Refer to the following partial ANOVA results from Excel (some information is missing). $$\begin{array} { | l | c | c | c | c | }
\hline \text { Source of Variation } & S S & d f & M S & F \
\hline \text { Between groups } & & & 210.2778 & \
\hline \text { Within groups } & 1483 & & 74.15 & \
\hline \text { Total } & 2113.833 & & & \
\hline
\end{array}$$ Degrees of freedom for between-groups variation are:
A)3.
B)4.
C)5.
D)Can't tell from given information.

Question

Refer to the following partial ANOVA results from Excel (some information is missing). $$\begin{array} { | l | c | c | c | c | } 
\hline \text { Source of Variation } & S S & d f & M S & F \
\hline \text { Between groups } & & & 210.2778 & \
\hline \text { Within groups } & 1483 & & 74.15 & \
\hline \text { Total } & 2113.833 & & & \
\hline
\end{array}$$ Degrees of freedom for between-groups variation are:
A)3.
B)4.
C)5.
D)Can't tell from given information.

Quizplus · Accepted Answer

The degrees of freedom for between-groups variation can be calculated using the formula: $df_{between} = k - 1$, where $k$ is the number of groups. The mean square (MS) for between groups is given as 210.2778, which is calculated as $SS_{between} / df_{between}$. Since the total sum of squares (SS) is 2113.833 and the within groups SS is 1483, the between groups SS is 2113.833 - 1483 = 630.833. Using the MS formula: 630.833 / df_{between} = 210.2778, solving for $df_{between}$ gives 3.

Refer to the Following Partial ANOVA Results from Excel (Some