> > I hope this is the appropriate place to post this
> > question. I am attempting
> > to show that an F-statistic is equal to a t-statistic
> > squared, or IOW F =
> > t^2. I am working with linear regression and extra
> > sum of squares tests. How
> > would one go about showing this? Or, can some kind
> > soul nudge me in the
> > right direction?
> The easiest way is to work with the canonical forms.
> F(m,n) = [X2(m)/m]/[X2(n)/n],
> where F(m,n) is an F with m and n degrees of freedom, X2(m) is a chi-square with m degrees of freedom, X2(n) is a chi-square with n degrees of freedom, and X2(m) and X2(n) are independent.
> In particular, for m = 1,
> F(1,n) = X2(1)/[X2(n)/n] = = z^2/[X2(n)/n] =
> [z/sqrt(X2(n)/n)]^2 = [t(n)]^2,
> where z ~ N(0,1) and t(n) is Student's t with n degrees of freedom.
> Thus an F with 1 and n degrees of freedom is the square of a t with n degrees of freedom.
> Jackwww.tomskystatistics.com
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