Answer
$z_0\gt -z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that $p_1\ne p_2$
Work Step by Step
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{451}{555}=0.8126$ and $p̂ _2=\frac{x_2}{n_2}=\frac{510}{600}=0.85$
Requirements:
$n_1p̂ _1(1-p̂ _1)=555\times0.8126(1-0.8126)=84.5\geq10$
$n_2p̂ _2(1-p̂ _2)=600\times0.85(1-0.85)=76.5\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{451+510}{555+600}=0.832$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.8126-0.85}{\sqrt {0.832(1-0.832)}\sqrt {\frac{1}{555}+\frac{1}{600}}}=-1.70$
Two-tailed test:
$z_{\frac{α}{2}}=z_{0.025}$
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, $-z_{\frac{α}{2}}=-1.96$
Since $z_0\gt -z_{\frac{α}{2}}$, we do not reject the null hypothesis.
