Question

5.5. Consumer finance. The data below show, for a consumer finance company operating in six cities, the number of competing loan companies operating in the city (X) and the number per thousand of the company's loans made in that city that are currently delinquent(Y): : X: Y:: 1 4 16 2 1 5 3 2 10 4 3 15 5 3 13

Please answer 5, 6, 7, 8, 9, 10.

Answer 1

We define the following matrices

$\begin{align*} Y=\begin{bmatrix}16\\5\\10\\15\\13\\22 \end{bmatrix} X=\begin{bmatrix}1&4\\1&1\\1&2\\1&3\\1&3\\1&4 \end{bmatrix} \epsilon=\begin{bmatrix}\epsilon_1\\\epsilon_2\\ ...\\\epsilon_6\\ \end{bmatrix} \beta=\begin{bmatrix}\beta_1\\\beta_2 \end{bmatrix} \end{align*}$

The regression equation that we want to estimate is

$\begin{align*} Y=X\beta+\epsilon \end{align*}$

5) The estimate of coefficients is

$\begin{align*} b&=(X^{'}X)^{-1}X^{'}Y\\ &=\left(\begin{bmatrix}1& 1& 1& 1& 1& 1\\ 4& 1& 2& 3& 3& 4 \end{bmatrix}\begin{bmatrix}1&4\\1&1\\1&2\\1&3\\1&3\\1&4 \end{bmatrix} \right )^{-1}\begin{bmatrix}1& 1& 1& 1& 1& 1\\ 4& 1& 2& 3& 3& 4 \end{bmatrix}\begin{bmatrix}16\\5\\10\\15\\13\\22 \end{bmatrix} \end{align*}$

We use the following R code to calculate this

#set Y
Y<-c(16,5,10,15,13,22)
#set x
x<-c(4,1,2,3,3,4)
#set the matrix X
X<-matrix(c(rep(1,6),x),ncol=2)

#part 4)
XX<-solve(t(X)%*%X)
#print
XX

#part 5)
b<-XX%*%(t(X)%*%Y)
#print
sprintf('The estimated intercept is %.4f',b[1])
sprintf('The estimated slope is %.4f',b[2])
sprintf('The estimated regression equation is yhat=%.4f+%.4fx',b[1],b[2])

#get this

sprintf ('The estimated intercept is 4f',b[l] [1] "The estimated intercept is 0.4390" sprintf ('The estimated slope is .4f',b[2]) [1"The estimated slope is 4.6098" sprintf ('The estimated regression equation is yhat-.4f+%. 4fx',b[1],b [2] ) 1"The estimated regression equation is yhat=0.4390+4.6098x"

6) The residual is calculated using

$\begin{align*}e=Y-\hat{Y}=Y-Xb\\ \end{align*}$

R code

#part 6)
e<-Y-X%*%b
#print
e

#get this

1,2.87804878 [2,0.04878049 [3, 0.34146341 [4,0.73170732 [5,- [6, -1.26829268 3.12195122

7) The hat matrix H is

$\begin{align*}H=X(X^{'}X)^{-1}X^{'} \end{align*}$

R code is

#part 7)
H<-X%*%XX%*%t(X)
#print
H

#get this

> H ,2 31 ,4 ,5] ,6] [1, 0.36585366 -0.1463415 0.02439024 0.1951220 0.1951220 0.36585366 [2,-0.14634146 0.6585366 0.39024390 0.1219512 0.1219512 -0.14634146 [3, 0.02439024 0.3902439 0.26829268 0.1463415 0.1463415 0.02439024 [4, 0.19512195 0.1219512 0.14634146 0.1707317 0.1707317 0.19512195 [5, 0.19512195 0.1219512 0.14634146 0.1707317 0.1707317 0.19512195 [6, 0.36585366 -0.1463415 0.02439024 0.1951220 0.1951220 0.36585366

8) The sum of square SSE is

$\begin{align*}SSE=e^{'}e \end{align*}$

where

$\begin{align*}e=Y-\hat{Y}=Y-Xb\\ \end{align*}$

R code

#part 8)
sse<-t(e)%*%e
#print
sprintf('SSE=%.4f',sse)

# get this

sprintf ('SSE=%.4I',Sse) [1] "SSE 20.2927"

9) the estimate of variance of regression is

$\begin{align*}\hat{\sigma}^2=MSE=\frac{SSE}{n-2} \end{align*}$

The variance-covariance matrix is

$\begin{align*}\hat{\sigma}^2(X^{'}X)^{-1} \end{align*}$

R code

#part 9)
#get the estimate of regression variance
mse<-sse/(6-2)
#get the estimate of variance-covariance
cvb<-drop(mse)*XX
#print
sprintf('Var(b0)=%.4f',cvb[1,1])
sprintf('Var(b1)=%.4f',cvb[2,2])

# get this

sprintf('Var (b0)=.4f',cvb[1,1]) [1] "Var (b0) =6.8055" sprintf('Var (bl) =% .4f', cvb [2, 2] ) [1 "Var (bl) =o.7424"

10) The variance of yhat at

$\begin{align*}x_h=\begin{bmatrix} 1\\2\end{bmatrix} \end{align*}$ is

$\begin{align*}s^2(\hat{y}_h)=\hat{\sigma}^2{x_h^{'}(X^{'}X)^{-1}x_h} \end{align*}$

R code

#part 10)
#set xh
xh<-c(1,2)
s2yhat<-drop(mse)*(t(xh)%*%XX%*%xh)
#print
sprintf('s2(yhat)=%.4f',s2yhat)

#get this

sprintf ('s2 (yhat)-.4f', s2yhat) 1 "s2 (yhat) =1.3611"

Code to all the parts together

#set Y
Y<-c(16,5,10,15,13,22)
#set x
x<-c(4,1,2,3,3,4)
#set the matrix X
X<-matrix(c(rep(1,6),x),ncol=2)

#part 4)
XX<-solve(t(X)%*%X)
#print
XX

#part 5)
b<-XX%*%(t(X)%*%Y)
#print
sprintf('The estimated intercept is %.4f',b[1])
sprintf('The estimated slope is %.4f',b[2])
sprintf('The estimated regression equation is yhat=%.4f+%.4fx',b[1],b[2])

#part 6)
e<-Y-X%*%b
#print
e

#part 7)
H<-X%*%XX%*%t(X)
#print
H

#part 8)
sse<-t(e)%*%e
#print
sprintf('SSE=%.4f',sse)

#part 9/10)
#get the estimate of regression variance
mse<-sse/(6-2)
#get the estimate of variance-covariance
cvb<-drop(mse)*XX
#print
sprintf('Var(b0)=%.4f',cvb[1,1])
sprintf('Var(b1)=%.4f',cvb[2,2])

#part 10)
#set xh
xh<-c(1,2)
s2yhat<-drop(mse)*(t(xh)%*%XX%*%xh)
#print
sprintf('s2(yhat)=%.4f',s2yhat)