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Week 5 E-Text Assignment Essay

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Week Five E-Text Assignment


Week Five E-Text Assignment
12.47 Regression analysis of free throws by 29 NBA teams during the 2002-2003 season revealed the fitted regression Y = 55.2 + .73X (R2 = 0.874, 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 2,382 (Golden State Warriors).
(a) Find the expected number of free throws made for a team that shoots 2,000 free throws.
Y=55.2+ .732000=1515.2
The expected number of successful free throws for a team that attempts 2,000 free throws is 1,515.
(b) Do you think that the intercept is meaningful?

I do not think the intercept is meaningful. When x = 0, y = 55.2 which is to say that when zero attempts are made, 55.2 attempts are successful. This is not possible since in order to have a successful free throw, players must attempt the free throw.
(c) Use the quick rule to make a 95 percent prediction interval for Y when X = 2, 000
Quick Rule = r>2n
1515 x 0.0724= ±109.69

Given the information above, we can say with 95% certainty that the number of shots made (Y) is probably between 1405.31 and 1624.69 when 2000 free throw attempts are made.
13.31 A sports enthusiast created an equation to predict Victories (the team’s number of victories in the National Basketball Association regular season play) using predictors FGP (team field goal percentage), FTP (team free throw percentage), Points = (team average points per game), Fouls (team average number of fouls per game), TrnOvr (team average number of turnovers per game), and Rbnds (tam average number of rebounds per game). The fitted regression was Victories = -281 + 523 FGP + 3.12 FTP + 0.781 Points – 2.90 Fouls + 1.60 TrnOvr +0.649 Rbnds (R2 = .802, F = 10.80, SE = 6.87). The strongest predictors were FGP (t = 4.35) and Fouls (t = 2.146). The other predictors were only marginally...

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