8-2 Testing a Claim About a Proportion 395 Solution: Confidence Interval Method The claim given in Section 8-1 Example 1, “Most Internet users utilize two-factor authentication to protect their online data,” can be tested with a 0.05 significance level by constructing a 90% confidence interval. (See Table 8-1 on page 376 to see why this right-tailed test with a 0.05 significance level corresponds to a 90% confidence interval.) The 90% confidence interval estimate of the population proportion p is found using the sample data consisting of n = 926 and x = 482 (found from pn = 0.52). Using the methods of Section 7-1 we get this confidence interval: 0.494 * p * 0.548 This confidence interval shows us that it is likely that the true value of the population proportion p can be anywhere between 0.494 and 0.548, so it could be less than 0.5 and it could be equal to 0.5. Therefore, we can’t conclude that p 7 0.5 and we can’t conclude that most Internet users utilize two-factor authentication. In this case, the conclusion is the same as with the P-value method and the critical value method, but that is not always the case. It is possible that a conclusion based on the confidence interval can be different from the conclusion based on the P-value method or critical value method. Alternative Methods: Resampling Methods of Bootstrapping and Randomization The claim given in the Chapter Problem, “Most Internet users utilize two-factor authentication to protect their online data,” can be tested by using the resampling methods of bootstrapping and randomization. See Section 8-5 “Resampling: Using Technology for Hypothesis Testing.” Left-Tailed Test The preceding discussions in this section illustrated the method of hypothesis testing for a claim made about a proportion, and those preceding discussions are based on a right-tailed test. The following example illustrates a left-tailed test. CP CP Fewer Than 30% of Adults Have Sleepwalked? EXAMPLE 1 A study of sleepwalking or “nocturnal wandering” was described in Neurology magazine, and it included information that 29.2% of 19,136 American adults have sleepwalked. Would a reporter be justified in stating that “fewer than 30% of adults have sleepwalked”? Let’s use a 0.05 significance level to test the claim that for the adult population, the proportion of those who have sleepwalked is less than 0.30. SOLUTION REQUIREMENT CHECK (1) The sample is a simple random sample. (2) There is a fixed number (19,136) of independent trials with two categories (a subject has sleepwalked or has not). (3) The requirements np Ú 5 and nq Ú 5 are both satisfied with n = 19,136 and p = 0.30. The three requirements are all satisfied. Step 1: The original claim is expressed in symbolic form as p 6 0.30. Step 2: The opposite of the original claim is p Ú 0.30. Step 3: Because p 6 0.30 does not contain equality, it becomes H1. We get H0: p = 0.30 1null hypothesis2 H1: p 6 0.30 1alternative hypothesis and original claim2 continued
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