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Testing Subhypotheses with Contrasts

A linear combination of the true means of \(k\) factor levels \(\mu_1,\mu_2,\cdots,\mu_k\) of the randomized-one-factor-design \(C=\displaystyle\sum_{j=1}^{k}c_j\mu_j\) is said to be a contrast if the sum of its coefficients \(\displaystyle\sum_{j=1}^{k}c_j=0\). Because \(\overline Y_{.j}\) is always an unbiased estimator for \(\mu_j\), we can use it to estimate C \(\hat C=\displaystyle\sum_{j=1}^{k}c_j\overline Y_{.j}\). Because \(Y_{ij}\) are normal, so \(\hat C\) is also normal. Then, \(E(\hat C)=\displaystyle\sum_{j=1}^{k}c_jE(\overline Y_{.j})=\displaystyle\sum_{j=1}^{k}c_j\mu_j=C\) and \(Var(\hat C)=\displaystyle\sum_{j=1}^{k}c_j^2Var(\overline Y_{.j})=\displaystyle\sum_{j=1}^{k}c_j^2\frac{\sigma^2}{n_j}=\sigma^2\displaystyle\sum_{j=1}^{k}\frac{c_j^2}{n_j}\). Replacing \(\sigma^2\) by its estimate \(MSE\) gives a formula for the estimated variance \(S_{\hat C}^2=MSE\displaystyle\sum_{j=1}^{k}\frac{c_j^2}{n_j}\). The Z transformation of \(\hat C\) is a standard normal: \[\frac{\hat C-E(\hat C)}{\sqrt{Var(\hat C)}}=\frac{\hat C-C}{\sqrt{Var(\hat C)}}\], then \[\Biggl[\frac{\hat C-C}{\sqrt{Var(\hat C)}}\Biggr]^2=\frac{(\hat C-C)^2}{\sigma^2\displaystyle\sum_{j=1}^{k}\frac{c_j^2}{n_j}}\] is a \(\chi^2\) random variable with \(1\) degree of freedom. When \(H_0:\mu_1 = \mu_2 = \cdots = \mu_k\) is true, \(C=0\) the ratio is \(\frac{\hat C^2}{\sigma^2\displaystyle\sum_{j=1}^{k}\frac{c_j^2}{n_j}}\).

For any contrast \(C_i=\displaystyle\sum_{j=1}^{k}c_{ij}\mu_j\) if we define the sum of squares associated with \(C_i\) is \[SS_{C_i}=\frac{\hat C_i^2}{\displaystyle\sum_{j=1}^{k}\frac{c_{ij}^2}{n_j}}=\frac{\Bigl[\displaystyle\sum_{j=1}^{k}c_{ij}\overline Y_{.j}\Bigr]^2}{\displaystyle\sum_{j=1}^{k}\frac{c_{ij}^2}{n_j}}\].

A set of \(n\) contrasts, \(\{C_i\}_{i=1}^{n}\) are said to be mutually orthogonal if \(\displaystyle\sum_{j=1}^{k}\frac{c_{aj}c_{bj}}{n_j}=0,\text{for all } a\ne b\). When \(\{C_{ij}\}_{j=1}^{k}\) are mutually orthogonal, \[SS_{C_i}=\frac{\hat C_i^2}{\displaystyle\sum_{j=1}^{k}\frac{c_{ij}^2}{n_j}}=\frac{\Bigl[\displaystyle\sum_{j=1}^{k}c_{ij}\overline Y_{.j}\Bigr]^2}{\displaystyle\sum_{j=1}^{k}\frac{c_{ij}^2}{n_j}}\]