Answer
$X^2\lt X_α^2$: the null hypothesis is not rejected.
There is not enough evidence to conclude that the die is loaded.
Work Step by Step
$H_0:$ the die is not loaded.
That is, $P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=\frac{1}{6}$
$H_1:$ the die is loaded.
Total:400 dice rolls.
Expected count of 1: $400\times\frac{1}{6}=66.667$
Expected count of 2: $400\times\frac{1}{6}=66.667$
Expected count of 3: $400\times\frac{1}{6}=66.667$
Expected count of 4: $400\times\frac{1}{6}=66.667$
Expected count of 5: $400\times\frac{1}{6}=66.667$
Expected count of 6: $400\times\frac{1}{6}=66.667$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(62-66.667)^2}{66.667}+\frac{(76-66.667)^2}{66.667}+\frac{(76-66.667)^2}{66.667}+\frac{(62-66.667)^2}{66.667}+\frac{(57-66.667)^2}{66.667}+\frac{(67-66.667)^2}{66.667}=4.670$
$k=6$. So, $d.f.=6-1=5$
$X_α^2=X_{0.01}^2=15.086$
(According to Table VII, for d.f. = 5 and area to the right of critical value = 0.01)
Since $X^2\lt X_α^2$, we do not reject the null hypothesis.
