Solved – Is it plausible to get a Cronbach’s alpha of .85 with only two Likert-type items

cronbachs-alphalikertpsychologyreliabilityscales

I was reading an article that reported a Cronbach's alpha of .85 for a two-item scale. The scale was measuring self-perceived competence. Each item was on a 7-point Likert-type scale.

Intuitively, that seemed like quite a high Cronbach's alpha for a scale with only two items.

  • Is it plausible to get an alpha that high on a two-item psychological scale based on Likert-type items?
  • How through equations or simulations can I better understand what degree of consistency would be required between the items to achieve a reliability coefficient of .85 with only two 7-point items?

Best Answer

I'm a bit confused by your question. Maybe I miss something in it? Alpha of .85 isn't too high. They say test designed for individual diagnostics / decision making (such as clinical psychological test on result of which important decision about patient are made) alpha should be .90 or higher. The data below shows alpha .855 (and standardized alpha, that is, for standardazed variables, .858). The data look quite realistic, isn't it?

1   1
2   1
2   3
3   3
3   3
4   2
4   3
5   7
5   4
5   7
5   6
5   4
5   4
6   6
6   6
6   5
6   3
6   5
6   5
7   6

As you know, alpha reflects the magnitude of mean inter-item covariance and variance, and standardized alpha - the magnitude of mean inter-item correlation. When there are only 2 items designed to correlate, it is plausible to expect correlation between them to be higher than the mean correlation between, say, 20 items designed to correlate. For the data above, r is .752 which gives st. alpha $\frac{2*.752}{1+(2-1)*.752}=.858$.

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