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Then we can reject H 0 : mi − mh = 0 in favor of H a : mi − mh ≠ 0 at level of significance a if either of the following equivalent conditions hold. | t | > ta / 2 or | yi − yh | > ta / 2 s (1 / ni ) + (1 / nh ) 2. p-value < a For example, in the gasoline mileage situation consider testing H 0 : mB − mA = 0 versus H a : mB − mA ≠ 0 . 17. 17. 17. 0018. 05 level of significance. 01 level of significance. This would be regarded as very strong evidence that mB and mA differ. Further examining the SAS output, we see this output gives the point estimates, standard errors of the estimates, test statistics, and p-values for testing H 0 : mA − mC = 0, H 0 : mB − mC = 0, and H 0 : mB − ( mC + mA ) / 2 = 0.

The mean of all of the treatment means is a mii = ∑∑ m ij i =1 j =1 a = b ab b ∑ ∑(m+a i =1 j =1 i + g j + qij ) ab a  b b  ab m + b ∑ a i + a ∑ g j + ∑  ∑ qij  i =1 j =1 i =1  j =1  = ab =m a This says that the overall mean m in the (a, g, q ) model is mii, the mean of all of the treatment means. The ith factor level mean, which we define to be the average of the treatment means when using level i of factor 1, where the averaging is done over the different levels of factor 2, is: b mi i = ∑m j =1 b = = ij b ∑( m + a + g j =1 i b j + qij ) b b j =1 j =1 b m + b ai + ∑ g j + ∑ qij = m + ai b This implies that ai , the main effect of the ith level of factor 1, is mi i − m , or mi i − mii, the difference between the ith factor level mean and the mean of all of the treatment means.

Recall, however, that the Tukey intervals are short enough to allow us to conclude with 95 percent confidence that mB is greater than mA and mC . We next suppose in the gasoline mileage situation that gasoline type B contains a chemical—Chemical XX—that is not contained in gasoline types A or C . To assess the effect of Chemical XX on gasoline mileage, we consider mB − mC + mA 2 This is the difference between the mean mileage obtained by using gasoline type B and the average of the mean mileages obtained by using gasoline types C and A.

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