Samples were drawn from three populations. The sample sizes were $n _ { 1 } = 5 , n _ { 2 } = 7$ and $n _ { 3 } = 5$ The sample means were $\bar { x } _ { 1 } = 1.43 , \bar { x } _ { 2 } = 1.85 , \text { and } \bar { x } _ { 3 } = 1.38$ The sample standard deviations were $s _ { 1 } = 0.31 , s _ { 2 } = 0.40 , \text { and } s _ { 3 } = 0.38$
The grand mean is $\overline { \bar { x } } = 1.588235$
Compute the mean squares MSE.
A) 0.8216
B) 0.4108
C) 0.1373
D) 1.9220

Question

Quizplus · Accepted Answer

The Mean Squares Error (MSE) is calculated using the formula $MSE = \frac{1}{df} \sum_{i=1}^{k} (n_i - 1)s_i^2$, where $df = n_1 + n_2 + n_3 - k$, $n_i$ is the sample size for group $i$, $s_i^2$ is the sample variance for group $i$, and $k$ is the number of groups. Here, $df = 5 + 7 + 5 - 3 = 14$, and the sum of squared deviations is $(5-1)0.31^2 + (7-1)0.40^2 + (5-1)0.38^2$. Calculating this gives $MSE = 0.1373$.

Samples Were Drawn from Three Populations n1=5,n2=7n _ { 1 } = 5 , n _ { 2 } = 7n1​=5,n2​=7

Samples Were Drawn from Three Populations $n _ { 1 } = 5 , n _ { 2 } = 7$