Answer
(a) $z_0\gt z_α$: null hypothesis is rejected.
(b) $P(z\gt z_0)\ltα$: null hypothesis is rejected.
Work Step by Step
$np_0(1-p_0)=200\times0.3(1-0.3)=42\gt10$
$p̂ =\frac{x}{n}=\frac{75}{200}=0.375$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.375-0.3}{\sqrt {\frac{0.3(1-0.3)}{200}}}=2.31$
(a) $z_α=z_{0.05}$
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.
(b) $P(z\gt z_0)=P(z\gt2.32)=1-P(z\lt2.31)=1-0.9896=0.0104$
Since $P(z\gt z_0)\ltα$, we reject the null hypothesis.
