# Based on factor loadings (in factor analysis) can we give unequal weights to Likert scale items?

After gathering data, we compute score of any Likert (summative) scale (previously identified as factor in factor analysis) by adding up its individual item scores (and maybe dividing the sum by the number of items to get the mean score). In that calculation, we assume that every item in the scale has equal weight. However we know from the factor analysis that some of the items had greater factor loadings than the other ones comprising that scale. They are thus explaining more of the variance. By using those factor loadings is it possible to give unequal weights to items? For example in a 6-item scale maybe item 4 is more effective on that scale score than other items.

Or, to restate my question: Although items of a Likert scale (construct) have not equal factor loadings (explaining variance of that factor) why do researchers generally use Likert scales with equally weighted items?

• I considerably reworded your question after I believed I got it correctly. Please check is it alright for you now. Jul 10, 2012 at 2:05
• See also a related question stats.stackexchange.com/q/191255/3277 Jan 19, 2016 at 11:16
• Jun 20, 2016 at 17:39

Yes, it is possible to supply each item with its own weight. This weight, however, cannot be the loading itself because - you might remember - loading is a regression coefficient of a factor in predicting an item, not vice versa. The weight you imply must be a regression coefficient of an item in predicting a factor. We obtain those weights when we compute factore scores; the weights $\mathbf{B}$ are estimated from inter-item correlation (or covariance) matrix $\mathbf{R}$ and loadings matrix $\mathbf{A}$ typically this way: $\mathbf{B}= \mathbf{R}^{-1} \mathbf{A}$. (If factors were obliquely rotated then in this formula factor structure matrix should replace $\mathbf{A}$.) See also, where coarse and refined methods are considered; coarse method permits using loadings as weights.
If so, then why do researchers generally use Likert scales with equally weighted items? In other words, why do they often prefer just binary weights 1 or 0 in place of the above computed fractional weights? There may be several reasons. To mention just three... First, the above weights $\mathbf{B}$ are not precise (unless we used PCA model rather than factor analysis model per se) due to the fact that the uniqness of an item is not known on the level of each case (respondent), and thereby computed factor scores are only approximation of true factor values. Second, computed weights $\mathbf{B}$ usually will vary from sample to sample and eventually they show not much better than simply 1 vs 0 weights. Third, the weighted-sum-model behind a summative (Likert) construct is a simplification in principle. It implies that the trait that is measured by the scale depends on all its items simultaneously whatever its pronouncedness. But we know that many traits behave differently. For instance, when a trait is weak, it may show only a subset of symptoms (i.e. items), but those expressed in full; as the trait grows stronger, more symptoms join in, some partly expressed, some expressed in full and even replacing those "older" symptoms. This dynamic and unpredictive internal growths of a trait cannot be modeled by weighted linear combination of its phenomena. In this situation, using fine fractional weights is in no way better than using binary 0-1 weights.