Question

9.72 G olf Scores In a professional golf tournament (c) Find a 95% interval to predict the average final e players participate in four rounds of golf and the player with the lowest score after all four rounds is th score of all golfers who shoot a +3 in the first round at the Masters. e champion. How well does a player's performance in the first round of the tournament predict the final score? Table 9.6 shows the first round score and final score for a random sample of 20 golfers who made the cut in a recent Masters tournament. The data are VariableNMean First Final StDev 20 0.550 3.154 20-0.65 5.82 also stored in MastersGolf The regression equation is Final 0.162 +1.48 First Computer output for a regression model to pre- dict the final score from the first-round score is Predictor Constant First Coef SE Coef T P 0.1617 0.8173 0.20 0.845 1.4758 0.2618 5.64 0.000 shown. Use values from this output to calculate and interpret the following. Show your work. S 3.59805 R-Sq-63.8% R-Sq(adj)-61.8% (a) Find a 95% interval to predict the average final score of all golfers who shoot aOon the first round Analysis of Variance Source Regression Residual Error Total DF MS 1 411.52 411.52 31.79 0.000 SS at the Masters. (b) Find a 95% interval to predict the final score of 18 233.03 12.95 19 644.55 a golfer who shoots a -5 in the first round at the Masters Table 9.6 Golf scores afrer the first and final rounds of the Masters First-4-4-1 Final 12 87 7 -5 -5 3 -2 -21 -1 0 4 1 -54 4 01 -3 -1 0 1 -1 0 -1 4 34-1 7 3 4 46 6 8 10

Answer 1

Theory:

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R-commands and outputs:

x=c(-4,-4,-1,-5,-4,-4,0,-1,-3,-1,0,1,-1,0,-1,4,3,4,-1,7)
y=c(-12,-8,-7,-7,-5,-5,-3,-2,-2,-1,-1,-1,0,3,4,4,6,6,8,10)

summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-5.00 -3.25 -1.00 -0.55 0.25 7.00
summary(y)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-12.00 -5.00 -1.00 -0.65 4.00 10.00
sd(x)
[1] 3.153528
sd(y)
[1] 5.824404

fit=lm(y~x)
fit
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
0.1617 1.4758

summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-6.2585 -2.1133 0.0762 1.3383 9.3141
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 0.1617 0.8173 0.198 0.845
x 1.4758 0.2618 5.638 2.38e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.598 on 18 degrees of freedom
Multiple R-squared: 0.6385, Adjusted R-squared: 0.6184
F-statistic: 31.79 on 1 and 18 DF, p-value: 2.385e-05

aov(fit)
Call:
aov(formula = fit)
Terms:
x Residuals
Sum of Squares 411.5233 233.0267
Deg. of Freedom 1 18
Residual standard error: 3.598045
Estimated effects may be unbalanced

# Some requirements to find intervals:
n=20
beta0=as.numeric(fit$coefficients[1])
beta0
[1] 0.161683
beta1=as.numeric(fit$coefficients[2])
beta1
[1] 1.475787
xbar=mean(x)
xbar
[1] -0.55
Sxx=sum((x-xbar)^2)
Sxx
[1] 188.95
sig=summary(fit)$sigma
MSRes=sig^2
MSRes
[1] 12.94593
alpha=0.05
tabt=qt(1-alpha/2,df=n-2)
tabt
[1] 2.100922

#(a)
# Find a 95 % interval to predict the average final score of all golfers who shoot a 0 on the first round at the Masters.
x0=0
muyx0=beta0+beta1*x0
muyx0
[1] 0.161683
(1/n)+((x0-xbar)^2/Sxx)
[1] 0.05160095
MSRes*((1/n)+((x0-xbar)^2/Sxx))
[1] 0.6680223
sqrt(MSRes*((1/n)+((x0-xbar)^2/Sxx)))
[1] 0.8173263
muyx0-tabt*sqrt(MSRes*((1/n)+((x0-xbar)^2/Sxx)))
[1] -1.555456
muyx0+tabt*sqrt(MSRes*((1/n)+((x0-xbar)^2/Sxx)))
[1] 1.878822
##Required 95% prediction Interval is [-1.555456, 1.878822]

#(b)
# Find a 95 % interval to predict the final score of a golfer who shoot a -5 in the first round at the Masters.
x0=-5
y0hat=beta0+beta1*x0
y0hat
[1] -7.217253
1+(1/n)+((x0-xbar)^2/Sxx)
[1] 1.154803
MSRes*(1+(1/n)+((x0-xbar)^2/Sxx))
[1] 14.95
sqrt(MSRes*(1+(1/n)+((x0-xbar)^2/Sxx)))
[1] 3.866522
y0hat-tabt*sqrt(MSRes*(1+(1/n)+((x0-xbar)^2/Sxx)))
[1] -15.34052
y0hat+tabt*sqrt(MSRes*(1+(1/n)+((x0-xbar)^2/Sxx)))
[1] 0.9060091
##Required 95% prediction Interval is [-15.34052, 0.9060091]

#(c)
# Find a 95 % interval to predict the average final score of all golfers who shoot a +3 in the first round at the Masters.
x0=3
muyx0=beta0+beta1*x0
muyx0
[1] 4.589045
(1/n)+((x0-xbar)^2/Sxx)
[1] 0.1166975
MSRes*((1/n)+((x0-xbar)^2/Sxx))
[1] 1.510758
sqrt(MSRes*((1/n)+((x0-xbar)^2/Sxx)))
[1] 1.229129
muyx0-tabt*sqrt(MSRes*((1/n)+((x0-xbar)^2/Sxx)))
[1] 2.006741
muyx0+tabt*sqrt(MSRes*((1/n)+((x0-xbar)^2/Sxx)))
[1] 7.171349
##Required 95% prediction Interval is [2.006741, 7.171349]