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I have an excel spreadsheet with

x    y
0   -1.5
100  1.6
200  0
300  -6.8
400  -19.8
500  -39.9

I want to find the values where x = 600 through 1500. I have tried making a graph and using the trend line and getting Polynomial 2, and it returns

y = -2.8857x2 + 12.686x - 11.7
R² = 0.999

So i plug this into my calculation using

=-2.8857*A110*A110+12686*A110-11.7

where A110 is the value 600, but it answers

6572736.3

I'm no math major, but in a trend of -6.8,-19.8,-39.9, the next number is not 6572736.3

Can someone please tell me how to figure out the equation so I can complete the series of numbers?

Matt Westlake
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    The equation you've provided fits your data very, very poorly and (relatedly) most certainly does not have an r-squared value of 0.999. The equation Excel gives me is: -0.0003x2 + 0.0691x - 1.9. I'd recommend looking again at how you came by that equation. Maybe try making a new sheet, entering all your data manually (no copy/paste to prevent incorrect type formatting) and trying again. – mkingston Jan 17 '13 at 03:57
  • +1 on the formatting issue – Matt Westlake Jan 21 '13 at 13:51

1 Answers1

3

I concur with @mkingston (see output below**).

I'd add two points:

1) I find it is always a good idea to plot the original data and the regression equation before doing anything with the equation. In this case, plotting @mkingston's result gives: Quadratic fit graph

... which shows that @mkingston's fitted results (shown by the lines) are, in fact, a good fit to the original data.

2) Extrapolation is always hazardous. If you already have a very good reason to believe that the underlying function is a quadratic of the form that we've fitted here, then the fit results below indicate the uncertainty in the parameters and hence can be used to estimate the uncertainty in the prediction (which may be quite substantial once you extrapolate to x = 1500). If, on the other hand, the quadratic equation that we've fitted is just a convenient shape that fits the data range that is available to us, then there are many alternative functions that could fit the available data roughly as well as this quadratic does, but would predict wildly different values for the range x = 600 to 1500. In this latter case, I'd descrbe any prediction at x = 600 as very uncertain and any prediction beyond that point as highly speculative, at best.

**The output I get from the Data | Data Analysis | Regression function of Excel 2007 is (after I've edited to change "X Variable" to "X" and "X Variable 2" to "X^2" for clarity):

SUMMARY OUTPUT                              

Regression Statistics                               
Multiple R  0.999516468                         
R Square    0.99903317                          
Adjusted R Square   0.998388617                         
Standard Error  0.647338875                         
Observations    6                           

ANOVA                               
            df  SS          MS          F           Significance F          
Regression  2   1299.01619  649.5080952 1549.9625   3.00625E-05         
Residual    3   1.257142857 0.419047619                 
Total       5   1300.273333                     

        Coefficients    Standard Error  t Stat          P-value     Lower 95%   Upper 95%   Lower 95.0% Upper 95.0%
Intercep   -1.9         0.586700679 -3.238448611    0.047907326 -3.767143409    -0.032856591    -3.767143409    -0.032856591
X       0.069142857     0.005518676 12.52888554     0.00109613  0.051579968 0.086705   746  0.051579968 0.086705746
X^2     -0.000288571    1.05946E-05 -27.23767444    0.000108607 -0.000322288    -0.000254855    -0.000322288    -0.000254855
Simon
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  • +1 for the dangers of extrapolation – chris neilsen Jan 17 '13 at 07:57
  • I appreciate this explanation. This is actually a ballistics chart I am trying to make for a friend. He wants to eventually shoot his gun out to over 1000yds, but all the data we can find is out to only 500yds. I was trying to trend the bullet drop out to a mile (1760 yds) – Matt Westlake Jan 21 '13 at 13:55