Answer
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
(b) $z_0=3.07$
(c) $z_α=z_{0.05}=1.645$
(d) $P$-value $=0.0011$
Work Step by Step
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
(b) $p̂ _1=\frac{x_1}{n_1}=\frac{368}{541}=0.680$ and $p̂ _2=\frac{x_2}{n_2}=\frac{351}{593}=0.592$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{368+351}{541+593}=0.634$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.680-0.592}{\sqrt {0.634(1-0.634)}\sqrt {\frac{1}{541}+\frac{1}{593}}}=3.07$
(c) $z_α=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
(d) $P$-value $=P(z\gt3.07)=1-P(z\lt3.07)=1-0.9989=0.0011$
Thus, we reject the null hypothesis. There is sufficient evidence that $p_1\gt p_2$.
