Answer
(a) $z_0\gt z_α$: null hypothesis is not rejected.
(b) $P(z\lt z_0)\gtα$: null hypothesis is not rejected.
Work Step by Step
$np_0(1-p_0)=400\times0.25(1-0.25)=75\gt10$
$p̂ =\frac{x}{n}=\frac{96}{400}=0.24$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.24-0.25}{\sqrt {\frac{0.25(1-0.25)}{400}}}=-0.46$
(a) $z_α=z_{0.1}$
If the area of the standard normal curve to the right of $z_{0.1}$ is 0.1, then the area of the standard normal curve to the left of $z_{0.1}$ is $1−0.1=0.9$
According to Table V, the z-score which gives the closest value to 0.9 is 1.28.
So, $-z_α=-1.28$
Since $z_0\gt -z_α$, we do not reject the null hypothesis.
(b) $P$-value $=P(z\lt z_0)=P(z\lt-0.46)=0.3228$
Since $P(z\lt z_0)\gtα$, we do not reject the null hypothesis.
