8

I am pretty new to Cronbach's Alpha, so I have a question about how to find it using R.

As far as I am aware, Cronbach's alpha is a measure of internal consistency, but to use the alpha() function in R (in the psych package) you need a dataframe or matrix - which has two+ variables. Why is this the case? Shouldn't you only need a list? Or am I understanding Cronbach's Alpha wrong?

Stephan Kolassa
  • 95,027
  • 13
  • 197
  • 357
japem
  • 81
  • 1
  • 1
  • 2
  • 1
    Note that if you are only asking about how R (/ an R function) works, that would be off-topic here. I think the idea is that there there is (or isn't) some level of consistency among several items. Thus you have columns w/ several items & rows w/ different raters. – gung - Reinstate Monica Dec 16 '14 at 00:55

1 Answers1

12

Cronbach's $\alpha$ is a measure of internal consistency of a questionnaire or test. It says how correlated the items are that are included in the scale. This is the reason why you need preferably much more than two items: you cannot correlate one item with itself and if you had only two items you could use a "traditional" correlation between the two items. So you need at least a few of them.

You also ask why a matrix or data.frame are needed instead of list. This is a broader topic about R's data types, however, questionnaire data that are used for calculation of Cronbach's $\alpha$ consist of several items, let's say $k$ and responses for those items by a group of $n$ individuals, so $n\times k$ matrix is a natural way of storing this kind of data. If you have this kind of data saved as a list (e.g., $k$ vectors of length $n$) you can always transform the list into a data.frame or a matrix.

What you have to remember with Cronbach's $\alpha$ is that it is a correlation measure, so you would get perfect $\alpha$ for a scale consisting of several items that are identical, while this would be a very poor questionnaire. So the general idea that correlated items are the best ones has its flaws and you have to remember about that. That is one of the reasons why you should rather not use $\alpha$ alone for psychometric analysis, but combine it with other methods, e.g. with Item Response Theory based methods being one of the most popular nowadays (check e.g. Comparison of Classical Test Theory and Item Response Theory and Their Applications to Test Development paper by Ronald K. Hambleton and Russell W. Jones ).

Check out also alpha documentation or the tutorial on personality-project.org (site by psych library developer with a great deal of information about psychometrics and R).

Tim
  • 108,699
  • 20
  • 212
  • 390
  • I know, it's impressive to me how well people do here. – gung - Reinstate Monica Dec 19 '14 at 21:12
  • With "list", OP meant what in R is called a vector. –  Nov 18 '15 at 18:07
  • @what to compute $\alpha$ you need at least two variables (OP mentions it himself/herself) so you cannot compute it with a vector, so it seems he/she meant R `list`. OP did not provide any clarification on this. – Tim Nov 18 '15 at 18:36
  • @Tim I agree that OP didn't clarify their question, but they specifically ask why you need an object with "two+ variables", instead of "only" a "list", and in the context of "two or more" versus "only" I can only understand "list" to contain "only" one variable, that is, a vector. Since OP also emphasizes "*internal* consistency", it seems to me that they were confused about why you need more than one variable to calculate the internal consistency of only one variable, because they did not understand that for cronbach's alpha to make sense a scale must consists of more than one item/variable. –  Nov 19 '15 at 06:18
  • @what even if so, the very first sentence in my answer is about what $\alpha$ is. – Tim Nov 19 '15 at 07:59
  • @Tim Yes, yes, wasn't saying you didn't answer that, just that maybe your answer was a bit beyond the basic misunderstanding of OP. When I started out with statistics, I remember thinking that each xi in var1 was an item (instead of an answer by a different person to the same item) and not understanding the difference between a scale (1 to 5) and a scale (Extroversion), and I can easily see how someone might not understand why you would need some outside comparison (other items or scales) to evaluate the validity of one scale = item = variable. Your answer was too complicated, I believe. –  Nov 19 '15 at 08:09
  • Which is why, I believe, your answer was upvoted by others but not accepted by OP, despite being correct. –  Nov 19 '15 at 08:13
  • @what this is a guessing game and I don't see any point in guessing "what could OP have possibly meant by his/her question?". OP did not provide any comments on the answer and my answer relates directly to to the question asked, it was upvoted few times so it seems somehow valid. I see no point in this discussion. – Tim Nov 19 '15 at 08:18
  • 1
    @StephanKolassa thanks! Unfortunatelly it is unavailable, but I was able to find it archived on archive.org – Tim May 17 '16 at 07:12