Answer
$n_1=201,229$
$n_2=200,745$
$\hat{p_1}=.999836$
$\hat{p_2}=.999427$
$\bar{p}=.999632$
$\bar{q}=.000368$
Work Step by Step
We know that n is the sample size, so we find that $n_1=201,229$ and $n_2=200,745$. We know that $\hat{p} $ is equal to the number of successes divided by the sample size. Thus:
$\hat{p_1}=\frac{201,229-33}{201,229}=.999836$
$\hat{p_2}=\frac{200,745-115}{200,745}=.999427$
We know that $\bar{p}$ is equal to $\frac{x_1+x_2}{n_1+n_2}$. Thus:
$\bar{p}=\frac{401,974-115-33}{401,974}=.999632$
We know that:
$\bar{q}=1-\bar{p}=.000368$
