Answer
(a) $H=0.875$
(b) Critical value: 5.6923
(c) $H\lt5.6923$: null hypothesis is not rejected.
There is not enough evidence to conclude that the distributions of the populations are different.
Work Step by Step
(a) See the picture.
$R_1=6.5+1.5+11+4.5=23.5$
$R_2=10+12+3+6.5=31.5$
$R_3=4.5+8+1.5+9=23$
$H=\frac{12}{N(N+1)}[\frac{R_1^2}{n_1}+\frac{R_2^2}{n_2}+\frac{R_3^2}{n_3}]-3(N+1)=\frac{12}{12(12+1)}[\frac{23.5^2}{4}+\frac{31.5^2}{4}+\frac{23^2}{4}]-3(12+1)=0.875$
(b) Small sample case:
Critical value: 5.6923
(According to Table XV, for $n_1=n_2=n_3=4$ and α = 0.049, the closest value to 0.05)
(c) Since $H\lt5.6923$, we do not reject the null hypothesis.
