Interpret the statistical results in managerial (real-world) terms and assess their practical importance
Interpret the statistical results in managerial (real-world) terms and assess their practical importance. Since we have rejected H0: μ = 50 in favor of Ha: μ > 50 by setting α equal to .05, we conclude (at an α of .05) that the mean breaking strength of the new trash bag exceeds 50 pounds. Furthermore, this conclusion has practical importance to the trash bag manufacturer because it means that the television network will approve running commercials claiming that the new trash bag is stronger than the former bag. Note, however, that the point estimate of μ, , indicates that μ is not much larger than 50. Therefore, the trash bag manufacturer can claim only that its new bag is slightly stronger than its former bag. Of course, this might be practically important to consumers who feel that, because the new bag is 25 percent less expensive and is more environmentally sound, it is definitely worth purchasing if it has any strength advantage. However, to customers who are looking only for a substantial increase in bag strength, the statistical results would not be practically important. This illustrates that, in general, a finding of statistical significance (that is, concluding that the alternative hypothesis is true) can be practically important to some people but not to others. Notice that the point estimate of the parameter involved in a hypothesis test can help us to assess practical importance. We can also use confidence intervals to help assess practical importance.
Considerations in setting α
We have reasoned in Section 9.1 that the television network has set α equal to .05 rather than .01 because doing so means that β, the probability of failing to advertise a true claim (a Type II error), will be smaller than it would be if α were set at .01. It is informative, however, to see what would have happened if the network had set α equal to .01. Figure 9.2 illustrates that as we decrease α from .05 to .01, the critical value zα increases from z.05 = 1.645 to z.01 = 2.33. Because the test statistic value z = 2.20 is less than z.01 = 2.33, we cannot reject H0: μ = 50 in favor of Ha: μ > 50 by setting α equal to .01. This illustrates the point that, the smaller we set α, the larger is the critical value, and thus the stronger is the statistical evidence that we are requiring to reject the null hypothesis H0. Some statisticians have concluded (somewhat subjectively) that (1) if we set α equal to .05, then we are requiring strong evidence to reject H0; and (2) if we set α equal to .01, then we are requiring very strong evidence to reject H0.
Figure 9.2: The Critical Values for Testing H0: μ = 50 versus Ha: μ > 50 by Setting α = .05 and .01
A p -value for testing a “greater than” alternative hypothesis
To decide whether to reject the null hypothesis H0 at level of significance α, steps 4, 5, and 6 of the seven-step hypoth esis testing procedure compare the test statistic value with a critical value. Another way to make this decision is to calculate a p -value, which measures the likelihood of the sample results if the null hypothesis H0 is true. Sample results that are not likely if H0 is true are evidence that H0 is not true. To test H0 by using a p-value, we use the following steps 4, 5, and 6: