Answer
$z_0\gt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that the proportion of Americans aged 15 years or older living alone today is greater than in 2000.
Work Step by Step
$H_0:~p=0.258$ versus $H_1:~p\gt0.258$
Requirement:
$np_0(1-p_0)=400\times0.258(1-0.258)=76.5744\gt10$
$p̂ =\frac{x}{n}=\frac{164}{400}=0.41$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.41-0.258}{\sqrt {\frac{0.258(1-0.258)}{400}}}=6.95$
Using the classical method:
$z_α=z_{0.05}$
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.
