Question: Let c1, c2,..., cI be numbers satisfying (ci = 0. Then (cii = c11 + ...+ cII is called a contrast in the i's. Notice

Let c1, c2,..., cI be numbers satisfying (ci = 0. Then (ciµi = c1µ1 + ...+ cIµI is called a contrast in the µi's. Notice that with c1 = 1, c2 = -1, c3 =... = cI = 0, (ciµi = µ1 - µ2 which implies that every pair-wise difference between µi's is a contrast (so is, e.g., µ1 - .5µ2 - .5µ3). A method attributed to Scheffé gives simultaneous CI's with simultaneous confidence level 100(1 - a) % for all possible contrasts (an infinite number of them!). The interval for (ciµi is
(cix̅i. ( ((c2i / Ji) 1/2, [(I - 1)] ( MSE ( Fa, J - I, n - I]1/2
Using the critical flicker frequency data of Exercise 42, calculate the Scheffé intervals for the contrasts µ1 - µ2, µ1 - µ3, µ2 - µ3, and .5 µ1 + .5µ2 - µ3 (this last contrast compares blue to the average of brown and green). Which contrasts appear to differ significantly from 0, and why?

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