1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.