0
$\begingroup$

I aim to test correlation for the following rank ordered ordinal variables :

  • 1st set of variables - Total 4 ( These are reduced variables using PCA from original 10 variables to 4, ranked on a scale of 1 to 10)
  • 2nd set of variables - Total 2 ( These are reduced variables using PCA from original 6 variables to 2, ranked on a scale of 1 to 6)
  • 3rd set of variables - Total 9 , ranked on a scale of 1 to 9

I believe Spearman Rank Correlation (SRC) would be the right technique to test correlation. However, the tricky part is all the above variables are ranked at different scales. All the SRC examples show that the variables tested for correlation are measured at same scale.

Is there any way to standardise the rank-ordered ordinal variables on the same scale to be eligible for Spearman Rank Correlation test though they are measured on different scales?

I would be grateful if you could advise on it.

Many Thanks.

Regards, - Tanuja

$\endgroup$
0
$\begingroup$

It is not necessary to have the same scales if you use Spearman's Rank Correlation.

The benefit of rank techniques is that the results do not change if you do monotonous transformations on the original data. Such transformations would take care of different scales.

Rank techniques basically replace the original values by their ranks, i.e. the number if the values were ordered. So if you transform monotonically the original data, their ranks do not change: E.g. the largest value will still be the largest value after the transformation. Consequently, statistics defined on the ranks (like Spearman's Rank Correlation or the Wilcoxon-Mann-Whitney-test) will be the same irrespective of the monotonous transformation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Is there any reference article or weblink which explains that SRC test need not have same scale variables? Secondly, what techniques are used to achieve transformation or does SRC take care of it? $\endgroup$ – Tanuja K Sep 16 '14 at 11:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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