Last sentence should be
Even better than pursuing a single figure-of-merit would be to plot
observed vs predicted and residuals vs predicted.
Nick
[email protected]
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Nick Cox
Sent: 22 May 2009 12:06
To: [email protected]
Subject: st: RE: Re: computation of R-squared with a non-linear model
I support this general idea. For another statement, see
How can I get an R-squared value when a Stata command does not supply
one?
http://www.stata.com/support/faqs/stat/rsquared.html
Even better than pursuing a single figure-of-merit would be to plot
observed vs predicted residuals vs predicted.
Nick
[email protected]
Paul Seed
There is a simple way to compute R-squared for any regression model,
if you do not believe the value given by Stata: Calculate the predicted
values and carry out your own correlation.
Using the auto data set:
**** Start Stata code *****
sysuse auto
regress weight price
predict pred_w
su weight pred_w
corr weight pred_w
di "R-squared = " r(rho)*r(rho)
**** End Stata code *****
Both ways giver a value of 0.2901023
In general, the use of weights and adjusted R-squared
makes things more complicated, and the last two lines
could be changed to allow for them;
but neither will alter a correltion of 1.0.
If Marcel Spijkerman uses this approach, he may find
a) Marcel is right - the second R-squared is different
from the first. (He does not say, but I assume that
both the adjusted and unadjusted R-squared are 1.0).
b) Martin Buis is right - the model has failed to converge,
and the predicted values are mostly or completely undefined.
c) Stata is right - both methods give R-squared = 1.0
d) Something else I haven't though to.
I'd be interested to know which.
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* http://www.ats.ucla.edu/stat/stata/
