Both the Schee procedure and the Tukey HSD procedure have exact experimentwise error rates. Why

is the number of pairs of treatments declared to be signicantly dierent by the Tukey HSD procedure

generally greater than the number of pairs declared by the Schee procedure?

A. HSD has a higher Type I error rate than the Type I error rate for Schee.

B. Schee considers all contrasts not just the contrasts dealing with pairwise dierences; whereas, HSD

is just considering pairwise dierences.

C. Schee fails to take into account all the factors; whereas HSD does.

D. HSD uses an estimate of the standard error which tends to be smaller than the value used by Schee.

E. None of the above are correct.