422 CHAPTER 8 Hypothesis Testing with Two Samples z 1 2 3 −1 −2 −3 0 α= 0.025 z0 = 1.96 −z0 = −1.96 z ≈ 1.85 1 2 α= 0.025 1 2 α 1 − = 0.95 Sample Statistics for Credit Card Debt Oklahoma North Carolina x1 = $5271 n1 = 250 x2 = $5121 n2 = 250 A Two-Sample z-Test for the Difference Between Means A credit card watchdog group claims that there is a difference in the mean credit card debts of people in Oklahoma and North Carolina. The results of a random survey of 250 people from each state are shown at the left. The two samples are independent. Assume that s1 = $960 for Oklahoma and s2 = $845 for North Carolina. Do the results support the group’s claim? Use a = 0.05. (Adapted from Experian) SOLUTION Note that s1 and s2 are known, the samples are random and independent, and both n1 and n2 are at least 30. So, you can use the z@test. The claim is “there is a difference in the mean credit card debts of people in Oklahoma and North Carolina.” So, the null and alternative hypotheses are H0: m1 = m2 and Ha: m1 ≠ m2. (Claim) Because the test is a two-tailed test and the level of significance is a = 0.05, the critical values are -z0 = -1.96 and z0 = 1.96. The rejection regions are z 6 -1.96 and z 7 1.96. The standardized test statistic is z = 1x1 - x22 - 1m1 - m22 Bs1 2 n1 + s2 2 n2 Use the z@test. = 15271 - 51212 - 0 B9602 250 + 8452 250 Assume m1 = m2, so m1 - m2 = 0. ≈ 1.85. Round to two decimal places. The figure at the left shows the location of the rejection regions and the standardized test statistic z. Because z is not in the rejection region, you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 5% level of significance to support the group’s claim that there is a difference in the mean credit card debts of people in Oklahoma and North Carolina. TRY IT YOURSELF 2 A survey indicates that the mean annual wages for forensic science technicians working for local and state governments are $63,560 and $62,070, respectively. The survey includes a randomly selected sample of size 100 from each government branch. Assume that the population standard deviations are $6200 (local) and $5575 (state). The two samples are independent. At a = 0.10, is there enough evidence to conclude that there is a difference in the mean annual wages? (Adapted from U.S. Bureau of Labor Statistics) Answer: Page A41 In Example 2, you can also use a P@value to perform the hypothesis test. For instance, the test is a two-tailed test, so the P@value is equal to twice the area to the right of z = 1.85, or P = 211 - 0.96782 = 210.03222 = 0.0644. Because 0.0644 7 0.05, you fail to reject H0. See TI-84 Plus steps on page 465. EXAMPLE 2
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