Assume Type 1 SS and no interaction. Under Model 1, your sums of squares (SS) is partitioned SS(M), SS(L|M), SS(E1|L,M). In Model 2 it is SS(L), SS(M|L), SS(E2|L,M). The total SS in both Model 1 & 2 are equal, and SS(E1|L,M) = SS(E2|L,M). [ If the design had been orthogonal then also SS(M)= SS(M|L) and SS(L)=SS(L|M) ]. In Model 3 it is SS(L), SS(E3|L). Now SS(E3|L) = SS(M|L)+ SS(E2|M,L).
If you want to test the _unconditional_ effect of Mother (ignoring Mother), you compare Model 1 to Model 3 (using drop1() for example). If you want to test the _conditional_ effect of Mother (Litter effect adjusted for Mother effect), you run Model 1 and test the main effect of Litter (=Litter|Mother). These are the same concepts as found in regression. Joe -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of michael watson (IAH-C) Sent: Thursday, April 21, 2005 3:51 AM To: Prof Brian Ripley Cc: r-help@stat.math.ethz.ch Subject: RE: [R] Anova - adjusted or sequential sums of squares? OK, I had no idea I was opening such a pandora's box, but thank you for all of your answers, it's been fascinating reading. This is how far I have got: I will fit the most complex model, that is the one that includes the interaction term. If the interaction term is significant, I will only interpret this term. If the interaction term is not significant, then it makes sense to test the effects of the factors on their own. This is where I get a little shaky... Using the example from the WNV paper, page 14. If I want to test for the effect of Litter, given that I have already decided that there is no interaction term, I can fit: Wt ~ Mother + Litter Wt ~ Litter + Mother Wt ~ Litter The latter tests for the effect of Litter ignoring the effect of Mother. The first two test for the effect of Litter eliminating the effect of Mother. Have I read that correct? However, it still remains that the top two give different results due to the non-orthogonal design. The way I see it I can do a variety of things when the interaction term is NOT significant and I have a non-orthogonal design: 1) Run both models "Wt ~ Mother + Litter" and "Wt ~ Litter + Mother" and take the consensus opinion. If that's the case, which p-values do I use in my paper? (that's not as flippant a remark as it should be...) 2) Run both models "Wt ~ Litter" and "Wt ~ Mother", and use those. Is that valid? 3) Believe Minitab, that I should use type III SS, change my contrast matrices to sum to zero and use drop1(model, .~., test="F") Many thanks Mick -----Original Message----- From: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Sent: 20 April 2005 16:35 To: michael watson (IAH-C) Cc: Liaw, Andy; r-help@stat.math.ethz.ch Subject: RE: [R] Anova - adjusted or sequential sums of squares? On Wed, 20 Apr 2005, michael watson (IAH-C) wrote: > I guess what I want to know is if I use the type I sequential SS, as > reported by R, on my factorial anova which is unbalanced, am I doing > something horribly wrong? I think the answer is no. Sort of. You really should test a hypothesis at a time. See Bill's examples in MASS. > I guess I could use drop1() to get from the type I to the type III in > R... Only if you respect marginality. The quote Doug gave is based on a longer paper available at http://www.stats.ox.ac.uk/pub/MASS3/Exegeses.pdf Do read it all. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
