91Ó°ÊÓ

Consider the ANOVA with \(a=2\) treatments. Show that the \(M S_{E}\) in this analysis is equal to the pooled variance estimate used in the two-sample \(t\) -test.

In a fixed-effects model, suppose that there are \(n\) observations for each of four treatments. Let \(Q_{1}^{2}, Q_{2}^{2}\) and \(Q_{3}^{2}\) be single-degree- of-freedom sums of squares for orthogonal contrasts. A contrast is a linear combination of the treatment means with coefficients that sum to zero. The coefficient vectors of orthogonal contrasts are orthogonal vectors. Prove that \(S S_{\text {Treatments }}=Q_{1}^{2}+Q_{2}^{2}+Q_{3}^{2}\).

Consider the random-effect model for the single-factor completely randomized design. Show that a \(100(1-\alpha) \%\) confidence interval on the ratio of variance components \(\sigma_{\tau}^{2} / \sigma^{2}\) is given by $$L \leq \frac{\sigma_{\tau}^{2}}{\sigma^{2}} \leq U$$ where $$L=\frac{1}{n}\left[\frac{M S_{\text {Treatments }}}{M S_{E}} \times\left(\frac{1}{f_{\alpha / 2, a-1, N-a}}\right)-1\right]$$ and$$U=\frac{1}{n}\left[\frac{M S_{\text {Treatments }}}{M S_{E}} \times\left(\frac{1}{f_{1-\alpha / 2, a-1, Na}}\right)-1\right]$$

Consider the fixed-effect model of the completely randomized single-factor design. The model parameters are restricted by the constraint \(\sum_{i=1}^{a} \tau_{i}=0\) (Actually, other restrictions could be used, but this one is simple and results in intuitively pleasing estimates for the model parameters.) For the case of unequal sample size \(n_{1}, n_{2}, \ldots, n_{a},\) the restriction is \(\sum_{i=1}^{a} n_{i} \tau_{i}=0\). Use this to show that $$E\left(M S_{\text {Treatments }}\right)=\sigma^{2}+\frac{\sum_{i=1}^{a} n_{i} \tau_{i}^{2}}{a-1}$$ Does this suggest that the null hypothesis in this model is \(H_{0}: n_{1} \tau_{1}=n_{2} \tau_{2}=\cdots=n_{a} \tau_{a}=0 ?\)