Answer
$z_0\gt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2.
Work Step by Step
$N_1,n_1~and~p_1$ refer to group 1 and $N_2,n_2~and~p_2$ refer to group 2.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{107}{710}=0.1507$ and $p̂ _2=\frac{x_2}{n_2}=\frac{67}{611}=0.1097$
Requirements:
$n_1p̂ _1(1-p̂ _1)=710\times0.1507(1-0.1507)=90.87\geq10$
$n_2p̂ _2(1-p̂ _2)=611\times0.1097(1-0.1097)=59.67\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{107+67}{710+611}=0.1317$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.1507-0.1097}{\sqrt {0.1317(1-0.1317)}\sqrt {\frac{1}{710}+\frac{1}{611}}}=2.20$
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.
