Mathematics, 17.12.2019 20:31, byejpg
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 winnings)i = β0 + β1(average number of putts per hole)i + εi,
where the deviations εi are assumed to be independent and normally distributed with a mean of 0 and a standard deviation of σ. this model was fit to the data using the method of least squares. the following results were obtained from statistical software.
r2 = 0.081
s = 281.8
variable
parameter estimate
standard error
constant
7897.2
3023.8
average putts
–4139.2
1698.4
the quantity s = 281.8 is an estimate of the standard deviation σ of the deviations in the simple linear regression model.
1)what are the degrees of freedom for this estimate?
2)what is the value of the intercept of the least-squares regression line?
3)suppose the researchers conducting this study wish to test the hypotheses h0: β1 = 0 versus ha: β1< 0. what is the value of the t statistic for this test?
4)what is an approximate 95% confidence interval for the slope β1?
5)if the researchers wish to estimate the mean winnings for all tour pros whose average number of putts per hole is 1.75, what would be a 95% confidence interval for the mean winnings?
6)if the researchers wish to estimate the winnings for a particular tour pro whose average number of putts per hole is 1.75, what would be a 95% prediction interval for the winnings?
7)what is the value of the sst, the total sum of squares?
8)what is the value of the f statistic for testing the hypotheses h0: β1 = 0 versusha: β1 ≠ 0?
Answers: 3
Mathematics, 21.06.2019 16:30, robclark128
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