SECTION 10.1 Goodness-of-Fit Test 529 Tax preparation method Observed frequency Expected frequency Accountant 60 72 By hand 43 60 Computer software 117 105 Friend or family 29 18 Tax preparation service 51 45 5 10 15 20 25 0 2χ 2χ = 13.277 ≈ 15.710 α= 0.01 Rejection region 2χ Performing a Chi-Square Goodness-of-Fit Test A retail trade association claims that the tax preparation methods of adults are distributed as shown in the table at the left below. A tax preparation company randomly selects 300 adults and asks them how they prepare their taxes. The results are shown in the table at the right below. At a = 0.01, test the association’s claim. (Adapted from National Retail Federation) Distribution of tax preparation methods Accountant 24% By hand 20% Computer software 35% Friend or family 6% Tax preparation service 15% Survey results 1n = 3002 Accountant 60 By hand 43 Computer software 117 Friend or family 29 Tax preparation service 51 SOLUTION The observed and expected frequencies are shown in the table at the left. The expected frequencies were calculated in Example 1. Because the observed frequencies were obtained using a random sample and each expected frequency is at least 5, you can use the chi-square goodness-of-fit test to test the proposed distribution. Here are the null and alternative hypotheses. H0: The expected distribution of tax preparation methods is 24% by accountant, 20% by hand, 35% by computer software, 6% by friend or family, and 15% by tax preparation service. (Claim) Ha: The distribution of tax preparation methods differs from the expected distribution. Because there are 5 categories, the chi-square distribution has d.f. = k - 1 = 5 - 1 = 4 degrees of freedom. With d.f. = 4 and a = 0.01, the critical value is x 2 0 = 13.277. The rejection region is x 2 7 13.277. Rejection region With the observed and expected frequencies, the chi-square test statistic is x 2 = Σ1O - E22 E = 160 - 7222 72 + 143 - 6022 60 + 1117 - 10522 105 + 129 - 1822 18 + 151 - 4522 45 ≈ 15.710. The figure at the left shows the location of the rejection region and the chi-square test statistic. Because x 2 is in the rejection region, you reject the null hypothesis. Interpretation There is enough evidence at the 1% level of significance to reject the claim that the distribution of tax preparation methods and the association’s expected distribution are the same. EXAMPLE 2
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