Problem 8.8

bailsofhay — Nov 25, 2013, 5:54 PM

###Problem 8.8 ######

data=read.table("http://www.stat.lsu.edu/exstweb/statlab/datasets/KNNLData/CH06PR18.txt")
names(data)=c("y","x1","x2","x3","x4")

### Part a #####

plot(data$x1, data$y, xlab="Property Age", ylab="Rental Rates")

x1=data$x1-mean(data$x1)
fit1=lm(data$y~x1+I(x1^2)+data$x2+data$x4)
fit1

Call:
lm(formula = data$y ~ x1 + I(x1^2) + data$x2 + data$x4)

Coefficients:
(Intercept)           x1      I(x1^2)      data$x2      data$x4  
   1.02e+01    -1.82e-01     1.41e-02     3.14e-01     8.05e-06  

coeff=fit1$coefficients
# yi= 10.20-.1818X1+.01415X1^2+.3140X2+.000008X4

plot(data$y, fit1$fitted.values,xlab="Rental Rates",ylab="Fitted Value")

# There is a much better fit than when the data was left with just x1,x2 and x4.The data 
# appears to be more linear than previously.


#### Part b###

r.square=summary(fit1)$adj.r.squared
# R^2 is:
r.square

[1] 0.5927

# R^2 measures the proportional reduction in the variation of the estimated Y by adding X1^2 into 
# linear model.

### part c ####

# Alternatives: Ho: B11=0, Ha: B11 is not equal to zero
# Decision Rule: If t* < or equal to t critical, conclude Ho. If t* is > than t critical
# conclude Ha

coeff=fit1$coefficients
b11=coeff[3]
summary=summary(fit1)$coefficients
ts=summary[3,3]
# t* is
ts

[1] 2.431

tc=qt(.975,76)
# t critical is:
tc

[1] 1.992

## Conclude Ha since t*>>t --> B11 is not equal to 0.

### part e ####

x1=data$x1
fit2=lm(y~x1+I(x1^2)+x2+x4, data=data)
fit2

Call:
lm(formula = y ~ x1 + I(x1^2) + x2 + x4, data = data)

Coefficients:
(Intercept)           x1      I(x1^2)           x2           x4  
   1.25e+01    -4.04e-01     1.41e-02     3.14e-01     8.05e-06  

# Yi= 12.49-.404X13+.01415X1^2+.3140X2+.000008X4

### part d ####
predict(fit2, newdata=data.frame(x1 = 8, x2 = 16, x4 = 250000),interval = c("confidence"), level = 0.95)

   fit   lwr   upr
1 17.2 16.46 17.94

# Yh= 17.2
# Confidence interval: [16.5,17.9]