Little by little, I think it's getting clearer...  

I have verified that ResMS is the same as the pooled variance for all
three samples, by calculating it manually.

However, using the SPM ANOVA model, the covariance of the parameter
estimate, is:

Covariance_of_BetaA = ResMS*Bcov = Pooled Variance/N_groupA

Then, the standard deviation is 

StandardDeviation_of_BetaA = sqrt(Pooled_Variance/N_groupA)

And thus

StandardError_of_BetaA =
[sqrt(Pooled_Variance/N_groupA)]/[sqrt(N_groupA)]

It's this extra division by sqrt(N_groupA) that results in the
discrepancy!!

If you want to know the standard error of the mean of group A, and let's
say I estimated it from the pooled variance, I'd use:

StandardError_of_meanGroupA = [sqrt(Pooled_variance)]/[sqrt(N_groupA)]

So, how do I interpret this extra factor?

Sorry Ged - you are going way above the call of duty in answering these
questions!!

A

-----Original Message-----
From: Ged Ridgway [mailto:[log in to unmask]] 
Sent: Thursday, August 23, 2007 9:10 AM
To: Nugent, Allison C. (NIH/NIMH) [E]
Cc: [log in to unmask]; Ferguson, Teresa (NIH/NIMH) [F]; Furey, Maura
(NIH/NIMH) [E]; Drevets, Wayne (NIH/NIMH) [E]
Subject: Re: Beta SE, revisited

Hello again,

> At least in a group-level ANOVA, with no non-sphericity estimation,
the
> covariance matrix of the betas, Bcov,

Remember that Bcov is slightly mis-named, for historical reasons. It 
isn't actually the covariance matrix *until* it's been multiplied by 
the voxel-wise ResMS term (see the comments in spm_spm.m). For 
full-rank designs, it's simply inv(X'*X).

> is a diagonal matrix, where the
> diagonal elements are simply equal to one over the number of data
points
> in the group

Sure, if X has just binary indicator variables for the groups, then 
X'*X just ends up summing up however many ones as there are in the 
each group; it's diagonal, and its inverse therefore has the 
reciprocal of each element on its diagonal, as you found.

> So, there is absolutely no information in the Bcov matrix concerning
the
> variance in the data for each group.  Thus, since the ResMS value is
the
> residual error from the whole model, the standard deviation of the
> parameter estimates calculated as:
> 
> sqrt(ResMS*Bcov) is really just equal to 
> 
> sqrt(ResMS/Na) for group A.

Right.

> My new question is this - if Bcov is only 1/Na (for group a), and
> therefore sqrt(ResMS*Bcov) contains only the variance contribution
from
> all groups' data left unexplained by the full model, can
sqrt(ResMS/Na)
> rightly be called the standard deviation of the parameter estimate, or
> would the standard deviation have to include the variance contribution
> from the individual groups' data?

Well, I think this just reflects the choice of model. With multiple 
groups together in an ANOVA design with equal variance, the best 
estimator of the individual group standard errors comes from 
proportionally distributing (via the elements you saw in the inv(X'*X) 
matrix) the pooled estimate of the total variance. In other words, it 
is the equal variance assumption that alters the concept of individual 
variances from what you expected.

I believe that in general statistics packages, if you specify unequal 
variances, then each group's variance would be the individual value 
that you originally expected. Now, you might think that you can do 
that in SPM (5 at least) via the unequal variance option... however, 
the complication here is that SPM estimates the variance components 
(AKA non-sphericity) by pooling together all the voxels which survive 
an initial main-effects F-test. So if you checked the data for an 
individual voxel in this case, you would not find the individual 
variance that you expected. How close it would be to this estimate 
would, I guess, depend on how much the non-sphericity actually varied 
among voxels, in relation to the constant/pooled structure estimate by 
SPM.

I hope that has helped more than it's further confused the issue!

Ged.