Consider the following partial ANOVA table: $$\begin{array} { | l | c c c c | }
\hline \text { Source of Variation } & \boldsymbol { SS } &df& \boldsymbol { M S } & \boldsymbol { F } \
\hline \text { Treatments } & 75 & * & 25 & 6.67 \
\text { Error } & 60 & * & 3.75 & \
\hline \text { Total } & 135 & 19 & & \
\hline
\end{array}$$ 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: $$\begin{array} { | l | c c c c | } 
\hline \text { Source of Variation }  & \boldsymbol { SS } &df& \boldsymbol { M S } & \boldsymbol { F } \
\hline \text { Treatments } & 75 & * & 25 & 6.67 \
\text { Error } & 60 & * & 3.75 & \
\hline \text { Total } & 135 & 19 & & \
\hline
\end{array}$$ 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 total degrees of freedom is 19. For treatments, $df = \frac{SS}{MS} = \frac{75}{25} = 3$. Thus, error $df = 19 - 3 = 16$.

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