## Elementary Statistics (12th Edition)

$n_1=201,229$ $n_2=200,745$ $\hat{p_1}=0.999836$ $\hat{p_2}=0.999427$ $\overline{p}=0.999632$ $\overline{q}=0.000368$
We know that $n$ is the sample size, hence we find that $n_1=201229$ and $n_2=200745$. We know that $\hat{p}$ is equal to the number of successes divided by the sample size. Thus: $\hat{p_1}=\frac{201,229−33}{201,229}=0.999836$ $\hat{p_2}=\frac{200,745−115}{200,745}=0.999427$ We know that $\overline{p}$ is equal to $\frac{x_1+x_2}{n_1+n_2}$. Thus: $\overline{p}=\frac{401,974−115−33}{401,974}=0.999632$ We know that: $\overline{q}=1−\overline{p}=0.000368$