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ANOVA Practice MA 223 1. (a) The ANOVA table is Source Factor Error Total df 3 16 19 SS 237.49 54.02 291.51 MS 79.16 3.38 F 23.45 p 0.000 (b) Soundly reject H0 : τ1 = τ2 = τ3 = τ4 = 0 at signiﬁcance level α = 0.05. We can use M SE as an estimate of σ 2 . For notational convenience let µk = µ + τk ; for each k (each column/treatment mean) the quantity x̄k,· − µk t= √ M SE /5 should follow a t-distribution with 16 df. Thus, for example, for the ﬁrst treatment mean (k = 1) we have x̄1,· = 46.08. This means a 95 percent conﬁdence interval is given by (note M SE /5 = 0.676) 46.08 − µ1 −2.12 < √ < 2.12 0.676 which leads to 44.3 < µ1 < 47.8. Similar computations lead to 49.2 < µ2 < 52.7, 40.4 < µ3 < 43.9, and 41.3 < µ4 < 44.8. (c) Prototypes 3 and 4 look best. In this case we should have that x̄3· − x̄4· − (µ3 − µ4 ) √ 2M SE /m t= follows a t-distribution with 16 df. Fill in the columns averages and M SE to ﬁnd that a 95 percent conﬁdence interval on the diﬀerence is given by −2.12 < −0.9 − (µ3 − µ4 ) < 2.12 1.163 leading to −3.37 < µ3 − µ4 < 1.57. Thus the two lowest kicking chains are not clearly diﬀerent. 2. (a) The null hypothesis is H0 : τ1 = τ2 = τ3 = 0, versus the alternative Ha : at least one τi ̸= 0. (b) We’ll use an F statistic with 2 df in the numerator and 9 df in the denominator. The critical value will be F > 4.26. (c) The ANOVA table looks like Source Factor Error Total df 2 9 11 SS 1.1816 0.305 1.487 MS 0.5908 0.0339 F 17.43 p ¡0.001 We reject the null hypothesis. (d) We assume that in the model xij = µ + τi + eij the eij are IID, normal with zero mean and common variance. 3. The ANOVA table from Minitab is Source Factor Error Total df 3 36 39 SS 5152 8348 13499 MS 1717 232 F 7.41 p ¡0.001 We reject H0 . The data easily passes the equality of variances test in Minitab (p around 0.95) and the normality test. 1 Steel wire made by four manufacturers is to be tested and compared. Ten random samples are drawn from a batch from each manufacturer and tested for strength. The results are in the ﬁle wire strength.mtw that I sent you. Test the hypothesis that there is no diﬀerence in the wire strength among the manufacturers. Also, assess the assumptions of normality and equality of variance in the data. 2