Answer
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
(b) $z_0=-0.35$
(c) $z_{\frac{α}{2}}=z_{0.025}=1.96$
(d) $P$-value $=0.7264$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, there is not 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) $p̂ _1=\frac{x_1}{n_1}=\frac{28}{254}=0.1102$ and $p̂ _2=\frac{x_2}{n_2}=\frac{36}{301}=0.1196$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{28+36}{254+301}=0.115$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.1102-0.1196}{\sqrt {0.115(1-0.115)}\sqrt {\frac{1}{254}+\frac{1}{301}}}=-0.35$
(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 $=2P(z\gt|z_0|)=2P(z\gt0.35)=2[1-P(z\lt0.35)]=2(1-0.6368)=0.7264$
We do not reject the null hypothesis.
