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For example, in column (1) we obtain for the fifth entry 5 = -(4) + 1, for the sixth entry 6 = -(-1) + 5, and so on. Column (2) is obtained from column (1) just as column (1) is obtained from the Response column. Column (3) is obtained from column (2) similarly. In general, for a 2k design we would construct k columns of this type. Column (3) [in general, column (k)] is the contrast for the effect designated at the beginning of the row. To obtain the estimate of the effect, we divide the entries in column (3) by n2k-1 (in our example, n2k-1 = 8).

For example, if the observations come from treatment 1, then x1j = 1 and x2j = 0 and the regression model is y1 j = β 0 + β 1 (1) + β 2 (0) + ε 1 j = β 0 + β1 + ε1j Since in the ANOVA model these observations are defined by y1 j = µ + τ 1 + ε 1 j , this implies that β 0 + β 1 = µ1 = µ + τ 1 Similarly, if the observations are from treatment 2, then y2 j = β 0 + β 1 (0) + β 2 (1) + ε 2 j = β0 + β2 + ε2 j and the relationship between the parameters is β 0 + β 2 = µ2 = µ + τ 2 Finally, consider observations from treatment 3, for which the regression model is y3 j = β 0 + β 1 (0) + β 2 (0) + ε 3 j = β0 + ε3j and we have β 0 = µ3 = µ + τ 3 Thus in the regression model formulation of the one-way ANOVA model, the regression coefficients describe comparisons of the first two treatment means with the third treatment mean; that is β 0 = µ3 β 1 = µ1 − µ 3 β 2 = µ2 − µ3 In general, if there are a treatments, the regression model will have a – 1 regressor variables, say yij = β 0 + β 1 x1 j + β 2 x2 j +"+ β a −1 xa −1 + ε ij RSi = 1,2,", a T j = 1,2,", n where xij = RS1 if observation j is from treatment i 0 otherwise T Since these regressor variables only take on the values 0 and 1, they are often called indicator variables.

1 we use the least squares approach to estimating the parameters in the single-factor model. Assuming a balanced experimental design, we fine the least squares normal equations as Equation 3-48, repeated below: a n anµ + nτ 1 + nτ 2 +"+ nτ a = ∑ ∑ yij i =1 j =1 n = ∑ y1 j nµ + nτ 1 j =1 nµ n = ∑ y2 j + nτ 2 j =1 # nµ n + nτ a = ∑ yaj j =1 where an = N is the total number of observations. As noted in the textbook, if we add the last a of these normal equations we obtain the first one. That is, the normal equations are not linearly independent and so they do not have a unique solution.

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