Answer
$z_0\gt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that the proportion of subjects taking Zoloft who experienced dry mouth is greater than the proportion of subjects taking the placebo who experienced dry mouth.
Work Step by Step
$N_1,n_1~and~p_1$ refer to experimental group and $N_2,n_2~and~p_2$ refer to control group.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{77}{553}=0.1392$ and $p̂ _2=\frac{x_2}{n_2}=\frac{34}{373}=0.0912$
Requirements:
$n_1p̂ _1(1-p̂ _1)=553\times0.1392(1-0.1392)=66.26\geq10$
$n_2p̂ _2(1-p̂ _2)=373\times0.0912(1-0.0912)=30.92\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{77+34}{553+373}=0.1199$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.1392-0.0912}{\sqrt {0.1199(1-0.1199)}\sqrt {\frac{1}{553}+\frac{1}{373}}}=2.21$
Right-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: $\frac{1.64+1.65}{2}=1.645$
Since $z_0\gt z_α$, we reject the null hypothesis.
