I'd like to check my understanding that there is an incorrect statement in my textbook. Then, I'd like to check my understanding on what is actually true.

Montgomery's Design of Experiments textbook (Design and Analysis of Experiments, edition 5, page 69), states the following in the section for standard one-way (fixed effects) ANOVA.

Because we have assumed that the errors $\epsilon_{ij}$ are normally and independently distributed with mean zero and variance $\sigma^2$, the observations $y_{ij}$ are normally and independently distributed with mean $\mu + \tau_i$ and variance $\sigma^2$. Thus, $SS_T$ [sum squares total] is a sum of squares in normally distributed random variables; consequently, it can be shown that $SS_{T}/{\sigma^2}$ is distributed chi-square with N-1 degrees of freedom.

I think this is false because it neglects to say "under the null hypothesis that $\tau_1 = \ldots = \tau_a = 0$." Right? I believe that $SS_{T}/{\sigma^2} \sim \chi^2(N-1)$ holds only under the null hypothesis.

Furthermore, in an attempt to understand this properly, I think $SS_{T}/{\sigma^2}$ should be noncentral chi-square. First, $SSE/{\sigma^2}$ is always distributed chi-square (with noncentrality parameter 0). Next, $SS_{Trt}/{\sigma^2}$ is chisquare with noncentrality 0 under null and noncentrality >0 in the alternative. Finally, $SSE$ and $SS_{Trt}$ are known to be independent in general. Thus $SS_{T}/{\sigma^2}$ is the sum of independent noncentral chi squared random variables, and thus noncentral chi-squared.

With this reassurance, it will not be hard for me to find the noncentrality parameters.

Thanks in advance!

--Ryan M.


1 Answer 1


You are right. If the null does not hold, $SS_T$ will follow noncentral Chi-square distribution. I'll leave it for you to find the noncentrality parameter.


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