3
Again, this means that for any constant
K
> 0,
= 1 and
lim
n
64
P
(
ˆ
F
>
K
)
= 1.
lim
n
64
P
(
ˆ
W
>
K
)
9
where
m
#
k
!
1, against the alternative hypothesis that the null hypothesis is false:
for at
β
i
…
0
least one index
i
#
m
. One possible way of testing this hypothesis
is to conduct
m
separate two
sided
t
tests for
i
= 1,.
.,
m
. However, the problem is that the lefthand side random variables in
(23) are in general not independent, hence under the null hypothesis
the test
β
1
'
....
'
β
m
'
0
statistics
are in general not independent. In particular,
it is impossible to select a critical
ˆ
t
1
,...,
ˆ
t
m
value
t
*
such that for a given significance level
α
×100%,
P
[
ˆ
t
1
>
t
(
,
ˆ
t
2
t
(
,...,
ˆ
t
m
t
(
]
'
α
,
because we do not know the joint distribution of
ˆ
t
1
,...,
ˆ
t
m
.
The solution of this problem is the following. Consider the restricted regression model
Y
j
'
β
m
%
1
X
m
%
1,
j
%
β
m
%
2
X
m
%
2,
j
%
...
%
β
k
&
1
X
k
&
1,
j
%
β
k
%
U
j
,
j
'
1,2,.
..,
n
.
(27)
Then it can be shown that:
Proposition 4
:
Under the null hypothesis
and the conditions of Proposition 2,
β
1
'
....
'
β
m
'
0
ˆ
F
'
(
SSR
0
&
SSR
)/
m
SSR
/(
n
&
k
)

F
m
,
n
&
k
,
(28)
and under the conditions of Proposition 2,
ˆ
W
'
m
.
ˆ
F
'
SSR
0
&
SSR
SSR
/(
n
&
k
)

χ
2
m
,
(29)
where SSR is the sum of squared residuals of the unrestricted model
(7)
and
SSR
0
is the sum of
squared residuals of the restricted model
(27).
Moreover, under the alternative hypothesis that
for at least one index i
#
m
,
the test statistics
converge in probability to
4
as
β
i
…
0,
ˆ
F and
ˆ
W
n
6
4
3
.
The test based on
is called, for obvious reasons, the
F
test, and the test based on
is called
ˆ
F
ˆ
W
the Wald test, named after the statistician with that name who proposed this test. The tests
involved are conducted rightsided. In particular in the case of the Wald test
the null hypothesis
involved is rejected at say the 5% significance level if
>
c
, where the critical value
c
is chosen
ˆ
W