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.
Sample plots of y_estim over y may be a good illustration as well.
Bob Hayden <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> ----- Forwarded message from Mike Stephenson -----
>
> I curve fit some data to the equation 1/y=a+bx using a set of x, y data.
I
> used the 1/y form in the curve fit because I could use a linear least
> squares approach. I treated 1/y as "Y" for the curve fit exercise. If I
> were to use it as y=f(x), the equation is no longer linear.
>
> Now, I need to calculate a correlation coefficient but am uncertain how to
> proceed. The curve is plotted as y vs. x and that is what I want the
> correlation coefficient to represent, because I'm comparing this curve fit
> to another more standard equation and would like to know which fits it
> better.
>
> ----- End of forwarded message from Mike Stephenson -----
>
> I would agree with most of the folks questioning whether this is a
> good idea, but if you really want to do this...
>
> You can compute the correlation of x and 1/y
>
> OR
>
> You can take the y-values (not 1/y values) predicted by your equation,
> subtract them from the observed y-values to get residuals, square the
> residuals and sum the squares, and then find the percentage reduction
> in the sum of squared residuals from the curve as compared to
> residuals from the mean. This will be a kind of "r^2" whose square
> root you can take (after converting to a decimal fraction).
>
> What if anything these numbers mean and how they might be interpreted
> I leave to others!-)
>
> _
> | | Robert W. Hayden
> | | Department of Mathematics
> / | Plymouth State College MSC#29
> | | Plymouth, New Hampshire 03264 USA
> | * | 82 River Street
> / | Ashland, NH 03217-9702
> | ) (603) 968-9914 (home)
> L_____/ fax (603) 535-2943 (work)
> [EMAIL PROTECTED]
> http://mathpc04.plymouth.edu
>
>
>
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