I am doing my thesis using a a non experimental descriptive correlation analysis with continuous ratio data. One of the variables is unevenly distributed. I have calculated Pearson's correlation coefficient ($-0.61$) and Spearman's correlation coefficient ($-0.48$), and I am now deciding which one to use. I have checked the distribution of each variable using The Kolmogorov-Smirnov test of normality and one of the variables is not normally distributed and the other is. Ideally, I would prefer to use Pearson's (as the correlation is stronger) but everything I have read has said to use Spearman's if unevenly distributed. Which should I use? Why is Pearson showing a stronger correlation?
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2$\begingroup$ Read topics with "Spearman Pearson" on the site. $\endgroup$– ttnphnsCommented Apr 6, 2021 at 7:27
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2 Answers
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Pearson's and Spearman's correlation coefficients have nothing to do with normal distributions (or with uneven distributions). Pearson's correlation coefficient measures the linear relationship between two variables, so if you were to check anything, it would be this.
Draw a scatter plot of your two variables and check if the relationship looks linear, if it doesn't, then Pearson's r is not appropriate. If the relationship is not linear but is monotonous then you can use Spearman's rho.
Pearson's r is a parametric measure, and as such has higher power (more easily detects effects) compared to non-parametric measures such as Spearman's rho, meaning if the relationship is truly linear then Pearson's r will have a higher (absolute) coefficient. However, this could also be for other unrelated reasons (such as outliers), so you cannot decide on which measure to use based solely on the coefficient.
answered Apr 6, 2021 at 5:19
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$\begingroup$ I have been unable to reconcile your initial statement that the Pearson correlation is unrelated to Normal distributions with your later statement that "Pearson's r is a parametric measure." Could you explain what you mean by the latter? What is being parameterized, if not some family of distributions, of which the Normal family would be the most natural in this context? $\endgroup$– whuber ♦Commented Jul 26, 2023 at 22:15
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$\begingroup$ @whuber I don't know. I guess the Pearson correlation coefficient can be parametric if used to estimate the parameters of a multivariate normal distributions, or it can non-parametric if making no assumptions about the data. Maybe the statement
Pearson's r is a parametric measureis wrong. $\endgroup$ Commented Jul 27, 2023 at 19:19 -
$\begingroup$ It's unclear what "parametric measure" even means. The term "parametric" is applied in situations where procedures assuming a finite-dimensional distribution family are used. The definition and interpretation of any "measure" (or statistic or metric or anything else like that) rarely, if at all, refers to any such distributional family. $\endgroup$– whuber ♦Commented Jul 27, 2023 at 19:36
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Have you read about Anscombe's quartet? Get that data and compute the Spearman's r and also Kendall's tau; the results are interesting. Bivariate correlation analysis is quite limited and can be misused. Do a scatter plot and let the reader decide.
Comparing Spearmans r and Pearsons r is like comparing apples and oranges: they have different meanings, so don't do it.
Also, it's relatively easy to achieve P < 0.05 (against the null hypothesis that r = 0) but large sample sizes are needed for narrow confidence intervals. If someone reports only r = 0.xx and P < 0.05 without any scatter plot, I can't accept that as evidence.
