Consider the following partial ANOVA table: $$\begin{array} { | l | c c c c | }
\hline \text { Source of Variation } & 5 & \boldsymbol { G } & \boldsymbol { M S } & \boldsymbol { F } \
\hline \text { Treatments } & 75 & * & 25 & 6.67 \
\text { Errar } & 60 & * & 3.75 & \
\hline \text { Total } & 135 & 19 & & \
\hline
\end{array}$$ The numerator and denominator degrees of freedom for the F-test (identified by asterisks)are
A) 4 and 15
B) 3 and 16
C) 15 and 4
D) 16 and 3

Question

Consider the following partial ANOVA table: $$\begin{array} { | l | c c c c | } 
\hline \text { Source of Variation } & 5 & \boldsymbol { G } & \boldsymbol { M S } & \boldsymbol { F } \
\hline \text { Treatments } & 75 & * & 25 & 6.67 \
\text { Errar } & 60 & * & 3.75 & \
\hline \text { Total } & 135 & 19 & & \
\hline
\end{array}$$ The numerator and denominator degrees of freedom for the F-test (identified by asterisks)are
A) 4 and 15
B) 3 and 16
C) 15 and 4
D) 16 and 3

Quizplus · Accepted Answer

The numerator degrees of freedom for the F-test is the number of treatments minus one, which is 4. The denominator degrees of freedom is the error degrees of freedom, which is total degrees of freedom minus the treatments degrees of freedom, which is 135-5=130. The error degrees of freedom is then divided among the number of groups minus one, which is 16. So the degrees of freedom are 3 and 16. Therefore, the answer is B.

Consider the Following Partial ANOVA Table The Numerator and Denominator Degrees of Freedom for the F-Test