SUGGESTED SOLUTIONS HW4
PAM 305
FALL 2006
Question 1.
4.6
(i) With
df
=
n
– 2 = 86, we obtain the 5% critical value from Table G.2 with
df
=
90.
Because each test is twotailed, the critical value is 1.987.
The
t
statistic for H
0
:
0
β
=
0 is about .89, which is much less than 1.987 in absolute value.
Therefore, we fail to
reject
0
β
= 0.
The
t
statistic for H
0
:
1
β
= 1 is (.976 – 1)/.049
≈
.49, which is even less
significant.
(Remember, we reject H
0
in favor of H
1
in this case only if 
t
 > 1.987.)
(ii) We use the SSR form of the
F
statistic.
We are testing
q
= 2 restrictions and the
df
in the unrestricted model is 86.
We are given SSR
r
= 209,448.99 and SSR
ur
=
165,644.51.
Therefore,
(209,448.99
165,644.51)
86
11.37,
165,644.51
2
F
−
=
⋅
≈
which is a strong rejection of H
0
:
from Table G.3c, the 1% critical value with 2 and 90
df
is 4.85.
(iii) We use the
R
squared form of the
F
statistic.
We are testing
q
= 3 restrictions
and there are 88 – 5 = 83
df
in the unrestricted model.
The
F
statistic is [(.829 –
.820)/(1 – .829)](83/3)
≈
1.46.
The 10% critical value (again using 90 denominator
df
in
Table G.3a) is 2.15, so we fail to reject H
0
at even the 10% level.
In fact, the
p
value is
about .23.
(iv) If heteroskedasticity were present, Assumption MLR.5 would be violated,
and the
F
statistic would not have an
F
distribution under the null hypothesis.
Therefore,
comparing the
F
statistic against the usual critical values, or obtaining the
p
value from
the
F
distribution, would not be especially meaningful.
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 Fall '06
 LUCARELLI
 Statistics, Critical Point, Regression Analysis, Variance, Chisquare distribution, Critical phenomena

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