Karen..off the top of my head, the VIF is the inverse of tolerance, hence,
if tolerance = (1 - r^2j), then VIF=
1/(1-r^2j)..[excuse the sloppiness of the notation, but r^2j would be the
percentage of variation accounted for by the predictors in predicting the
other predictor..ie., the linear combination of x1 and x2 in predicting x3];
anyway, as with any cutoff value there can be an element of arbitrariness,
though some have registered concern if VIF > 10.0; my personal (possibly
misinformed!) opinion is that the aforementioned cutoff value is way too
liberal; for VIF to equal 10.0 then 1/=(1 - .9) entails a multiple R of
.9486!!!; for me it is a stretch to conceive that collinearity only becomes
problematic when R = .9486...I'll be interested to see what others
think............

Dale Glaser, Ph.D.
Senior Statistician, Pacific Science and Engineering Group
Adjunct faculty/lecturer, SDSU/USD/CSPP
San Diego, CA.



-----Original Message-----
From:   [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
On Behalf Of Karen Scheltema
Sent:   Tuesday, May 30, 2000 1:52 PM
To:     [EMAIL PROTECTED]
Subject:        VIF

What is the usual cutoff for saying the VIF is too high?

Karen Scheltema, M.A., M.S.
Statistician
HealthEast
1700 University Ave W
St. Paul, MN 55104
(651) 232-5212   fax: (651) 641-0683

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