Answer
The results are ambiguous. We can conclude that $µ_2\ne µ_3$, but we could not conclude that $µ_1\ne µ_2$ or $µ_1\ne µ_3$. There is at least one Type II error (when we do not reject $H_0$ when we should have rejected).
Work Step by Step
$q_{α,ν,k}=q_{0.05,12,3}=3.773$
(According to table IX, for $ν=12$, $k=3$ and α = 0.05)
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\ne µ_2$
$q_0=\frac{x ̅_1-x ̅_2}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{9.5-9.1}{\sqrt {\frac{26.2}{2}(\frac{1}{5}+\frac{1}{5})}}=0.175$
Since $q_0\lt q_{α,ν,k}$, do not reject the null hypothesis.
$H_0:~µ_1=µ_3$ versus $H_1:~µ_1\ne µ_3$
$q_0=\frac{x ̅_3-x ̅_1}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_3})}}=\frac{18.1-9.5}{\sqrt {\frac{26.2}{2}(\frac{1}{5}+\frac{1}{5})}}=3.757$
Since $q_0\lt q_{α,ν,k}$, do not reject the null hypothesis.
$H_0:~µ_2=µ_3$ versus $H_1:~µ_2\ne µ_3$
$q_0=\frac{x ̅_3-x ̅_2}{\sqrt {\frac{s^2}{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{18.1-9.1}{\sqrt {\frac{26.2}{2}(\frac{1}{5}+\frac{1}{5})}}=3.932$
Since $q_0\gt q_{α,ν,k}$, reject the null hypothesis.
