Question

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There is an old saying in golf: "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 the total winnings (in thousands of dollars) using the simple linear regression model: (1993 winningsPo B1(average number of putts per hole)i Eiv where the deviations E's are assumed to be independent and Normally distributed with a mean of 0 and a standard deviation of ơ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software R20.081,s-281.8 Parameter estimate 7897.2 -4139.2 Variable Constant Average putts Standard error 3023.8 1698.4 Determine whether each of the following statements is true or false: 1. The explanatory variable in this study is the average number of putts per hole. The value of the correlation between 1993 winnings and average number of putts per hole is 0.285. If the average number of putts per hole increases by 1, we estimate the winnings in 1993 to decrease by $4,139.20. We predict the 1993 winnings for golf pro with an average of 1.75 putts per hole to be $653,600.00 a. b. c. d. 2. what are the degrees of freedom for s, the estimate of the standard deviation of σ? 3. What is the value of the intercept of the least-squares regression line? What is the value of the slope of the least-squares regression line? We would like to show that the higher the putts are, the lower the earnings are. 4. 5. a. What is are the hypotheses? b. What is the test statistic, its distribution and its value? c. What is your conclusion? what is a 95% confidence interval for ß? 6, 7. The following results were obtained from a statistical software for the case the average putts per game is 1.75: 95.0% C.I 530.6, 776.6) Predicted winnings Standard error 61.62 95.0% P.1 77.7, 1229.4 653.6 What is the predicted winnings for a player who averages 1.75 putts per game? what is the 95% confidence interval for the mean winnings for that player? a. b. The following is the partial ANOVA table for the above simple linear regression: Sum of squares 471,605 5,319,690 DF Source Model Error 67 8. Which of the following statements are correct? a. The ej are assumed to be Normally distributed with a mean of 0 and a standard deviation of σ. The parameters of this model are β0, β1, and σ. b. + βιχί is the mean response when x-xi. The εί, 1-1, 2, , n are independent. All of the above are true. c. d. e. Which of the following statements about simple linear regression is/are FALSE? 9. The term "residual" refers to the difference between the observed response and the predicted response using the regression equation. a. b. If bo and b1 are the intercept and slope, respectively, of the least-squares line, ỹ - bo + bix" is an unbiased estimator of the mean response when x = x*. The estimate of σ is given by s C. d. The e sum to zero. 10. What is the value of the SST, the total sum of squares 11. What is the value of F statistic and its distribution? 12. What is the hypothesis that the test statistic is defined for? 13. What is the estimate of the standard deviation of the error term?

Answer 1

Solution:

Question 1

Part a

Answer: True

For the given regression model, the response variable is given as 1993 winnings and explanatory variable is given as the average number of putts per hole.

Part b

Answer: True

We are given R^2 = 0.081

So, correlation coefficient = r = sqrt(0.081) = 0.284605

r = 0.285

Part c

Answer: True

We are given the slope as -4139.2 which indicate a decrement of -4139.2 in response variable as per increase of one unit in explanatory variable.

Part d

Answer: True

The regression equation is given as below:

1993 winnings = 7897.2 – 4139.2*Average putts

We are given average putts = 1.75

1993 winnings = 7897.2 – 4139.2*1.75

1993 winnings = 653.6 thousand dollars

1993 winnings = $653,600.00

Question 2

We are given n = 69

df = n – 1 = 69 – 1 = 68

Question 3

The value of the intercept of the least squares regression line is given as 7897.2.

Question 4

The value of the slope of the least squares regression line is given as - 4139.2.