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My statistics book points out that when using sample mean and sample SD for numerical variables the confidence interval is calculated using the Student's t distribution instead of the normalized normal distribution (Z value). However, in the formula for the confidence limit for proportions the book uses the Z value both for populations (population proportion known) and for samples (only sample proportion known). I would expect that the the Student's t distribution should also be used calculating the confidence intervals for proportions based on a sample?

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If the proportion $p$ is not close to 0 or 1 and the number of observations $N$ is large, the normal distribution describes with a good approximation the distribution of $p$ as estimated from the sample. If one wants to be more precise, then it is true that the proportion is a mean, but it is not a mean of real (continuous) numbers, it is a mean of integers (0,1). So the Student's distribution would not be the correct one. You should use instead the Binomial proportion confidence interval, there are some different approximations that you can use and that keep into account that the proportion estimation from the sample is discretized.

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  • $\begingroup$ My course taught that we should have both n*p>=15 and n*(1-p)>=15 ..... it seems like this guidance accords with your advice, and that it takes into account sufficient population size and effects from being close to zero or one? Unfortunately, the course did not proffer an explanation of why Student's T should not be used, or that an alternative is the Binomial proportion CI - so thanks for providing that clarification. $\endgroup$
    – James
    Jul 17, 2018 at 22:29

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