I have the following quantity $$ Z = \frac{\sqrt{n}(\hat{p}-0.2)}{\sqrt{0.16}}. $$ I have already proven by the WLLN that $\hat{p} \stackrel{P}{\longrightarrow} p = 0.2$. Given that, can I claim that $Z \stackrel{P}{\longrightarrow} 0$? What is troubling me is that, in any limiting process, this $\sqrt{n}$ would blow up. So even though ($\hat{p} - 0.2$) converges to $0$, I'm afraid the $\sqrt{n}$ would "ruin" the converge.
For some context, we have that $\hat{p} = (1/n)\sum_{i=1}^{n}X_i = \bar{X}_n,$ where $X_i \sim Bernoulli(p)$. We can assume that the real value of $p$ is $p=0.2$
This is central limit theorem. In general, for i.i.d. random variables with finite second moment, we have $$ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\longrightarrow Z\sim N(0,1). $$ In this case, each $X$ is a Bernoulli r.v. so $\mu=p$ and $\sigma=\sqrt{p(1-p)}$.
It's really in the form $$ Z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}} $$