Problem 3. The Athletic Department of Leland University is considering whether t

Question

Problem 3. The Athletic Department of Leland University is considering whether to hold an extensive campaign next year to raise funds for a new athletic field. The response to the campaign depends heavily upon the success of the football team this fall. In the past, the football team has had winning seasons 60 percent of the time. If the football team has a winning season (W) this fall, then many of the alumnae and alumni will contribute and the cam paign will raise $3 million. If the team has a losing season (L), few will contribute and the campaign will lose $2 million. If no campaign is undertaken, no costs are incurred. On September 1, just before the football season begins, the Athletic Department needs to make its decision about whether to hold the campaign next year. A famous football guru, William Walsh, has offered his services to help evaluate whether the team will have a winning season. For $100,000, he will carefully evaluate the teanm throughout spring practice and then throughout preseason workouts. William then will provide his prediction on September 1 regarding what kind of season, W or L, the team will have. In similar situations in the past when evaluating teams that have winning seasons 50 percent of the time, his predictions have been correct 75 percent of the time. (a) Determine the optimal policy regarding whether to hire William and whether to un- dertake the campaign. (b) What is the maximum price that the team should pay for William's services? (c) Assuming that an analysis more accurate than William's can be conducted by a differ- ent expert, what is the maximum cost of such an analysis for the team?

Explanation / Answer

a) Expected value without hiring = 0.6*3-0.4*2= 1 million

Expected value with hiring= 0.6*0.75*3-0.6*0.25*2=1.05

Expected value including cost of hiring = 1.05-0.1 = 0.95 million

We should undertake the campaign as it is.

b) Maximum price= 1.05-1=0.05 million= 5000 dollars

c) If accuracy=100%, Expected value = 0.6*3=1.8 million

Cost of analysis = 1.8-1=0.8 million

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