chapter 1

Chapter 14: The Analysis of Categorical Datac. We will test to see if the average score on a controlled word association test is the samefor soccer and non-soccer athletes. H 0 : µ 1 = µ 2 vs H a : µ 1 ? µ 2 . We’ll use test statistic( x − x )21 2 s1 t = . With = 3. 2062 2s1s m2+m n( 37.50 − 39.63)t == −.95. The df =3.206 + 1.8542snand 2= 1. 854 ,( 3.206 + 1.854)3.2062521.854+5522≈ 56. The p-value willbe > .10, so we do not reject H 0 and conclude that there is no difference in the averagescore on the test for the two groups of athletes.d. Our hypotheses for ANOVA are H 0 : all means are equal vs H a : not all means are equal.MSTrf = .MSE22SSTr = 91(.30 − .35) + 96(.49 −.35)+ 53(.19 −.35)3.4659MSTr = = 1.732952222= 90(.67) + 95(.87) + 52(.48) = 124.2873124.28731.73295MSE = = .5244 . Now, f = = 3. 30237.5244The test statistic isSSE and2= 3.4659. Using df 2,200 fromtable A.9, the p value is between .01 and .05. At significance level .05, we reject the nullhypothesis. There is sufficient evidence to conclude that there is a difference in theaverage number of prior non-soccer concussions between the three groups.48.a. H o : p 0 = p 1 = … = p 9 = .10 vs H a : at least one p i ? .10, with df = 9.b. H o : p ij = .01 for I and j= 1,2,…,9 vs H a : at least one p ij ? 0, with df = 99.c. For this test, the number of p’s in the Hypothesis would be 10 5 = 100,000 (the number ofpossible combinations of 5 digits). Using only the first 100,000 digits in the expansion,the number of non-overlapping groups of 5 is only 20,000. We need a much largersample size!d. Based on these p-values, we could conclude that the digits of p behave as though theywere randomly generated.456