the value of the test statistic. Despite this, you have enough information to determine the pair of \(p\) -values associated with the following alternative hypotheses. i. \(H_{1}: \mu<15\) ii. \(H_{1}: \mu>15\) Note that you will need more information to determine which \(p\) -value goes with which alternative. Determine the pair of \(p\) -values. Here the value of the sample mean is the same in both cases. b. Suppose the statistician tells you that the value of the test statistic is negative. Match the \(p\) -values with the alternative hypotheses. Note that the result for one of the two alternatives implies that the sample mean is not on the same side of \(\mu=15\) as the rejection region. Although we have not discussed this scenario in the book, it is important to recognize that there are many real-world scenarios in which this type of situation does occur. For example, suppose the EPA is to test whether or not a company is exceeding a specific pollution level. If the average discharge level obtained from the sample falls below the threshold (mentioned in the null hypothesis), then there would be no need to perform the hypothesis test.

Short Answer

Expert verified
The p-values for the one-tailed tests are each .2273. Given that the test statistic is negative, the p-values would be matched as follows: \(p=.2273\) for \(H_{1}: \mu<15\) and \(p=.2273\) for \(H_{1}: \mu>15\).

Step by step solution

01

Understand the p-value

The provided p-value (.4546) is for a two-tailed test. As we are shifting to one-tailed tests, we understand that the p-value of a one-tailed test is half of the two-tailed test. This is because the two-tailed test distributes the significance level (伪) over two areas on the distribution (both tails), while a one-tailed test puts all the significance level in one tail. Therefore, we divide the given p-value by 2 to get the p-value for each of the one-tailed tests.
02

Calculate the p-values

Divide the given two-tailed p-value by 2: \(.4546 / 2 = .2273\). Therefore, the p-values for the one-tailed tests are each .2273.
03

Interpret the test statistic

The test statistic is negative. This suggests that the sample mean is less than the hypothesized population mean under the null hypothesis. Therefore, the alternative hypothesis that corresponds to this case is \(H_{1}: \mu<15\).
04

Match the p-values with the alternative hypotheses

Given the negative test statistic, it is unlikely that the sample mean is greater than the hypothesized population mean (\(H_{1}: \mu>15\)). So, the matched pairs would be \(p=.2273\) for \(H_{1}: \mu<15\), and \(p=.2273\) for \(H_{1}: \mu>15\).

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