We are doing a lab report comparing attachment scores and their effect on trust scores. We have been told to do correlations followed by linear regression to get rid of confounders.

However, we have found that trust, the dependent variable, strongly deviates from normality kurtosis = 2.02 and q-q plot looks abnormal but the KS test does not suggest abnormality.

I am not sure whether to report Pearson's or Spearman's considering the independent variable is normally distributed and confused as to why we have been asked to do regression if trust is not normally distributed.

Spearman's is not significant whereas Pearson's is.

A simplified answer would be preferred as I am an undergraduate doing my first research project.

  • 5
    $\begingroup$ There's no assumption about the marginal distribution of either dependent or independent variables for regression, whether you're looking at hypothesis tests, confidence intervals or whatever. $\endgroup$
    – Glen_b
    Commented Jan 11, 2015 at 12:27
  • $\begingroup$ On that aspect of your question, see here, here or here $\endgroup$
    – Glen_b
    Commented Jan 11, 2015 at 13:08

1 Answer 1


Focusing on the correlation issue (Pearson vs. Spearman), based on what you've told us, it sounds like the Pearson correlation will be adequate for this situation.

The KS test tends to be overly sensitive to departures from normality, so that non-significant KS test, along with the modest kurtosis, suggests that normality isn't grossly violated. The Pearson correlation will suffer from increased Type I and II errors, but only with much more extreme departures from normality. For example, some simulation studies find inflated Type I and II errors when excess kurtosis is as high as 12, but I have not seen such results for excess kurtosis of 2. See this previous post for a reference and more details.

Some caveats and other considerations: 1. The traditional significance test of the Pearson correlation (the one you probably learned in intro stats) assumes that both variables are approximately normal. In contrast, the typical significance tests for regression assume normal residuals (that is, that the datapoints' vertical distances relative to the regression line are normally distributed). 2. The fact that you have very different results for the Pearson versus Spearman correlation suggests that you should examine a scatterplot carefully. Ask yourself whether a single outlier is driving the pattern. If so, the Spearman correlation might actually make more sense, as it is less sensitive to outliers.

  • 1
    $\begingroup$ @Glen_b has pointed out in comments that testing the dependent variable for normality is irrelevant. What, then, justifies drawing any conclusion from a KS test result (whether positive or negative)? $\endgroup$
    – whuber
    Commented Jan 11, 2015 at 22:24
  • 1
    $\begingroup$ @Glen_b was referring to regression, where the traditional significance test relies on an assumption of normal residuals, not marginals. For a significance test of the Pearson correlation, there is an assumption of bivariate normality. $\endgroup$
    – Anthony
    Commented Jan 11, 2015 at 22:46
  • 2
    $\begingroup$ Thank you. Some explicit explanation of your point of view and assumptions appear necessary in the answer itself, in light of the fact that the question appears to be about regression and not correlation: after all, it distinguishes a "dependent variable," which is not an aspect of a correlation calculation. Be aware that many beginners do not distinguish Pearson correlation-based tests from regression-based tests and often use "correlation" interchangeably with "regression coefficient." $\endgroup$
    – whuber
    Commented Jan 11, 2015 at 22:53
  • $\begingroup$ I've tried to incorporate comments into the revised answer. I still think that the question's focus is on correlation rather than regression, so I've mainly focused on correlation. $\endgroup$
    – Anthony
    Commented Jan 12, 2015 at 19:40
  • $\begingroup$ Thank you so much for this reply Anthony, it has really helped my understanding! $\endgroup$
    – Mary
    Commented Jan 12, 2015 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.