Refer to the following partial ANOVA results from Excel (some information is missing). $$\begin{array} { | l | c | c | c | c | }
\hline \text { Source of Variation } & S S & d f & M S & F \
\hline \text { Nozzle setting } & 3.46722 & 1 & 3.46722 & 4.87198 \
\hline \text { Pressure level } & 8.07444 & & 4.03722 & 5.67291 \
\hline \text { Interaction } & 2.80779 & 2 & 1.40389 & 1.97268 \
\hline \text { Error } & 8.54000 & & 0.711667 & \
\hline \text { Total } & 22.8894 & & & \
\hline
\end{array}$$ How many pressure levels were observed?
A)4
B)3
C)2
D)1

Question

Refer to the following partial ANOVA results from Excel (some information is missing). $$\begin{array} { | l | c | c | c | c | } 
\hline \text { Source of Variation } & S S & d f & M S & F \
\hline \text { Nozzle setting } & 3.46722 & 1 & 3.46722 & 4.87198 \
\hline \text { Pressure level } & 8.07444 & & 4.03722 & 5.67291 \
\hline \text { Interaction } & 2.80779 & 2 & 1.40389 & 1.97268 \
\hline \text { Error } & 8.54000 & & 0.711667 & \
\hline \text { Total } & 22.8894 & & & \
\hline
\end{array}$$ How many pressure levels were observed?
A)4
B)3
C)2
D)1

Quizplus · Accepted Answer

The degree of freedom (df) for pressure level is missing from the table, but we can calculate it using the formula df = total df - df for nozzle setting - df for interaction - 1. Since there is only one nozzle setting and two factors in the interaction, df for pressure level must be 12 - 1 - 2 - 1 = 8. With a df of 8, and given that the F-value for pressure level is 5.67291, we can conclude that there were 3 pressure levels observed, since the numerator df for this factor is 2 and the denominator df is 8. Therefore, the answer is B.

Refer to the Following Partial ANOVA Results from Excel (Some