Answer
${{\mu }_{1}}$ and ${{\mu }_{2}}$, ${{\mu }_{1}}$ and ${{\mu }_{4}}$, ${{\mu }_{3}}$ and ${{\mu }_{4}}$, and ${{\mu }_{2}}$ and ${{\mu }_{4}}$ are significantly different.
Work Step by Step
The null hypothesis is
\[{{H}_{0}}:\ {{\mu }_{1}}={{\mu }_{2}}={{\mu }_{3}}={{\mu }_{4}}\]
The arranged values of sample means $\bar{x}_{1}, \bar{x}_2, \bar{x}_3,$ and $\bar{x}_4$ are 78, 70, 64, and 47, respectively. There are a total of four means, which are compared in this study. The value of n is 31. The degree of freedom is
\[\begin{align}
& v=n-k \\
& =31-4 \\
& =27
\end{align}\]
The value of \[{{q}_{\left( 0.05,\,27,\,4 \right)}}\]is 3.685.
Now, take a difference of
\[\begin{align}
& {{{\bar{x}}}_{1}}-{{{\bar{x}}}_{3}}=78-70 \\
& =8
\end{align}\]
The Tukey’s test statistic is
\[\begin{align}
& {{q}_{{{\mu }_{1}},\,{{\mu }_{3}}}}=\frac{{{{\bar{x}}}_{1}}-{{{\bar{x}}}_{3}}}{\sqrt{\frac{{{s}^{2}}}{2}\left( \frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{3}}} \right)}} \\
& =\frac{8}{\sqrt{\frac{373}{2}\left( \frac{1}{31}+\frac{1}{31} \right)}} \\
& =2.306
\end{align}\]
Hence, the hypothesis is not rejected because $${{q}_{{{\mu }_{1}},\,{{\mu }_{3}}}}<
{{q}_{\left( 0.05,\,27,\,4 \right)}}$$
Now, take a difference of
\[\begin{align}
& {{{\bar{x}}}_{1}}-{{{\bar{x}}}_{2}}=78-64 \\
& =14
\end{align}\]
The Tukey’s test statistic is
\[\begin{align}
& {{q}_{{{\mu }_{1}},\,{{\mu }_{2}}}}=\frac{{{{\bar{x}}}_{1}}-{{{\bar{x}}}_{2}}}{\sqrt{\frac{{{s}^{2}}}{2}\left( \frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}} \right)}} \\
& =\frac{14}{\sqrt{\frac{373}{2}\left( \frac{1}{31}+\frac{1}{31} \right)}} \\
& =4.036
\end{align}\]
Hence, the hypothesis is rejected because $${{q}_{{{\mu }_{1}},\,{{\mu }_{2}}}}\gt {{q}_{\left( 0.05,\,27,\,4 \right)}}$$
Now, take a difference of
\[\begin{align}
& {{{\bar{x}}}_{1}}-{{{\bar{x}}}_{4}}=78-47 \\
& =31
\end{align}\]
The Tukey’s test statistic is
\[\begin{align}
& {{q}_{{{\mu }_{1}},\,{{\mu }_{4}}}}=\frac{{{{\bar{x}}}_{1}}-{{{\bar{x}}}_{4}}}{\sqrt{\frac{{{s}^{2}}}{2}\left( \frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{4}}} \right)}} \\
& =\frac{31}{\sqrt{\frac{373}{2}\left( \frac{1}{31}+\frac{1}{31} \right)}} \\
& =8.936
\end{align}\]
Hence, the hypothesis is rejected because $${{q}_{{{\mu }_{1}},\,{{\mu }_{4}}}}\gt{{q}_{\left( 0.05,\,27,\,4 \right)}}$$
Now, take a difference of
\[\begin{align}
& {{{\bar{x}}}_{3}}-{{{\bar{x}}}_{2}}=70-64 \\
& =6
\end{align}\]
The Tukey’s test statistic is
\[\begin{align}
& {{q}_{{{\mu }_{3}},\,{{\mu }_{2}}}}=\frac{{{{\bar{x}}}_{3}}-{{{\bar{x}}}_{2}}}{\sqrt{\frac{{{s}^{2}}}{2}\left( \frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{4}}} \right)}} \\
& =\frac{6}{\sqrt{\frac{373}{2}\left( \frac{1}{31}+\frac{1}{31} \right)}} \\
& =1.7297
\end{align}\]
Hence, the hypothesis is not rejected because
$${{q}_{{{\mu }_{3}},\,{{\mu }_{2}}}}\lt{{q}_{\left( 0.05,\,27,\,4 \right)}}$$
Now, take a difference of
\[\begin{align}
& {{{\bar{x}}}_{2}}-{{{\bar{x}}}_{4}}=64-47 \\
& =17
\end{align}\]
The Tukey’s test statistic is
\[\begin{align}
& {{q}_{{{\mu }_{2}},\,{{\mu }_{4}}}}=\frac{{{{\bar{x}}}_{2}}-{{{\bar{x}}}_{4}}}{\sqrt{\frac{{{s}^{2}}}{2}\left( \frac{1}{{{n}_{2}}}+\frac{1}{{{n}_{4}}} \right)}} \\
& =\frac{17}{\sqrt{\frac{373}{2}\left( \frac{1}{31}+\frac{1}{31} \right)}} \\
& =4.490
\end{align}\]
Hence, the hypothesis is rejected because $${{q}_{{{\mu }_{2}},\,{{\mu }_{4}}}}\gt{{q}_{\left( 0.05,\,27,\,4 \right)}}$$
From the solutions, it can be concluded that Tukey’s test is indefinite.
