Solved – How to calculate confidence intervals for a population proportion (ordinal variable)

How can I compute confidence intervals for an ordinal variable (or Likert-type scale)?

My sample size is 430, and the proportion of each category is as follows: 5 = 24%, 4 = 29%, 3 = 27%, 2 = 15%, and 1 = 5%.

Is it possible to calculate the 95% CI of, say, category 5 (24%)? I tried the following:

Margin of error: $sqrtfrac{p(1-p)}{n} = sqrtfrac{0.24(1-0.24)}{430} = 0.0206$, CI: $0.24 pm 1.96*0.0206$

But I'm not sure if it's correct. Can anyone help?

The fact that it's ordered makes no difference. Under the usual the usual assumptions for a Bernoulli process*, the proportion in a given category is just a binomial proportion (an observation is either in the category of interest, or it isn't), so any of the methods for binomial proportion confidence intervals will work just as well as they do in any other binomial case.

In the case where $n$ is sufficiently large$^dagger$, the asymptotic normal interval you mention should be fine.

* Independence is probably reasonable (/approximately true); the assumption that the probability is the same for every observation is less likely to be true (however, it's not likely to be much of an issue even if it isn't very close to true. Even a fair bit of variation about the mean proportion will still tend to lead to very similar answers)

$^dagger$ i.e. large enough that the interval is sufficiently accurate for your purposes; it should be fine for proportions like 0.24, but may not work so well for much smaller proportions depending on how picky you are about coverage.

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