Samples were drawn from three populations. The sample sizes were $n _ { 1 } = 9 , n _ { 2 } = 6$, and $n _ { 3 } = 6$. The sample means were $\bar { x } _ { 1 } = 1.57 , \bar { x } _ { 2 } = 1.64$, and $\bar { x } _ { 3 } = 1.36$. The sample standard deviations were $s _ { 1 } = 0.42 , s _ { 2 } = 0.29$, and $s _ { 3 } = 0.40$. The grand mean is $\bar { x } = 1.53$. Compute the sum of squares SSTr.
A) 2.6317
B) 0.2604
C) 0.1302
D) 0.8905

Question

Quizplus · Accepted Answer

First, calculate the total sum of squares SST:
$$SST = \sum_{i=1}^{3}\sum_{j=1}^{n_i}(x_{ij} - \bar{x})^2 = \sum_{i=1}^{3}\sum_{j=1}^{n_i}(x_{ij} - \bar{x}_i + \bar{x}_i - \bar{x})^2$$
$$= \sum_{i=1}^{3}\sum_{j=1}^{n_i}(x_{ij} - \bar{x}_i)^2 + \sum_{i=1}^{3}n_i(\bar{x}_i - \bar{x})^2$$
Using the given values:
$$SST = (8	imes0.42^2) + (5	imes 0.29^2) + (5 	imes 0.4^2) + 9(1.57-1.53)^2 + 6(1.64-1.53)^2 + 6(1.36-1.53)^2 = 1.7916$$
Next, calculate the sum of squares due to treatment (or regression) SSR:
$$SSR = \sum_{i=1}^3 n_i(\bar{x}_i - \bar{x})^2 = 8(1.57 - 1.53)^2 + 5(1.64-1.53)^2 + 5(1.36-1.53)^2 = 0.2604$$
Finally, calculate SSTr by subtracting SSR from SST:
$$SSTr = SST - SSR = 1.7916 - 0.2604 = 1.5312$$
Therefore, the answer is B.

Samples Were Drawn from Three Populations n1=9,n2=6n _ { 1 } = 9 , n _ { 2 } = 6n1​=9,n2​=6

Samples Were Drawn from Three Populations $n _ { 1 } = 9 , n _ { 2 } = 6$