Hi. Here's an interesting reference:

Chart trendline formula is inaccurate in Excel
http://support.microsoft.com/kb/211967

I don't have the original post, but with this data...

809920.89 1688.5
811026.35 1685.9
812370.50 1680.8
813553.47 1673.8
814357.32 1670

I get:

= 5163.22295426061-0.00428863203403701*x
or
=-317931.918702048+0.791376677921689*x-4.8985630175318E-07*x^2

Assuming of course that these are the appropriate equations.

--
Dana DeLouis


"Gary''s Student" wrote in message
...
I don't understand how your r-squared can be so good if the variation at a
single point is so large. Using the simple linear fit I posted, here are
the
x-values, y-values, fitted y-values and the variation-squared at each
point:

809920.89 1688.5 1686.815012 2.839185485
811026.35 1685.9 1683.471295 5.898609287
812370.5 1680.8 1679.405605 1.944336986
813553.47 1673.8 1675.827441 4.110518666
814357.32 1670 1673.396013 11.53290394

--
Gary''s Student - gsnu200735


"tom r" wrote:



i just added a trendline, using a 2nd order polynomial fit to better
match
the curve of the data. r^2 = .994.