Answer
(a) $H=1.223$
(b) Critical value: 5.5985
(c) $H\lt5.5985$: 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.
$R_1=7.5+4+1.5=13$
$R_2=10+4+9+1.5=24.5$
$R_3=4+6+11+7.5=28.5$
$H=\frac{12}{N(N+1)}[\frac{R_1^2}{n_1}+\frac{R_2^2}{n_2}+\frac{R_3^2}{n_3}]-3(N+1)=\frac{12}{11(11+1)}[\frac{13^2}{3}+\frac{24.5^2}{4}+\frac{28.5^2}{4}]-3(11+1)=1.223$
(b) Small sample case:
Critical value: 5.5985
(According to Table XV, for $n_1=3$ and $n_2=n_3=4$ and α = 0.049, the closest value to 0.05)
Notice that the option $n_1=3$ and $n_2=n_3=4$ does not exist in Table XV, but given that the order of the populations to be tested does not matter, we use $n_1=n_2=4$ and $n_3=3$
(c) Since $H\lt5.5985$, we do not reject the null hypothesis.
