My study is about the correlation of the ratings of the respondents to the product to their willingness to purchase it. I also made use of likert scale. My values are ordinal and i also have outliers after making use of scatter plot. But it still showed linearity.

  • $\begingroup$ check the distributions if they don't look normal go for a spearman. if you are overly concerned in having ordinal data go for a spearman. $\endgroup$ Feb 25, 2020 at 10:09
  • $\begingroup$ I'm sure there are better discussions on this site, but I made a few comments on Pearson and Spearman here: stats.stackexchange.com/questions/451269/… $\endgroup$ Feb 26, 2020 at 16:28
  • $\begingroup$ Welcome to CV, Alyssa Lopez! Are you trying to measure linear association, or monotonic association? $\endgroup$
    – Alexis
    Jul 27, 2020 at 16:50

1 Answer 1


If you assume linearity behind your ordinal values (linearly assign values to the ranks), then Pearson and Spearman have identical coefficients!

Spearman is used to diminish the effect of outliers (which are not part of your assumed normal-distribution of the observable) by assigning ordinary values to them. Since your data is already in ordinary scale, it doesn't do anything other than Pearson.

edit: (an example in Python)

Imagine the flowing ordinal values were used to describe your product:

purchase_willingness = [low, rather_low, high, rather_high]
rating = [rather_bad, bad, rather_good, good]

And linearly assigned values by their intrinsic order (that all ordinal scales have):

   low, rather_low, rather_high, high = 11 ,13, 15, 17
   bad, rather_bad, rather_good, good = 31, 33, 35, 37

Then here is the result of Spearman vs Pearson:

scipy.stats.spearmanr(purchase_willingness, rating)
SpearmanrResult(correlation=0.6000000000000001, pvalue=0.3999999999999999)

scipy.stats.pearsonr(purchase_willingness, rating)
(0.6, 0.4)

As you can see, the result is identical. Also note that it is enough to linearly assign values, you don't need to assign ranks or whatever was meant with the comment below.

  • $\begingroup$ This answer is incorrect: it is based on confusing ordinal data with the data ranks. The two needn't be the same (and usually are not). $\endgroup$
    – whuber
    Jul 4, 2020 at 14:53
  • $\begingroup$ I added an example for clarification. $\endgroup$
    – KaPy3141
    Jul 27, 2020 at 16:49
  • 1
    $\begingroup$ Thank you: that shows us what you have in mind. I think what was bothering me was the lack of justification for the linearity assumption, but since you have explained what that is, there's nothing incorrect about your conclusion that the Pearson and Spearman coefficients are the same: this is a consequence of the fact that "linearly assigned" values are an affine transformation of the ranks. $\endgroup$
    – whuber
    Jul 27, 2020 at 17:18
  • $\begingroup$ I appreciate your comment! Sorry for overreacting earlier! $\endgroup$
    – KaPy3141
    Jul 28, 2020 at 9:34

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