430 CHAPTER 8 Hypothesis Testing with Two Samples 1 − = 0.90 α= 0.05 1 2 α= 0.05 1 2 t 3 4 0 1 −3 −4 −t0 = −1.895 t ≈ 0.922 t0 = 1.895 −1 −2 α Sample Statistics for State Mathematics Test Scores Teacher 1 Teacher 2 x1 = 473 s1 = 39.7 n1 = 8 x2 = 459 s2 = 24.5 n2 = 18 Sample Statistics for Annual Earnings High school diploma Associate’s degree x1 = $41,775 s1 = $6050 n1 = 25 x2 = $51,650 s2 = $9580 n2 = 16 A Two-Sample t-Test for the Difference Between Means The results of a state mathematics test for random samples of students taught by two different teachers at the same school are shown at the left. Can you conclude that there is a difference in the mean mathematics test scores for the students of the two teachers? Use a = 0.10. Assume the populations are normally distributed and the population variances are not equal. SOLUTION Note that s1 and s2 are unknown, the samples are random and independent, and the populations are normally distributed. So, you can use the t@test. The claim is “there is a difference in the mean mathematics test scores for the students of the two teachers.” So, the null and alternative hypotheses are H0: m1 = m2 and Ha: m1 ≠ m2. (Claim) Because the population variances are not equal and the smaller sample size is 8, use d.f. = 8 - 1 = 7. The test is a two-tailed test with d.f. = 7 and a = 0.10, so the critical values are -t0 = -1.895 and t0 = 1.895. The rejection regions are t 6 -1.895 and t 7 1.895. The standardized test statistic is t = 1x1 - x22 - 1m1 - m22 Bs1 2 n1 + s2 2 n2 Use the t@test (variances are not equal). = 1473 - 4592 - 0 B139.722 8 + 124.522 18 Assume m1 = m2, so m1 - m2 = 0. ≈ 0.922. Round to three decimal places. The figure at the left shows the location of the rejection regions and the standardized test statistic t. Because t is not in the rejection region, you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 10% level of significance to support the claim that the mean mathematics test scores for the students of the two teachers are different. TRY IT YOURSELF 1 The results of a survey on the annual earnings of random samples of people with a high school diploma and with an associate’s degree are shown at the left. Can you conclude that there is a difference in the mean annual earnings based on level of education? Use a = 0.05. Assume the populations are normally distributed and the population variances are not equal. (Adapted from U.S. Census Bureau) Answer: Page A41 You can also use technology and a P@value to perform a hypothesis test for the difference between means. For instance, in Example 1, you can enter the statistics in a TI-84 Plus, as shown at the left, and find P ≈ 0.379. Because P 7 a, you fail to reject the null hypothesis. Note that when using technology, the number of degrees of freedom for the t@test is often determined by the formula d.f. = 1s2 1 n1 + s 2 2 n22 2 1s2 1 n12 2 1n 1 - 12 + 1s 2 2 n22 2 1n 2 - 12 . This formula will not be used in the text. See Minitab steps on page 464. EXAMPLE 1 TI-84 PLUS μ1≠μ2 t=0.9224141169 p=0.37924039 df=9.458685946 x1=473 x2=459 2-SampTTest
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