I have been trying to use this test recently, following the text from this link: http://books.google.com/books?id=1eTyuMDND94C&pg=PA145&lpg=PA145&dq=nonparam#v=onepage&q&f=false
I ordered my data based on ranks, and ran a type III ANOVA from the car package - something like Anova(lm(var1~var2*var3,contrasts=list(var2='contr.sum',var3='contr.sum')),type='III'). Then I calculate SStotal (sum of the SS for the factors, interaction, and residual). I calculate MStotal, which is the SStotal divided by the degrees of freedom (add up the DF for factors, interaction, and residual). Then, calculate SS/MStotal for each factor and combination of factors. The p value for each is calculated as follows: 1-pchisq(the SS/MStotal, the degrees of freedom). I'm pretty sure there is an error in the text, as the first example they give calculates the SS as 1496, which includes the intercept-SS, and their math doesn't work out then (1496/16 is not = to 22). The second example makes more sense, and they don't include the intercept-SS. Anyhow, this seems like a useful test, but I think it should be used with caution. Hopefully, this helps, and if I'm doing something wrong here, that would be great to know (: -- View this message in context: http://r.789695.n4.nabble.com/Scheirer-Ray-Hare-tp3818476p3891860.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.
