SECTION 8.1 Testing the Difference Between Means (Independent Samples, s1 and s2 Known) 423 EXAMPLE 3 Sample Statistics for Monthly Rent for a One-Bedroom Apartment Seattle, WA Washington, DC x1 = $2166 n1 = 400 x2 = $2181 n2 = 300 Sample Statistics for Monthly Rent for a One-Bedroom Apartment Atlanta, GA Nashville, TN x1 = $1610 n1 = 200 x2 = $1603 n2 = 300 Using Technology to Perform a Two-Sample z-Test A rental market researcher claims that the average monthly rent for a one-bedroom apartment in Seattle, WA, is less than the average monthly rent for a one-bedroom apartment in Washington, DC. The table at the left shows the results of a random survey of people who rent apartments in each city. The two samples are independent. Assume that s1 = $190 for Seattle and s2 = $210 for Washington, and that both populations are normally distributed. At a = 0.01, is there enough evidence to support the claim? (Adapted from RentPath Holdings, Inc.) SOLUTION Note that s1 and s2 are known, the samples are random and independent, and the populations are normally distributed. So, you can use the z@test. The claim is “the average monthly rent for a one-bedroom apartment in Seattle, WA, is less than the average monthly rent for a one-bedroom apartment in Washington, DC.” So, the null and alternative hypotheses are H0: m1 Ú m2 and Ha: m1 6 m2 (claim). The left display shows how to set up the hypothesis test using a TI-84 Plus. The right displays show the results of selecting Calculate or Draw. Because the test is a left-tailed test and a = 0.01, the rejection region is z 6 -2.33. The standardized test statistic z ≈ -0.97 is not in the rejection region, so you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 1% level of significance to support the rental market researcher’s claim. TRY IT YOURSELF 3 A rental market researcher claims that the average monthly rent for a one-bedroom apartment in Atlanta, GA, is greater than the average monthly rent for a one-bedroom apartment in Nashville, TN. The table at the left shows the results of a random survey of people who rent apartments in each city. The two samples are independent. Assume that s1 = $210 for Atlanta and s2 = $190 for Nashville, and that both populations are normally distributed. At a = 0.05, is there enough evidence to support the claim? (Adapted from RentPath Holdings, Inc.) Answer: Page A41 Tech Tip Note that the TI-84 Plus displays P ≈ 0.1651. Because P 7 a, you fail to reject the null hypothesis. TI-84 PLUS 2-SampZTest Inpt:Data Stats s1:190 s2:210 x1:2166 n1:400 x2:2181 n2:300 μ1:≠μ2 <μ2 >μ2 Calculate Draw TI-84 PLUS 2-SampZTest μ1<μ2 z=-.9738412097 p=.1650676893 x1=2166 x2=2181 n1=400 TI-84 PLUS z=-0.9738 p=0.1651
RkJQdWJsaXNoZXIy NjM5ODQ=