 # 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 Line 2020-10-20 Answered
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 $M{S}_{E}=1.75$ and you reject the null hypothesis. The sample means you obtain for each lightbulb are below:
Type 1: 20
Type 2: 30
Type 3: 28
Type 4: 22
Which pairs are significantly different at the $5\mathrm{%}$ significance level? Select all that apply.
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Step 1
From the provided data we have,
$M{S}_{E}=1.75$

$k=$ number of groups $=4$
$\begin{array}{|cccc|}\hline \text{Type 1}& \text{Type 2}& \text{Type 3}& \text{Tepe 4}\\ 20& 30& 28& 22\\ \hline\end{array}$
Step 2
Using Tukey’s HSD:
We first compute the Tukey’s HSD as below:

(from studentized q - table)
$=2.393646$
(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 $5\mathrm{%}$ 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.