Answer
$X^2\gt X_α^2$: 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 $H_1:~p\gt0.258$
$np(1-p)=400\times0.258(1-0.258)=76.5744\gt10$
Total: 400 Americans
From item (a):
$E(15~years~or~older~living~alone)=103.2$
$E(15~years~or~older~not~living~alone)=296.8$
Observed number of 15 years or older living alone: 164
Observed number of 15 years or older not living alone: $400-164=236$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(164-103.2)^2}{103.2}+\frac{(236-296.8)^2}{296.8}=48.28$
$k=2$. So, $d.f.=2-1=1$
$X_α^2=X_{0.05}^2=3.841$
(According to Table VII, for d.f. = 1 and area to the right of critical value = 0.05)
Since $X^2\gt X_α^2$, we reject the null hypothesis.
