Dear Marko,
If your design matrix is just a column of n 1's ie a one sample t-test
on n scans then we have
X=ones(n,1)
The betas are then given by
beta=pinv(X)*y
where y are the scans
Here, pinv(X) will give you a matrix equivalent to ones(1/n,1).
This is a long way of saying that the beta will be the average of the
data points. Therefore the contrast '1' on the beta will also be the
average.
This means two con images can be easily compared.
The same principle applies to more complex designs - the beta's will be
the response per event (ie the average).
Best,
Will.
Marko Wilke wrote:
> Dear All,
>
> quick question: I am manually comparing values from con-images from
> different designs and am unsure whether or not different group sizes are
> already reflected in these estimates. As far as I understand it, df (und
> thus, n) are only taken into account later but I may easily be wrong.
> Any hints are appreciated.
>
> Thanks,
> Marko
--
William D. Penny
Wellcome Trust Centre for Neuroimaging
University College London
12 Queen Square
London WC1N 3BG
Tel: 020 7833 7475
FAX: 020 7813 1420
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URL: http://www.fil.ion.ucl.ac.uk/~wpenny/
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