Table 17.21 is incomplete. The data were actually obtained from a Graeco-

Latin square that incorporates four different operators as well as the four different machines. The Table 17.21: Hydrostatic pressure tests: Machine, yield.
2 4 3 1 39.0 39.0 41.0 41.0 1 3 4 2 36.5 42.5 40.5 38.5 4 2 1 3 40.0 39.0 41.5 41.5 3 1 2 4 41.5 39.5 39.0 44.0 Table 17.22: Hydrostatic pressure tests: Operator, machine.
B,2 A,4 D,3 C,1 A,1 B,3 C,4 D,2 D,4 C,2 B,1 A,3 C,3 D,1 A,2 B,4 Operators are A, B, C, D.
Machines are 1, 2, 3, 4.
correct design is given in Table 17.22. Note that this is a Latin square for machines when we ignore the operators and a Latin square for operators when we ignore the machines. Moreover, every operator works once with every machine. Give the new analysis of variance table. How do the results on machines change?What evidence is there for differences among operators?Was the analysis for machines given earlier incorrect or merely inefficient?