Let's just say that the 4 items making one construct are not well correlating (for whatever reason) and their Cronbach is 0.62 despite loading into 1 factor. Is there an alternative way to proceed? Can I skip Cronbach/validity issues/factor analysis by just summing all 4 items and then getting their mean by dividing the total sum by 4? After that, I can use the new mean (variable) as a dependent one in my regression analysis. Has this been done before? Is it an acceptable practice?

  • $\begingroup$ By "loading into 1 factor" do you mean that you performed a factor analysis, that factor analysis had a one factor solution, and all items loaded on that one factor? Or simply that you believe the four items should all be indicators of the same latent construct? $\endgroup$
    – Ian_Fin
    Oct 13, 2016 at 13:33
  • $\begingroup$ They all actually loaded on 1 factor but their Cronbach reliability is low (0.62). Therefore, if I proceed with this, it will be a mess of reliability/validity. I am asking if I can skip all of this and go directly to the alternative suggested above. $\endgroup$
    – R. AS.
    Oct 13, 2016 at 13:34
  • $\begingroup$ How many observations did you have for the factor analysis, and were the variables ordinal or some continuous type? $\endgroup$
    – Ian_Fin
    Oct 13, 2016 at 13:36
  • $\begingroup$ Around 180. All are Likert-scale items (back to the debate of ordinal vs continuous). However, the low reliability is bothering me. I tested a regression model with both the factor scores and the summed mean; both results were approximately similar (and significant). I just want to skip reliability/validity of this problematic construct since its Cronbach is low. $\endgroup$
    – R. AS.
    Oct 13, 2016 at 13:39
  • $\begingroup$ You may want to read around formative measurement models where less emphasis is placed on reliability. However, if you're certain that your items come from a formative measurement model then you may just have to accept that your final scores are likely to be unreliable. $\endgroup$
    – Ian_Fin
    Oct 13, 2016 at 14:02


Your Answer

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

Browse other questions tagged or ask your own question.