RE: [R] Anova - adjusted or sequential sums of squares?

From: Lucke, Joseph F <LUCKE_at_uthscsa.edu>
Date: Fri 22 Apr 2005 - 00:28:55 EST


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: r-help-bounces@stat.math.ethz.ch
[mailto:r-help-bounces@stat.math.ethz.ch] 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:ripley@stats.ox.ac.uk] 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,                  ripley@stats.ox.ac.uk
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

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Received on Fri Apr 22 00:41:36 2005

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