To calculate the confidence-interval (CI) for mean with unknown population standard deviation (sd) we estimate the population standard deviation by employing the t-distribution.
Notably, $CI=\bar{X} \pm Z_{95\% }\sigma_{\bar X}$ where $\sigma_{\bar X} = \frac{\sigma}{\sqrt n}$. But because, we do not have point estimate of the standard deviation of the population, we estimate through the approximation $CI=\bar{X} \pm t_{95\% }(se)$ where $se = \frac{s}{\sqrt n}$
Contrastingly, for population proportion, to calculate the CI, we approximate as $CI = \hat{p} \pm Z_{95\% }(se)$ where $se = \sqrt\frac{\hat{p}(1-\hat{p})}{n}$ provided $n \hat{p} \ge 15$ and $n(1-\hat{p}) \ge 15$
My question is, why are we complacent with standard distribution for population proportion?
Best Answer
Both the standard Normal and Student t distributions are rather poor approximations to the distribution of
$$Z = \frac{\hat p - p}{\sqrt{\hat p(1-\hat p)/n}}$$
for small $n,$ so poor that the error dwarfs the differences between these two distributions.
Here is a comparison of all three distributions (omitting the cases where $\hat p$ or $1-\hat p$ are zero, where the ratio is undefined) for $n=10, p=1/2:$
The "empirical" distribution is that of $Z,$ which must be discrete because the estimates $\hat p$ are limited to the finite set $\{0, 1/n, 2/n, \ldots, n/n\}.$
The $t$ distribution appears to do a better job of approximation.
For $n=30$ and $p=1/2,$ you can see the difference between the standard Normal and Student t distributions is completely negligible:
Because the Student t distribution is more complicated than the standard Normal (it's really an entire family of distributions indexed by the "degrees of freedom," formerly requiring entire chapters of tables rather than a single page), the standard Normal is used for almost all approximations.