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STAT3008 Assignment 2 Solutions
(2010 – 2011 Semester 2) Q1. Problem 2.1 2.1.1 R Codes: library(alr3) data(htwt) plot(htwt)
With only 10 points, it is hard to determine whether a simple linear regression is enough or not. 2.1.2 R Codes: library(alr3) data(htwt) plot(htwt) fit=lm(Wt~Ht,data=htwt) #Fit the regression line abline(fit) average=mean(htwt) #Compute the means s=(10-1)*cov(htwt) #cov() computes the sample covariance matrix s #Display Sxx,Syy,Sxy average #Display the means
summary(fit) #Display the summary of the fitted model
Using R, we have 165.52x = , 59.47y = , 472.076SXX = , 731.961SYY = ,
274.786SXY = . 0 36.8759β = − and 1 0.5821β = .
2.1.3 R Codes: summary(fit) #Display the summary of the fitted model vcov(fit) #Display the variance-covariance matrix of the estimates
Using the above commands, we get
0( ) 64.4728se β = ,
1( ) 0.3892se β = and
2 28.456 71.502σ = = . Also,
0
1
4156.7419 25.070025.0700 0.1515
Covβ
β
− = −
.
2.1.4 R Codes: anova(fit) #Display the anova table of the fitted model We can get from the results that 2.237F statistic− = and 0.1731p value− = . Therefore, there is no strong evidence that the slope is significant.
Q2. 2.3 2.3.1
α is the value of ( | )E Y X x= .
2.3.2 2
1 1
2 2 21 1
( , ) ( ( ))
( ) 2 ( )( ) ( )i i
i i i i
RSS y x x
y y x x x x
α β α β
α β α β
= − − −
= − − − − + −
∑∑ ∑ ∑
And also, ( )( ) ( ) ( )0
i i i i iy x x y x x x xSXY SXY
α α− − = − − −
= − =∑ ∑ ∑
Therefore, 2 21 1 1( , ) ( ) 2iRSS y SXY SXXα β α β β= − − +∑
2 ( ) 0iRSS y
yα α
αα
α=
∂= − − =
∂
=
∑
1 1
11
1
2 2 0RSS SXY SXX
SXYSXX
β β
ββ
β
=
∂= − + =
∂
=
2.3.3
2
( ) ( )Var Var ynσα = = ,
2
1( ) ( )SXYVar VarSXX SXX
σβ = =
Q3. 2.4.1 R Codes: library(alr3) data(heights) model1=lm(Dheight~Mheight,data=heights) summary(model1) anova(model1)
From the above commands, we can get 2 22.27 5.1529σ = = , 0 29.917β = ,
1 0.542β = ,
0( ) 1.623se β = and
1( ) 0.026se β = . 2 0.241R = means 0.241 of the
variability of the data is explained by the model. The ANOVA table shows that
435F statistic− = with 162 10p value −− =< × . F-test suggests that 1β is significant
or 1 0β ≠ . Q4. a)
Since 0 1i i iy x eβ β= + + ……(1)
0 1i i iy n x eβ β= + +∑ ∑ ∑
Then 0 1y x eβ β= + + ……(2)
(1)-(2), 0 1 0 1( ) ( )i i iy y x e x eβ β β β− = + + − + +
As a result, 1( ) ( )i i iy y x x e eβ− = − + −
b)
ie ’s are i.i.d. with ( ) 0iE e = and 2( )iVar e σ= . Thus, 2 2( )iE e σ= and
2 2( )iE e nσ=∑ .
Also, ( ) 0E e = and 2( ) /Var e nσ= . Thus, 2 2( ) /E e nσ= .
2 2 2 21 1( ) ( ) ( ) 2 ( )( )i i i i iy y x x e e e e x xβ β− = − + − + − −
2 2 2 21 1( ) ( ) ( ) 2 ( )( )i i i i iy y x x e e e e x xβ β− = − + − + − −∑ ∑ ∑ ∑
2 2 2 21 1[ ( ) ] ( ) ( ) 2 ( )[ ( )]i i i i iE y y x x E e e x x E e eβ β− = − + − + − −∑ ∑ ∑ ∑
( ) 0iE e e− =
22 2
22
222
22 2
( ) ( 2 )
( ) ( ) 2 ( )
2 ( )
1
i i i
i i
i
E e e E e e ee
E e E e E eeeE
n nn
n n
σσ
σσ σ
− = + −
= + −
= + −
−= − =
Therefore, 2 2 2 21[ ( ) ] ( ) ( 1)i iE y y x x nβ σ− = − + −∑ ∑
c)
0 1 iiy xβ β= +
1 1( ) iiy y x xβ β= − +
1 1( )i i iiy y y y x xβ β− = − − −
1( ) ( )i i iiy y y y x xβ− = − − −
22 2 21 1( ) ( ) ( ) 2 ( )( )i i i i iiy y y y x x x x y yβ β− = − + − − − −
22 2 21 1( ) ( ) ( ) 2 ( )( )i i i i iiy y y y x x x x y yβ β− = − + − − − −∑ ∑ ∑ ∑
22 2 2 21 1 2
( )( )( ) ( ) ( ) 2 ( )
( )i i
i i i iii
x x y yy y y y x x x x
x xβ β
− −− = − + − − × −
−∑∑ ∑ ∑ ∑∑
2 22 2 2 21 1( ) ( ) ( ) 2 ( )i i i iiy y y y x x x xβ β− = − + − − −∑ ∑ ∑ ∑
22 2 21( ) ( ) ( )i i iiy y y y x xβ− = − − −∑ ∑ ∑
d) From the results of (c), we get
22 2 21[ ( ) ] [ ( ) ] ( ) ( )i i iiE y y E y y E x xβ− = − − −∑ ∑ ∑
From the results of (b) and
22 2 2 21 1 1 12
( ) ( ) [ ( )]( )i
E Var Ex xσβ β β β= + = +−∑
,
22 2 2 2 2 2
1 12[ ( ) ] [ ( ) ( 1) ] [ ] ( )
( )i i iii
E y y x x n x xx xσβ σ β− = − + − − + −−∑ ∑ ∑∑
2 2 2 2 2 2 21 1[ ( ) ] ( ) ( 1) ( )i i iiE y y x x n x xβ σ σ β− = − + − − − −∑ ∑ ∑
2 2[ ( ) ] ( 2)i iE y y n σ− = −∑
Therefore,
22 2( )
[ ]2
i iy yE
nσ σ
−= =
−∑ .
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