Why is the z statistic of a binomial proportion test normally distributed? Suppose there is single sample $X_1\dots X_n$ of binary variable $X\sim Bern(p_1)$ and the following hypothesis is being tested:
$$H_0: p_1 = p $$
$$H_1: p_1 \not= p $$
Let us use z-statistic without continuity correction:
$$z = \frac{\hat p -p}{\sqrt{\hat p \hat q / n}}, $$
where $\hat p = \frac 1n \sum X_i,$ $\hat q = 1-\hat p$.
Fleiss in "Statistical Methods for Rates and Proportions" on page 26 states that $z$ is normal just "thanks to the CLT". But this is not obvious to me, because the denominator is also a random variable and we can't use the CLT directly.
Please tell me where can I read about this fact?

The statistic should be $z = \frac{\hat p -p}{\sqrt{ p  q / n}}$ which follows a scaled binomial distribution. This approaches the normal distribution.

But also, the expression $z = \frac{\hat p -p}{\sqrt{\hat p \hat q / n}},$ will be approximately equal to $ \frac{\hat p -p}{\sqrt{ p  q / n}}$, it is the first term in Taylor series expansion around $p$.
