I would like to know if there are any rules to determine if Pearson's r correlation values are similar to Spearman's rho?
For example, if r = -.207 and rho = -.282, are these similar enough to just report r?
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$\begingroup$ Thank you. The reason I ask is because my data have violated the assumptions of normality (as measured by Shapiro Wilks statistic). I know that one way to determine if the violation is excessively non-normal is to take a look at how similar the correlations between r and rho are. I am just having difficulty determining what constitutes "similar"!! $\endgroup$– AnonCommented Sep 16, 2011 at 7:09
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$\begingroup$ "... way to determine if the violation is excessively non-normal is to take a look at how similar the correlations between r and rho are". This is wrong statement. Magnitude of both coefficients has nothing to do with notion of normality (still magnitude of Pearson has something to do with shape of distribution) $\endgroup$– ttnphnsCommented Sep 16, 2011 at 7:17
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$\begingroup$ Oops. I was going to make the argument that if r and rho are similar then I can keep my outliers?? (the dataset with my outliers violated normality... I thought that if r and rho were similar then I could keep my outliers). Also, would a way to check whether r and rho are similar be to check how small, mod, and large their correlations are using Cohen's recommendations? That is, if both r and rho suggest "small" relationships then they are both similar? $\endgroup$– AnonCommented Sep 16, 2011 at 9:41
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$\begingroup$ Pearson r is strongly sensitive to outliers, as you know. Sometimes they exaggerate, sometimes they diminish r - depending on their spatial location. Spearman, by definition, is insensitive to outliers. But there is much more than just outliers that can distinguish r and rho. So, your idea "if r and rho were similar then I could keep my outliers" goes adrift. $\endgroup$– ttnphnsCommented Sep 16, 2011 at 11:43
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2 Answers
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In case of large sample sizes, the significance test is the same for both indexes:
$t = \sqrt{\frac{(n-2)}{(1-r^2)}}$
So you can generally treat both indexes the same way and test their difference for significance.
However, things might be a bit complicated, as they are non-independent correlations (i.e., both are derived from the same sample). There have been articles by Steiger (1980, [1]) and Meng et al. (1992, [2]) which treat this issue. In the cases covered there, however, it is always a correlation between one variable $x$ and two other variables $y$ and $z$ (i.e., comparing $r_{xy}$ with $r_{xz}$), which is not exactly your case.
[1] Steiger, J. H. (1980). Tests for comparing elements of a correlation matrix. Psychological Bulletin, 87, 245-251.
[2] Meng, X. L., Rosenthal, R., & Rubin, D. B. (1992). Comparing correlated correlation coefficients. Psychological Bulletin, 111, 172-175.
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$\begingroup$ In what way bivariate normality can show in ranked data? $\endgroup$– ttnphnsCommented Sep 16, 2011 at 6:49
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$\begingroup$ @ttnphns: Good point - that was wrong. I should have written (and correct my answer to): "with large sample sizes". Usually one would use a permutation test for rho in small samples; my idea was to put both Pearson and Spearman coefficients onto the same metric to make them comparable. $\endgroup$– Felix SCommented Sep 16, 2011 at 7:19
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Because Spearman rho is the same formula as Pearson r, only applied to ranked rather than to raw data, the two coefficients can and may be directly compared in magnitude. (In contrast, for example you cannot directly compare Spearman and Kendall correlations.). So, in your case, as -.282 is somewhat greater then -.207, you could conclude that the "true" association between the variables is not so much linear but rather monotonic.
See this for more particulars.
