Can I run PCA (principal components analysis) if my Likert variables are measured differently? E.g. one Likert variable is in the range 0-6, another one is in the range 0-10 and yet another one is in the range 4-12. These are all risk variables. I want to keep them this way because the risk if you answered a 4 on one scale (first) would be equivalent to the risk if you answered 2 on another scale.

  • $\begingroup$ Why would answering 4 on a scale 0-6 be equivalent to answering 2 on a scale 0-10 ? $\endgroup$ – amoeba says Reinstate Monica Jul 21 '16 at 23:13

Usually, data is normalized before performing a PCA, as this post explains extensively. "Standardizing is usually done when the variables on which the PCA is performed are not measured on the same scale" and this quote seems to fit your problem very well.

If you don't normalize your data, variance will be dominated by the variables with larger ranges. In fact, not normalizing is like weighting your variables with their variances - that is, giving more importance to the variable with range 1-10 than the variable with range 1-6.

Furthermore, if you need to go back to your original scales, you can always denormalize any result.

  • $\begingroup$ What if I want the variables to be weighted (so that certain variables are more important)? Eg I want to keep the fact that certain Likert scales go to 6 while others go to 12. If I run pca based on a correlation matrix (which standardizes the data) does this get rid of this weighting? $\endgroup$ – maya Jul 22 '16 at 23:58
  • $\begingroup$ When I run the pca when all variables range 0-6 and then run again with some variables have been manipulated to range from 0-6 and others 0-12 I get two different models with different interpretations (eg number of components to retain and loading loadings) are different. this makes me believe that even with standardization that happens with the correlation matrix, the range of the Likert scale still matters (which I want). $\endgroup$ – maya Jul 22 '16 at 23:59

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.