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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?

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    $\begingroup$ Strictly speaking, it is not a t-statistic nor a z-statistic. Even if $p_0$ were known exactly, then $T$ is not exactly normal even though its distribution converges to the normal distribution by force of the CLT for $n\to\infty$. So that is a first approximation. A second approximation is replacing $p_0$ by $\hat{p}$ in the denominator in order to get a statistic that you can actually compute from the sample under the null. For any finite $n$, $T$ is not normal or $t$-distributed, but there is convergence ("almost surely") to the normal distribution asymptotically for $n\to\infty$. $\endgroup$ Nov 3 '20 at 19:22
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I agree with you. When null is true (the true p is p0), the test statistic T is indeed a Z statistic, since when the true p is p0, the true variance is p0(1-p0)/n.

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