George
> Why is the numerator degrees of freedom equal to K?
One way of thinking about the f-test that we perform on the GLM is as it
being a test of the ratio of the variance explained by the contrasts in
the f-test (the numerator) to the variance in the residuals (the
denominator). The numerator is related to a sum of the squares over the K
contrasts, hence is chi-square distributed with K dof. The denominator is
chi-square distributed with T-P dof.
> In a statistics book I
> in an F-test the numerator degrees of freedom is (P - 1), but maybe this is
> contrast matrix that is a single vector.
With this f-test, if you have a contrast matrix with a single t-contrast
then (numerator dof)=K=1, and F=T^2 and the F-test is equivalent to a
two-tailed t-test.
The discrepancy between this and the f-test in your
stats book is possibly due to them performing a test for a different
variance ratio to the one discussed above. Different numerator variance
(sum of squares over a different set of components) leads to a different
numerator dof.
Mark Woolrich.
Oxford University Centre for Functional MRI of the Brain (FMRIB),
John Radcliffe Hospital, Headington, Oxford OX3 9DU, UK.
Work: +44-(0)-1865-222713, Mobile: +44-(0)-7808-727745
Email: [log in to unmask]
On Fri, 10 Oct 2003, George Tourtellot wrote:
> I have another question about the F-statistic formula:
>
> Mark Woolrich wrote:
> if:
> length of beta vector is P
> number of contrast vectors is K
> length of time points is T
>
> then the f-stat is distributed with degrees of freedom K (numerator) and
> (T-P) (denominator). Using (T-P) assumes known autocorrelation with
> prewhitening, or that there is no autocorrelation.
>
> -------------
> my question:
>
> Why is the numerator degrees of freedom equal to K? In a statistics book I read that
> in an F-test the numerator degrees of freedom is (P - 1), but maybe this is just for a
> contrast matrix that is a single vector. Could someone clarify this issue for me?
>
> thanks,
> George
>
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