Mike Stephenson wrote in message ...
>Thanks again for everyone's input. It definitely helped me. I'm an
>engineer, not a statistician and sometimes it shows. :)
>
>I did get a response on another NG, which I think will give me an
acceptable
>way to calculate a correlation coefficient that might have a little more
>meaning as how good the fit is. Here it is:
>
>If your true sample values are (x,y) and your model calculates an
>estimated value y_estim = m(x), then the empiric correlation factor R
>between y and y_estim might help you. R^2 is used as a "goodness of fit"
>measure for linear models. It can be interpreted as a ratio of variances:
>R^2 = the variance of y_estim divided by the variance of y. The variance
>of y_estim is the part of the variance of y that is "explained" by the
>model.
>
** It will NOT help.
It is trivial to find a set of y_estim which will give an excellent R^2
(e.g.
R^2 = 100%), and yet a very bad fit.
e.g. y_estim = 0.001*y + 1000000
FORGET about correlations.
Look at the residuals - as everyone else is telling you.
Also, why don't you find the non-linear least squares fit?
You could use the parameters from your 1/Y fot as
starting values.
>Sample plots of y_estim over y may be a good illustration as well.
>
--
Alan Miller, Retired Scientist (Statistician)
CSIRO Mathematical & Information Sciences
Alan.Miller -at- vic.cmis.csiro.au
http://www.ozemail.com.au/~milleraj
http://users.bigpond.net.au/amiller/
>
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