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I have heard that Cronbach's alpha is used to measure reliability of a test (to check if a set of items/question are measuring the same construct). How does the Cronbach's alpha compare with the average of Item-Sum correlation or the Item-(Sum minus Item) correlation of all the items?

I have a set of testing items and I am interested in understanding if they are measuring different concepts/constructs or are measuring the same construct.

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The standardized alpha coefficient is more intuitive so let's start there:

$$\alpha_{standardized}=\frac{K\bar{r}}{1+(K-1)\bar{r}}$$

where $K$ is the number of items and $\bar{r}$ is the mean of the non-redundant correlations between items.

The non-standardized alpha coefficient just needs a few changes to the denominator.

$$\alpha=\frac{K\bar{c}}{\bar{v}+(K-1)\bar{c}}$$

where $K$ is the number of items, $\bar{v}$ is the average variance of each item, and $\bar{c}$ is the mean of the non-redundant covariances between items.

Here is some sample data in subject-by-item format (with four items A, B, C, and D):

A   B   C   D
-------------
5   5   5   5
3   1   3   1
5   5   5   5
3   1   3   1
5   5   4   5
1   2   3   3
3   1   1   1
1   1   1   3
3   3   4   4
1   3   1   1
5   5   5   5
1   1   1   1
1   1   1   1
1   1   1   1
3   3   4   3
1   3   3   4
4   4   5   5
5   5   5   5
5   3   3   3
3   2   3   3
5   3   5   5
3   3   3   4
1   1   1   1
1   1   1   1
1   1   3   3
3   3   3   1
3   3   1   1
3   3   1   1
3   3   2   5

To calculate the standardized alpha, we need the inter-item correlations ($r$):

$r_{AB}=.776$
$r_{AC}=.756$
$r_{AD}=.630$
$r_{BC}=.730$
$r_{BD}=.750$
$r_{CD}=.810$

Which makes for a mean inter-item correlation of $\bar{r}=.742$.

$$\alpha_{standardized}=\frac{4(.742)}{1+(4-1).742}=.920$$

To calculate alpha, we need the variances ($v$) and covariances ($c$):

$v_A=2.433$
$v_B=2.101$
$v_C=2.384$
$v_D=2.933$
$c_{AB}=1.754$
$c_{AC}=1.820$
$c_{AD}=1.683$
$c_{BC}=1.633$
$c_{BD}=1.861$
$c_{CD}=2.142$

which makes for a mean variance of $\bar{v}=2.463$ and mean covariance of $\bar{c}=1.816$.

$$\alpha=\frac{4(1.816)}{2.463+(4-1)1.816}=.918$$

Now let's compare these results to your idea of the average item-sum correlation.

$$S=A+B+C+D$$

$r_{AS}=.877$
$r_{BS}=.901$
$r_{CS}=.919$
$r_{DS}=.895$

which makes for a mean item-sum correlation of $\bar{r}_{XS}=.898$.

So your result is similar to but lower than the alpha coefficient(s). Perhaps someone else who is better at algebra than I am can put together an equation for calculating one from the other.

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