Answer
$z_0\gt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that the student-centered model has a higher pass rate than the traditional model.
Work Step by Step
$N_1,n_1~and~p_1$ refer to the student-centered model and $N_2,n_2~and~p_2$ refer to the traditional model.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\gt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{335}{567}=0.5908$ and $p̂ _2=\frac{x_2}{n_2}=\frac{364}{743}=0.4899$
Requirements:
$n_1p̂ _1(1-p̂ _1)=567\times0.5908(1-0.5908)=137.1\geq10$
$n_2p̂ _2(1-p̂ _2)=743\times0.4899(1-0.4899)=185.7\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{335+364}{567+743}=0.5336$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.5908-0.4899}{\sqrt {0.5336(1-0.5336)}\sqrt {\frac{1}{567}+\frac{1}{743}}}=3.63$
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.
