STAT 645 HOMEWORK SOLUTION FOR CHANPTER 7 AND 8
Chapter 7
Problem 7.4
a. The AVNOVA table is:
Source
X1
X3|X1
X2|X1, X3
Error
Total
DF
1
1
1
48
51
Seq SS
MS
136366
136366
2033565
2033565
6675
6675
985530 20531.88
3162136 62002.67
b. H0: β2 = 0
HA: β2 ≠ 0
SSR(X 2 | X1 , X 3 ) SSE(X1 , X 2 , X 3 ) 6675 985530
F* =
÷
=
÷
= 0.3251
1
n−4
1
48
F(0.95, 1, 48) = 4.0427
F* < F(0.95, 1, 48), we conclude H0, that is, X2 can be dropped.
P-value = 0.5713
c. SSR(X1) + SSR(X2|X1) = 136366 + 5726 = 142092
SSR(X2) + SSR(X1|X2) = 11395 + 130697 = 142092
So they are equal to each other, and the value is actually SSR(X1, X2).
Problem 7.7
a. The AVNOVA table is:
Source
DF
X4
1
X1|X4
1
X2|X4, X1
1
X3|X4, X1, X2
1
Error
76
Total
80
Seq SS
MS
67.775
67.775
42.275
42.275
27.857
27.857
0.42
0.42
98.231 1.292513
236.557 2.956963
b. H0: β3 = 0
HA: β3 ≠ 0
SSR(X 3 | X 4 , X1 , X 2 ) SSE(X1 , X 2 , X 3 , X 4 ) 0.42 98.231
F* =
÷
=
÷
= 0.3249
1
n −5
1
76
F(0.99, 1, 76) = 6.9806
F* < F(0.99, 1, 76), we conclude H0, that is, X3 can be dropped.
P-value = 0.5704
Problem 7.8
H0: β2 = 0 and β3 = 0
HA: β2 ≠ 0 or β3 ≠ 0
SSR(X 2 , X 3 | X 4 , X1 ) SSE(X1 , X 2 , X 3 , X 4 ) 138.327 - 110.05 98.231
÷
=
÷
= 10.9388
2
n −5
2
76
F(0.99, 2, 76) = 4.8958
F* > F(0.99, 1, 76), we conclude HA, that is, X2 and X3 can be dropped at the same
time.
P-value < 0.001
F* =
Problem 7.14
a. The R values are:
SSR(X1 )
8275.4
2
R Y1 =
=
= 0.6190
SSTO(X 1 ) 13369.3
2
R Y1|2 =
2
R Y1|23 =
SSR(X1 | X 2 ) 3896.0
=
= 0.4579
SSE(X 2 )
8509.0
SSR(X1 | X 2 , X 3 ) 2857.6
=
= 0.4021
SSE(X 2 , X 3 )
7106.4
Problem 7.22
The significance of the predictors depends on other variables in the model. When we
add four additional variables, if they are related to the three already in the model it is
possible for none of the new variables to be significant and those previously in the
model can go from significant to not significant. This does not mean the model is
worse.
Problem 7.25
a. The regression equation is: Y = 4080 + 0.000935 X1
b. In 6.10, the regression function is: Y = 4150 + 0.000787 X1 - 13.2 X2 + 624 X3.
What we can see from here is that if we add the factor X2 and X3, there will be a
large impact on the coefficient for X1.
c. SSR(X1) = 136366, SSR(X1|X2) = 130697, they are not equal each other.
d. r12 = 0.08490
Problem 7.29
a. RHS = SSR(X1) + SSR(X2, X3|X1) + SSR(X4| X1, X2, X3)
= SSR(X1) + [SSR(X1, X2, X3) – SSR(X1)]
+ [SSR(X1, X2, X3, X4) – SSR(X1, X2, X3)]
= SSR(X1, X2, X3, X4) = LHS
b. RHS = SSR(X2, X3) + SSR(X1|X2, X3) + SSR(X4| X1, X2, X3)
= SSR(X2, X3) + [ SSR(X1, X2, X3) – SSR(X2, X3)]
+ [SSR(X1, X2, X3, X4) – SSR(X1, X2, X3)]
= SSR(X1, X2, X3, X4) = LHS
Chapter 8
Problem 8.6
a. The regression equation is: Y = - 26.3 + 4.87 X - 0.118 X2.
The graph for the regression function is:
It seems to be a good fit, and R2 = SSR / SSTO = 1046.27 / 1284.81 = 0.8143.
b. The null and alternative hypothesis are:
H0: β1=β11=0
HA: β1≠0 or β11≠ 0
F* = MSR / MSE = 52.63; F(0.99; 2, 24) = 5.61359
F* > F(0.99; 2, 24), we reject H0, which implies that there is a regression
relationship. The p-value is less than 0.001.
c. We know that the prediction intervals could be Yˆh ± Ss{ pred } or Yˆh ± Bs{ pred } .
In this problem, S2 = gF(1-α;g, n-3) = 3F(0.99; 3, 24) = 14.1542, S = 3.7622, B =
t(1-α/6; 24) = 3.25838.
Therefore, for Xh = 10, 15, 20
Xh = 10, fitted value for Yh = 10.52
s2{pred} = 9.939 (1/27 + (10 - 15.7778)2 / 30.256380 ) = 11.3342, s = 3.3666.
The prediction interval is (12.6103, 34.5497)
Xh = 15, fitted value for Yh = 20.02
s2{pred} = 9.939 (1/27 + (15 - 15.7778)2 / 30.256380 ) = 0.5668, s = 0.7529.
The prediction interval is (21.1268, 26.0332)
Xh = 20, fitted value for Yh = 23.58
s2{pred} = 9.939 (1/27 + (20 - 15.7778)2 / 30.256380 ) = 6.2241, s = 2.4948.
The prediction interval is (15.4509, 31.7091).
d. Xh = 15, fitted value for Yh = 20.02, B = t(1-α/6; 24) = 3.25838
s2{pred} = 9.939 ( 1 + 1/27 + (15 - 15.7778)2 / 30.256380 ) = 10.5058, s=3.2431.
The prediction interval is (9.4587, 30.5813)
e.
H0: β11=0
HA: β11≠ 0
SSR(X 2 | X) SSE(X, X 2 ) 252.99 238.54
F* =
÷
=
÷
= 25.4538 ;
1
n−2
1
24
F(0.99; 1, 24) = 7.82287
F* > F(0.99; 1, 24), we reject H0, which implies that the quadratic term can not be
dropped. The p-value is less than 0.001.
Problem 8.10
a. E{Y} = 14 + 7X1 + 5X2 – 4X1X2
If X1 = 1, then E{Y} = 14 + 7 + 5X2 – 4X2 = 21 + X2
If X1 = 4, then E{Y} = 14 + 28 + 5X2 – 16X2 = 44 - 11X2
The graph looks like:
If the effects for X1 and X2 are additive, then the two lines should be parallel to
each other. But from the graph, they have very different slope, so the effect are not
additive.
Problem 8.21
a. Hard hat: E{Y} = (β0 +β2) + β1X1
Bump cap: E{Y} = (β0 +β3) + β1X1
None: E{Y} = β0 + β1X1
b. When X1 fixed, test if wearing a bump cap reduced the expected severity of
injury.
H0: β3 ≥ 0
HA: β3 < 0
When X1 is fixed, test if the expected severity of injury the same when
wearing a hard hat as when wearing a bump cap.
H0: β2 = β3
HA: β2 ≠ β3
Problem 8.23
It means summer and winter are about the same.
Problem 8.24
a. The plot looks as the following:
Scatterplot of Y vs X1
100
X2
0
1
Y
90
80
70
60
70.0
72.5
75.0
X1
77.5
80.0
The regression relations seem to be different.
b. The regression equation is: Y = - 127 + 2.78 X1 + 76.0 X2 - 1.11 X1X2
To test if there is any regression relation, the null and alternative hypotheses
are:
H0: β2 = β3 = 0
HA: β2 ≠ 0 or β3 ≠ 0
To
test
it,
F*=
SSR( X 2 , X 3 | X 1 , X 2 , X 3 ) SSE ( X 1 , X 2 , X 3 ) 4237.1 - 3670.9 909.1
÷
=
÷
= 18.62
p−q
n− p
2
60
F(0.95; 2, 60) = 3.15041
F* > F(0.95; 2, 60), we reject H0, which implies that there is a regression
relation between Y and X2.
c. The estimated regression functions for two populations are:
Y = −126.9052 + 2.7759X1 for noncorner lots
Y = −50.8836 + 1.6684X1 for corner lots
From the graph, we can see that the prices for houses not located at the corners are
higher than the prices for houses located at the corner.
Problem 8.34
a. The linear regression model is: Y = β0 + β1X1 + β2X2 + β3X3
b. The response functions for three types of banks are:
Commercial bank: Y = (β0 + β2) + β1X1
Mutual savings: Y = (β0 + β3) + β1X1
Savings and loan: Y = (β0 – β2 – β3) + β1X1
c. β2, β3, –β2–β3 indicates the adjustments for profit or loss of the three different
types of bank.