Read in the data

golf=read.table('http://www.stat.ufl.edu/~winner/data/pgalpga2008.dat')
colnames(golf) <- c('drive_distance', 'accuracy', 'gender')
golf_female <- subset(golf,gender==1)
golf_male <- subset(golf,gender==2)

Plots

We fit a linear regression model to the female golfer data.

golf_female_lm = lm(accuracy~drive_distance,data=golf_female)
summary(golf_female_lm)
## 
## Call:
## lm(formula = accuracy ~ drive_distance, data = golf_female)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -23.6777  -2.6583   0.9829   3.6346  10.2339 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    130.89331   10.92765  11.978  < 2e-16 ***
## drive_distance  -0.25649    0.04424  -5.797 3.66e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.246 on 155 degrees of freedom
## Multiple R-squared:  0.1782, Adjusted R-squared:  0.1729 
## F-statistic: 33.61 on 1 and 155 DF,  p-value: 3.662e-08

Based on the linear regression, determine the accuracy based on a drive_distance of 260 yards.

Therefore, accuracy=coef(golf_female_lm)[1] + coef(golf_female_lm)[2]*260

coef(golf_female_lm)[1] + coef(golf_female_lm)[2]*260 
## (Intercept) 
##    64.20573

Determine the 95% posterior predictive interval for the accuracy of new female golfer whose average driving distance is 260 yards.

predict(golf_female_lm,data.frame(drive_distance=260),interval="predict")
##        fit      lwr      upr
## 1 64.20573 53.74528 74.66619