One-way ANOVA is performed on independent samples taken from three normally distributed populations with equal variances.The following summary statistics were calculated: $$\begin{array} { l l l }
n _ { 1 } = 6 & \bar { x } _ { 1 } = 50 & s _ { 1 } = 5.2 \
n _ { 2 } = 8 & \bar { x } _ { 2 } = 55 & s _ { 2 } = 4.9 \
n _ { 3 } = 6 & \bar { x } _ { 3 } = 51 & s _ { 3 } = 5.4
\end{array}$$ The grand mean equals:
A) 50.0.
B) 52.0.
C) 52.3.
D) 53.0.

Question

One-way ANOVA is performed on independent samples taken from three normally distributed populations with equal variances.The following summary statistics were calculated: $$\begin{array} { l l l } 
n _ { 1 } = 6 & \bar { x } _ { 1 } = 50 & s _ { 1 } = 5.2 \
n _ { 2 } = 8 & \bar { x } _ { 2 } = 55 & s _ { 2 } = 4.9 \
n _ { 3 } = 6 & \bar { x } _ { 3 } = 51 & s _ { 3 } = 5.4
\end{array}$$ The grand mean equals:
A) 50.0.
B) 52.0.
C) 52.3.
D) 53.0.

Quizplus · Accepted Answer

The grand mean can be calculated as the weighted average of the sample means, where the weights are the respective sample sizes. The calculation is as follows:

Grand mean = (6/20) * 50 + (8/20) * 55 + (6/20) * 51 
= 3 * 10 + 4 * 5.5 + 3 * 17 
= 30 + 22 + 51 
= 103 
So, the grand mean is 103/20 = 5.15 or 52.3 (rounded to one decimal place). Therefore, the correct answer is C.

One-Way ANOVA Is Performed on Independent Samples Taken from Three