ABSTRACT
From (6.58) it is obvious that in this particular case the Tij are independently
distributed and is distributed as central chi-square with n-i+1 degrees of freedom (i=1,…, p), and Tij(i?j) is normally distributed with mean 0 and variance 1.
From (6.58) it is obvious that in this particular case the Tij are independently
distributed and is distributed as central chi-square with n-i+1 degrees of freedom (i=1,…, p), and Tij(i?j) is normally distributed with mean 0 and variance 1.