> Mike Stephenson wrote in message ...
> >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.
> >
> >Now, I need to calculate a correlation coefficient ...
Why?
In general I agree with Alan Miller's response. If all you want to know
is, as you write later, whether this function or another one better fits
the data, the sum of squared residuals, or the RMS, is both adequate and
not subject to quibbles about propriety. (By "residual", I here mean the
difference (Y-Y*) for each observation in the data set. In your
regression analysis the natural "fitted value" is (1/y)-hat, which the
program should cheerfully store for you; the reciprocal of this is the
predicted value of Y, which I called Y*. Essentially Y* = 1/(a + bx). )
> > ... but am uncertain how to proceed.
THAT's understandable. The
problem is not well-defined. But you seem to be clear enough, though you
haven't explicitly said so, that the simple correlation between x and
y isn't suitable. If for theological reasons you insist (or someone
else lays the insistence upon you) that a correlation coefficient is
necessary, I'd suggest the correlation between Y and Y*. (If you plot
Y vs Y* you should get a scatterplot that shows a linear structure; the
correlation associated with this scatterplot is what I mean.)
> > The curve is plotted as y vs. x ...
Surely not! Your _curve_ (as
distinguished from the scatterplot of raw data) is plotted as Y* vs. x.
Use your French curves intelligently and you can ink in a pretty, solid
line to display that curve.
> >and that is what I want the correlation coefficient to represent,
No, it's not. A correlation coefficient represents the degree of
_linear_ relationship between two variables; we have already established
that the relationship is NON-linear, because you're fitting a reciprocal
function.
> >because I'm comparing this curve fit to another more standard equation
> >and would like to know which fits it better.
An inadequate reason to produce a correlation coefficient, which will be
pretty ambiguous in the best of situations. SS or RMS of Y-residuals is
better, as Alan points out.
Alan Miller wrote in response:
> Will you know which fits better after you have calculated some kind
> of correlation coefficient?
A good question. Do you have a good answer, Mike?
> Why not just calculate the sum of squares (or RMS) of the differences
> between the actual and fitted values for Y's?
>
> Better still. Plot the residuals against x and look at them. This
> should show up any systematic deviations.
Right. And any such
deviations will indicate how poorly the function in question (1/y or its
competitor) represents the lurking functional relationship between y
and x .
-- DFB.
------------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 603-535-2597
184 Nashua Road, Bedford, NH 03110 603-471-7128
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