Answer
$z_0\gt -z_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that $p̂ _1\lt p̂ _2$.
Work Step by Step
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\lt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{156}{650}=0.24$ and $p̂ _2=\frac{x_2}{n_2}=\frac{143}{550}=0.26$
Requirements:
$n_1p̂ _1(1-p̂ _1)=650\times0.24(1-0.24)=118.56\geq10$
$n_2p̂ _2(1-p̂ _2)=550\times0.26(1-0.26)=105.82\geq10$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{156+143}{650+550}=0.2492$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.24-0.26}{\sqrt {0.2492(1-0.2492)}\sqrt {\frac{1}{650}+\frac{1}{550}}}=-0.80$
Left-tailed test:
$z_α=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
So, $-z_α=-1.645$
Since $z_0\gt -z_α$, we do not reject the null hypothesis.
