Basic Concepts of ANOVA – Advanced

Property 1: If a sample is made as described in Definition 1 of Basic Concepts of Anova, with the x_ij independently and normally distributed and with all σ_j² equal, then

Proof: The first two of these results from Property 3 of Chi-square Distribution. Also, by this property

where MS_j = SS_j / df_j and df_j = n_j – 1. Thus,

If all the σ_j² are equal, the third assertion follows by Property 2 of Chi-square Distribution.

Property 2:

Proof: The assertion about the degrees of freedom is clear. The proof of the assertion about the SS is as follows:

Property 3:

Proof: Since ē  = 0 and e_ij = x_ij – x̄_j, by Definition 1, we have