Suppose we're testing whether more than $100p_0$% Bernoulli trials are successful at the $\alpha$% significance level. We take a sample of $n$ Bernoulli trials and find that $\hat p$ trials are successful.
Our hypothesis test is:
$$H_0: p \leq p_0 \text{ Vs. }H_1:p>p_0$$
Then our test statistic is:
$$T=\frac{\hat p -p_0}{\sqrt{p_{0}(1-p_{0})/n}}$$
My professor says $T$ has a t-distribution with $(n-1)$ degrees of freedom, i.e. it is a t-statistic.
I am not sure what to think of this. Reflecting on the z-statistic for a sample mean, $\frac{\bar X-\mu}{\sigma/\sqrt{n}}$, if we did not know $\sigma$, we would estimate this with the sample standard deviation, $s$, and swap $\sigma$ for $s$. Therefore, this becomes a t-statistic, $\frac{\bar X-\mu}{s/\sqrt{n}}$.
In the case above, we have computed $\sqrt{p_{0}(1-p_{0})/n}$. We assume to know the value of $p_0$ under $H_0$. Therefore, we have nothing to estimate. So why isn't this a z-statistic rather than a t-statistic, as my professor claims?
Can someone more experienced comment on what distribution this test statistic has and why?
