SOLUTION: Regression analysis of free throws by 29 NBA teams during the 2002-2003 season revealed the fitted regression Y = 55.2 + .73X (R2 = .874, Syx = 53.2) where Y = total free throws ma

Question 181941: Regression analysis of free throws by 29 NBA teams during the 2002-2003 season revealed the fitted regression Y = 55.2 + .73X (R2 = .874, Syx = 53.2) where Y = total free throws made and X = total free throws attempted. The observed range of X was from 1, 620 (New York Knicks) to 2,383 (Golden State Warriors) (a.) Find the expected number of free throws made for a team that shoots 2,000 free throws. (b.) Do you think that the intercept is meaningful? Hint: Make a scatter plot and let Excel fit the line (c.) Use the quick rule to make a 95 percent predicition interval for Y when X = 2,000.
FGP FTP Points Fouls TrnOvr
FGP 1.000
FTP - 0.039 1.000
Points 0.475 0.242 1.000
Fouls - 0.014 0.211 0.054 1.000
TrnOvr 0.276 0.028 0.033 0.340 1.000
Rbnds 0.436 0.137 0.767 - 0.032 0.202

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Regression analysis of free throws by 29 NBA teams during the 2002-2003 season revealed the fitted regression Y = 55.2 + .73X
(R2 = .874, Syx = 53.2) where Y = total free throws made and X = total free throws attempted.
The observed range of X was from 1, 620 (New York Knicks) to 2,383 (Golden State Warriors)
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(a.) Find the expected number of free throws made for a team that shoots 2,000 free throws.
Ans: Y(2000) = 55.2 + 0.73*2000 = 1515
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(b.) Do you think that the intercept is meaningful? Hint: Make a scatter plot and let Excel fit the line
I'll leave that to you.
-----------------------------
(c.) Use the quick rule to make a 95 percent predicition interval for Y when X = 2,000.
I am not familiar with your Quick Rule.
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Cheers,
Stan H.
================================
FGP FTP Points Fouls TrnOvr
FGP 1.000
FTP - 0.039 1.000
Points 0.475 0.242 1.000
Fouls - 0.014 0.211 0.054 1.000
TrnOvr 0.276 0.028 0.033 0.340 1.000
Rbnds 0.436 0.137 0.767 - 0.032 0.202

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