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From: Yun Zhao on 9 Mar 2010 06:17
Thanks for your reply. But remember, the set of data values is 20 data
points, and my computed distribution of n(t) is a function of t, so
thousands and thousands of points. When I tried to use

Correlation[data1, n(t)]

I get the error "Correlation::vctmat: The arguments to Correlation are not a
pair of vectors or a pair of matrices of equal length." Please tell me what
I did wrong. Thank you.

Correlation::vctmat: "

StyleBox[\"\"\", \"MT\"] The arguments to Correlation are not a pair of
vectors or a pair of matrices of equal length


On Mon, Mar 8, 2010 at 5:13 AM, Sjoerd C. de Vries <
sjoerd.c.devries(a)gmail.com> wrote:

> Surprisingly, you can use the function Correlation to calculate the
> correlation between two lists. In your case you should probably use
> the set of predicted values and the sat of actual data values.
>
> Cheers -- Sjoerd
>
> On Mar 7, 11:05 am, Yun Zhao <yun.m.z...(a)gmail.com> wrote:
> > Hi everyone,
> >
> > I solved a differential equation, and got a solution n(t). Now I have
> > collected 8 data points at 8 different times. I plotted the solution of
> > n(t), and the curve intersect the 8 data points quite well on a graph of
> > n(t) vs. t. How do I use Mathematica 7.0 to calculate the correlation
> > coefficient R^2 value of how well the n(t) solution fit the data points?
> > Thank you very much.
>
>
>


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