Psych 610 Handout #24, p. 1Prof. MooreContrasts, Comparisons and Simple Effects on Repeated Measures DesignsI. Alternate method for computing single df contrasts on repeated measuresData A1 A2 A3Sl 13 24 22S2 6 30 29S3 25 13 23S4 20 16 25S5 25 37 16S6 19 30 12A. Suppose we want to compare A1 and A3 __ __1. First compute the contrast value ψ = (-1)(YA1) + (1)(YA3) for each subject. ∧ψS1 9 2. Compute [ψ] = ∑ψ2/1 = 769S2 23S3 -2 3. Compute [T] = 192/6 = 60.17S4 5S5 -9 4. Compute [ψ] – [T] = 708.83S6 -7∑ψ = 192. Put together as “one-group anova.” This partitions the total variance into the grandmean and error. Then we test H0: µ = 0.Source df EQ SS MS F Mean 1 [T] 60.17 60.17 .42Error 5 [ψ] – [T] 708.83 141.77 --Note that this procedure gives the correct F-ratio for Acomp , but the SSMean (60.17) andthe SSError (141.77) are on a different scale from the original analysis. If you need theactual values of the SS, you will need to divide each of these by n*∑cj2, where n* is thenumber of observations in the score to which the coefficients have been applied.Ordinarily, this will be 1, i..e., one observation per subject per cell. However, n* can begreater if the scores in the analysis are totals pooled over replications or pooled oversome other factors.For illustration, the normalized Ss for the present problem are:Handout #24, p. 2SSAcomp = SSmean/2 = 30.09SSSxAcomp = SSerror/2 = 354.42B. What about Keppel’s more “complex” contrast, A1 & A3 versus A2 ? Just use the sameprocedure exactly. __ __ __1. First compute (-1)(YA1) + (-2)(YA2) + (1)(YA3) for each subject. ∧ψ'S1 -13 2. Compute [ψ'] = ∑ψ2/1 = 3377S2 -25S3 22 3. Compute [T'] = -652/6 = 704.17S4 13S5 -33 4. Compute [ψ'] – [T'] = 2672.83S6 -29∑ψ' = -652. Put together as one-group anova and test H0: µ = 0.Source df EQ SS MS F Mean 1 [T'] 704.17 704.17 1.32Error 5 [ψ'] – [T'] 2672.83 534.57 --Note, again, the F ratio is correct, but to get back to the original sums of squares weneed to normalize by dividing by n*∑cj2.SSAcomp = 704.17/6 = 117.36SSSxAcomp = 2672.83/6 = 445.47C. Since Acomp and Acomp′ are a complete partition of A (and SxAcomp and SxAcomp′, are acomplete partition of SxA) we can check that the sum of the SS’s is what it should be:SSA = 30.09 + 117.36 = 147.45SSSxA = 354.42 + 445.47 = 799.89Handout #24, p. 3II. Two-way (and higher-way) contrasts on repeated measures.Consider Lopes’ (1976, JEP: General) poker experiment. The subject rates (scale of 1-10) theprobability of beating a pair of opposing stimulus hands (Hand 1 x Hand 2 design) with a pair ofsevens. Look at H1LINEAR x H2LINEAR.Hand 2StrongWeakStrongP1P2P3P4P512112P1P2P3P4P512212P1P2P3P4P52322333323Hand 1P1P2P3P4P512212223234543576667DataTableWeakP1P2P3P4P51212234433767871099109X-3-113-131-1-3000001-3-113Table ofCoefficientsof L x L ∧1. Make a table of ψ s:ψS1 24 2. Compute [ψ] = ∑ψ2/1 = 2444S2 19S3 21 3. Compute [T] = 1102/5 = 2420S4 25S5 21 4. Compute [ψ] – [T] = 24∑ψ = 110Handout #24, p. 45. Source df EQ SS MS F Mean 1 [T] 2420 2420 403.33Error 4 [ψ] – [T] 24 6If you want you may normalize by dividing the SSs by n*∑cj2 = 40.III. Partial Interactions for Repeated MeasuresNote – A new set of data begins here.A1A2A3A4B11211234433767871099109B224432744638754634546Suppose we wish to test the hypothesis that the linear effect for factor A differs at the two levelsof B (i.e., to test the ALINEAR x B partial interaction). C1 C2 Because there are only 2 levels B, this is same asψ@B1 ψ@B2 ALINEAR x BLINEAR.S1 31 4 Coefficients are: -3, -1, 1, 3S2 23 3S3 27 4 Notice that we have now a simple one-way designS4 32 1 with repeated measures. If there is a significantS5 25 15 partial interaction then there will be a significant138 27 difference between the two columns when testedagainst the S x Column interaction.Because this is a one df contrast, it can be doneas a one group anova by taking C1 – C2.1. Compute the basic ratios[T] = 1652/10 = 2722.5[SC] = 4135/1 = 4135.0[C] = 19773/5 = 3954.6Handout #24, p. 5[S] = 5551/ 2 = 2775.52. SSC = 3954.6 – 2722.5 – 1232.1SSSxC = 4135.0 – 3954.6 – 2775.5 + 2722.5 = 127.43. MSC = 1232.1/1 = 1232.1MSSxC = 127.4/4 = 31.854.68.3885.31/1.1232 ===xBACLINFFTo normalize, divide by n*∑cj2 = 20.V. Simple Main Effects and Simple Interaction EffectsSimply divide the data set according to the factor or factors being ignored, and run the regularrepeated measures ANOVA. This will give the correct analysis including error terms.For example, to compute the F ratio for A at B1, simply separate out the data for B1 and testfactor A as though it were a one-way design.Handout #24, p. 6Poker ExampleH2 @ H1 StrongSource df SS Partitions:____________________________________A = A + AxBMean 1 84.05 AxS = AxS + AxBxSSubjects 4 3.20A 3 6.55A x S 12 1.20 Mean: Mean + B____________________________________ S: S + BxSH2 @ H1 MediumSource df SS____________________________________Mean 1 266.45Subjects 4 3.30A 3 68.15A x S 12 3.10____________________________________H2 @ H1 WeakSource df SS____________________________________Mean 1 572.45Subjects 4 .80A 3 184.95A x S 12