Statistics 500 – Fall 2009
Solutions to Homework 8
1
1. Linerboard and regression tests
My fitted model was:
)
log(
34
.
)
log(
286
.
)
log(
543
.
)
log(
126
.
762
.
1
)
log(
labor
deprec
energy
rawmat
product
+
+
+
+
−
=
()
( ) ( ) ( )
34
.
0
286
.
0
543
.
0
126
.
0
)
762
.
1
exp(
labor
deprec
energy
rawmat
product
−
=
Variable
DF
Estimate
Std Error
t Value
Pr > t
Intercept
1
1.76206
0.79507
2.22
0.0384
lograwmat
1
0.12634
0.04316
2.93
0.0083
logenergy
1
0.54267
0.07886
6.88
<.0001
logdeprec
1
0.28626
0.06516
4.39
0.0003
loglabor
1
0.33994
0.14459
2.35
0.0291
(a) This is the Ftest of the entire regression model:
Sum of
Mean
Source
DF
Squares
Square
F Value
Pr > F
Model
4
4.83013
1.20753
647.29
<.0001
Error
20
0.03731
0.00187
Corrected Total
24
4.86744
The test statistic is F=647, with p < 0.0001. There is very strong evidence that at least one
regression slope differs from zero.
* Again, remember the overall test does not tell you that all slopes are nonzero, just at least one.
(b) Test
1
:
4
3
2
1
0
=
+
+
+
β
H
You can test this hypothesis three ways.
Each gives you the same pvalue and conclusion, but the
test statistics may not be the same.
P < 0.017.
There is evidence that the sum of the coefficients is
not 1.
In econometric terms, there is not constant return to scale.
1) Estimate
4
3
2
1
+
+
+
and its standard error, then construct a ttest:
T = (estimate of sum  1)/se = (1.2951)/0.114 = 2.59.
You can get the estimate and s.e. from proc glm; … estimate rawmat 1 energy 1 deprec 1 labor 1;
Parameter
Estimate
Error
t Value
Pr > t
sum1
1.29521599
0.11377588
11.38
<.0001
* The ttest provided by SAS tests the wrong hypothesis (sum = 0).
2) Use a test statement in proc reg to construct the F statistic for a linear constraint.
Source
DF
Mean Sq
F Value
Pr > F
Numerator
1
0.01256
6.73
0.0173
Denominator
20
0.00187
3) Construct a reduced model corresponding to the null hypothesis.
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View Full DocumentStatistics 500 – Fall 2009
Solutions to Homework 8
2
Full model:
Model in (a)
Reduced model:
( )
ε
β
+
−
−
−
+
+
+
+
=
4
3
2
1
3
3
2
2
1
1
0
log
1
log
log
log
)
log(
X
X
X
X
Y
Or:
[]
[
][
]
+
−
+
−
+
−
+
=
−
4
3
3
4
2
2
4
1
1
0
4
log
log
log
log
log
log
)
log(
)
log(
X
X
X
X
X
X
X
Y
(c) You need the s.e. of the sum, which you can get from the estimate (above) command or by
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 Fall '08
 Staff
 Statistics, Normal Distribution, Regression Analysis, proc reg

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