According to the statistical test procedure described in Section 6.2, the rejection region and, correspondingly, the value of $\alpha$ are selected prior to conducting the test, and the conclusions are stated in terms of rejecting or not rejecting the null hypothesis. A second method of presenting the results of a statistical test is one that reports the extent to which the test statistic disagrees with the null hypothesis and leaves to the reader the task of deciding whether to reject the null hypothesis. This measure of disagreement is called the observed significance level (or p-value) for the test.

Recall testing $H_0:\mu =2,400$ versus $H_{\mathrm{a}}:\mu =2,400$, where $\mu$ is the mean breaking strength of sewer pipe (Section 6.1). The value of the test statistic computed for the sample of $n=50$ sections of sewer pipe was $z=\mathrm{2.12.}$ Because the test is one-tailed—that is, the alternative (research) hypothesis of interest is $H_{\mathrm{a}}:\mu >2,400$—values of the test statistic even more contradictory to $H_0$ than the one observed would be values larger than $z=\mathrm{2.12.}$ Therefore, the observed significance level (p-value) for this test is

or, equivalently, the area under the standard normal curve to the right of $z=2.12$ (see Figure 6.6).

The area A in Figure 6.6 is given in Table II in Appendix B as .4830. Therefore, the upper-tail area corresponding to $z=2.12$ is

Figure 6.6

Consequently, we say that these test results are “very significant” (i.e., they disagree rather strongly with the null hypothesis, $H_0:\mu =2,400,$ and favor $H_{\mathrm{a}}:\mu >2,400$). The probability of observing a z-value as large as 2.12 is only .0170, if in fact the true value of $\mu$ is 2,400.

If you are inclined to select $\alpha =.05$ for this test, then you would reject the null hypothesis because the p-value for the test, .0170, is less than .05. In contrast, if you choose $\alpha =.01,$ you would not reject the null hypothesis because the p-value for the test is larger than .01. Thus, the use of the observed significance level is identical to the test procedure described in the preceding sections except that the choice of $\alpha$ is left to you.

The steps for calculating the p-value corresponding to a test statistic for a population mean are given in the next box.

Figure 6.7

Figure 6.8

Example 6.4 Comparing Rejection Regions to p-Values

Problem

1. Consider the one-tailed test of hypothesis, $H_0:\mu =100$ versus $H_{\text{a}}:\mu >100$.

   1. Suppose the test statistic is $z=1.44$. Find the p-value of the test and the rejection region for the test when $\alpha =.05$. Then show that the conclusion using the rejection region approach will be identical to the conclusion based on the p-value.

   2. Now suppose the test statistic is $z=3.01$; find the p-value and rejection region for the test when $\alpha =.05$. Again, show that the conclusion using the rejection region approach will be identical to the conclusion based on the p-value.

Solution

1. The p-value for the test is the probability of observing a test statistic more contradictory to the null hypothesis (i.e., more supportive of the alternative hypothesis) than the value $z=1.44$. Since we are conducting an upper-tailed test $(H_{\text{a}}:\mu >100)$, the probability we seek is:

   $$p-\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}=P(z>1.44)=1-P(z<1.44)$$

   The latter probability in the expression is a cumulative probability that can be obtained from the standard normal table (Table II, Appendix B) or by using statistical software. The Minitab printout giving this probability is shown in Figure 6.9.

   

   Figure 6.9

   Therefore, we compute

   $$p-\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}=P(z>1.44)=1-P(z<1.44)=1-.925=.075$$

   This p-value is shown on Figure 6.10.

   Since $\alpha =.05$ and the test is upper-tailed, the rejection region for the test is $z>1.645$ (see Table 6.2). This rejection region is also shown in Figure 6.10. Observe that the test statistic $(z=1.44)$ falls outside the rejection region, implying that we fail to reject $H_0$. Also, $\alpha =.05$ is less than $p-\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}=.075$. This result also implies that we should fail to reject $H_0$. Consequently, both decision rules agree—there is insufficient evidence to reject $H_0$.

   

   Figure 6.10

2. For $z=3.01$, the observed significance level of the test is:

   $$p-\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}=P(z>3.01)=1-P(z<3.01)$$

   A MINITAB printout giving the cumulative probability, $P(z<3.01)$, is shown in Figure 6.11.

   

   Figure 6.11

   Thus, we have

   $$p-\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}=P(z>3.01)=1-P(z<3.01)=1-.9987=.0013.$$

   This p-value is shown on Figure 6.12.

   Again, for $\alpha =.05$ and an upper-tailed test, the rejection region is $z>1.645$. This rejection region is also shown in Figure 6.12. Now the test statistic $(z=3.01)$ falls within the rejection region, leading us to reject $H_0$. And, $\alpha =.05$ now exceeds the p-value (.0013), which also implies that we should reject $H_0$. Once again, both decision rules agree—and they always will if the same value of $\alpha$ is used to make the decision.

Figure 6.12

Look Ahead

Since p-values are easily computed using statistical software, most analysts and researchers utilize the p-value approach to hypothesis testing. In future examples, we will adopt this approach generally, although we’ll continue to show the appropriate rejection regions where necessary for illustrative purposes.

Now Work Exercise 6.25

When publishing the results of a statistical test of hypothesis in journals, case studies, reports, and so on, many researchers make use of p-values. Instead of selecting $\alpha$ beforehand and then conducting a test, as outlined in this chapter, the researcher computes (usually with the aid of a statistical software package) and reports the value of the appropriate test statistic and its associated p-value. It is left to the reader of the report to judge the significance of the result (i.e., the reader must determine whether to reject the null hypothesis in favor of the alternative hypothesis, based on the reported p-value). Usually, the null hypothesis is rejected if the observed significance level is less than the fixed significance level, $\alpha$, chosen by the reader. The inherent advantage of reporting test results in this manner is twofold: (1) Readers are permitted to select the maximum value of $\alpha$ that they would be willing to tolerate if they actually carried out a standard test of hypothesis in the manner outlined in this chapter, and (2) a measure of the degree of significance of the result (i.e., the p-value) is provided.