Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is no significant evidence to conclude that the proportion of females aged 15 years of age and older living alone today has changed since 2000.
Work Step by Step
$H_0:~p=0.58$ versus $H_1:~p\ne0.58$
Requirement:
$np_0(1-p_0)=500\times0.58(1-0.58)=121.8\gt10$
$p̂ =\frac{x}{n}=\frac{285}{500}=0.57$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.57-0.58}{\sqrt {\frac{0.58(1-0.58)}{500}}}=-0.45$
Using the classical method:
$z_{\frac{α}{2}}=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
Also, $-z_{\frac{α}{2}}=-z_{0.05}=-1.645$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
