Answer
$µ_1=µ_2=µ_3\neµ_4$
Work Step by Step
$q_{α,ν,k}=q_{0.05,20,4}=5.018$
(According to table IX, for $ν=12$, $k=3$ and α = 0.05)
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\ne µ_2$
$q_0=\frac{x ̅_2-x ̅_1}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{49.1-42.6}{\sqrt {\frac{26.2}{2}(\frac{1}{6}+\frac{1}{6})}}=3.111$
Since $q_0\lt q_{α,ν,k}$, do not reject the null hypothesis.
$H_0:~µ_1=µ_3$ versus $H_1:~µ_1\ne µ_3$
$q_0=\frac{x ̅_3-x ̅_1}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_3})}}=\frac{46.8-42.6}{\sqrt {\frac{26.2}{2}(\frac{1}{6}+\frac{1}{6})}}=2.010$
Since $q_0\lt q_{α,ν,k}$, do not reject the null hypothesis.
$H_0:~µ_1=µ_4$ versus $H_1:~µ_1\ne µ_4$
$q_0=\frac{x ̅_4-x ̅_1}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{63.7-42.6}{\sqrt {\frac{26.2}{2}(\frac{1}{6}+\frac{1}{6})}}=10.097$
Since $q_0\gt q_{α,ν,k}$, reject the null hypothesis.
$H_0:~µ_2=µ_3$ versus $H_1:~µ_2\ne µ_3$
$q_0=\frac{x ̅_2-x ̅_3}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{49.1-46.8}{\sqrt {\frac{26.2}{2}(\frac{1}{6}+\frac{1}{6})}}=1.101$
Since $q_0\lt q_{α,ν,k}$, do not reject the null hypothesis.
$H_0:~µ_2=µ_4$ versus $H_1:~µ_2\ne µ_4$
$q_0=\frac{x ̅_4-x ̅_2}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{63.7-49.1}{\sqrt {\frac{26.2}{2}(\frac{1}{6}+\frac{1}{6})}}=6.987$
Since $q_0\gt q_{α,ν,k}$, reject the null hypothesis.
$H_0:~µ_3=µ_4$ versus $H_3:~µ_3\ne µ_4$
$q_0=\frac{x ̅_2-x ̅_3}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{63.7-46.8}{\sqrt {\frac{26.2}{2}(\frac{1}{6}+\frac{1}{6})}}=8.087$
Since $q_0\gt q_{α,ν,k}$, reject the null hypothesis.
