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I have a question about correlations. Let's say I want to measure the associations between a number of variables and all of them are measured on a ratio scale. I know I can use Pearson's $r$ correlation to do this, assuming the assumptions to run Pearson's correlation are met (i.e., normality, linearity, no outliers, homoscedasticity, etc.).

However, let's say I just wanted to save time and not have to check all of these assumptions and instead calculate Spearman's $\rho$ correlations (as there are far fewer assumptions to check – all one really has to check for is that relationships are monotonic).

This of course begs the question: If I do this, will it in any way give less precise/less accurate results? I've heard that this is something that isn't a big deal, but I thought I'd ask.

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  • $\begingroup$ But you can check normality, linearity, no outliers simply by plotting the data. Then applying whatever sufficient transform to make the data normal and linear. You can also visually get a rough idea on homoscedasticity. I don't see how you save time by skipping doing that plot. $\endgroup$
    – smci
    Commented Feb 19, 2023 at 22:42

2 Answers 2

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Pearson's correlation coefficient ($\boldsymbol{r}$) provides a measure of linear association between paired variables.

Spearman's correlation coefficient ($\boldsymbol{r_{\bf{S}}}$) provides a measure of monotonic association between paired variables.

The latter is different than the former. A linear relationship is just one kind of monotonic relationship between two variables. For example, an exponential curve, a logistic curve, and a stair step function are also all examples of monotonic functions between $x$ and $y$. Because of this, it is possible to observe $r > r_{\text{S}}$, $r \approx r_{\text{S}}$ and $r < r_{\text{S}}$ to a significant degree depending on what's going on in your data. These two measures can even have different signs.

As a consequence, without knowing the specifics of your data and sample distributions ahead of time, it is probably best to respect that $\boldsymbol{r}$ and $\boldsymbol{r_{\bf{S}}}$ are measures of substantively different things.

Aside: Spearman's correlation can also be thought of as a measure of linear association between ranks of paired variables..

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I think a couple valuable journal articles on this subject are de Winter et al., 2016 and Bishara & Hittner, 2015. I highlight the main points of both below, focusing primarily on Pearson and Spearman coefficients since that is what your question asks.

Bishara & Hittner, 2015

In this simulation study, the authors mainly sought to test how non-normality affected correlation coefficients since so much of psychological research has non-normal data anyway. Their conclusions generally were the following:

  1. Non-normality causes exaggerated point estimates for Pearson's $r$, with particular issues present when kurtosis is present. Therefore, in these cases, something like Spearman's $\rho$ can be better, but that of course is also dependent on the shape of the data distribution.
  2. Pearson's allow for much easier interpretation when the relationship is very linear. It does not require a rank-transformation explanation of why the data is weak or strong.
  3. RMSE (re: error) for Pearson and Spearman coefficients is generally the same for both at low estimates, but RMSE decreases as the point estimates for Spearman become larger.

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  1. For a variety of distributions, the differences in RMSE is minimal while in other cases it is more drastic. These differences also vary across sample size but not by much.

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de Winter et al., 2016

This study used a large pool of real world data, and showed some interesting similarities and differences.

  1. Normally distributed variables show similar values for $r$ and $\rho$, but $\rho$ seemed to be less consistent when the correlation was strong and prone to lower values, which meant that Pearson coefficients are more consistent for normal data.
  2. Reflecting what the other simulation study stipulated, kurtosis seemed to affect Pearson coefficients more. Outliers also unsurprisingly affected Pearson values more.
  3. Pearson coefficients perform much better for light-tailed distributions, whereas Spearman values performed better for heavy-tailed distributions.

So looking at these results, we can say that in some cases, each correlation coefficient approximates bivariate relationships in a more straightforward fashion when meeting the particular class of association it is intended for (strictly linear vs monotonic), whether the data is normal (the tail shape, etc.), how large the sample size is, and other relevant factors. By the way, both of these papers have some good material that explains past lit on the properties of correlations if you feel you would like to explore this topic further.

Citations

  • Bishara, A. J., & Hittner, J. B. (2015). Reducing bias and error in the correlation coefficient due to nonnormality. Educational and Psychological Measurement, 75(5), 785–804. https://doi.org/10.1177/0013164414557639

  • de Winter, J. C. F., Gosling, S. D., & Potter, J. (2016). Comparing the Pearson and Spearman correlation coefficients across distributions and sample sizes: A tutorial using simulations and empirical data. Psychological Methods, 21(3), 273–290. https://doi.org/10.1037/met0000079

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  • $\begingroup$ Nice! +1 Can you clarify what you mean by "behave better"? Is that shorthand for "is more sensitive to violations of normality" and similar assumptions? $\endgroup$
    – Alexis
    Commented Feb 19, 2023 at 3:24
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    $\begingroup$ Eh... "true population correlation" ignores that linear association and monotonic association are actually different things. $\endgroup$
    – Alexis
    Commented Feb 19, 2023 at 3:26
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    $\begingroup$ Hmm...perhaps what I mean to say is that when meeting the particular class of association (strictly linear vs monotonic), the Pearson case approximates the relationship in a more straightforward fashion than Spearman in normal data because it is more consistent in measurement. Would you agree? $\endgroup$ Commented Feb 19, 2023 at 3:28
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    $\begingroup$ These comments help me learn more so I appreciate your input. I've edited that into my answer to better clarify that point. $\endgroup$ Commented Feb 19, 2023 at 3:40
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    $\begingroup$ Thank you so much for this Shawn!! So much here. Very helpful and very clear. $\endgroup$ Commented Feb 19, 2023 at 3:54

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