1st Yr Statistics Question: Create an approximate $\alpha$ level test of $H_0 : p_1 = p_2$

Let $$X_1$$ and $$X_2$$ be binomial random variables with respective parameters $$n_1, p_1$$ and $$n_2, p_2$$. Show that when $$n_1$$ and $$n_2$$ are large, an approximate level $$\alpha$$ test of $$H_0 : p_1 = p_2$$ versus $$H_1 : p_1 \neq p_2$$ is as follows, reject $$H_0$$ if $$\frac{|X_1/n_1-X_2/n_2|}{\sqrt{\frac{X_1+X_2}{n_1+n_2} \left( 1 - \frac{X_1+X_2}{n_1+n_2}\right) \left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} > z_{\alpha/2}$$ Hint below in screenshot.

My attempt

Given $$H_0$$ is true I can say that $$p = p_1 = p_2$$ and since $$n_1$$ and $$n_2$$ are large, I can use the normal approximation to the binomial to say that

$$V = \frac{X_1 - n_1 p}{\sqrt{n_1 p q}} = \frac{ \frac{X_1}{n_1} - p}{\sqrt{\frac{p q}{n_1}}} \, \dot\sim \, N(0,1)$$

$$W = \frac{X_2 - n_2 p}{\sqrt{n_2 p q}} = \frac{ \frac{X_2}{n_2} - p}{\sqrt{\frac{p q}{n_2}}} \, \dot\sim \, N(0,1)$$

Then we have $$\frac{V-W}{\sqrt{2}} \dot\sim N(0,1)$$ and we can build a two sided hypothesis using the fact that

$$P \left( -z_{\alpha/2} \le \frac{V-W}{\sqrt{2}} \le z_{\alpha/2} \right) = 1-\alpha$$

The book answer seems superior because you don't need to know $$p_1$$ or $$p_2$$. However I'm having difficulty getting rid of those two values. Thanks for your help and patience!

Book problem with hint Define $$\hat p_i=X_i/n_i$$ as the observed binomial proportion, $$i=1,2$$.

Since $$n_1,n_2$$ are large, by CLT $$\frac{\sqrt{n_i}(\hat p_i-p_i)}{\sqrt{p_i(1-p_i)}}\stackrel{L}\longrightarrow N(0,1)\quad,\,i=1,2$$

For a formal derivation of the result, suppose $$n=n_1+n_2$$ and that $$\min(n_1,n_2)\to\infty$$ such that $$n_1/n\to\lambda \in(0,1)$$ (which implies $$n_2/n\to1-\lambda$$). Hence assuming $$X_1$$ and $$X_2$$ are independent,

$$\frac{\sqrt{n}\left((\hat p_1-\hat p_2)-(p_1-p_2)\right)}{\sqrt{\frac{p_1(1-p_1)}{\lambda}+\frac{p_2(1-p_2)}{1-\lambda}}}\stackrel{L}\longrightarrow N(0,1)$$

If $$p$$ is the common value of $$p_1$$ and $$p_2$$ under $$H_0$$, then

$$\frac{\sqrt{n}(\hat p_1-\hat p_2)}{\sqrt{p(1-p)\left(\frac{1}{\lambda}+\frac{1}{1-\lambda}\right)}}\stackrel{L}\longrightarrow N(0,1)\tag{1}$$

Let $$\hat\lambda=n_1/n$$ and define $$\hat p=\hat\lambda \hat p_1+(1-\hat\lambda)\hat p_2=\frac{1}{n}(X_1+X_2)$$

Now note that $$\hat p\stackrel{P}\longrightarrow p$$

Therefore, $$\frac{1}{\sqrt{\hat p(1-\hat p)}}\stackrel{P}\longrightarrow\frac{1}{\sqrt{p(1-p)}}\qquad,\,\hat p\ne 0,1$$

Or, $$\frac{\sqrt{p(1-p)}}{\sqrt{\hat p(1-\hat p)}}\stackrel{P}\longrightarrow 1\tag{2}$$

Applying Slutsky's theorem on $$(1)$$ and $$(2)$$, we get under $$H_0$$,

$$\frac{\sqrt{n}(\hat p_1-\hat p_2)}{\sqrt{p(1-p)\left(\frac{1}{\lambda}+\frac{1}{1-\lambda}\right)}}\times \frac{\sqrt{p(1-p)}}{\sqrt{\hat p(1-\hat p)}}\stackrel{L}\longrightarrow N(0,1)$$

That is, the test statistic under $$H_0$$ is given by $$\color{blue}{T=\frac{\sqrt{n}(\hat p_1-\hat p_2)}{\sqrt{\hat p(1-\hat p)\left(\frac{1}{\hat\lambda}+\frac{1}{1-\hat\lambda}\right)}}\stackrel{L}\longrightarrow N(0,1)}$$

(The expression above is the same as the one given in your question.)

We reject $$H_0$$ approximately at level $$\alpha$$ if $$|\text{observed }T|>z_{\alpha/2}$$.

• Excellent answer, thank you. – HJ_beginner May 22 at 21:57
• Quick notation question - what does the L over the arrow stand for? Does that mean in the limit as n goes to infinity the distribution approaches the same distribution as as unit normal? Something like that? Thanks. – HJ_beginner May 24 at 18:34
• Yes, $L$ denotes convergence in law/convergence in distribution. – StubbornAtom May 24 at 18:37
• Thanks for confirming. You rock! – HJ_beginner May 24 at 19:06

Interpret $$X_i$$ as a sum of $$n_i$$ Bernoulli rv's with parameter $$p_i$$ so the LLN shows that $$\frac {X_i}{n_i} \approx p_i$$ for large $$n_i$$. Now the CLT gives the hint and the denominator of the test statistic can be found by substituting $$X_i/n_i=p_i=p$$.

• Ah, thank you that's a good point, another implication of $n_1$ and $n_2$ large is you can use the LLN for $p_i$ – HJ_beginner May 22 at 2:42