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I used my TI-83 to find the quadratic regression of two data columns. The accuracy wasn't close at all. So I tried cubic and then finally quartic regression. The accuracy still isn't close enough. Is there any way I can improve the accuracy? I can't add any more data, unfortunetly.

Edit 1:

To address Henry's comment, my plotted data looks similar to y=x^(1/3)

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    It is difficult to comment without seeing the data. You could try plotting the points to see what the pattern looks like and whether that suggests anything.2011-05-20
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    If it is in fact a cube root, that explains why polynomials don't fit well. The cube root gets too flat for a polynomial. As I suggested below, taking the logarithm of all the data will check this. y=x^(1/3) implies log(y)=(log(x))/3 and you should get a good linear fit.2011-05-20

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This may mean your data is poorly fit by a polynomial. Three ways it can fail is either to be exponential, to have poles, or be like a sine wave. Plotting the data with the fits you have overlayed may give you some ideas. If it is exponential, taking the logarithm of one will render it linear (or close to polynomial). If it has poles, using a rational function will help. Procedures are given in chapter 3.2 of Numerical Recipes. The obsolete versions are free. If it is a sine wave, an FFT can help-chapter 12 of NR.

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    What do you mean 'take the logarithm of one'? Log the value of the equation?2011-05-20
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    Take the logarithm of the data. If the data fits $y=exp(ax)$ your polynomial fitters will not do well, because polynomials can't increase that fast. But then $\log y=ax$ will be linear.2011-05-20
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    I'm really sorry, I still don't follow. Say my function is y = -1.5E-8*x^4 + 1.8E-5*x^3 + .007*x^2 + 1.7*x + -15. Intead of raising each, I would lof each?2011-05-20
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    I wasn't talking about the function, but the data itself. Instead of doing a regression of x and y, do one on log(x) and log(y). See if that helps.2011-05-20
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    Ok thanks, that makes more sense. How do you predict something with that though? If I understood you right you take the log of both x and y, make a data chart and then do regression?2011-05-21
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    Right. That gives you an equation connecting log x and log y. In your case you might get $\log y = (\log x)/3 +a$. Now if you get an $x$ value, you take its log, calculate log y, and exponentiate that to get y.2011-05-21
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    Ok, thanks for bearing with me. Sorry, for being difficult.2011-05-21
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    You're welcome. Having found the equation in log x and log y, you may be able to exponentiate both sides and wind up with x and y. It depends upon the degree you use.2011-05-21
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    Ok, I'm sorry to reopen this. I did what you said, and the graph looks great. But I'm still stuck on finding y. What am I exponentiating to find y?2011-05-23
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    Normally when you do a regression you get an equation that looks like y=f(x) where x is a polynomial. Your tool is probably giving you that, but you need to remember that you took the log of all your y values. So the output of the equation is the log of the y you should predict. Just exponentiate that y.2011-05-23
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If, as you say, it looks like $y=x^\frac{1}{3}$, then you might try a power regression, which will give you a formula of the form $y=ax^b$.