a. Calculate the mean response for each treatment b. The MINITAB ANOVA printout is shown here. Test for interaction at the $\alpha = 0.05$ leve of significance. Analysis of variance for response. $$\begin{array} { l l l l l }
\text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F } \
\hline \text { A } & 1 & 0.53777 & 0.53777 & 0.11851 \
\text { B } & 2 & 5.02708 & 2.51334 & 0.55391 \
\text { AB } & 2 & 13.49334 & 6.74667 & 1.48678 \
\text { Error } & 6 & 27.22667 & 4.53778 & \
\hline \text { Total } & 11 & 46.28486 & & \
\hline
\end{array}$$ c. Does the result warrant tests of the two factor mean effects?

Question

a. Calculate the mean response for each treatment b. The MINITAB ANOVA printout is shown here. Test for interaction at the $\alpha = 0.05$ leve of significance. Analysis of variance for response. $$\begin{array} { l l l l l } 
\text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F } \
\hline \text { A } & 1 & 0.53777 & 0.53777 & 0.11851 \
\text { B } & 2 & 5.02708 & 2.51334 & 0.55391 \
\text { AB } & 2 & 13.49334 & 6.74667 & 1.48678 \
\text { Error } & 6 & 27.22667 & 4.53778 & \
\hline \text { Total } & 11 & 46.28486 & & \
\hline
\end{array}$$ c. Does the result warrant tests of the two factor mean effects?

Quizplus · Accepted Answer

The Answer of  a. Calculate the mean response for each treatment b. The MINITAB ANOVA printout is shown here. Test for interaction at the $\alpha = 0.05$ leve of significance. Analysis of variance for response. $$\begin{array} { l l l l l } 
\text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F } \
\hline \text { A } & 1 & 0.53777 & 0.53777 & 0.11851 \
\text { B } & 2 & 5.02708 & 2.51334 & 0.55391 \
\text { AB } & 2 & 13.49334 & 6.74667 & 1.48678 \
\text { Error } & 6 & 27.22667 & 4.53778 & \
\hline \text { Total } & 11 & 46.28486 & & \
\hline
\end{array}$$ c. Does the result warrant tests of the two factor mean effects?

A Calculate the Mean Response for Each Treatment B α=0.05\alpha = 0.05α=0.05

A Calculate the Mean Response for Each Treatment B $\alpha = 0.05$