# When is a poor reliability index acceptable?

I have tested this particular scale and it turns out that it has low reliability index (a=4.5). I have already deleted two items but it only increased a bit by a=.67. My sample size is 25 as part of pilot testing. Does adding more participants increase the reliability index?

I wanted the reliability index to reach a=0.80 without deleting any items as there are only about 7 items originally and if I push thru with deleting another item to increase reliability, I may run short again on reaching the acceptable cutoff value.

What should i do?

• The title of the question does not quite correspond to the body. Jul 30, 2014 at 10:15
• In general, deleting items just because they reduce reliability should be done with caution. Make sure you look closely at the item wording to make sure you understand why they could be reducing reliability. It could indicate you have a multidimensional construct. Jan 9, 2015 at 9:06

Regarding the title question: If there is really no better scale of the construct you wish to measure, a low reliability may be better than nothing, but it might be an indication that the thing you are trying to measure isn't a thing. If you told us what you are trying to measure and what the questions are, we might be able to help more.

Regarding the questions in the body: Adding more subjects does not, in general, increase or decrease the reliability estimate, but it will increase its precision.

As to what you should do - examine each question and see what's going on. Look at the full correlation matrix of all the questions. Look for reverse coded questions. Consider adding questions.

I'm assuming you are using Cronbach's $$\alpha$$ to measure reliability. Adding on to @Peter's comment regarding sample size and reliability, look at the formula for $$\alpha$$:

$$\alpha=\frac{K}{K-1} \left(1-\frac{\Sigma^K_i\sigma^2_Y{_i}}{\sigma^2_X=1}\right)$$

Where K is the number of items. Notice $$n$$ doesn't fit in there anywhere. The classic definition of reliability is:

$$\text{Reliability}=\frac{\text{True Score Variance}}{\text{Observed Score Variance}}$$

Again, $$n$$ isn't there.