In one-way ANOVA, suppose that there are five treatments with $$n _ { 1 } = n _ { 2 } = n _ { 3 } = 5$$ and $$n _ { 4 } = n _ { 5 } = 7$$ . Then the mean square for error, MSE, equals: $\begin{array}{|l|l|}
\hline \text { A. } & \mathrm{SSE} / 4 . \
\hline \text { B. } & \mathrm{SSE} / 29 . \
\hline \text { C. } & \mathrm{SSE} / 24 \
\hline \text { D. } & \mathrm{SSE} / 5 . \
\hline
\end{array}$

Question

Quizplus · Accepted Answer

The Answer of In one-way ANOVA, suppose that there are five treatments with $$n _ { 1 } = n _ { 2 } = n _ { 3 } = 5$$ and $$n _ { 4 } = n _ { 5 } = 7$$ . Then the mean square for error, MSE, equals: $\begin{array}{|l|l|}
\hline \text { A. } & \mathrm{SSE} / 4 . \
\hline \text { B. } & \mathrm{SSE} / 29 . \
\hline \text { C. } & \mathrm{SSE} / 24 \
\hline \text { D. } & \mathrm{SSE} / 5 . \
\hline
\end{array}$

In One-Way ANOVA, Suppose That There Are Five Treatments With n