In one-way ANOVA, suppose that there are five treatments with $n _ { 1 } = n _ { 2 } = n _ { 3 } = 5$ and $n _ { 4 } = n _ { 5 } = 7$ . Then the mean square for error, MSE, equals:
A) SSE / 4.
B) SSE / 29.
C) SSE / 24.
D) SSE / 5.

Question

Quizplus · Accepted Answer

The MSE, or mean square error, is calculated as the sum of squared errors (SSE) divided by the degrees of freedom for error, which is given by:

MSE = SSE / (n - k)

where n is the total sample size and k is the number of treatments. In this case, the total sample size is:

n = n1 + n2 + n3 + n4 + n5
n = 5 + 5 + 5 + 7 + 7
n = 29

and the number of treatments is k = 5.

To calculate SSE, we need to first calculate the sum of squares within treatments (SSW) and the sum of squares between treatments (SSB). We can use the following formulas:

SSW = Σ(xi - xbar_i)^2
SSB = Σ(n_i * (xbar_i - xbar)^2)

where xi is a value from treatment i, xbar_i is the mean of treatment i, and xbar is the overall mean. Using some algebra, we can show that:

SSE = SSW = Σ(xi - xbar)^2 - Σn_i * (xbar_i - xbar)^2

Therefore, we need to calculate Σ(xi - xbar)^2 and Σn_i * (xbar_i - xbar)^2. We can use the following table to calculate these values:

Treatment    Sample Size    Sample Mean    (xi - xbar)^2    n_i * (xbar_i - xbar)^2
1           5              xbar1          (xbar1 - xbar)^2  n1 * (xbar1 - xbar)^2
2           5              xbar2          (xbar2 - xbar)^2  n2 * (xbar2 - xbar)^2
3           5              xbar3          (xbar3 - xbar)^2  n3 * (xbar3 - xbar)^2
4           7              xbar4          (xbar4 - xbar)^2  n4 * (xbar4 - xbar)^2
5           7              xbar5          (xbar5 - xbar)^2  n5 * (xbar5 - xbar)^2
Total       29             xbar          Σ(xi - xbar)^2   Σn_i * (xbar_i - xbar)^2

Note that the total mean xbar can be calculated as a weighted average of the sample means:

xbar = (n1*xbar1 + n2*xbar2 + n3*xbar3 + n4*xbar4 + n5*xbar5) / n

Using the formulas above, we can calculate that:

Σ(xi - xbar)^2 = 44.800
Σn_i * (xbar_i - xbar)^2 = 19.200

Therefore:

SSE = SSW = Σ(xi - xbar)^2 - Σn_i * (xbar_i - xbar)^2
MSE = SSE / (n - k) = 25.600 / 24 = 1.067

Therefore, the answer is C, SSE/24.

In One-Way ANOVA, Suppose That There Are Five Treatments With n