SECTION 10.2 Independence 539 Performing a Chi-Square Independence Test In Words In Symbols 1. Verify that the observed frequencies were obtained from a random sample and each expected frequency is at least 5. 2. Identify the claim. State the null and State H0 and Ha. alternative hypotheses. 3. Specify the level of significance. Identify a. 4. Determine the degrees of freedom. d.f. = 1r - 121c - 12 5. Determine the critical value. Use Table 6 in Appendix B. 6. Determine the rejection region. 7. Find the test statistic and sketch the x 2 = Σ1O - E22 E sampling distribution. 8. Make a decision to reject or fail to If x 2 is in the rejection reject the null hypothesis. region, then reject H0. Otherwise, fail to reject H0. 9. Interpret the decision in the context of the original claim. GUIDELINES Performing a Chi-Square Independence Test The contingency table shows the results of a random sample of 1984 undergraduate students classified by the student’s living arrangement and family college experience. The expected frequencies, calculated in Example 1, are displayed in parentheses. At a = 0.01, can you conclude that the variables student’s living arrangement and family college experience are related? Student’s living arrangement Family college experience With parents, rent free With parents, pay rent On campus Off campus, w/others Off campus, alone Total First in family 190 (163.622) 39 (28.794) 97 (102.142) 43 (76.266) 17 (15.175) 386 2nd generation 651 (677.378) 109 (119.206) 428 (422.858) 349 (315.734) 61 (62.825) 1598 Total 841 148 525 392 78 1984 SOLUTION Because each expected frequency is at least 5 and the adults were randomly selected, you can use the chi-square independence test to determine whether the variables are independent. Here are the null and alternative hypotheses. H0: The student’s living arrangement is independent of family college experience. Ha: The student’s living arrangement depends on family college experience. (Claim) EXAMPLE 2 Study Tip For instance, a contingency table with three rows and four columns will have 13 - 1214 - 12 = 122132 = 6 d.f.
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