Statistics Lab #7 ANOVA Part 2 & ANCOVA
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1 Statistics Lab #7 ANOVA Part 2 & ANCOVA PSYCH Initialize R Initialize R by entering the following commands at the prompt. You must type the commands exactly as shown. options(contrasts=c("contr.sum","contr.poly") ) # set definition of contrasts load(url(" ) load(url(" ) goats <- read.csv(file=url(" ) closeallconnections() 7.1 ab data An experiment was done to measure the effects of treatment A and treatment B on a dependent variable, y, using a crossed-factorial design. The data are stored in the dataframe L5.dat.1. Tasks: 1. Conduct an ANOVA that evaluates the effects of A and B on y. Explain your results. 2. Interpreting interactions: Example: Before computing the simple main effects, it might be helpful to graph the data. The following command created Figure 1. with(l5.dat.1,interaction.plot(b,a,y)) Example: The following code calculates the simple main effect of A at b1: MS.resid < # from main anova df.resid <- 30 # from main anova levels(l5.dat.1$b) # levels of B [1] "b1" "b2" "b3" B.b1 <- subset(l5.dat.1,b=="b1") # get subset of data aov.a.at.b1 <- aov(y~a,data=b.b1); # one-way anova of A at b1 summary(aov.a.at.b1) # anova table Df Sum Sq Mean Sq F value Pr(>F) A Residuals (F.A.b1 < /MS.resid) # recalculate F 1
2 A mean of y a1 a2 b1 b2 b3 B Figure 1: Interaction plot of the AB data. 2
3 [1] (1-pf(F.A.b1, df1=1, df2=df.resid)) # recalculate p [1] Task: Calculate the simple main effect of B at each level of A. 3. Pairwise comparisons: Example: The following code shows how to use TukeyHSD to do pairwise comparisons of each level on factor B. Note that the functions computes 90% adjusted confidence intervals, and the familywise α therefore is 0.1: ab.aov.01 <- aov(y~a + B + A:B,data=L5.dat.1) # the anova TukeyHSD(ab.aov.01,which="B",conf.level=0.90) Tukey multiple comparisons of means 90% family-wise confidence level Fit: aov(formula = y ~ A + B + A:B, data = L5.dat.1) $B diff lwr upr p adj b2-b b3-b b3-b Task: Use TukeyHSD to evaluate all pairwise comparisons of cell means while maintaining a familywise α = Are all of these comparisons equally interesting? 4. Calculate Cohen s f for the A B interaction. (For more information about calculation Cohen s f in a factorial design, see Section 7.8 in the notes.) 7.2 cd data This section, in which we analyze data from an unbalanced design, draws on material in Sections in the course notes. An experiment was done to measure the effects of treatment C and treatment D on a dependent variable, y, using a crossed-factorial design. Six subjects were assigned randomly to each condition, however the data from two subjects in one of the conditions were lost. The data are stored in the dataframe L5.dat Verify that the CD data are unbalanced. 2. Verify that the results of the two-way ANOVA depend on the order of the terms in the full linear model. 3. Your book defines Type I sums of squares as the sums of squares associated with one main effect when all other variables are ignored. According to this definition, what are the Type I sums of squares for C and D? What null hypotheses about the main effects are being evaluated with these Type I sums of squares? 3
4 4. What are the Type II sums of squares for C and D? Use Type II sums of squares to evaluate the main effects of C and D. 5. Use drop1 to compute the Type III sums of squares for C and D. Verify that these sums of squares do not depend on the order of terms in the model. What null hypotheses about the main effects are being evaluated with these Type III sums of squares? (See Section in the course notes for an example of how to use drop1.) 7.3 goats The following text was taken from material posted at Experiments were carried out on six commercial goat farms to determine whether the standard worm drenching program was adequate. Forty goats were used in each experiment. Twenty of these, chosen completely at random, were drenched according to the standard program, while the remaining twenty were drenched more frequently. The goats were individually tagged, and weighed at the start and end of the year-long study. For the first farm in the study the resulting liveweight gains are given along with the initial liveweights. In each experiment the main interest was in the comparison of the liveweight gains between the two treatments. The data from one of these experiments was taken from and is stored in the data frame goats. Values of baseline (i.e., pre-treatment) weight and weight gain are stored in the variables wt and gain, respectively. Type of worm drenching standard and intensive is stored in the variable treatment linear regression The following code uses linear regression to evaluate the linear relation between gain and wt: goats.lm.01 <- lm(gain~wt,data=goats) summary(goats.lm.01) Call: lm(formula = gain ~ wt, data = goats) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-09 *** wt e-05 *** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: on 38 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 38 DF, p-value: 6.681e-05 The summary function prints a regression table, which the regression coefficient, or β, for wt as -0.35, which differs significantly from zero (t(38) = 4.47, p <.0001). This regression coefficient means that for every increase of 1 in wt, the value of gain decreases (on average) by The intercept is 14.39, and the t test 4
5 shows that it differs significantly from zero. Together, the intercept and regression coefficient define a straight line gain = intercept wt (1) The overall fit of the regression model, as indexed by R 2 = 0.34, also is significant, F (1, 38) = 20.05, p < R 2 is evaluated by noting the change in the goodness-of-fit that occurs when all of the parameters except the intercept are set to zero. In this case, there is only one parameter the regression coefficient for wt and so you would think that the overall F test should be related to the t test for wt s β value... and you would be correct. In the case of a regression model that has only one regression coefficient, the overall F equals t 2 : F = t 2 = = The following code uses plot to create a scatter plot of gain vs. wt and then uses the function abline to add the regression line defined by Equation 1. The resulting graph is shown in Figure 2. with(goats,plot(wt,gain,"p",xlab="wt",ylab="gain")) abline(goats.lm.01) gain wt Figure 2: Plot of weight gain as a function of pre-treatment weight. The solid line shows the regression line ANOVA & ANCOVA In this section we will use ANOVA and ANCOVA to evaluate the effect of treatment. 5
6 1. Conduct an ANOVA to evaluate the effect of treatment on gain. Calculate the strength of association between treatment and gain. 2. Conduct an analysis of covariance (ANCOVA) that evaluates the effect of treatment after controlling for the linear association between wt and gain. 3. Use the function coef to examine the coefficients, or parameters, of the ANCOVA model. What are the values of the parameters (i.e., the α s) for the two levels of treatment. Explain what the coefficients for the ANCOVA model mean. 4. Evaluate the ANCOVA s homogeneity of slopes assumption. 5. Calculate the adjusted means for the two treatment conditions. 6. Use TukeyHSD to evaluate each pairwise difference between adjusted means. (N.B. There are only two groups, so a Tukey test obviously is unnecessary in this case. Our purpose here is to show that the command works with an ANCOVA model.) 7. Calculate the strength of association between treatment and gain using omega-squared and partial omega-squared. (See Section 9.3 of the course notes.) (a) What is the difference between these two measures of association strength? (b) How do these association strengths compare to the association strength estimated from the ANOVA in question 1? 6
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