Answer
$X^2\gt X_α^2$: the null hypothesis is rejected. There is sufficient evidence to conclude that the distribution of M&Ms is different from what is stated.
Work Step by Step
Total plain M&Ms: $61+64+54+61+96+64=400$
Expected count of Brown: $400\times0.13=52$
Expected count of Yellow: $400\times0.14=56$
Expected count of Red: $400\times0.13=52$
Expected count of Blue: $400\times0.24=96$
Expected count of Orange: $400\times0.20=80$
Expected count of Green: $400\times0.16=64$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(61-52)^2}{52}+\frac{(64-56)^2}{56}+\frac{(54-52)^2}{52}+\frac{(61-96)^2}{96}+\frac{(96-80)^2}{80}+\frac{(64-64)^2}{64}=18.738$
$k=6$. 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 $X^2\gt X_α^2$, we reject the null hypothesis.
