Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that the proportion of cardiovascular events in each treatment group is different.
$z_0=-1.44$
Work Step by Step
$n_1~and~p_1$ refer to the aspirin group and $n_2~and~p_2$ refer to the placebo group.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{477}{19,934}=0.02393$ and $p̂ _2=\frac{x_2}{n_2}=\frac{522}{19,942}=0.02618$
Requirements:
$n_1p̂ _1(1-p̂ _1)=19,934\times0.02393(1-0.02393)=465.6\geq10$
$n_2p̂ _2(1-p̂ _2)=19,942\times0.02618(1-0.02618)=508.3\geq10$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{477+522}{19,934+19,942}=0.02505$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.02393-0.02618}{\sqrt {0.02505(1-0.02505)}\sqrt {\frac{1}{19,934}+\frac{1}{19,942}}}=-1.44$
Two-tailed test:
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
Also, $-z_{\frac{α}{2}}=-z_{0.025}=-1.96$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
