the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and 2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data.
I created a dummy data set to explain my situation: df1 <- data.frame( Observation = c(rep("CTRL",15), rep("TREAT_1",13), rep("TREAT_2", 12)), COV1 = c(rep("A1", 30), rep("A2", 10)), COV2 = c(rep("B1", 5), rep("B2", 5), rep("B3", 10), rep("B1", 5), rep("B2", 5), rep("B3", 10)), Variable = c(3944133, 3632461, 3351754, 3655975, 3487722, 3644783, 3491138, 3328894, 3654507, 3465627, 3511446, 3507249, 3373233, 3432867, 3640888, 3677593, 3585096, 3441775, 3608574, 3669114, 4000812, 3503511, 3423968, 3647391, 3584604, 3548256, 3505411, 3665138, 4049955, 3425512, 3834061, 3639699, 3522208, 3711928, 3576597, 3786781, 3591042, 3995802, 3493091, 3674475) ) plot(Variable ~ Observation, data = df1) As you can see from the plot there is a linear relationship between the control and the treatment groups. To check if this linear effect is statistical significant I change the contrasts using the contr.poly() function and fit a linear model like this: contrasts(df1$Observation) <- contr.poly(levels(df1$Observation)) lm1 <- lm(log(Variable) ~ Observation, data = df1) summary.lm(lm1) >From the summary we can see that the linear effect is statistically >significant: Observation.L 0.029141 0.012377 2.355 0.024 * Observation.Q 0.002233 0.012482 0.179 0.859 However, this first model does not include any of the two covariates. Including them results in a non-significant p-value for the linear relationship: lm2 <- lm(log(Variable) ~ Observation + COV1 + COV2, data = df1) summary.lm(lm2) Observation.L 0.04116 0.02624 1.568 0.126 Observation.Q 0.01003 0.01894 0.530 0.600 COV1A2 -0.01203 0.04202 -0.286 0.776 COV2B2 -0.02071 0.02202 -0.941 0.354 COV2B3 -0.02083 0.02066 -1.008 0.320 So far so good. However, I have been told to conduct a Type II Anova rather than a Type I. To conduct a Type II Anova I used the Anova() function from the car package. Anova(lm2, type="II") Anova Table (Type II tests) Response: log(Variable) Sum Sq Df F value Pr(>F) Observation 0.006253 2 1.4651 0.2453 COV1 0.000175 1 0.0820 0.7763 COV2 0.002768 2 0.6485 0.5292 Residuals 0.072555 34 The problem here with using Type II is that you do not get a p-value for the linear and quadratic effect. So I do not know if the treatment effect is statistically linear and or quadratic. I found out that the following code produces the same p-value for Observation as the Anova() function. However, the result also does not include any p-values for the linear or quadratic effect: lm2 <- lm(log(Variable) ~ Observation + COV1 + COV2, data = df1) lm3 <- lm(log(Variable) ~ COV1 + COV2, data = df1) anova(lm2, lm3) Does anybody know how to conduct a Type II anova and the contrasts function to obtain the p-values for the linear and quadratic effects? Help would be very much appreciated. Best Peter [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.