Answer
(a) $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
(b) $P$-value $\gt α$: null hypothesis is not rejected.
There is not enough evidence to conclude that $p\ne0.35$
Work Step by Step
Requirement:
$np_0(1-p_0)=420\times0.35(1-0.35)=95.55\gt10$
$p̂ =\frac{x}{n}=\frac{138}{420}=0.3286$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.3286-0.35}{\sqrt {\frac{0.35(1-0.35)}{420}}}=-0.92$
(a) $z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
Also, $-z_{\frac{α}{2}}=-1.96$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
(b) $P$-value $=2P(z\gt |z_0|)=2P(z\gt0.92)=2[1-P(z\lt0.92)]=2(1-0.8212)=0.3576$
Since $P$-value $\gt α$, we do not reject the null hypothesis.
