Hi

On Fri, 26 Mar 2004, David Jones wrote:

> jim clark wrote:
> > No, the two ways of computing r^2 should agree, no matter how
> > many predictors you have.
> 
> Unfortunately/importantly, this is not true. They only agree if the
> predictor is of a specific form:
>   Y2 = a + b*Y3
> where Y3 is another predictor, and a and b are effectively fitted by
> least squares (possibly at the same time Y3 is fitted). If the form of
> the predictor Y2 is not such  as to have the "free" parameters a and b
> of this type, then Pearson's r^2 will give a different answer because
> it essentially does allow for these extra free parameters (because it
> is the correlation).
> 
> If you do use  Pearson's r^2  as a measure of "fit" you are
> effectively allowing for further adjustment of your predictor, beyond
> what you originally fitted ... thus this r^2 (correlation) will be at
> least as large as R^2 (calculated from the errors).

Below is what I wrote in its entirety.  I would really like to
see an example where the correlation between the original y and
the predicted y does NOT equal the multiple R defined as
sqrt(SSpred/SStotal).

-----------------------------------------------------------
No, the two ways of computing r^2 should agree, no matter how
many predictors you have.  Just to be clear,

if y^ = b0 + b1*x1 + b2*x2 ....

, where one or more xs could be polynomial predictors,

then

R^2 = SSy^ / SSy  (just another variation of your formula)

will equal r(yy^)^2
-----------------------------------------------------------

Best wishes
Jim

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James M. Clark                          (204) 786-9757
Department of Psychology                (204) 774-4134 Fax
University of Winnipeg                  4L05D
Winnipeg, Manitoba  R3B 2E9             [EMAIL PROTECTED]
CANADA                                  http://www.uwinnipeg.ca/~clark
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