Question

Suppose that independent samples of sizes n1,n2,...,nk are taken from each of k normally distributed populations with means µ1,µ2,...,µn and common variances, all equal to s2. Let Yij denote the jth obervations from population i, for j=1,2,...,ni and i= 1,2,...,k, and let n=n1+n2+...+nk.

a) Argue that SSE/s2 has a X2(chi-square) distribution with (n1-1)+(n2-1)+...+(nk-1)= n-k df

Answer 1

The total degrees of freedom for SSE, when divided by the common variance s², follow a chi-square distribution with (n1 - 1) + (n2 - 1) + ... + (nk - 1) degrees of freedom, which is equivalent to n - k degrees of freedom, where n is the total number of observations and k is the number of groups.

In the context of analysis of variance (ANOVA), SSE (Sum of Squares Error) represents the variation within groups. The formula for SSE is the sum of squares of the deviations of individual data points from their respective group means.

Given that independent samples are taken from k normally distributed populations, and assuming that the populations have a common variance (s²), we can argue that SSE divided by the common variance (s²) follows a chi-square distribution with degrees of freedom equal to the total number of observations minus the number of groups.

The total degrees of freedom in the SSE term can be computed by adding up the degrees of freedom from each group. For each group, the degrees of freedom associated with the SSE term is equal to the sample size minus 1 (ni - 1, where ni is the sample size of group i).

Therefore, the total degrees of freedom for SSE, when divided by the common variance s², follow a chi-square distribution with (n1 - 1) + (n2 - 1) + ... + (nk - 1) degrees of freedom, which is equivalent to n - k degrees of freedom, where n is the total number of observations and k is the number of groups.