# two-sample proportions hypothesis test: 10 successes and failures rule

I asked this question on mathematics stack exchange, and reposting here as it did not get resolved.

Suppose the sample sizes and number of successes are ($$n_1, y_1$$) and ($$n_2, y_2$$), for the two samples, respectively. Let the true proportions of successes be $$p_1, p_2$$.

Null hypothesis $$H_0$$: $$p_1-p_2 = 0$$

Alternative hypothesis $$H_a$$: $$p_1 - p_2 \ne 0$$

Everywhere I have seen, it is required that both samples need to have at least $$10$$ successes and failures. I understand that a binomial to be approximated by a normal distribution needs to have that condition met for a single distribution.

Here, the null hypothesis is that $$p_1 = p_2$$. Then the estimate for the true proportion $$p$$ under that is $$\hat p = \frac{y_1 + y_2}{n_1 + n_2}$$. Is it not enough that the number of combined successes and failures meet $$y_1 + y_2 > 10$$ and $$n_1 + n_2 - y_1 - y_2 > 10$$, if in addition $$\hat p n_1, (1 - \hat p) n_1, \hat p n_2, (1 - \hat p) n_2 > 10$$? Under the null hypothesis, then, would that imply that the individual samples are drawn from approximately normal distributions. This would then further imply that $$\hat p_1 - \hat p_2$$ is approximately normally distributed.

An example is as follows. Suppose $$n_1=n_2=50, y_1=7,y_2=13$$. Then $$\hat p =(7+13)/100=0.2$$. So $$\hat p n_1=\hat p n_2=10$$, and $$(1− \hat p)n_1=(1−\hat p)n_2=40$$.