# Applying CLT to difference in sample proportions?

I am reading a derivation that has the following statement about a two-sample proportion test: $$\frac{\hat p_1 - \hat p_2}{\sqrt{(\frac{1}{n_1}+\frac{1}{n_2})\hat p(1-\hat p)}} \stackrel{d}{\rightarrow} N(0,1)$$ where this is under a null hypothesis $H_0: p_1=p_2$ and the estimates in the above expression are the regular sample proportions.

The text justifies this convergence using the Central Limit Theorem, but I am somewhat confused as to why it works here. From what I understand, the CLT applies to sums of iid random variables; while $\hat p_1$ and $\hat p_2$ fit this criteria, their difference does not. How does one prove the above convergence?

• newonlinecourses.science.psu.edu/stat414/node/268 – HelloWorld Jun 1 '18 at 3:11
• "Recall that $\hat p_1 - \hat p_2$ is approximately normally distributed..." How does one show this? As stated in the question, I do not see how the CLT applies here, since this expression is not a sum of iid random variables – mai Jun 1 '18 at 3:17
• – HelloWorld Jun 1 '18 at 3:22
• That link shows that $\hat p$ converges to a normal, which I understand is a basic application of the CLT... but it does not say anything about the difference of two sample proportions – mai Jun 1 '18 at 3:25
• I know that $\hat p_1 \stackrel{d}{\rightarrow} Z_1$ and $\hat p_2 \stackrel{d}{\rightarrow} Z_2$ via a basic application of the CLT (where $Z_1$ and $Z_2$ are normally distributed r.v.s). However, I don't think that it is generally true that convergence is "additive" in the sense that $\hat p_1 - \hat p_2 \stackrel{d}{\rightarrow} Z_1 - Z_2$. – mai Jun 1 '18 at 4:00