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I have a data set expressed as in the figure:

data

Here $y$ is some measured quantity with known error and 'fit' is some attempt to fit a function with zero error.

In order to evaluate the quality of the fit (and thereby rank different fits) it is possible to calculate a Pearson Correlation Coefficient, in this case approx $0.8$.

My question is that although I can calculate a correlation coefficient is it possible to also calculate an error for the Pearson's correlation coefficient? i.e. $0.8\pm 0.05$?

If a Pearson Correlation Coefficient is not the best way to score the fit I would also be interested in alternatives.

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  • $\begingroup$ Just as an aside, there are many ways to evaluate fits. No single scoring method is ever "best", and in order to provide informative, truthful results to the readers of your work, you need to usually apply a wide array of model checks and fit scores. One place to start reading is Chapter 6 of the excellent resource book "Bayesian Data Analysis" by Gelman, Carlin, Stern, and Rubin. $\endgroup$
    – ely
    Mar 23, 2012 at 1:44
  • $\begingroup$ very true. I did my best to search the web for alternatives but my problem generally suffers from low expected counts. I will have a look at the book. $\endgroup$
    – MarcF
    Mar 23, 2012 at 14:03

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If I recall right, then the function $\small \tanh^{-1}(r) $ where r is the pearson correlation coefficient is distributed normally if the correlated data are normally distributed too, so you can compute confidence intervals based on this (the coefficient r has range $\small -1 \ldots 1 $ and this range gets stretched to $\small -\infty \ldots \infty $ by the $\small \tanh^{-1}$ - transformation). I've seen this been discussed much intensely by James Steiger, but it's long time ago and I cannot give a reference at the moment (surely this should also be mentioned in the wikipedia). HTH anyway.

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Probably a more rigorous scoring statistic is chi squared. The error in chi squared can then be derived directly. This has been used and results are very promising.

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