Answer
$X^2\lt X_α^2$: the null hypothesis is not rejected. There is not enough evidence to conclude that the distribution of M&Ms is different from what is stated.
Work Step by Step
Total plain M&Ms: $53+66+38+96+88+59=400$
Expected count of Brown: $400\times0.12=48$
Expected count of Yellow: $400\times0.15=60$
Expected count of Red: $400\times0.12=48$
Expected count of Blue: $400\times0.23=92$
Expected count of Orange: $400\times0.23=92$
Expected count of Green: $400\times0.15=60$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(53-48)^2}{48}+\frac{(66-60)^2}{60}+\frac{(38-48)^2}{48}+\frac{(96-92)^2}{92}+\frac{(88-92)^2}{92}+\frac{(59-60)^2}{60}=3.569$
$k=6$. So, $d.f.=6-1=5$
$X_α^2=X_{0.05}^2=11.070$
(According to Table VII, for d.f. = 5 and area to the right of critical value = 0.05)
Since $X^2\lt X_α^2$, we do not reject the null hypothesis.
