Answer
(a) $-z_\frac{α}{2}\lt z_0\lt z_\frac{α}{2}$: the null hypothesis is not rejected.
(b) $P$-value $\gtα$: the null hypothesis is not rejected.
Work Step by Step
$np_0(1-p_0)=1000\times0.4(1-0.4)=240\gt10$
$p̂ =\frac{x}{n}=\frac{420}{1000}=0.42$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.42-0.4}{\sqrt {\frac{0.4(1-0.4)}{1000}}}=1.29$
(a) $z_{\frac{α}{2}}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
Also, $-z_\frac{α}{2}=-2.575$
Since $-z_\frac{α}{2}\lt z_0\lt z_\frac{α}{2}$, we do not reject the null hypothesis.
(b) $P$-value $=2P(z\gt|z_0|)=2P(z\gt1.29)=2(1-0.9015)=0.197$
Since $P$-value $\gtα$, we do not reject the null hypothesis.
