Answer
$z_0\gt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that the new procedure cures a higher percentage of patients.
Work Step by Step
$N_1,n_1~and~p_1$ refer to the new procedure and $N_2,n_2~and~p_2$ refer to the standard treatment.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{697}{2115}=0.3296$ and $p̂ _2=\frac{x_2}{n_2}=\frac{642}{2105}=0.3050$
Requirements:
$n_1p̂ _1(1-p̂ _1)=2115\times0.3296(1-0.3296)=467\geq10$
$n_2p̂ _2(1-p̂ _2)=2105\times0.3050(1-0.3050)=446\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{697+642}{2115+2105}=0.3173$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.3296-0.3050}{\sqrt {0.3173(1-0.3173)}\sqrt {\frac{1}{2105}+\frac{1}{2115}}}=1.72$
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.
