Answer
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
(b) $z_0=-2.99$
(c) $z_{\frac{α}{2}}=z_{0.025}=1.96$
(d) $P$-value $=0.0028$
Since $z_0\lt -z_{\frac{α}{2}}$, there enough evidence to conclude that $p̂ _1\ne p̂ _2$.
Work Step by Step
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
(b) $p̂ _1=\frac{x_1}{n_1}=\frac{804}{874}=0.9199$ and $p̂ _2=\frac{x_2}{n_2}=\frac{902}{954}=0.9455$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{804+902}{874+954}=0.9333$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.9199-0.9455}{\sqrt {0.9333(1-0.9333)}\sqrt {\frac{1}{874}+\frac{1}{954}}}=-2.19$
(c) $z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
Also, $-z_{\frac{α}{2}}=-z_{0.025}=-1.96$
(d) $P$-value $=2P(z\gt|z_0|)=2P(z\gt2.19)=2[1-P(z\lt0.35)]=2(1-0.9986)=0.0028$
Thus, we reject the null hypothesis.
