Answer
(a) $z_0\gt z_α$: the null hypothesis is not rejected.
(b) $P(z\lt z_0)\gtα$: the null hypothesis is not rejected.
Work Step by Step
$np_0(1-p_0)=150\times0.55(1-0.55)=37.125\gt10$
$p̂ =\frac{x}{n}=\frac{78}{150}=0.52$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.52-0.55}{\sqrt {\frac{0.55(1-0.55)}{150}}}=-0.74$
(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 $=P(z\lt z_0)=P(z\lt-0.74)=0.2296$
Since $P(z\lt z_0)\gtα$, we do not reject the null hypothesis.
