SECTION 8.2 Testing the Difference Between Means (Independent Samples, s1 and s2 Unknown) 431 1 − = 0.95 α= 0.05 α t 0 1 2 3 −2 −3 t ≈ −1.930 t0 = −1.671 −1 Sample Statistics for Sedan Driving Costs Manufacturer Competitor x1 = $0.52 mi s1 = $0.05 mi n1 = 30 x2 = $0.55 mi s2 = $0.07 mi n2 = 32 A Two-Sample t-Test for the Difference Between Means An automobile manufacturer claims that the mean driving cost per mile of its sedans is less than that of its leading competitor. You conduct a study using 30 randomly selected sedans from the manufacturer and 32 from the leading competitor. The results are shown at the left. At a = 0.05, can you support the manufacturer’s claim? Assume the population variances are equal. (Adapted from American Automobile Association) SOLUTION Note that s1 and s2 are unknown, the samples are random and independent, and both n1 and n2 are at least 30. So, you can use the t@test. The claim is “the mean driving cost per mile of the manufacturer’s sedans is less than that of its leading competitor.” So, the null and alternative hypotheses are H0: m1 Ú m2 and Ha: m1 6 m2. (Claim) The population variances are equal, so d.f. = n1 + n2 - 2 = 30 + 32 - 2 = 60. Because the test is a left-tailed test with d.f. = 60 and a = 0.05, the critical value is t0 = -1.671. The rejection region is t 6 -1.671. To make the calculation of the standardized test statistic easier, first find the standard error. sx 1 -x2 = B1n1 - 12s1 2 + 1n 2 - 12s2 2 n1 + n2 - 2 # A1 n1 + 1 n2 = B130 - 1210.0522 + 132 - 1210.0722 30 + 32 - 2 # A1 30 + 1 32 ≈ 0.0155416 The standardized test statistic is t = 1x1 - x22 - 1m1 - m22 sx 1 -x2 Use the t@test (variances are equal). ≈ 10.52 - 0.552 - 0 0.0155416 Assume m1 = m2, so m1 - m2 = 0. ≈ -1.930. Round to three decimal places. The figure at the left shows the location of the rejection region and the standardized test statistic t. Because t is in the rejection region, you reject the null hypothesis. Interpretation There is enough evidence at the 5% level of significance to support the manufacturer’s claim that the mean driving cost per mile of its sedans is less than that of its competitor’s. TRY IT YOURSELF 2 An automobile manufacturer claims that the mean driving cost per mile of its minivans is less than that of its leading competitor. You conduct a study using 34 randomly selected minivans from the manufacturer and 38 from the leading competitor. The results are shown at the right. At a = 0.10, can you support the manufacturer’s claim? Assume the population variances are equal. (Adapted from American Automobile Association) Answer: Page A41 See TI-84 Plus steps on page 465. EXAMPLE 2 Sample Statistics for Minivan Driving Costs Manufacturer Competitor x1 = $0.57 mi s1 = $0.09 mi n1 = 34 x2 = $0.59 mi s2 = $0.08 mi n2 = 38 Tech Tip It is important to note that when using a TI-84 Plus for the two-sample t-test, select the Pooled: Yes input option when the variances are equal.
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