Example 92
The decision rule is to reject H0 in favor of Ht if
For a 5%-level test, a = .05, and za!2 = 2.025 = 1.96. Then, since —2.50 is less than -1.96, the null hypothesis is rejected at the 5% significance level.
In fact, according to the decision rule, the null hypothesis will be rejected for any significance level a for which — zaj2 is bigger than —2.50. From Table 3 in the Appendix, we see that when za/2 is 2.50, a/2 is equal to .0062. Hence, a — .0124. This is the p-value of the test, implying that the null hypothesis can be rejected against the two-sided alternative at any level of significance greater than 1.24%. This certainly casts substantial doubt on the hypothesis that the drill is functioning correctly. What we have found is that if the null hypothesis were true, the probability would be only .0124 of finding a sample mean this far or farther from two inches.
Until now we have dealt only with the (generally unrealistic) case where the population variance is known. However, if the available number of sample observations is large, the tests can readily be modified to deal with an important class of practical problems. As in Section 8.2, we are indebted to the central limit theorem, which allows us to conclude that for large samples, the sampling distribution of the sample mean will be approximately normal, even though the population distribution is not normal.
Tests for the Mean: Large Sample Sizes
Suppose that we have a random sample of n observations from a population with mean (jl and variance a2. If the sample size n is large,2 the test procedures developed for the case where the population variance is known can be employed when it is unknown, replacing a1 by the observed sample variance s,2. Moreover, these procedures remain approximately valid even if the population distribution is not normal.
A random sample of 541 consumers was asked to respond on a scale from one (strongly disagree) to five (strongly agree) to the assertion that a limit should be placed on the amount of punitive damages awarded for product liability.1 The sample
2 This approximation is generally satisfactory for samples of thirty or more observations.
3 J. DeConinck and J. Kopf, "Consumers' attitudes toward product liability reform," American Business Review, 10, no. 2 (1992), 78-83.
mean response was 3.68 and the sample standard deviation was 1.21. Suppose that a population mean response of 3.75 or more is taken as broad general support for the assertion. Test the null hypothesis that the population mean is at least 3.75 against the alternative that it is less than 3.75.
We want to test the null hypothesis
against the alternative hi. (jl< 3.75
The decision rule is to reject Ho in favor of H, if
Here we have x = 3.68 /if, = 3.75 ¿v — 1.21 « = 541
According to the decision rule, the null hypothesis is rejected for any significance level a for which — z„ is bigger than — 1.35. From Table 3 of the Appendix, we see that when is 1.35, a is equal to .0885. Therefore, the probability of observing a sample mean of 3.68 or less, if the population mean is 3.75, would be .0885. The null hypothesis can be rejected at significance levels above 8.85%, so there is moderately strong evidence against that hypothesis. The conclusion of the test is illustrated in Figure 9.6, which shows the distribution of the decision rule criterion when the popu-
FIGURE 9.6 Conclusion of the test in Example 9.3: the null hypothesis H0: ¡x > fia is rejected against the alternative Hù ¡J. < (JU) at significance levels greater than .0885.
FIGURE 9.6 Conclusion of the test in Example 9.3: the null hypothesis H0: ¡x > fia is rejected against the alternative Hù ¡J. < (JU) at significance levels greater than .0885.
lation mean is 3.75. Also illustrated is the lower-tail area probability corresponding to the observed value, —1.35, of the test statistic.
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