Suppose that $ (\Omega,\Sigma,\mathsf{P}) $ is a probability space and that $ (X_{k})_{k \in \mathbb{N}} $ is a sequence of i.i.d. Bernoulli trials on $ (\Omega,\Sigma,\mathsf{P}) $, each with probability of success $ p \in (0,1) $. If we define another sequence $ (\hat{P}_{n})_{n \in \mathbb{N}} $ of random variables on $ (\Omega,\Sigma,\mathsf{P}) $ by $$ \forall n \in \mathbb{N}: \qquad \hat{P}_{n} \stackrel{\text{df}}{=} \frac{1}{n} \sum_{k = 1}^{n} X_{k}, $$ then according to the Central Limit Theorem, we have $$ \forall z \in \mathbb{R}: \qquad \lim_{n \to \infty} \mathsf{P} \! \left( \frac{\hat{P}_{n} - p}{\sqrt{p (1 - p) / n}} \leq z \right) = \Phi(z), $$ where $ \Phi $ denotes the standard normal c.d.f. For each $ n \in \mathbb{N} $, we call $ \hat{P}_{n} $ a sample proportion for a sample of size $ n $.
When most statistics textbooks discuss confidence intervals for a sample proportion, they implicitly claim that $$ \frac{\hat{P}_{n} - p}{\sqrt{\hat{P}_{n} (1 - \hat{P}_{n}) / n}} \stackrel{\text{d}}{\longrightarrow} \operatorname{N}(0,1), $$ which is the same as saying that $$ \forall z \in \mathbb{R}: \qquad \lim_{n \to \infty} \mathsf{P} \! \left( \frac{\hat{P}_{n} - p}{\sqrt{\hat{P}_{n} (1 - \hat{P}_{n}) / n}} \leq z \right) = \Phi(z). $$ However, I was unable to rigorously establish this claim using the Central Limit Theorem.
Could anyone kindly provide references? Thanks!
The most straightforward proof of this result requires knowledge of Slutsky's theorem, which in turn requires the concept of convergence in probability. Write $$\frac{\hat{P}_{n} - p}{\sqrt{\hat{P}_{n} (1 - \hat{P}_{n}) / n}}= \frac{\hat{P}_{n} - p}{\sqrt{p(1-p) / n}} \cdot \sqrt{ \frac{p(1-p)}{\hat P_n(1-\hat P_n)}}, $$ a product of two factors. The first factor converges in distribution to the standard normal, by the central limit theorem. The second factor converges almost surely to the constant value $1$, by the law of large numbers. Now apply Slutsky's theorem, since convergence a.s. implies convergence in probability.