HronnE wrote on 09/27/2011 06:27:38 AM: > > Hi all > > This is probably a simple problem but somehow I am having much trouble with > finding a solution, so I seek your help! > > I have a data-set with continuous response variables. The explanatory > variably is 4xpH treatments (so 8.08, 7.94, 7.81 and 7.71) so also > continuous and not technically factorial. > > However I have decided to do Anova's (as well as regression) to explore the > effect of pH. > > What I don't understand is why the anova done in such a way: > > summary(aov(BioMass~pH)) > > ... gives me completely different p-values if I define the pH as factor or > not. And what would be the correct approach? > > Help on this subject would be much appreciated! > > Hronn
If pH is a numeric variable (see ?class) then your formula is instructing aov() to include it as a linear term with one degree of freedom, just like lm() does if you're fitting a regression. If you want to treat pH as a categorical variable, then you must make sure that it is included in the model as a factor. Then you will see that it uses three degrees of freedom. BioMass <- rnorm(100) pH <- sample(c(8.08, 7.94, 7.81, 7.71), size=100, replace=TRUE) pH.f <- as.factor(pH) summary(aov(BioMass ~ pH)) summary(lm(BioMass ~ pH)) summary(aov(BioMass ~ pH.f)) summary(lm(BioMass ~ pH.f)) Jean [[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.