Well, you bring up very good points. I agree that it is better to determine
if it is a good fit by one or more of the methods suggested. I also agree
that a correlation coefficient may lead one astray - it may be a "good fit"
based on corr. coeff. , but could be terrible for predicting intermediate
values, which would be clear from a graph of the curve.
I was just curious as to how this is calculated for this scenario more than
anything else. I probably wasn't very clear in my original post.
Donald F. Burrill <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > 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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