Answer
$z_0\lt -z_α$: null hypothesis is rejected.
There is enough evidence to conclude that $p_1\lt p_2$.
Work Step by Step
$H_0:~p_1=p_2$ versus $H_1:~p_1\lt p_2$
$p̂ _1=\frac{x_1}{n_1}=\frac{40}{135}=0.2963$ and $p̂ _2=\frac{x_2}{n_2}=\frac{60}{150}=0.4$
Requirements:
$n_1p̂ _1(1-p̂ _1)=135\times0.2963(1-0.2963)=28.1\geq10$
$n_2p̂ _2(1-p̂ _2)=150\times0.4(1-0.4)=36\geq10$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{40+60}{135+150}=0.3509$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.2963-0.4}{\sqrt {0.3509(1-0.3509)}\sqrt {\frac{1}{135}+\frac{1}{150}}}=-1.83$
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 $z_0\lt -z_α$, we reject the null hypothesis.
