Answer
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\lt p̂ _2$
(b) $z_0=-0.36$
(c) $z_α=z_{0.05}=1.645$
(d) $P$-value $=0.3594$
Work Step by Step
(a) $H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\lt p̂ _2$
(b) $p̂ _1=\frac{x_1}{n_1}=\frac{109}{475}=0.229$ and $p̂ _2=\frac{x_2}{n_2}=\frac{78}{325}=0.24$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{109+78}{475+325}=0.23375$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.229-0.24}{\sqrt {0.23375(1-0.23375)}\sqrt {\frac{1}{475}+\frac{1}{325}}}=-0.36$
(c) $z_α=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
(d) $P$-value $=P(z\lt-0.36)=0.3594$
We do not reject the null hypothesis.
