A pro golfer collects data on her clubhead speeds in mph and distance in yards for 20 games. Clubhead Speed (mph) Distance (yards) Clubhead Speed (mph) Distance (yards) 79 211 95 255 83 214 96 248 85 221 97 263 88 225 98 273 89 225 98 285 90 228 100 280 90 239 101 286 92 240 104 289 93 250 106 294 95 247 106 299 (a) The golfer would like to find the least squares regression line predicting distance traveled (y) from clubhead speed (x) for the 20 games she played. Create a scatterplot of the data. Does a linear regression model make sense for these data? Yes/No (b) Calculate the slope. (Round your answer to four decimal places.) Calculate the y-intercept. (Round your answer to four decimal places.) (c) For every 1-mph increase in swing speed, the distance traveled by the golf ball increases by 3.6963 yards. True/False
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Refer to The Sport Journal (Winter 2007) study of a new method for ranking the total driving performance of golfers on the Professional Golf Association (PGA) tour, presented in Exercise 2.66 (p. 90). Recall that the method computes a driving performance index based on a golfer's average driving distance (yards) and driving accuracy (percent of drives that land in the fairway). Data for the top 40 PGA golfers (as ranked by the new method) are saved in the PGA file. (The first five and last five observations are listed in the next table.) a. Write the equation of a straight-line model relating driving accuracy ( $y$ ) to diving distance $(x)$. b. Use simple linear regression to fit the model you found in part a to the data. Give the least squares prediction equation. c. Interpret the estimated $y$ -intercept of the line. d. Interpret the estimated slope of the line. e. A professional golfer, practicing a new swing to increase his average driving distance, is concerned that his driving accuracy will be lower. Which of the two estimates, $y$ -intercept or slope, will help you determine whether the golfer's concern is a valid one? Explain.
Develop an estimated regression equation that can be used to predict the average number of yards per drive given the ball speed and the launch angle. Y = 81.596 + 1.093X1 + 1.646X2, where Y = average number of yards per drive, X1 = ball speed, X2 = launch angle. Adjusted R2 = 0.826. Suppose a new member of the PGA Tour for 2013 has a ball speed of 170 miles per hour and a launch angle of 11 degrees. Use the estimated regression equation in part (d) to predict the average number of yards per drive for this player. The ball speed of the new player = 170 mph launch angle = 11 degree. The predicted value of the average number of yards per drive is Y = 81.596 + 1.093*170 + 1.646*11 = 285.512. In part (d) of question-2, data contained in the DATAfile PGADrivingDistance (PGA Tour website, November 1, 2012) was used to develop an estimated regression equation to predict the average number of yards per drive given the ball speed and the launch angle. Help with these two questions below: 1. Does the estimated regression equation provide a good fit to the data? Explain. 2. In part (b) of question-2, an estimated regression equation was developed using only ball speed to predict the average number of yards per drive. Compare the fit obtained using just ball speed to the fit obtained using ball speed and the launch angle.
Adi S.
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 2013 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: winnings = ̢̣̣̑̑ + ̢̣̣̑̑ (average number of putts per hole) This model was fit to the data using the method of least squares. The following results were obtained from statistical software: Variable Parameter Estimate SE of Parameter Estimate Intercept 7,897,179 3,023,782 Avg. Putts -4,139,198 1,698,371 A 95% confidence interval for the slope in the simple linear regression model is (approximately): 7,897,179 6,047,564 7,897,179 3,023,782 -4,139,198 3,396,742 -4,139,198 1,698,371
David N.
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