Lab 4 ANCOVA and Two-Way ANOVA_SR_rev

1 1. Submit answers to the Google form by 11:20a.m. on Mar. 17, 2022: https://forms.gle/mrotGVoHofCtGHpU8 LAB 4 ANCOVA and Two-Way ANOVA Review ANCOVAmodel : y ij = μ + α i + b x ij + ϵ ij . Represented using regression, the model is: y = θ 0 + θ 1 d 1 + ⋯ + θ a − 1 d a − 1 + θ k x + ϵ 1 . In the regression model, the intercept, θ 0 , costs 1 degree of freedom. The covariate, x, costs 1 degree of freedom to estimate θ k . Factor A costs a-1 degree of freedom. This is because if there are “a” groups, there will be a-1 dummy codes. Consequently, we need to estimate a-1 regression coefficients for factor A, and the degree of freedom for factor A is a-1. Finally, the degree freedom for error, ϵ , is N-1-(a-1)-1= N-a-1. For sum of squares, it is easy to calculate the sum of squares of regression residual (variance of the residual *(N-1)), which equals to the within group sum of squares. To calculate sum of squares for Factor A, the most intuitive way is to first estimate the residual sum of squares for this regression: y = γ 0 + γ 1 x + ϵ 2 . Then, SS A = SS ϵ 2 − SS ϵ 1 . This is the intuitive way to calculate sum of squares, it is mathematically equivalent to the formula you learned during the lecture. Type 3 sum of squares is also calculated in a similar way. The two-way ANOVA model is: Two way ANOVA model : y ij = μ + α + β + αβ + ε . In regression terms, suppose factor A has 3 groups and factor B has 3 groups, the model looks like: y = c 0 + c 1 da 1 + c 2 da 2 + c 3 db 1 + c 4 db 2 + c 5 da 1 db 1 + c 6 da 1 db 2 + c 7 da 2 db 1 + c 8 da 2 db 2 + ε . For degree of freedom, the intercept, c 0 , costs 1 degree of freedom. Factor A costs 3-1=2 degree of freedom to estimate c 1 and c 2 . Similarly, factor B costs 2 degree of freedom to estimate c 3 and c 4 . The interaction costs 2*2=4 degree of freedom. The error costs N-1-2-2-4 degree of freedoms. More generally, Factor A costs a-1 df, Factor B costs b-1 df, interaction costs (a-1)(b-1) df, residual costs N-1- (a-1)-(b-1)-(a-1)(b-1)=N-1-a+1-b+1-ab+a+b-1=N-ab. Sum of squares for factor A, B and interaction can be calculated in a similar way like in ANCOVA.

2 ANCOVA Exercise A researcher was interested in studying the effect of a drug on dementia patients’ memory. Thirty participants were randomly assigned to three groups: placebo, low dose, and high dose. Participants’ memory was measured using a memory test that scored from 0 to 10. The test was administered before and after the drug treatment. The data is presented in ‘ ANCOVA data.sav ’. Using what you have learned in the previous lab, conduct an ANOVA test with dose as the independent variable, and pre memory score as the dependent variable. A. Report the results. Does dose have a significant effect on pre memory score? F(2,27)=1.979m p=.158. No pre-memory score is not significant Conduct another ANOVA with dose as the independent variable, and post memory score as the dependent variable, report the results. B. Does dose have a significant effect on post memory score? What can you conclude? F(2,27)=2.416, p=.108. No, also not significant (p> .05) C. Why is this approach problematic? Because now it seems both the pre-test and post- test do not have a significant relationship with the independent variable, dosage Now, conduct an ANCOVA test by clicking Analyze -> General Linear Model -> Univariate, put post_score into the dependent variable box, put dose in the fixed factor(s) box, and put pre_score into the Covariate(s) box, click EM Means, put dose to the “Display Means for” box, check compare main effects, select LSD(none), Continue, Options, check Descriptive statistics, parameter estimates, Homogeneity tests, Continue, OK. [PASTE Levene’s Test of Equality of Error Variances here] D. Based on the Levene’s test, is the homogeneity of variance assumption satisfied?

4 *. The mean difference is significant at the .05 level. b. Adjustment for multiple comparisons: Least Significant Difference (equivalent to no adjustments). G. Based on pairwise comparisons , which groups are significantly different? Placebo & Low Dose, Placebo & High Dose ANCOVA also assumes Homogeneity of regression, which suggests that the regression slope between covariate and dependent variables are the same across groups. To check this assumption, click analyze, general linear model, Univariate. Since we just conducted an ANCOVA, we can keep the rest of the settings, click Model, Build custom terms, move/drag dose and pre_score to the model box, then use the down arrow to move Dose into Build Term, click By *, select pre_score under Factors & Covariates, move pre_score into Build Term, and move them both together to the model box by clicking “Add.” This will create the interaction term . Continue, OK. (*Hint: See screenshots on the next page.)

5 Tests of Between-Subjects Effects Dependent Variable: post_score Source Type III Sum of Squares df Mean Square F Sig. Corrected Model 52.346 a 5 10.469 4.286 .006 Intercept 53.542 1 53.542 21.921 .000 Dose 36.558 2 18.279 7.484 .003 pre_score 17.182 1 17.182 7.035 .014 Dose * pre_score 20.427 2 10.213 4.181 .028 Error 58.621 24 2.443 Total 683.000 30 Corrected Total 110.967 29 a. R Squared = .472 (Adjusted R Squared = .362) If there is a significant interaction between dose and pre_score, it suggests that the assumption of homogeneity of regression is violated. H. Report the results.

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8 ANOVA a Model Sum of Squares df Mean Square F Sig. 1 Regression 6.734 1 6.734 1.809 .189 b Residual 104.232 28 3.723 Total 110.967 29 2 Regression 31.920 3 10.640 3.500 .030 c Residual 79.047 26 3.040 Total 110.967 29 a. Dependent Variable: post_score b. Predictors: (Constant), pre_score c. Predictors: (Constant), pre_score, d2, d1 J. What are the sum of squares residuals for model 1 and model 2? What is their difference? What does it equal/represent in the ANCOVA model? Model 1: R2=104.232, F(1,29)=1.809, p=.189. Model 2: 79.047, F(3,29)=3.500, p=.030. Difference = 25.185.

10 [Dose=2] -.439 .811 -.541 .593 -2.107 1.228 [Dose=3] 0 a . . . . . pre_scor e .416 .187 2.227 .035 .032 .800 a. This parameter is set to zero because it is redundant. Two-Way ANOVA Exercise An anthropologist was interested in the effects of alcohol on mate selection at nightclubs. Her rationale was that after alcohol had been consumed, subjective perceptions of physical attractiveness would become more inaccurate (the well-known ‘beer-goggles effect’). She was also interested in whether this effect was different for men and women. She picked 48 students: 24 male and 24 female. She then took groups of eight participants to a night-club and gave them no alcohol (participants received placebo drinks of alcoholfree lager), 2 pints of strong lager, or 4 pints of strong lager. At the end of the evening she took a photograph of the person that the participant was chatting up. She then got a pool of independent judges to assess the attractiveness of the person in each photograph (out of 100). The data are in Table 12.1 and Goggles.sav To conduct Factorial ANOVA, click Analyze, general linear model, univariate, put attractiveness into the dependent variable box, put gender and alcohol into the fixed factor box, click plots, move gender to the separate lines box, move alcohol to the horizontal axis box, add, continue, click EM Means, move all variables to the display means for box, check compare main effects, Continue, Options, check descriptive statistics, estimates of effect size, parameter estimates, and homogeneity tests box, continue, Paste. In the syntax window, after the line /EMMEANS=TABLES(Gender*Alcohol), paste COMPARE(Gender) ADJ(LSD) as shown below. The file should look like this: UNIANOVA Attractiveness BY Gender Alcohol /METHOD=SSTYPE(3) /INTERCEPT=INCLUDE /PLOT=PROFILE(Alcohol*Gender) /EMMEANS=TABLES(Gender) COMPARE ADJ(LSD) /EMMEANS=TABLES(Alcohol) COMPARE ADJ(LSD) /EMMEANS=TABLES(Gender*Alcohol) COMPARE(Gender) ADJ(LSD) /PRINT=ETASQ DESCRIPTIVE PARAMETER HOMOGENEITY /CRITERIA=ALPHA(.05) /DESIGN=Gender Alcohol Gender*Alcohol.

11 This modification will allow SPSS to do post hoc comparison for interaction by gender. Select (highlight with your mouse) all the above syntax. Click the green triangle icon to run the syntax. Assume that Levene’s test is not significant, F(5,42)=1.527, p=.202, so the assumption of homogeneity of variance was not violated. Tests of Between-Subjects Effects Dependent Variable: Attractiveness of Date Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared Corrected Model 5479.167 a 5 1095.833 13.197 .000 .611 Intercept 163333.333 1 163333.333 1967.025 .000 .979 Gender 168.750 1 168.750 2.032 .161 .046 Alcohol 3332.292 2 1666.146 20.065 .000 .489 Gender * Alcohol 1978.125 2 989.062 11.911 .000 .362 Error 3487.500 42 83.036 Total 172300.000 48 Corrected Total 8966.667 47 a. R Squared = .611 (Adjusted R Squared = .565) L. Report the factorial ANOVA results. Which effects were significant? Alcohol, F(2,42)=20.065, p<.001). Gender*Alcohol, F(2,42)=11.911, p<.001) Parameter B Std. Error t Sig. Partial Eta Squared Intercept 57.500 3.222 17.848 .000 .884 [Gender=0] -21.875 4.556 -4.801 .000 .354 [Gender=1] 0 a . . . . [Alcohol=1] 3.125 4.556 .686 .497 .011 [Alcohol=2] 5.000 4.556 1.097 .279 .028 [Alcohol=3] 0 a . . . . [Gender=0] * [Alcohol=1] 28.125 6.443 4.365 .000 .312 [Gender=0] * [Alcohol=2] 26.250 6.443 4.074 .000 .283 [Gender=0] * [Alcohol=3] 0 a . . . . [Gender=1] * [Alcohol=1] 0 a . . . .

13 Since the interaction effect was significant, main effects were less meaningful. N. Interpret the interaction effect based on plot and the post hoc comparison. There was a significant interaction effect