SECTION 10.4 Analysis of Variance 561 Performing a One-Way ANOVA Test A medical researcher wants to determine whether there is a difference in the mean lengths of time it takes three types of pain relievers to provide relief from headache pain. Several headache sufferers are randomly selected and given one of the three medications. Each headache sufferer records the time (in minutes) it takes the medication to begin working. The results are shown in the table. At a = 0.01, can you conclude that at least one mean time is different from the others? Assume that each population of relief times is normally distributed and that the population variances are equal. Medication 1 Medication 2 Medication 3 12 16 14 15 14 17 17 21 20 12 15 15 19 n1 = 4 n2 = 5 n3 = 4 x1 = 56 4 = 14 x2 = 85 5 = 17 x3 = 66 4 = 16.5 s2 1 = 6 s 2 2 = 8.5 s 2 3 = 7 SOLUTION The null and alternative hypotheses are as follows. H0: m1 = m2 = m3 Ha: At least one mean is different from the others. (Claim) Because there are k = 3 samples, d.f.N = k - 1 = 3 - 1 = 2. The sum of the sample sizes is N = n1 + n2 + n3 = 4 + 5 + 4 = 13. So, d.f.D = N - k = 13 - 3 = 10. Using d.f.N = 2, d.f.D = 10, and a = 0.01, the critical value is F0 = 7.56. The rejection region is F 7 7.56. To find the test statistic, first calculate x, MSB, and MSW. x = Σx N = 56 + 85 + 66 13 ≈ 15.92 MSB = SSB d.f.N = Σni(xi - x)2 k - 1 ≈ 4114 - 15.9222 + 5117 - 15.9222 + 4116.5 - 15.9222 3 - 1 = 21.9232 2 = 10.9616 MSW = SSW d.f.D = Σ1ni - 12s 2 i N - k = 14 - 12162 + 15 - 1218.52 + 14 - 12172 13 - 3 = 73 10 = 7.3 EXAMPLE 1
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