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I find that a polynomial trend line gives a better $r^2$ value. Is Pearson's Correlation Coefficient $r$ still a good indicator in this scenarios between the strength of correlation between my two variables?

If not - is there a better indicator for strength of relationship between two variables when using a polynomial trend line?

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The square root of $r^2$ is in effect the correlation between observed values and those predicted by the polynomial. (That's why it is denoted $r^2$.)

Whether the polynomial is a better fit than a straight line is a more challenging question. A polynomial with quadratic and possibly other terms can wiggle much more than a straight line but whether it is a better idea overall is harder to say.

Positively, a polynomial can capture important features of a relationship that a straight line just can't match, such as a turning point. Negatively, and this is the advantage reversed, a polynomial can wiggle in ways that make no substantive sense and/or would be treacherous if taken literally or extrapolated incautiously.

What you should always do includes

  1. Comparing $r^2$ from a polynomial fit with the square of plain Pearson's $r$ to see how much improvement is implied by the polynomial.

  2. Plotting the polynomial fit together with the original data to gauge the improvement.

  3. Thinking about the substantive interpretation of the polynomial fit, especially whether explicit or implicit turning points, inflections and axis crossings make sense.

(I find it hard to imagine that a Bayesian would say anything different in spirit.)

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  • $\begingroup$ Thank you so much! re:1 Are you recommending comparing a linear r-squared vs a polynomial r-sqaured and seeing the improvement through the delta? $\endgroup$ – luckee.13 Jan 4 at 17:08
  • $\begingroup$ Yes; that is #1 of my suggestions (now edited). Equivalently compare the root of $r^2$ with your Pearson $r$. $\endgroup$ – Nick Cox Jan 4 at 17:38
  • $\begingroup$ Thank you Nick for taking the time to educate me and in explaining the material thoroughly. I appreciate it. $\endgroup$ – luckee.13 Jan 5 at 7:11

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