Exhibit 10-14
The following is part of an ANOVA table that was obtained from data regarding three treatments and a total of 15 observations.
$$\begin{array} { l c c }
\begin{array} { l }
\text { Source of } \
\text { Variation }
\end{array} & \begin{array} { c }
\text { Sum of } \
\text { Squares }
\end{array} & \begin{array} { c }
\text { Degrees of } \
\text { Freedom }
\end{array} \
\text { Between Treatments } & 64 & \
\text { Error Within Treatments) } & 96 &
\end{array}$$
-Refer to Exhibit 10-14. The computed test statistics is
A) 32
B) 8
C) 0.667
D) 4

Question

Exhibit 10-14
The following is part of an ANOVA table that was obtained from data regarding three treatments and a total of 15 observations.
 $$\begin{array} { l c c } 
\begin{array} { l } 
\text { Source of } \
\text { Variation }
\end{array} & \begin{array} { c } 
\text { Sum of } \
\text { Squares }
\end{array} & \begin{array} { c } 
\text { Degrees of } \
\text { Freedom }
\end{array} \
\text { Between Treatments } & 64 & \
\text { Error Within Treatments) } & 96 &
\end{array}$$
-Refer to Exhibit 10-14. The computed test statistics is
A) 32
B) 8
C) 0.667
D) 4

Quizplus · Accepted Answer

The F-statistic is calculated as the ratio of the mean square between treatments to the mean square within treatments. The mean square between treatments is 64/2 = 32 (since there are 3 treatments, degrees of freedom = 3-1 = 2). The mean square within treatments is 96/12 = 8 (since there are 15 observations, degrees of freedom = 15-3 = 12). Therefore, the F-statistic is 32/8 = 4.

Exhibit 10-14 the Following Is Part of an ANOVA Table

Exhibit 10-14
the Following Is Part of an ANOVA Table