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UW-Madison PSYCH 610 - PSYCH 610 Handout 30

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Psych 610 Handout #30, p. 1 Prof. Moore 1. Example from Kirk, pp. 353 ff. Latin Square “Crossover Design” A S Totals _ YPA A1 A2 A3 S1 B1 7 B2 14 B3 12 33 11.0 S2 B1 3 B3 11 B2 5 19 6.33 S3 B2 7 B3 11 B1 6 24 8.0 S4 B3 9 B1 12 B2 13 34 11.33 S5 B2 9 B1 7 B3 8 24 8.0 S6 B3 9 B2 13 B1 8 30 10.0 Totals A = 44 68 52 T=164 _ YT=9.11 _ YAj 7.33 11.33 8.67 [Y] = ∑Y2 = 1652 [A] = 9264/6 = 1544 [T] = 1642/18 = 1494.22 [B] = 9170/6 = 1528.33 [S] = 4658/3 = 1552.67 B1 B2 B3 7 3 12 7 6 8 7 9 14 13 5 13 9 9 11 11 12 8 B = 43 61 60 _ YBk 7.17 10.17 10.00Handout #30, p. 2 ANOVA Table Source df SS MS F Mean 1 [T] = 1494.22 1494.22 773.99 A a-1 = 2 [A] – [T] = 49.78 24.89 12.89 p < .01 B b-1 = 2 [B] – [T] = 34.11 17.06 8.83 p < .01 Subjects n-1 = 5 [S] – [T] = 58.44 11.69 Error (ℓ-1)(n-2) = 15.44 1.93 = 8 or “residual [Y] - [A] - [B] - [S] df” + 2[T], or “residual” Total Total [Y] = 1652 obs = 18 ℓn = (3)(6) Let ℓ = # levels in the Latin Square In a Latin Square, a = b = ℓHandout #30, p. 3 Linear model: Population: Yijk = µ + αj + βk + πi + εijk where αj = µj - µ and ∑αj = 0 βk = µk - µ and ∑βk = 0 πi = µi - µ and ~ N(0, σπ2) εijk = Yijk - µ - αj – βk – πi εijk ~ N(0, σe2) Sample: ! Yijk= YT+ (YAj" YT) + (YBk" YT) + (YPi" YT) +#ijk ! ˆ " j= YAj# YT ! ˆ " k = YBk# YT ! "ijk= Yijk# YAj# YBk# YPi+


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