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When variance is unknown, and sample size is big enough, we will use sample variance and t-distribution as substitutes when we are constructing a confidence interval for population mean. But when we are constructing confidence intervals for population proportion (for example the p in Bernouli(p)), we will simply use sample variance without subsitituting the normal distribution with t distribution. Why should we do this? Shouldn't proportion be a special case of mean? So why don't we use t-distribution when we are constructing CI for bernouli distribution?

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The t-distribution is used when there are two unknown population parameters, the mean and the variance, which in our case is $p$ and $p(1-p)$, respectively.

A Bernoulli Random Variable only really has one parameter to estimate. If we know $p$, which is considered the population mean, then we know the variance since this is just $p(1-p)$. The sample proportion $\hat{p}$ gives us estimates of both the mean and variance.

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