I count how many days within a month people have executed a certain activity. The maximum is 30 (i.e., the person executed the activity each day), the minimum is 0 (i.e., the person never executed the activity). I get the following distribution:

enter image description here

I want to calculate the 95% confidence intervals for three proportions, namely for low (0-9), medium (10-19), and high (20-30) activity. Total n = 683, the first group (low) contains 311 cases. From this question I understand that the binomial proportion confidence intervals apply and:

$\hat p \pm 1.96 \sqrt{\frac{\hat p (1 - \hat p)}{n}}$

However, my data seems to be skewed towards the edges, especially having excessive zeros. My question: Is the described approach still valid for calculating 95% CI in my scenario?

  • 1
    $\begingroup$ Assuming this is a sample from a much larger population, you can simply put $\hat p =\frac{311}{683}$ onto your expression. The uncertainty is related to who you sampled rather the pattern of answers. $\endgroup$
    – Henry
    Nov 25 '21 at 23:24

Once you would categorize your ordinal variable into low, medium, and high the skewness will not matter since the probability mass will be reduced to three proportions.


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