STAT 511
Solutions to Homework 3
Spring 2004
1. > X <- matrix(c(1,1,rep(c(rep(0,6),1),3),1),6,4)
> Y <- c(2,1,4,6,3,5)
> V1 <- c(1,4,4,1,1,4)
> fit <- lm(Y~X-1,weights=1/V1)
> fit$coefficients
X1 X2 X3 X4
1.8 4.0 6.0 3.4
2. > homes <- read.table("homes.txt", header=T)
> dim(homes)
# The dimension of the matrix is: 88 rows and 15 columns
[1] 88 15
> Y <- as.matrix(homes[,1])
> X <- as.matrix(homes[,c(2,5,10,11,13)])
> X0 <- rep(1,length(Y))
> X <- cbind(X0,X)
> library(lattice)
> splom(~homes[,c(1,2,5,10,11,13)],aspect="fill")
The first plot in the last row suggests that size might be the best single predictor of price. This scatterplot matrix
shows no clear evidence of multicollinearity.
10000 15000
15000
10000
Land
5000 10000
1200
600800
1000
1200
1000
800
600
600
FinishedBsmt
400
200
0 200400600
1200
8001000
1200
1000
800
Basement
600
400
200400600
5
3
4
200
5
4
3
BedRooms
3
2
1
2
3
1
1500 2000
2000
1500
Size
1500
1000
1000 1500
150000
2e+05
2e+05
150000Price150000
1e+05
1e+05
150000
Scatter Plot Matrix
Figure 1: Scatterplot matrix for y, x1 , x2 , . . . , x5
1
0
10000
5000
> round(cor(homes[,c(1,2,5,10,11,13)]),4)
Price
Size BedRooms Basement FinishedBsmt
Land
Price
1.0000 0.6649
0.2974
0.3597
0.3152 0.4353
Size
0.6649 1.0000
0.4647
0.4028
0.2044 0.1975
BedRooms
0.2974 0.4647
1.0000
0.1794
-0.0268 -0.0240
Basement
0.3597 0.4028
0.1794
1.0000
0.3153 -0.0157
FinishedBsmt 0.3152 0.2044 -0.0268
0.3153
1.0000 0.0854
Land
0.4353 0.1975 -0.0240 -0.0157
0.0854 1.0000
> qr(X)$rank
[1] 6
> b <- solve(t(X)%*%X)%*%t(X)%*%Y
> yhat <- X%*%b
> e <- Y-yhat
20000
−40000
−20000
0
Residual
40000
60000
80000
Residual Plot
60000
80000
100000
120000
140000
160000
180000
200000
Predicted Y
Figure 2: Residuals versus Fitted values
To answer what does fin=c(6.0,6.0) inside the command par() do, one can type help(par) or type help.start()
which opens a browser in ”C:\Program Files\R\rw1080\doc\html\rwin.html” and then type par in the Search box.
’fin’ A numerical vector of the form ’c(x, y)’ which gives the
size of the figure region in inches.
> MSE <- (t(e)%*%e)/(dim(X)[1]-qr(X)$rank)
> MSE
[1,] 624915133
> cov.b <- as.numeric(MSE)*solve(t(X)%*%X)
> labels <- c("Intercept","Size","BedRooms","Basement","FinishedBsmt","Land")
> results <- round(cbind(b,sqrt(diag(cov.b))),4)
> tmp <- cbind(labels,results)
> colnames(results) <- c("Estimate","Std Error")
> colnames(tmp) <- c("
","Estimate ","Std Error")
> library(MASS)
> write.matrix(tmp,"hw03.out")
Estimate Std Error
#hw03.out
Intercept
-20167.0441 16406.6402
Size
56.4467
10.2605
BedRooms
2888.6011
4081.838
Basement
21.446
16.7706
FinishedBsmt 21.4798
10.7385
Land
3.847
0.8769
2
1500
2000
1
2
3
4
5
200
400
600
800
1000
Residual Plot
Residual Plot
Normal Probability Plot
400
600
800
FinishedBsmt
Sample Quantiles
−40000 −20000
0
Residual
0
−40000 −20000
0
200
20000 40000 60000 80000
Basement
20000 40000 60000 80000
BedRooms
20000 40000 60000 80000
Size
−40000 −20000
0
20000 40000 60000 80000
−40000 −20000
0
Residual
20000 40000 60000 80000
−40000 −20000
−40000 −20000
1000
Residual
Residual Plot
0
Residual
20000 40000 60000 80000
Residual Plot
0
Residual
Residual Plot
4000
8000
12000
16000
−2
Land
−1
0
1
2
Theoretical Quantiles
Figure 3: Residual plots and Normal probability plot
When we change the smoothing parameter in loess() from 0.9 to 0.5, we obtain a less smooth line.
3. > fit <- lm(Price~Size+BedRooms+Basement+FinishedBsmt+Land,data=homes)
> summary(fit)
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.017e+04 1.641e+04 -1.229
0.2225
Size
5.645e+01 1.026e+01
5.501 4.18e-07 ***
BedRooms
2.889e+03 4.082e+03
0.708
0.4812
Basement
2.145e+01 1.677e+01
1.279
0.2046
FinishedBsmt 2.148e+01 1.074e+01
2.000
0.0488 *
Land
3.847e+00 8.769e-01
4.387 3.39e-05 ***
Residual standard error: 25000 on 82 degrees of freedom
Multiple R-Squared: 0.5778,
Adjusted R-squared: 0.5521
F-statistic: 22.45 on 5 and 82 DF, p-value: 4.105e-14
3