Consider the following ANOVA table. $\begin{array}{llll}
\begin{array}{l}
\text { Source } \
\text { of Variation }
\end{array} & \begin{array}{l}
\text { Sum } \
\text { of Squares }
\end{array} & \begin{array}{l}
\text { Degrees } \
\text { of Freed om }
\end{array} & \begin{array}{l}
\text { Mean } \
\text { Square }
\end{array}&F \
\text { Between Treatments } & 2073.6 & 4 & \
\text { Between Blocks } & 6000 & 5 & 1200 \
\text { Error } & & 20 & 288 \
\text { Total } & & 29 &
\end{array}$
The sum of squares due to error equals
A) 13833.6.
B) 2073.6.
C) 5760.
D) 6000.

Question

Consider the following ANOVA table.   $\begin{array}{llll}
\begin{array}{l}
\text { Source } \
\text { of Variation }
\end{array} & \begin{array}{l}
\text { Sum } \
\text { of Squares }
\end{array} & \begin{array}{l}
\text { Degrees } \
\text { of Freed om }
\end{array} & \begin{array}{l}
\text { Mean } \
\text { Square }
\end{array}&F \
\text { Between Treatments } & 2073.6 & 4 & \
\text { Between Blocks } & 6000 & 5 & 1200 \
\text { Error } & & 20 & 288 \
\text { Total } & & 29 &
\end{array}$ 
The sum of squares due to error equals
A) 13833.6.
B) 2073.6.
C) 5760.
D) 6000.

Quizplus · Accepted Answer

The sum of squares due to error is the total sum of squares minus the sum of squares due to treatments and the sum of squares due to blocks. Therefore, the sum of squares due to error is 29000 - 2073.6 - 6000 = 5760.

Consider the Following ANOVA Table The Sum of Squares Due to Error Equals A) 13833

Consider the Following ANOVA Table The Sum of Squares Due to Error Equals
A) 13833