Damián I asked a similar question a few months ago (3 August 2004):
> temp.aov <- aov(S~rep+trt1*trt2*trt3, data=dummy.data) > model.tables(temp.aov, type='mean', se=T) > > Returns the means, but states "Design is unbalanced - use se.contrasts > for se's" which is a little surprising since the design is balanced. To which Prof Ripley replied: If you used the default treatment contrasts, it is not. Try Helmert contrasts with aov(). If I recall correctly, following Prof Ripley's suggestion led aov() to accept the design was balanced, but model.tables() still did not (but that could have been my error). However, se.contrast() worked. Cheers ........ Peter Alspach >>> Damián Cirelli <[EMAIL PROTECTED]> 02/12/04 09:50:13 >>> Hi all, I'm new to R and have the following problem: I have a 2 factor design (a has 2 levels, b has 3 levels). I have an object kidney.aov which is an aov(y ~ a*b), and when I ask for model.tables(kidney.avo, se=T) I get the following message along with the table of effects: Design is unbalanced - use se.contrast() for se's but the design is NOT unbalanced... each fator level combination has the same n I' d appreciate any help. Thanks. ______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ______________________________________________________ The contents of this e-mail are privileged and/or confidenti...{{dropped}} ______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
