Of the six candidates for three managerial positions, three are female and three are male. Denote the females by F1, F2, F3 and the males by M1, M2, M3.

The result of choosing the managers is (F2, M1, M3).

a. Identify the 20 possible samples that could have been selected, and construct the contingency table for the sample actually obtained.

b. Let p1 denote the sample proportion of males selected and p2 the sample proportion of females. For the observed table, p1 − p2 = 1/3. Of the 20 possible samples, show that 10 have p1 − p2 ≥ 1/3. Thus, if the three managers were randomly selected, the probability would equal 10/20 = 0.50 of obtaining p1 − p2 ≥ 1/3. This reasoning provides the P-value for Fisher’s exact test with Ha: π1 > π2.