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I am attempting to conduct a logistic regression on the average of items in a Likert scale. The scale is: strongly oppose, somewhat oppose, neutral, somewhat support, strongly support and is coded 1 to 5 respectively. I'm rounding the average to reduce the number of classes. One practice I noticed in the literature is that people performed principal components analysis on the items in the Likert scale, confirmed that there is one loading that explains most of the variance before they did the average. Is it necessary to perform PCA and notice one loading with the majority of the variance to take an average of a likert scale?

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  • $\begingroup$ PCA is surely not mandatory and whether this helps at all will depend on your data. $\endgroup$ – Lewian Jan 7 '20 at 15:14
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First, why are you doing logistic regression on the average? Once you take the average, the result is no longer discrete and you could try linear regression.

Second, whether you can average Likert scale items at all isn't completely agreed. Technically, you can't. But a lot of people do it and it seems to "work" for some not-too-precise definition of "work".

Third, while nothing prohibits you from averaging any numbers you like, first doing PCA (or otherwise checking that your variables are all related) is a good idea. You don't want to average numbers that represent different things. E.g. you could average the temperature in degrees C, the wind speed in KPH and the humidity - but the result would be mush.

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    $\begingroup$ Further, rounding makes no sense. If an average makes sense, it doesn't deserve or require rounding. If it's dubious, then making it more nearly discrete won't make it less dubious. Yet further, make sure you reverse whatever answers need reversing before averaging. $\endgroup$ – Nick Cox Jan 7 '20 at 13:43
  • $\begingroup$ I decided to switch to linear regression on the unrounded average. I was rounding to force logistic regression but I am abandoning that idea. The numbers I am averaging are all on the same exact likert scale, and cronbach's alpha is high (>.9). $\endgroup$ – Bayesian1701 Jan 7 '20 at 16:57

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