8-3 Testing a Claim About a Mean 409 Confidence Interval Method Critical Value Method a5 0.05 m5 7 or t 5 0 Critical Value: t 521.796 Sample mean: x 5 6.833 hours or t 520.290 FIGURE 8-8 t Test: Critical Value Method Adult Sleep: Critical Value Method EXAMPLE 3 Example 1 is a left-tailed test with test statistic t = -0.290 (rounded). The sample size is n = 12, so the number of degrees of freedom is df = n - 1 = 11. Given the significance level of a = 0.05, refer to the row of Table A-3 corresponding to 11 degrees of freedom, and refer to the column identifying an “area in one tail” of 0.05 (the significance level). The intersection of the row and column yields the critical value of t = 1.796, but this test is left-tailed, so the actual critical value is t = -1.796. Figure 8-8 shows that the test statistic of t = -0.290 does not fall within the critical region bounded by the critical value t = -1.796, so we fail to reject the null hypothesis. The conclusions are the same as those given in Example 1. Adult Sleep: Confidence Interval Method EXAMPLE 4 Example 1 is a left-tailed test with significance level a = 0.05, so we should use 90% as the confidence level (as indicated by Table 8-1 on page 376). For the sample data given in Example 1, here is the 90% confidence interval estimate of m: 5.8 hours 6 m 6 7.9 hours. In testing the claim that m 6 7 hours, we use H0: m = 7 hours, but the assumed value of m = 7 hours is contained within the confidence interval limits, so the confidence interval is telling us that 7 hours could be the value of m. We don’t have sufficient evidence to reject H0: m = 7 hours, so we fail to reject this null hypothesis and we get the same conclusions given in Example 1. Alternative Methods Used When Population Is Not Normal and n " 30 The methods of this section include two requirements: (1) The sample is a simple random sample; (2) either the population is normally distributed or n 7 30. If we have sample data that are not collected in an appropriate way, such as a voluntary response sample, it is likely that there is nothing that can be done to salvage the data, and the methods of this section should not be used. If the data are a simple random Practical Significance Merck conducted clinical trials of its sleep-aid drug Belsomra. Some key findings: (1) Subjects taking Belsomra fell asleep six minutes faster than those taking a placebo; (2) Belsomra subjects slept 16 minutes longer than the placebo subjects; (3) the preceding two results were found to have statistical significance. Merck proceeded to spend $96 million in one year for the promotion of Belsomra, and Merck expected to eventually have more Belsomra sales than Ambien and Lunesta, the current market leaders. Belsomra costs around $12 per pill and was met with negative reviews and poor sales given the high cost and potential side effects. Given the relatively small effects on sleep, it appears that Belsomra has statistical significance, but not practical significance.
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