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I would like to calculate simple correlations. Since my sample size is quite small and data are not normally distributed I thought about Spearman’s correlations. Then I read about bootstrapping as a distribution free method and, considering my small sample size, I decided to use it with Pearson’s correlation coefficient. However, recently I’ve found bootstrapping for Spearman correlations and got confused. If bootstrapping solves the problem with the data distribution and small sample size (i.e. one can use Pearson’s method) why I should bother with bootstrapped Spearman. Where does the problem lie? What method should I use? Are there any rules?

just adding more details:

The whole problem is as follows: I’d like to test 14 multiple correlations (1 vs. 14). A sample size is fairly small (approximately n=50) and the data are not normally distributed. So, I have two problems one with correlations and the second with the Type I error. My natural choice was to look at Spearman’s correlations and in order to check if they are robust (effect size and significance) I wanted to bootstrap them. I’d followed this comment: http://www.methodspace.com/profiles/blogs/bonferroni-correcting-lots-of-correlations. I don’t know if assumed relationship is linear – from the scatterplot it looks like linear and without evident outliers. Then, using bootstrapped p-values I was thinking about Bonferroni-Holm’s correction for repeated measurements or/and calculating q-value (to control for False Discovery Rate)

So the basic questions are:

1) If this approach is correct? 2) How can I justify (e.g. for a reviewer) that I’d chosen Spearman’s or Pearson’s correlations (results are different - I’ve just checked)

thanks!

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  • $\begingroup$ Bootstrapping does not solve any problems with distribution of your data - it is a method for estimating confidence intervals, errors etc. for different methods. You would use bootstrap the same way with Pearson or Spearman correlation. What do you need bootstrap for? How do you want to use it? Please clarify since your question is unclear. $\endgroup$
    – Tim
    Nov 2, 2015 at 14:39
  • $\begingroup$ as Tim writes, you should clarify your question/your process. Bootstrapping is usually used for estimating standard errors, not correlations themselves. I.e., you'd bootstrap the error of your correlation (pearson or spearman) to ascertain its significance... $\endgroup$
    – metjush
    Nov 2, 2015 at 14:44
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    $\begingroup$ It's not normality that is an ideal for Pearson correlations, but linearity. $\endgroup$
    – Nick Cox
    Nov 2, 2015 at 15:03
  • $\begingroup$ thanks for your replies. I've just provided more details $\endgroup$
    – mcdomell
    Nov 2, 2015 at 15:39
  • $\begingroup$ It sounds like your problem does not relate to bootstrap at all but to choosing between Pearsons and Spearmans correlations, for this see stats.stackexchange.com/questions/8071/… Bootstrap procedure would work the same in both cases. $\endgroup$
    – Tim
    Nov 2, 2015 at 15:46

1 Answer 1

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Do you mean 1 target variable (Y) correlated against 14 X variables? In case you want to test if two variables, say the target $Y$ and the predicted $\hat{Y} $ are correlated or not in a nonparametric way, you need to use bootstrap. What you'll do is compute the correlation (spearman/ pearson) of the $Y,\hat{Y}$ values several times (100/500) across different resamples and then create a distribution of the computed values. If $0$ is included in the $[q_{\alpha/2},q_{100-\alpha/2}]$ interval where $q_p$ denotes the $p$th percentile of the computed distribution, then you can say that the correlation coeff is not significantly different from $0$ at level $\alpha$, usually alpha is taken as 5%. Also Pearson correlation $0$ indicates absence of linear dependence, while Spearman correlation $0$ indicates absence of monotonic dependence.

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  • $\begingroup$ Thanks. That’s right, One Y variable against X1 then: Y vs. X2; Y vs.X3; … up to: Y vs. X14. Sorry for being a layman but saying ‘predicted Y^’ you mean subsequent X variables? If so, everything is clear. $\endgroup$
    – mcdomell
    Nov 2, 2015 at 16:24
  • $\begingroup$ Yup, in that case, you need to use the same approach 14 times. Btw, if you like my answer, please accept it as an answer or upvote it. $\endgroup$ Nov 2, 2015 at 16:48

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