I have a vector with 3000 data points consisting of values on a likert scale from 1 to 9. It is ordinal data without normal distribution. Now I would like to calculate the correlation of this vector to three other vectors each of size 3000 as follows:

  1. Correlation to a vector with continuous values.

  2. Correlation to a vector containing 21 different continuous values between 0.25 and 0.8.

  3. Correlation to a vector containing only 4 different values (-2,2,-0.2,0.2).

Is it correct if I'm using for all three cases the Kendal's tau correlation?


Kendal tau is not just for ordinal variables, but rather for the case where the correlation is ordinal. For instance, imagine that the true relationship is $y=\ln x+\varepsilon$, but you don't know it. You got your sample $(x_i,y_i)$, and run a Pearson correlation analysis $cor[x_i,y_i]$. The problem is that although both variables are not just ordinal but cardinal, their relationship is not linear. That's why Pearson correlation is not appropriate here. If you'd know the true relationship, you do look at Pearson correlation of $\ln x$ and $y$, but in this case you don't know this. In such a case, you could run Kendall $\tau$ correlation because it's ordinal and $\ln x$ happens to be a monotonic function, which is often the case.

Hence, Kendall $\tau$ analysis will work in a wider set of circumstances compared to Pearson correlation. Particularly, when the relationship between variables is monotonic it will work. Linear is monotonic, so it will work wherever Pearson work, but for linear relationship Pearson would be a better test.

  • $\begingroup$ Thanks for your explanation. I think I don't know if the correlation is ordinal. As said, one vector contains user ratings on a likert scale from 1 to 9. The other vectors contain remaining available time for completion of a tasks (point 1 above), task difficulty (point 2) and money given to users (point 3). Do you think that this could be ordinal correlations? $\endgroup$
    – BlackHawk
    Sep 7 '18 at 15:39
  • $\begingroup$ If not, should I use spearmen or pearson? $\endgroup$
    – BlackHawk
    Sep 7 '18 at 15:39
  • $\begingroup$ @BlackHawk, I was trying to say that Kendall application is Ok, wherever Pearson works, Kendal will work too. in case where the relationship is linear Pearson will work better though $\endgroup$
    – Aksakal
    Sep 7 '18 at 15:41
  • $\begingroup$ Alright, because I don't know if the relationship is linear I should probably use Kendall. Although I'm getting stronger correlation values when using Pearson. $\endgroup$
    – BlackHawk
    Sep 7 '18 at 15:43
  • $\begingroup$ I am unclear on what you mean by "the correlation is ordinal" here. $\endgroup$
    – Glen_b
    Sep 8 '18 at 2:48

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