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A Goodness-of-Fit Test: Specified Probabilities A goodness-of-fit test with specified probabilites, of significance level a, of H, against the alternative that the specified probabilities are not correct is based on the decision rule K (O; - E;) reject H, if Xx-1,0 X6 E; where xx-1, is the number for which P(xx-1 Xx-1,0) = a and the random variable xk-1 follows a chi-square distribution with K - 1 degrees of freedom. i=1 > To illustrate this test, consider again the data of Table 14.3 on brand selection. The null ypothesis is that the probabilities are the same for the three categories. The test of this ypothesis is based on the following: 3 (O; E;) (75 100) x = (110 100) (115 100) = 9.50 E; 100 100 100 here are three (K = 3) categories, so K - 1 = 2 degrees of freedom are associated with ne chi-square distribution. From Appendix Table 7, = 9.210 + + i=1 X2,0.01 herefore, according to our decision rule, the null hypothesis is rejected at the 1% signifi- cance level. These data contain strong evidence against the hypothesis that a randomly hosen subject is equally likely to select any of the three soft drink brands. Example 14.1 Do Customers Have a Preference for Any of Four Hershey Chocolate Bars? (Chi-Square) Suppose that the Hershey Company wants to determine if customers have a preference for any of the following four candy bars: A, Mr. Goodbar; B, Hershey's Milk Chocolate; C, Hershey's Special Dark Mildly Sweet Chocolate; or D, Krackel. From a random sam- ple of 200 people, it was found that 43 preferred Mr. Goodbar; 53 preferred Hershey's Milk Chocolate; 60 preferred Hershey's Special Dark Mildly Sweet Chocolate, and the remainder preferred Krackel. Test the null hypothesis that customers have no prefer- ence for any of the four candy bars against the alternative hypothesis that customers have a preference for at least one of the candy bars