Answer
There is enough evidence to conclude that the vaccine was effective.
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 $H_1:~p̂ _1\lt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{33}{200,000}=0.000165$ and $p̂ _2=\frac{x_2}{n_2}=\frac{115}{200,000}=0.000575$
Requirements:
$n_1p̂ _1(1-p̂ _1)=200,000\times0.000165(1-0.000165)=32.99\geq10$
$n_2p̂ _2(1-p̂ _2)=200,000\times0.000575(1-0.000575)=114.93\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{33+115}{200,000+200,000}=0.00037$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.000165-0.000575}{\sqrt {0.00037(1-0.00037)}\sqrt {\frac{1}{200,000}+\frac{1}{200,000}}}=-6.74$
Left-tailed test:
Let's use $α=0.01$ level of significance.
$z_α=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33.
So, $-z_α=-2.33$
Since $z_0\lt -z_α$, we reject the null hypothesis.
Notice that the null hypothesis would be rejected even for the $α=0.05$ or $α=0.10$ level of significance. They would provide a higher value of $-z_α$.
