----- 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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