Stat 5303 (Oehlert): Tukey One Degree of Freedom 1

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1 Stat 5303 (Oehlert): Tukey One Degree of Freedom 1 > catch<-c(19.1,50.1,123,23.4,166.1,407.4, 29.5,223.9,398.1,23.4,58.9,229.1,16.6,64.6,251.2) These data are average numbers of insects trapped for 3 kinds of traps used in 5 periods. Data from Snedecor and Cochran. > trap<-factor(rep(1:3,5)) Factors to indicate traps and periods. Replication is just n=1. > period<-factor(rep(1:5,each=3)) > fit1 <- lm(catch period+trap); anova(fit1) Here is the standard ANOVA. Because we have only a single replication, we have no estimate of pure error. Our error here is really interaction, so we might want to check to see if some of this could be reduced via transformation. Response: catch period trap *** Residuals > plot(fit1,which=1) OK, here is trouble. If the interaction can truly be ignored, this plot should just look like random noise. It doesn t. This curved shape suggests that a transformation could help. Residuals vs Fitted 1 Residuals Fitted values lm.default(catch ~ period + trap) > preds<-predict(fit1) To do Tukey One DF, start by getting the predicted values.

2 Stat 5303 (Oehlert): Tukey One Degree of Freedom 2 > coef(fit1) (Intercept) period1 period2 period3 period4 trap trap > rspv<-predsˆ2/2/coef(fit1)[1] Get the squares of the predicted values and rescale them by dividing by twice the mean. (The first coefficient is ˆµ.) > fit2 <- lm(catch period+trap+rspv);anova(fit2) Get the Tukey 1df test by doing an ANOVA with the squared predicted values as a third term after rows and columns. This is the simplest way to get a test for Tukey nonadditivity. In this case, preds2 is highly significant, so the Tukey test is indicating that interaction can be reduced through a transformation. Response: catch period ** trap e-06 *** rspv ** Residuals > plot(fit2,which=1) The residual plot is improved, but still not great. Residuals vs Fitted 8 Residuals Fitted values lm.default(catch ~ period + trap + rspv)

3 Stat 5303 (Oehlert): Tukey One Degree of Freedom 3 > summary(fit2) The coefficient of preds2 is 0.88, with a standard error of.17. Thus, a reasonable range for the coefficient is.54 to So a reasonable range for 1 minus the coefficient is -.22 up to.46. So log is in the range, as is a fourth root, (almost) a square root, and various others. The fact of the matter is that this is not always the best way to find the transformation power. A better way is to try a bunch of different powers and then take the one that minimizes the Tukey F-test. Here, the power that minimizes things is about -.5 (not shown). Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) period period period period trap trap rspv ** Residual standard error: on 7 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 7 and 7 DF, p-value: 4.122e-05 > boxcox(fit1) Box Cox also suggests a reciprocal square root. 95% log Likelihood λ

4 Stat 5303 (Oehlert): Tukey One Degree of Freedom 4 > inverseresponseplot(fit1) The inverse response plot is also supposed to transform to a better fit. It likes the regular square root. lambda RSS λ^: yhat catch > # So we now have three suggested transformations ranging from square root to reciprocal square root. Which one do we use? The proof of the pudding is in the tasting, and the proof of the transformation is in doing the transformation and then looking at the residuals. > fit3 <- lm(log(catch) period+trap);anova(fit3) Here we look at the log. Note that the error, which is also the interaction, is about 4% of total variation (it was about 12% on the original scale). Response: log(catch) period * trap e-06 *** Residuals

5 Stat 5303 (Oehlert): Tukey One Degree of Freedom 5 > plot(fit3,which=1) This is better than the original scale, but in no way could it be called good. Compare this to the residuals when we include rspv. Residuals vs Fitted Residuals Fitted values lm.default(log(catch) ~ period + trap) > fit4 <- lm(catchˆ-.5 period+trap);anova(fit4) Repeat with reciprocal square root. Now the error is about 0.3% of total variation, so it s the best one yet. Response: catchˆ-0.5 period * trap e-07 *** Residuals

6 Stat 5303 (Oehlert): Tukey One Degree of Freedom 6 > plot(fit4,which=1) This is the best we ve seen, but still not very good. Residuals vs Fitted Residuals Fitted values lm.default(catch^ 0.5 ~ period + trap) > fit5 <- lm(sqrt(catch) period+trap);anova(fit5) Repeat with square root. Now the error is about 7% of total variation, so not so good. Response: sqrt(catch) period * trap e-05 *** Residuals > # Let s look at what a Tukey style interaction looks like in residuals. We work from the original model. What these steps will do is to put the residuals in a matrix, and then order the rows and columns so that the row and column effects are in increasing order. > m <- matrix(residuals(fit1),3);m Here is the matrix of residuals. Note that this worked because the data were entered in a systematic order. If things were randomized we would need to work harder to associate treatments with their residuals in a matrix form. [,1] [,2] [,3] [,4] [,5] [1,] [2,] [3,] > p <- model.effects(fit1,"period");p

7 Stat 5303 (Oehlert): Tukey One Degree of Freedom 7 These are the period effects > op <- order(p);op This is the order in which we will need to sort the columns to put column effects in increasing order. [1] > t<-model.effects(fit1,"trap");t Trap effects > ot <- order(t);ot Order to put trap effects into increasing order. Well, they were already in increasing order, so this doesn t change anything. [1] > m[ot,op] Here are the residuals reordered into increasing row and column effect orders. Note the pattern of negatives and positives in opposite corners. This is the Tukey 1 df interaction pattern in residuals. It looks a lot like a linear by linear interaction (once the rows and columns have been properly ordered). [,1] [,2] [,3] [,4] [,5] [1,] [2,] [3,] > # What we re going to do now is to redo the Tukey using comparison values (in preparation for the Mandel models). > p <- model.effects(fit1,"period");allp <- p[period] As above, but now repeat these values as appropriate for levels of period. > t <- model.effects(fit,"trap");allt <- t[trap] Similarly for trap effects. > compvals <- allp*allt/coef(fit1)[1] These are the comparison values. > fit2b <- lm(catch period+trap+compvals) Refit the Tukey model with comparison values.

8 Stat 5303 (Oehlert): Tukey One Degree of Freedom 8 > summary(fit2b) Two points. First, the estimated coefficient using compvals is the same as using rspv. However, when you use compvals the coefficients for period and trap are the same as they were in fit1 (this is not true when using rspv). Call: lm.default(formula = catch period + trap + compvals) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-07 *** period ** period ** period ** period trap e-05 *** trap * compvals ** Residual standard error: on 7 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 7 and 7 DF, p-value: 4.122e-05 > fit.rowmodel <- lm(catch trap+period+period*allt) The distinction of rows versus columns is arbitrary, but here we we fit a different slope to trap effects separately for each period. > fit.rowmodel The ally will always be non-estimable, and the trap and period effects are the same as in the base model. Call: lm.default(formula = catch trap + period + period * allt) Coefficients: (Intercept) trap1 trap2 period1 period period3 period4 allt period1:allt period2:allt NA period3:allt period4:allt

9 Stat 5303 (Oehlert): Tukey One Degree of Freedom 9 > model.effects(fit.rowmodel,"period:allt") The slope adjustments sum to 0. allt > anova(fit.rowmodel) The row model is borderline significantly better than the additive model. Response: catch trap *** period * period:allt Residuals > anova(fit1,fit2b,fit.rowmodel) Tukey is significantly better than additive, but the row model does not significantly improve on Tukey. Model 1: catch trap + period Model 2: catch period + trap + compvals Model 3: catch trap + period + period * allt Res.Df RSS Df Sum of Sq F Pr(>F) ** > fit.colmodel <- lm(catch trap+period+trap*allp) Here we fit different period slopes for each trap. > anova(fit.colmodel) Column model is significantly better than additive. Response: catch trap e-06 *** period ** trap:allp ** Residuals

10 Stat 5303 (Oehlert): Tukey One Degree of Freedom 10 > anova(fit1,fit2b,fit.colmodel) But the column model is not really significantly better than the Tukey model. Model 1: catch trap + period Model 2: catch period + trap + compvals Model 3: catch trap + period + trap * allp Res.Df RSS Df Sum of Sq F Pr(>F) *** > fit.slpmodel <- lm(catch trap+period+trap*allp+period*allt) We can fit the model with both the slopes. > anova(fit.slpmodel)... but it is not significantly better than period slopes model. Response: catch trap ** period * trap:allp * period:allt Residuals > # Note that we get the same predicted values but different individual terms estimated in the model depending on which order we enter the rows and column slope effects.