# st: RE: RE: Re: computation of R-squared with a non-linear model

 From "Nick Cox" To Subject st: RE: RE: Re: computation of R-squared with a non-linear model Date Fri, 22 May 2009 12:14:11 +0100

```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
n.j.cox@durham.ac.uk

-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Nick Cox
Sent: 22 May 2009 12:06
To: statalist@hsphsun2.harvard.edu
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
n.j.cox@durham.ac.uk

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

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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