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According to my EFA, items q1-q5 of my questionnaire load on factor1 (loyalty), and q6-10 load on factor2 (satisfaction). Factor loadings are not similar between items within each factor.

I would like to calculate factor scores which retain the scale of my items, but instead of simply averaging the raw scores for q1-q5 for instance (which, despite solving my need for a score that retains my items scale, would only give me a factor-based score), I would like to weight each item by their factor loadings.

I know SPSS can compute regression-based factor scores for each participant, but they're standardized to have a mean of 0 and SD of 1. How would I "un-standardize" them back to the original scale? Or alternatively, is there a better method to achieve my aims of calculating a score for each latent factor?

The_Dude
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  • A hint. What is the variance accounted by the factor? – ttnphns Sep 04 '18 at 15:52
  • Would that be the sum of the squared factor loadings divided by the number of variables in that factor? I'm probably misunderstanding your hint, but if the factor only accounts for 45% of the variance for instance, are you saying it's erroneous to calculate a score for that factor as I intend? – The_Dude Sep 04 '18 at 16:34
  • Please go to https://stats.stackexchange.com/a/126985/3277 and read this paragraph "The scores computed via B are scaled: they have variances equal to or close to..." (with further link there) and also this "In FA (not PCA), regressionally computed factor scores will appear... – ttnphns Sep 04 '18 at 16:41
  • Thanks. Some clarity please: is the (true) variance accounted for by the factor simply equal to the sum of the squared loadings, and not further divided by the number of variables as I first thought? Also, the page you linked states in order to get "factor scores with the true factor's variance, multiply the scores (having standardized them to st.dev. 1) by the sq. root of that variance". However, since the variance of scores only approximates 1 does this mean I need to first truly standardize the scores to have a variance of 1 before multiplying by the sq. root of the "true" variance? – The_Dude Sep 04 '18 at 17:34
  • 1. Yes. 2. Yes. – ttnphns Sep 04 '18 at 17:37

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