Answer
$z_0\lt -z_{\frac{α}{2}}$: null hypothesis is rejected.
There is significant evidence to conclude that parents of children in high school feel differently today than they did in 1994. It is a serious problem that high school students are not being taught enough math and science.
Work Step by Step
$H_0:~p=0.52$ versus $H_1:~p\ne0.52$
Requirement:
$np_0(1-p_0)=800\times0.52(1-0.52)=199.68\gt10$
$p̂ =\frac{x}{n}=\frac{256}{800}=0.32$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.32-0.52}{\sqrt {\frac{0.52(1-0.52)}{800}}}=-11.32$
Using the classical method:
$z_{\frac{α}{2}}=z_{0.025}$
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, $-z_{\frac{α}{2}}=-1.96$
Since $z_0\lt -z_{\frac{α}{2}}$, we reject the null hypothesis.
