0
$\begingroup$

How can I calculate Cronbach Alpha for two tests that I have administered to 165 people? One test contains 110 items and the other contains 67 items. The scoring is done on the hard copies and the scores of each subscale rather than individual items are entered for further calculations. There are 11 sub-scales in the one with 110 items, and four sub-scales in the one with 67 items. Presumably, the subscales are uni-dimensional in construct, so should I be calculating alpha for them. I mean do the results become invalid if I do not find alpha for my data. Is there any other more practical way of ensuring the internal consistency, or do the original alpha values given by the developers of the scales suffice? The task of putting scores of 165 people for 177 items is not practical. My tables show scores of each subscale (not items) for all the participants. Please help me understand what is the best way to go about it.

$\endgroup$
  • $\begingroup$ My concern is about the amount of time and effort that would be required to do such work. I would like to know a more practical alternative. $\endgroup$ – Hadia Pasha Jun 6 '14 at 12:09
1
$\begingroup$

Cronbach's alpha is applicable only to unidimensional scales. Do you really have many dozens of items taping only a single dimension? What does principal component analysis (or factor analysis) say about the dimensionality of your data?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sorry, I did not make myself clear. The test with 110 items has 11 sub-scales and the test with 67 items has four sub-scales and a full scale score comprising of the total number of items. These are standardized tests and I have bought them from publishers to use them in my research. The manuals give information about the reliability, validity and correlations of the subscales with each other. I am concerned because my supervisor was telling me that alpha values for my sample would be required. How am I supposed to enter scores for each item (110 + 67 = 177 items) for 165 people. $\endgroup$ – Hadia Pasha Jun 6 '14 at 13:15
  • $\begingroup$ I repeat: Cronbach's alpha is only appropriate for unidimensional scales. Alpha values are meaningless otherwise. $\endgroup$ – Alexis Jun 6 '14 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.