Answer
$X^2\lt X_α^2$: the null hypothesis is rejected.
There is sufficient evidence to conclude that the grade distribution of home-schooled children is different in her district from the national grade distribution.
Work Step by Step
$H_0:$ the grade distribution of home-schooled children is equal in her district to the national grade distribution.
$H_1:$ the grade distribution of home-schooled children is different in her district from the national grade distribution.
Total: 25 home-schooled children.
Expected count of K-3: $25\times0.329=8.225$
Expected count of 4-8: $25\times0.393=9.825$
Expected count of 9-12: $25\times0.278=6.95$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(15-8.225)^2}{8.225}+\frac{(7-9.825)^2}{9.825}+\frac{(3-6.95)^2}{6.95}=8.638$
$k=3$. So, $d.f.=3-1=2$
$X_α^2=X_{0.05}^2=5.991$
(According to Table VII, for d.f. = 3 and area to the right of critical value = 0.05)
Since $X^2\lt X_α^2$, we reject the null hypothesis.
