I have 21 items that should load onto four factors. Several of these items had been reverse-coded. However, in a factor analysis, all the reverse coded items load onto one factor (when they actually should load onto four separate ones). I had reverse coded those items prior to running the CFA, but a previous question did suggest I am correct to do this, so what might be the issue here?


1 Answer 1


This is quite common and I have seen this many times before. The reverse-coded items share commonalities because they all share a similar methodological detail. If you measured the same construct using a self-report questionnaire and a physiological measure, for example, you would find that the self-report and physiological indices load on different factors because they are different methods, despite measuring the same construct. I would simply explain the reverse-worded factor as it is, a factor comprised of the items that are reverse-worded because these items share a similar response pattern.

Edit: I will also mention that reverse-coding items should not matter. If the items are reverse-worded, while the rest of the items are worded straightforwardly, the items that are reverse-worded will likely correlate strongly simply due to their reverse-worded nature.

  • $\begingroup$ People sometimes talk about trait factors (which you want) and method factors (which are artifacts). In a confirmatory factor analysis framework you can add a method factor and then have the traits that you are interested in, but you can't do this in EFA. $\endgroup$ Commented May 29, 2013 at 21:38
  • $\begingroup$ In a CFA you could correlate the errors of the different "methods" and such, but indeed, in an EFA you can only explain the factor. I don't see this as a problem if it is explained clearly. $\endgroup$
    – Behacad
    Commented May 29, 2013 at 22:02
  • $\begingroup$ In EFA, you won't get a complex factor structure like that to emerge though, if you've a method artifact, because it tries to rotate to simple structure (and it ain't). $\endgroup$ Commented May 29, 2013 at 22:25

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