Answer
(a) $z_0\lt z_α$: null hypothesis is rejected.
(b) $P(z\lt z_0)\ltα$: null hypothesis is rejected.
Work Step by Step
$np_0(1-p_0)=250\times0.6(1-0.6)=60\gt10$
$p̂ =\frac{x}{n}=\frac{124}{250}=0.496$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.496-0.6}{\sqrt {\frac{0.6(1-0.6)}{250}}}=-3.36$
(a) $z_α=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33.
So, $-z_α=-2.33$
Since $z_0\lt z_α$, we reject the null hypothesis.
(b) $P$-value $=P(z\lt z_0)=P(z\lt-3.36)=0.0004$
Since $P(z\lt z_0)\ltα$, we reject the null hypothesis.
