446 CHAPTER 9 Inferences from Two Samples Critical Value Method The critical value method of testing hypotheses (see Figure 8-1 on page 376) can also be used for Example 1. In Step 6, instead of finding the P-value, find the critical values. With a significance level of a = 0.05 in a two-tailed test based on the normal distribution, we refer to Table A-2 and find that an area of a = 0.05 divided equally between the two tails corresponds to the critical values of z = {1.96. In Figure 9-1(b) on the preceding page we can see that the test statistic of z = 3.51 falls within the critical region beyond the critical value of 1.96. We again reject the null hypothesis. The conclusions are the same as in Example 1. Confidence Intervals Using the format given in the preceding Key Elements box, we can construct a confidence interval estimate of the difference between population proportions 1p1 - p22. If a confidence interval estimate of p1 - p2 does not include 0, we have evidence suggesting that p1 and p2 have different values. The confidence interval uses a standard deviation based on estimated values of the population proportions, whereas a hypothesis test uses a standard deviation based on the assumption that the two population proportions are equal. Consequently, a conclusion based on a confidence interval might be different from a conclusion based on a hypothesis test. See the caution that follows Example 2. Use the sample data given in Example 1 to construct a 95% confidence interval estimate of the difference between the two population proportions. What does the result suggest about the claim that there is no difference in success rates between the two treatment groups? CP EXAMPLE 2 Confidence Interval for Claim About Two Treatment Groups SOLUTION REQUIREMENT CHECK We are using the same data from Example 1, and the same requirement check applies here, so the requirements are satisfied. The confidence interval can be found using technology; see the preceding Statdisk display. If not using technology, proceed as follows. With a 95% confidence level, za>2 = 1.96 (from Table A-2). We first calculate the value of the margin of error E as shown here. E = za>2Bpn 1q n 1 n1 + pn 2q n 2 n2 = 1.96Ta 79 438ba 359 438b 438 + a 44 446ba 402 446b 446 = 0.04541521 With pn 1 = 79>438 = 0.18036530 and p n 2 = 44>446 = 0.09865471, we get pn 1 - p n 2 = 0.08171059. With p n 1 - p n 2 = 0.08171059 and E = 0.04541521, the confidence interval is evaluated as follows, with the confidence interval limits rounded to three significant digits. 1pn 1 - p n 22 - E 6 p1 - p2 6 1p n 1 - p n 22 + E 0.08171059 - 0.04541521 6 p1 - p2 6 0.08171059 + 0.04541521 0.0363 6 p1 - p2 6 0.127 See the preceding Statdisk display in Example 1 showing that after rounding, we get the same confidence interval obtained here. C T al c Polio Experiment In 1954 an experiment was conducted to test the effectiveness of the Salk vaccine as protection against the devastating effects of polio. Approximately 200,000 children were injected with an ineffective salt solution, and 200,000 other children were injected with the vaccine. The experiment was “double blind” because the children being injected didn’t know whether they were given the real vaccine or the placebo, and the doctors giving the injections and evaluating the results didn’t know either. Only 33 of the 200,000 vaccinated children later developed paralytic polio, whereas 115 of the 200,000 injected with the salt solution later developed paralytic polio. Statistical analysis of these and other results led to the conclusion that the Salk vaccine was indeed effective against paralytic polio. In e w to e o v
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