# Question

A pool of six candidates for three managerial positions includes three females and three males. Denote the three females by F1, F2, F3 and the three males by M1, M2, M3. The result of choosing three individuals for the managerial positions is (F2, M1, M3).

a. Identify the 20 possible samples that could have been selected. Explain why the contingency table relating gender to whether chosen for the observed sample is

b. Let 1 denote the sample proportion of males selected and 2 the sample proportion of females. For the observed table, 1 – 2 = (2/3) - (1/3) = 1/3. Of the 20 possible samples, show that 10 have 1 – 2 ≥ 1/3. (If the three managers were randomly selected, the probability would equal 10/20 = 0.50 of obtaining 1 – 2 ≥ 1/3. This is the reasoning that provides the one-sided P-value for Fisher’s exact test with Ha: p1 > p2.)

a. Identify the 20 possible samples that could have been selected. Explain why the contingency table relating gender to whether chosen for the observed sample is

b. Let 1 denote the sample proportion of males selected and 2 the sample proportion of females. For the observed table, 1 – 2 = (2/3) - (1/3) = 1/3. Of the 20 possible samples, show that 10 have 1 – 2 ≥ 1/3. (If the three managers were randomly selected, the probability would equal 10/20 = 0.50 of obtaining 1 – 2 ≥ 1/3. This is the reasoning that provides the one-sided P-value for Fisher’s exact test with Ha: p1 > p2.)

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