Answer
$X^2\gt X_α^2$: the null hypothesis is rejected.
The given sample evidence indicates that the distribution of residents in the United States has changed since 2000.
Work Step by Step
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(900-950)^2}{950}+\frac{(1089-1145)^2}{1145}+\frac{(1846-1780)^2}{1780}+\frac{(1165-1125)^2}{1125}=9.240$
$k=4$. So, $d.f.=4-1=3$
$X_α^2=X_{0.05}^2=7.815$
(According to Table VII, for d.f. = 3 and area to the right of critical value = 0.05)
Since $X^2\gt X_α^2$, we reject the null hypothesis.
