A vice president for marketing recruits 20 college graduates for management trai

Question

A vice president for marketing recruits 20 college graduates for management training. Each of the 20 individuals is randomly assigned, 10 each, to one of two groups. A "traditional" method of training (T) is used in one group, and an "experimental" method(E) is used in the other. After the graduates spend six months on the job, the vice president ranks them on the basis of their performance, from 1 (worst) to 20 (best), with the following results
(stored in the le TestRank): T: 1 2 3 5 9 10 12 13 14 15 E: 4 6 7 8 11 16 17 18 19 20
Is there evidence of a dierence in the median performance between the two methods? ( Use ? = 0.05) A vice president for marketing recruits 20 college graduates for management training. Each of the 20 individuals is randomly assigned, 10 each, to one of two groups. A "traditional" method of training (T) is used in one group, and an "experimental" method(E) is used in the other. After the graduates spend six months on the job, the vice president ranks them on the basis of their performance, from 1 (worst) to 20 (best), with the following results
(stored in the le TestRank): T: 1 2 3 5 9 10 12 13 14 15 E: 4 6 7 8 11 16 17 18 19 20
Is there evidence of a dierence in the median performance between the two methods? ( Use ? = 0.05)

Explanation / Answer

We will use Mann-Whitney test:

T1= sum of rank of R = 84

T2 = sum of rank of E =126

Selecting highest of these two rank:

Tx = 126

Number of member in group which give T1=10

Number of member in group which give T2=10

Nx =Number of member in group which give Tx = 6

Critical value :

U = N1*N2+Nx*(Nx+1)/2 -TX

=6*6+(6*7/2)-128

=-71

Critical Region = U(10,10,0.05) =23

AS -71<23.SO, we reject the claim of equality.

Yes, there is evidence of difference.

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