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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?

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  • $\begingroup$ 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

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