Answer
(a) $z_0\gt z_α$: null hypothesis is rejected.
(b) $P$-value $\lt α$: null hypothesis is rejected.
There is enough evidence to conclude that $p\gt0.6$
Work Step by Step
Requirement:
$np_0(1-p_0)=250\times0.6(1-0.6)=60\gt10$
$p̂ =\frac{x}{n}=\frac{165}{250}=0.66$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.66-0.6}{\sqrt {\frac{0.6(1-0.6)}{250}}}=1.94$
(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) $P$-value $=P(z\gt z_0)=P(z\gt1.94)=1-P(z\lt1.94)=1-0.9738=0.0262$
Since $P$-value $\ltα$, we reject the null hypothesis.
