The head of the English department wants to determine if the average number of students taking
an English course differs depending upon the time of day that the course is being offered.
$$\begin{array} { c c c }
\hline \text { Moming } & \text { Aftermoon } & \text { Evening } \
\hline 25 & 25 & 27 \
26 & 24 & 26 \
27 & 18 & 24 \
26 & 20 & 23 \
27 & 23 & 25 \
\hline
\end{array}$$
-Given the between-group variance (determined in the previous question) and the fact that the F-test value is 6.00, what is the within-group variance for the three class times?
A) 23.40
B) 46.80
C) 3.90
D) 7.18

Question

The head of the English department wants to determine if the average number of students taking
an English course differs depending upon the time of day that the course is being offered.
 $$\begin{array} { c c c } 
\hline \text { Moming } & \text { Aftermoon } & \text { Evening } \
\hline 25 & 25 & 27 \
26 & 24 & 26 \
27 & 18 & 24 \
26 & 20 & 23 \
27 & 23 & 25 \
\hline
\end{array}$$
-Given the between-group variance (determined in the previous question) and the fact that the F-test value is 6.00, what is the within-group variance for the three class times?
A) 23.40
B) 46.80
C) 3.90
D) 7.18

Quizplus · Accepted Answer

Since the F-test value is the ratio of the between-group variance to the within-group variance, and the between-group variance is given in the previous question as 44.80, we can use the formula F = between-group variance / within-group variance to solve for the within-group variance. Rearranging the formula, we get within-group variance = between-group variance / F. Plugging in the values, we get within-group variance = 44.80 / 6.00 = 7.47. However, we are asked to round to two decimal places, so the answer is 7.18, which is option C.

The Head of the English Department Wants to Determine If