SECTION 10.2 Independence 541 Using Technology for a Chi-Square Independence Test A health club manager wants to determine whether the number of days per week that adults exercise is related to demographic cohort, specifically generation. A random sample of 275 adults is selected and the results are classified as shown in the table. At a = 0.05, is there enough evidence to conclude that the number of days an adult exercises per week is related to generation? Number of days of exercise per week Generation 0 –1 2–3 4 –5 6 –7 Total Gen Z (ages 18–24) 40 53 26 6 125 Millennial (ages 25– 40) 34 68 37 11 150 Total 74 121 63 17 275 SOLUTION Here are the null and alternative hypotheses. H0: The number of days of exercise per week is independent of generation. Ha: The number of days of exercise per week depends on generation. (Claim) Because d.f. = 3 and a = 0.05, the critical value is x 2 0 = 7.815. So, the rejection region is x 2 7 7.815. Using Minitab (see below), the test statistic is x 2 ≈ 3.493. Because x 2 ≈ 3.493 is not in the rejection region, you fail to reject the null hypothesis. MINITAB Chi-Square Test for Association: Generation, Number of days of exercise Rows: Generation Columns: Number of days of exercise 0 to 1 2 to 3 4 to 5 6 to 7 All Gen Z 40 53 26 6 125 Millennial 34 68 37 11 150 All 74 121 63 17 275 Cell Contents Chi-Square DF P-Value Count Pearson 3.493 3 0.322 Interpretation There is not enough evidence to conclude that the number of days a student exercises per week is related to generation. TRY IT YOURSELF 3 A researcher wants to determine whether age is related to whether or not a tax credit would influence an adult to make a charitable donation. A random sample of 1250 adults is selected and the results are classified as shown in the table. At a = 0.01, is there enough evidence to conclude that age is related to the response? Age Response 18 –34 35 –54 55 and older Total Yes 257 189 143 589 No 218 261 182 661 Total 475 450 325 1250 Answer: Page A42 EXAMPLE 3 Study Tip You can also use a P@value to perform a chi-square independence test. For instance, in Example 3, note that Minitab displays P = 0.322. Because P 7 a, you fail to reject the null hypothesis.
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