Suppose you are testing whether the lifetime of 4 different types of lightbulbs. You run 5 experiments each. You get a mean square error of MS_{E}=1.7

Step 1
From the provided data we have,
${\displaystyle M{S}_E=1.75}$
${\displaystyle d{f}_t=4\times 5=20}$
${\displaystyle d{f}_w=20-4=16}$
${\displaystyle k=}$ number of groups ${\displaystyle =4}$
$\begin{array}{|cccc|}\hline \text{Type 1}& \text{Type 2}& \text{Type 3}& \text{Tepe 4}\\ 20& 30& 28& 22\\ \hline\end{array}$
Step 2
Using Tukey’s HSD:
We first compute the Tukey’s HSD as below:
${\displaystyle D=q_{\alpha ,k,d{f}_w}\sqrt{\frac{M{S}_E}{n}}}$
${\displaystyle =4.046\sqrt{\frac{1.75}{5}}}$ (from studentized q - table)
${\displaystyle =2.393646}$
(Where n represent the number of observations in each group).
The absolute differences in group means is computed as below:
$\begin{array}{|cc|}\hline \text{Groups}& \text{|Differences in group means|}\\ 1-2& 10\\ 1-3& 8\\ 1-4& 2\\ 2-3& 2\\ 2-4& 8\\ 3-4& 6\\ \hline\end{array}$
We say that the groups are significantly only when the absolute difference in group mean is larger than the HSD value.
Thus, this difference is larger for Group 1 and 2, Group 1 and 3, Group 2 and 4 and Group 3 and 4
Thus, the pairs that are significantly different at the ${\displaystyle 5\mathrm{\%}}$ significance level are,
$\begin{array}{|cccc|}\hline \text{Type 1-2}& \text{Type 1-3}& \text{Type 2-4}& \text{Type 3-4}\\ \hline\end{array}$
and other groups are not significant.