BA 275 Quizzes
Winter 2007
Quiz #6
Name (please print) / Answer KeySection (circle one) / 12noon – 1:50pm / 2:00 – 3:50pm / 4:00 – 5:50pm
Question 1. An old saying in golf is “you drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data on the top 69 money winners on the PGA tour in 1993 are examined. The average number of putts per hole for each player is used to predict their total winnings using the simple linear regression model 1993 winnings = b0 + b1 (average number of putts per hole) + e. Partial regression reports were obtained.
Parameter / Estimate / Standard Error / T Statistic / P-ValueCONSTANT / 7,897,179 / 3,023,782 / ? / ?
Avg. Putts / –4,139,198 / 1,698,371 / ? / ?
Correlation Coefficient = –0.285
Standard Error of Est. = 281,777
A. Suppose the researchers test the hypotheses H0: b1 = 0 vs. Ha: b1 < 0. What is the value of the T statistic for this test?
ANSWER
/ T = estimate / (Standard Error) = –4139198 / 1698371 = –2.43716B. What should be the degrees of freedom used in the above test?
ANSWER
/ n – p – 1 = 69 – 1 – 1 = 67C. Given a = 5%, what is the conclusion of the above test?
ANSWER
/ Rejection region is: Reject H0 if t < –1.671 (if 60 degrees of freedom are used) or –1.664 (if 80 degrees of freedom are used.) Given that the t statistic (from A) is –2.43716, we would reject H0 at a = 5%.D. Construct a 95% confidence interval for the slope parameter b1. Report your answer in the following format: ( quick estimate ) ± ( margin of error )
ANSWER
/ If 60 degrees of freedom are used: –4139198 ± (2.00) × (1698371)If 80 degrees of freedom are used: –4139198 ± (1.99) × (1698371)
E. What percentage of variation in the dependent variable has been explained by the independent variable?
ANSWER
/ R-Squared = = r2 = the percentage of total variation in Y explained by X =(–0.285)2 = 0.08123
Question 2. A linear regression analysis has produced the following equation relating profits to hours of managerial time spent developing the past year’s projects at a firm:
Profits = -- $957 + $85 × ( Number of Hours )
A. According to this estimated relationship, how large would the profits (or losses) be if only 10 hours were spent in planning?
ANSWER
/ Estimated profits = –957 + 85 × 10 = –107 (i.e., a loss of $107.)B. On the average, how large an increase in profits would have been resulted last year if additional 8 hours were spent on planning?
ANSWER
/ The estimated slope coefficient $85 means the amount of profits will change with an additional hour of planning. Hence, for additional 8 hours, the profits are expected to increase by $85 × 8 = $680C. If the sample correlation is r = 0.351, what percentage of the total variation in profits has been explained by the time spent?
ANSWER
/ R-Squared = = r2 = the percentage of total variation in Y explained by X = (0.351)2 = 0.1232D. What percentage of the total variation in profits is left unexplained by the number of hours spent?
ANSWER
/ 1 – r2 = 1 – 0.1232 = 0.8768Hsieh, P-H 1