Cronbachs Alpha

(Formula)

Relationship between Cronbachs Alpha and inter-item correlation over varying number of items (k)

Internal Consistency Reliability

What are reasons for cronbachs alpha being low? (4)

Test is short (see Figure 1, slide 3)

Average item correlation is low (see Figure 1, slide 3)

Low variability in your total scores or small range of ability in the sample you are testing. (Recall that correlational or covariance-based measures, respectively, are variance dependent)

Test only contains either very easy items or very hard dichotomous items

• This is because item variance is a function of item difficulty p (i.e., proportion of people who answered the item correctly): Var(X) = pq, where q = 1-p

Caveat about Cronbach’s Alpha

What is Cronbach’s Alpha sensitive to?

A high alpha may mean that the items in the test are highly correlated. However, α is also sensitive to the number of items in a test: A larger number of items can result in a larger α, a smaller number of items in a smaller α

=> If you have (even weakly) positively correlated items but many of them, you will get a high Cronbach’s alpha

What about unidimensionality?

A common misunderstanding: Alpha is a measure of unidimensionality: High Alpha —> test is unidimensional (i.e., measures primarily one thing)

If you have (even weakly) positively correlated items but many of them, you will get a high Cronbach’s alpha

• But this doesn’t necessarily imply that your test is one- dimensional!

If you have a one-dimensional test, Cronbach’s Alpha can be interpreted unambiguously and indicates the amount of systematic variance in your data

If you have a multidimensional test, an “overall” Cronbach Alpha doesn’t make sense

• Cronbach’s Alpha for the entire Big Five inventory?

A proper interpretation of Alpha

Internal consistency: Usually assessed by split-half reliability

Randomly divide all items into two sets of equal size

Calculate total scores for each half and correlate the scores

Problem: This split-half reliability coefficient is based on alternate forms that have only one-half the number of items that the full test has

• Reliability is also a function of the number of items but we have effectively halved the number of itemsàAdjust the calculated correlation to estimate the reliability of a scale that is twice the length, using the Spearman Brown formula:

You might see a problem here because you just picked two halves at random —> Likely get different reliabilities from different random splits

Random variation in split-half reliabilities stems from the fact that any estimate of split-half reliability that one gets depends, to some extent, on the particular manner in which one chooses to split the test: First half vs. second half; easy vs. harder items; odd- even split-half method (dividing the test into odd-even numbered items, with odd-numbered items (1, 3, 5 etc.) forming the first group while even-numbered items (2, 4, 6 etc.) forms the second group); etc.)

Wouldn’t it be better to take all possible split-halves into account and to calculate an average (i.e., typical) split-half reliability?

That’s exactly the idea of Cronbach’s Alpha: Average split-half reliability of all possible split-halves

– The formal proof of Cronbach’s Alpha as the average split-half reliability of all possible split-halves can be found in Cronbach (1951, p. 305) or in Lord and Novic (1968, p. 93)

• What we need to determine: Number of ways of picking k unordered outcomes from n possibilities

• This is known as the binomial coefficient or choice number and read "n choose k“

Example: With 8 items, there are 8!/(4!(8 - 4 )!) = 70 possible split-half combinations

– To avoid possible confusion: No, it‘s not 8!/(2!(8-2)!) because we need split-halves, hence 8/2 = 4

Spearman Brown Formula

Reliability is also a function of the number of items but we have effectively halved the number of items —> Adjust the calculated correlation to estimate the reliability of a scale that is twice the length, using the Spearman Brown formula:

Appendix:

What if a test has an odd number of items, so that any split would have one more item in one split versus the other?

As Warrens (2015) showed, alpha is approximately identical to the mean of all split-half reliabilities, if a test consists of an odd number of items and has at least eleven items

Last changed16 days ago