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

[Lucke, Joseph F]

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 at stat.math.ethz.ch
[mailto:r-help-bounces at 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 at 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 at stats.ox.ac.uk] 
Sent: 20 April 2005 16:35
To: michael watson (IAH-C)
Cc: Liaw, Andy; r-help at 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 at 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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