5. Tests of hypotheses and
condence intervals
An alternative hypothesis (Ha)
gives a set of alternaties to the null
hypothesis
Consider the linear model with
E (Y) = X and V ar(Y) = We may test
H0 : C = d vs Ha : C 6= d
where
C is an m k matrix of constants
d is an m 1 vector of constants
This can also be expressed as
Y = X + where E () = 0 and V ar() = .
Typical null hypothesis (H0)
species the values for one or more elements of species the values for some linear functions of the elements of The null hypothesis is rejected if it is shown to
be suciently incompatible with the observed
data.
288
287
Failing to reject H0 is not the same as proving
H0 is true.
too little data to accurately estimate
C
relatively large variation in (or Y)
if H0 : C = d is false, C ; d may be
\small"
You can never be completely sure that you
made the correct decision
Type I error (signicance level)
Type II error
289
Basic considerations in specifying a null
hypothesis H0 : C = d
(i) C should be estimable
(ii) Inconsistencies should be avoided,
i.e., C = d should be a consistent
set of equations
(iii) Redundancies should be eliminated,
i.e., in C = d we should have
rank(C ) = number of rows in C
290
Example 5.1 Eects model from Example 3.2
Yij = + i + ij i = 1; 2; 3
j = 1; : : : ; ni
In this case
2Y
66 Y
66 Y
64 Y
Y
11
12
21
31
32
Y33
3 21
77 66 1
77 = 66 1
75 64 1
1
1
1
0
0
0
1 0
0
0
1
0
0
0
0
0
0
1
1
1
3
77
77
75
2
3 6 75 + 666 64 2 64 11
12
1
21
2
31
3
33
32
3
77
77
75
291
+ 1 = cT H0 : + 1 = 60 seconds
against
HA : + 1 6= 60 seconds
(two-sided alternative)
Or we can test
H0 : + 1 = 60 seconds
against
HA : = 1 < 60 seconds
(one-sided alternative)
By denition
E (Yij ) = + i is estimable.
In this case
We can test
213
6 7
where c = 664 10 775
0
Note that this quantity is estimable, e.g.,
0 0 0 0)Y):
cT = + 1 = E (( 12 21
Then, any solution b = (X T X );X T Y to the
normal equations X T X b = X T Y yields the
same value cT b and it is the unique blue for
cT .
We will reject H0 : cT = 60 if
cT b = cT (X T X );X T Y
is too far away from 60.
293
292
(ii) Dierence between the mean response for
two treatments is estimable
1 ; 3 = ( + 1) ; ( + 3)
= 12 21 0 ;31 ;31 ;31 E (Y)
ans we can test
H0 : 1 ; 3 = 0 vs. HA : 1 ; 3 6= 0
The unique blue for
1 ; 3 = (0 1 0 ; 1) = cT is
cT b for any b = (X T X );X T Y
Reject H0 : 1 ; 3 = cT = 0 if cT b is too
far from 0.
294
In Example 3.2 (on Page 3.7) we found several
solutions to the normal equations:
(iii) It would not make much sense to attempt to test
H0 : 1 = 3 vs. HA : 1 6= 3
because 1 = [0 1 0 0] = cT is not
estimable
Although E (Y1j ) = + 1 neither nor 1
has a clear interpretation.
Dierent solutions to the normal equations
produce dierent values for
^1 = CT b = CT (X T X );X T Y
The data alone do not provide information
about whether or not 1 = 3.
2
66 00
b = 66
40
0 0
1 0
2
0 11
0 0 0
0 37 2 Y:: 3 2 0 3
0 77 666 Y1: 777 = 666 61 777
0 75 4 Y2: 5 4 71 5
1
Y3:
60
3
2 1 ;1 ;1 0 3 2 Y 3 2 69 3
::
1 6 ;1 2:5 1 0 777 666 Y1: 777 = 666 ;8 777
b = 664
3 ;1 1 4 0 5 4 Y2: 5 4 2 5
0 0 0 0 Y3:
0
2 2 ;1 ;1
66 61 61 6
0
6;
b = 66 16 2 1
64 ; 6 0 1
; 16 0
;1 3 2
6 7 Y::
0 777 66 Y1:
6
0 775 4 Y2:
0 31 Y3:
295
2
3
0 1 0 ;1 7
6
(iv) For C = 4 1 1 0 0 5
1 0 0 1
consider testing
2 3
2 3
;3 7
;3
6
H0 : C = 4 60 5 vs. HA : C 6= 64 60 75
70
70
In this case C is estimable, but
there is an inconsistancy. If the null
hypothesis is true,
2
3 2 3
;
;3
1
3
C = 64 + 1 75 = 64 60 75
70
+ 3
Then + 1 = 60 and + 3 = 70
implies
(1 ; 3) = ( + 1) ; ( + 3)
= 60 ; 70 = ;10
Such inconsistancies should be avoided.
297
3 2 66:66 3
77 666 ;5:66 777
75 = 66
7
4 4:33 75
2:33
296
2
3
0 1 0 ;1 7
6
(v) For C = 4 1 1 0 0 5
1 0 0 1
consider testing
2
3
2
3
;10 7
;10 7
6
6
H0 : C = 4 60 5 vs. HA : C 6= 4 60 5
70
70
In this case C is estimable and the equations
specied by the null hypothesis are consistent.
There is a redundancy
[1 1 0 0] = + 1 = 60
[1 0 0 1] = + 3 = 70
imply that
[0 1 0 ; 1] = 1 ; 3
= ( + 1) ; ( + 3)
= 60 ; 70 = ;10
298
The rows of C are not linearly independent,
i.e., rank(C ) < number of rows in C .
There are many equivalent ways to remove a
redundancy:
"
#
" #
H0 : 11 10 00 01 = 60
70
"
#
"
H0 : 01 11 00 ;01 = ;10
60
"
#
"
H0 : 01 10 00 ;11 = ;10
70
#
In each case:
The two rows of C are linearly independent
and
rank(C ) = 2
= number of rows in C
The two rows of C are a basis for the
same 2-dimensional subspace of R4.
This is the 2-dimensional space spanned by
the rows of
2
3
0
1
0
;
1
C = 64 1 1 0 0 75
1 0 0 1
#
"
#
"
#
50
H0 : 12 21 00 ;11 = 130
are all equivalent.
We will only consider null hypotheses of the
form H0 : C = d where rank(C ) = number of
rows in C .
299
This leads to the following concept of a
\testable" hypothesis.
Defn 5.1: Consider a linear model E (Y) = X where V ar(Y) = and X is an n k matrix.
For an m k matrix of constants C and an
m 1 vector of constants d, we will say that
H0 : C = d
is testable if
(i) C is estimable
(ii) rank(C ) = m = number of rows in C
301
300
To test H0 : C = d
(i) Use the data to estimate C .
(ii) Reject H0 : C = d if the estimate of C is to far away from d.
How much of the deviation of the
estimate of C from d can be attributed to random errors?
{ measurement error
{ sampling variation
Need a probability distribution for
the estimate of C Need a probability distribution for a
test statistic
302
Normal theory Gauss-Markov model
2 3
Y1
Y = 64 .. 75 N (X ; 2I )
Yn
Result 5.1.
A least squares estimate b for minimizes
(Y ; X b)T (Y ; X b)
For a testable null hypothesis
H0 : C = d
the OLS estimator for C is
C b = C ( X T X ); X T Y
(i) Since C is estimable, C b is invariant to
the choice of (X T X );. (Result 3.10).
(ii) C b is the unique b.l.u.e. for C . (Result
3.11).
Any solution to the normal equations
(X T X )b = X T Y
has the form
b = (X T X );X T Y
303
(iii)
E (C b) ; d) = C ; d
V ar(C b ; d) = V ar(C b)
= 2 C ( X T X ); C T
This follows from V ar(Y) = 2I , because
V ar(C b) = V ar(C (X T X );X T Y)
= C (X T X );X T V ar(Y) X [(X T X );]T C T
= C (X T X );X T (2I )X [(C T X );]T C T
= 2C (X T X );X T X [(X T X );]T C T
Since C is estimable, C = AX for some A and
V ar(C b) = 2AX (X T X );X T [X (X T X );X T ]T AT
= 2AX (X T X );X T AT
= 2C (X T X );C T
305
304
(iv) C b ; d N (C ; d, 2C (X T X );C T )
This follows from the normal theory
Gauss-Markov assumption
Y N (X ; 2I );
property (iii) and Result 4.1.
(v) When H0 : C = d is true,
C b ; d N (0; 2C (X T X );C T )
(vi) Dene
SSH0 = (C b ; d)T [C (X T X );C T ];1(C b ; d)
then
1
2 2
2 SSH0 m( )
where
2 = 12 (C ; d)T [C (X T X );C T ];1(C ; d)
306
This follows from Result 4.7 using
C b ; d N (C ; d; 2C (X T X );C T )
A = 12 [C (X T X );C T ];1
= V ar(C b ; d) = 2C (X T X );C T
Clearly, A = I is idempodent.
We also need the estimability of C and
rank(C ) = m = number of rows in C
to ensure that C (X T X );C T is positive denite
and (C (X T X );C T );1 exists. Then,
d.f. = rank(A) = rank(C (X T X );C T )
=m
Since C (X T X );C T is positive denite, we have
2 = 212 (C ; d)T [C (X T X );C T ];1(C ; d)
> 0
unless C ; d = 0. Hence 2 = 0 if and only if
H0 : C = d is true.
Consequently, from Result 4.7, we have
(vii) 12 SSH0 2m
if and only if
H0 : C ; d is true.
308
307
Then,
SSresiduals = eT e
= YT (I ; PX )Y
is invariant to the choice of (X T X ); used to
obtain PX = X (X T X );X T and b.
To obtain an estimate of
V ar(C b ; d) = 2C (X T X );C T
we need an estimate of 2.
(viii)
Since E (Y) = X is estimable,
^ = X b = X (X T X );X T Y = PX Y
Y
is the unique b.l.u.e. for X .
where
Consequently, the residual vector
^ = (I ; PX )Y
e=Y;Y
is invariant to the choice of (X T X ); used to
obtain PX = X (X T X );X:T
and
E (SSresiduals) = (n ; k)2
k = rank(X) = rank(PX)
n ; k = rank(I ; PX)
and it follows that
MSresiduals = SSnresiduals
;k
is an unbiased estimator of 2.
309
310
Result (viii) is obtained by applying Result 4.7
to
SSresidual = YT(I ; PX)Y
(ix) 12 SSresiduals 2n;k
To show this use the assumption that
Y N (X ; 2I ) and apply Result 4.7 to
1
T 1
2 SSresiduals = Y 2 (I ; PX ) Y
E (YT AY) =
=
T A + tr(A)
T X T (I ; PX )X + tr((I ; PX )2I )
"
this is a zero matrix
= 2tr(I ; PX )
= 2(n ; k)
where k = rank(X).
This used the assumption of a Gauss-Markov
model, but does not use the normality assumption.
%
E (Y) = X = V ar(Y) = 2I = Clearly A = 12 (I ; PX )2I = I ; PX is idempodent and the noncentrality parameter is
2 = T A
= 12 T X T (I ; PX )X = 0
%
this is a matrix
of zeros
312
311
(x) SSH0 and SSresiduals are independently
distributed.
To show this note that SSH0 is a
function of
Then, by Result 4.4, C b and e are independent
because
Cov(C b; e) =
=
=
=
C b = C (X T X );X T Y
and SSresiduals is a function of
e = (I ; X )Y
By Result 4.1,
" # "
#
C b = C (X T X ); X T Y
e
I ; PX
has a multivariate normal distribution
because Y N (X ; 2I ).
313
this is A
COV (C (X T X );X T Y; (I ; PX )Y)
C (X T X );X T (V ar(Y))(I ; PX )T
C (X T X );X T (2I )(I ; PX )
2C (X T X ); X T (I ; PX ) = 0
%
This is a matrix of
zeros since it is
the transpose of
(I ; PX )X = X ; X = 0
Consequently, SSH0 is independent of
SSresiduals, and it follows that
314
(xi)
SSH 0
2
F = SSm residuals
2
=
(n;k)
SSH0
m
SSresiduals
n;k
Perform the test by rejecting H0 : C = if
Fm;n;k(2)
F > F(m;n;k);
with noncentrality parameter
2 = 12 (C ; d)T [C (X T X );C T ];1(C ; d)
0
where is a specied signicance level (Type
I error level) for the test.
and 2 = 0 if and only if H0 : C = d is true.
315
316
Type I error level:
n
o
= Pr F > Fm;n;k; j H0 is true
%
when H0 is true,
H0
F = MSMS
residuals
has a central F distribution
with (m; n ; k) d.f.
This is the probability of incorrectly rejecting
a null hypothesis that is true.
317
318
Type II error level:
0.8
Central F Distribution with (4,20) df
= Prffail to reject H0 H0 is falseg
0.4
0.6
= PrfF < Fm;n;k; H0 is falseg
%
when H0 is false;
F = MSMSH0
0.2
f(x;n)
= PrfType II errorg
0.0
residuals
0
1
2
3
4
5
has a noncentral F distribution
with (m; n ; k) d:f : and
noncentrality parameter > 0:
x
319
Power of a test:
power = 1 ; = PrfF > Fm;n;k; H0 is falseg
%
this determines the value
of the noncentrality
parameter.
320
321
0.8
Central and Noncentral
F Distributions with (4,20) df
Noncentral
Noncentral
F(1)
F(1)
0.4
Noncentral
Noncentral
F(3)
F(3)
Noncentral
Noncentral
F(5)
F(5)
0.0
0.2
f(x;v,w)
0.6
Central
Central
FF
0
1
2
3
4
5
x
322
0.6
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 1.5
Central F
0.4
0.2
0.0
f(x;v,w)
Noncentral F
0
1
2
3
4
5
x
323
0.6
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 3
Central F
0.4
0.0
0.2
f(x;v,w)
Noncentral F
0
1
2
3
4
5
x
324
0.6
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 10
Central F
0.4
0.2
0.0
f(x;v,w)
Noncentral F
0
1
2
3
4
5
x
325
0.6
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 20
For a xed type I error level (signicance level)
, the power of the test increases as the
noncentrality parameter increases.
Central F
2 = 12 (C ; d)T [C (X T X );C T ];1(C ; d)
"
"
0.4
size of
how much
the error the actual
variance value of
C
deviates
from the
hpothesized
value d
"
the design
of the experiment
(Note: the number
of observations
also aects
degrees of
freedom.)
0.0
0.2
f(x;v,w)
Noncentral F
0
1
2
3
4
5
x
326
Example 5.2 Eects of three diets on blood
coagulation times in rats.
Diet factor: Diet 1, Diet 2, Diet 3
Response: blood coagulation time
Model for a completely randomized experiment
with ni rats assigned to the i-th diet.
Yij = + i + ij
where
ij NID(0; 2)
for i = 1; 2; 3 and j = 1; 2; :::; ni.
Here, E (Yij ) = + i is the mean coagulation
time for rats fed the i-th diet.
327
Test the null hypothesis that the mean blood
coagulation time is the same for all three diets
H0 : + 1 = + 2 = + 3
against the general alternative that at least two
diets have dierent mean coagulation times
HA : ( + i) 6= ( + j ) for some i 6= j :
Equivalent ways to express the null hypothesis
are:
H0 : 1 = 2 = 3
"
#
0
1
;
1
0
H0 : C = 0 0 1 ;1
2 3
66 1 77 " 0 #
64 75 = 0
2
3
328
2 3
# 6 7 " #
H0 : C = 00 10 01 ;11 664 1 775 = 00
2
3
"
"
this is d
Obtain the OLS estimator for C C b where b = (X T X );X T Y
and evaluate
SSH0 = (C b ; 0)T [C (X T X );C T ];1(C b ; 0)
3
X
=
ni(Yi: ; Y::)2
i=1
SSH0
2 on 3 ; 1 = 2 d:f :
ni
3 X
X
SSresiduals = YT (I ; PX )Y =
(Yij ; Yi:)2
MSH0 =
SSresiduals on
MSresiduals = X
3
(ni ; 1)
i=1
i=1 j =1
3
X
(ni ; 1) d:f :
i=1
Reject H0 in favor of Ha if
F = MSMSH0 > F(2;P3i=1(ni;1));
residuals
How many observations (in this case rats) are
needed? Suppose we are willing to specify:
(i)
(ii)
(iii)
(iv)
= type I error level = .05
n1 = n2 = n3 = n
power .90 to detect
a specic alternative
( + 1) ; ( + 3) = 0:5
( + 2) ; ( + 3) = 330
329
For this particular alternative
2 3
"
#6 7 "
#
;1 66 1 77 = :5
C = 00 01 10 ;
1 4 2 5
3
and the power of the F-test is
n
o
power = Pr F(2;3n;3)(2) > F(2;3n;3);:05
where
"
# " #!T
2 = 12 :5 ; 00
h T ; T i;1 " :5 # " 0 #!
C (X X ) C
" ;# 0
h
i
= [:5 1] C (X T X );C T ;1 :15
%
X has n rows of (1 1 0 0)
n rows of (1 0 1 0)
n rows of (1 0 0 1)
331
In this case,
2 3n
6
X T X = 664 nn
n
n n n 37
n 0 0 77
0 n 05
0 0 n
and a generalized inverse is
2
3
66 00 n0;1 00 00 77
(X T X ); = 66 0 0 n;1 0 77
4
5
0 0 0 n;1
Then,
"
#
C (X T X );C T = n1 21 12
and
" #
2
T
;
T
;
1
= [:5 1][C (X X ) C ] :51
= n2
Choose n to achieve
:90 = power
= Pr F(2;3n;3) n2 > F(2;3n;3):05
332
0.6
0.4
0.0
0.2
Power
0.8
1.0
Power versus Sample Size
F-test for equal means
0
10
20
30
40
Sample Size
333
For testing
H0 : ( + 1) = ( + 2) = = ( + k )
against
HA : ( + 1) 6= ( + j )for somei 6= j
# This file is stored in power.spl
#===========================================
use
n <- 2:50
power <- 1 - pf(qf(.95,2,3*n-3),2,3*n-3,n/2)
#
#
F = MSMSH0
residuals
Unix users show insert the motif( )
function here to open a graphics window
where
par(fin=c(7.5,8.5),cex=1.2,mex=1.5,lwd=4,
mar=c(5.1,4.1,4.1,2.1))
plot(c(0,50), c(0,1), type="n",
xlab="Sample Size",ylab="Power",
main="Power versus Sample Size\n
F-test for equal means")
lines(n, power, type="l",lty=1)
F(k;1;Pki=1(ni;1))(2)
k
X
2 = 12 ni(i ; :)2
i=1
with
#============================================
k
X
nii
: = i=1k
X
i=1
334
ni
335
50
To obtain the formula for the noncentrality parameter, write the null hypothesis as
31 2 3
0H20 : 0 = C = 3 2
0
0 1 0 0
B
6 0 0 1 0 7 6 0
B
@64 .. 0 . . . . 75 ; 64 ..
0 0 0 1
Use
0
2
66 11
66 1
66
61
X = 666 1
66 ..
66 ..
64 .
1
1
1
0
0.
...
..
1 0
n1
n
..
..
n
:
1
n
:
0 0
0
1
1.
...
..
0 n
n
k
:
77CC 66 5A 64 ..
1
2
n
n
k
:
k
77
75
3
07
0 77
0 77
0 777
0. 77
. 77
1 777
15
1
2
66 nn: nn1
1 1
X T X = 666 n2
64 ..
nk
20 0
66 0 n;1
1
T
;
(X X ) = 6666 0 0
4 ..
3
n2 nk 7
77
7
n2
. . . 775
nk
0 0 n;2 1 ...
0 0 0
0
0
n;k 1
3
77
77
77 :
5
Then
2 = 12 (C ; 0)T [C (X T X );C T ];1(C ; 0)
k
X
= 12 ni(1 ; :)2
i=1
336
Consider minimum power of the test that rejects H0 when
F = MSMSH0
residuals
For n1 = n2 = = nk = n we have
2 n w2
Then, the power of the test is at least
> F(k;1;Pki=1(ni;1));
against alternatives where
ji ; :j w for at lease one i and a specied w.
Then,
337
power Pr F(k;1;k(n;1)) (n w2) > F(k;1;k(n;1);
%
%
noncentral F
percentile
random
of a
variable
central F
distribution
F = MSMS
H0
residuals
k
X
2 = 12 ni(i ; :)2
i=1
w2 minfn1; : : : ; nkg
The following graph shows minimum power for
= :01; k = 4; w = :50 as a function of the
common sample size n.
338
339
0.6
0.4
0.0
0.2
Power
0.8
1.0
Minimum Power vs. Sample Size
At least one difference exceeds sigma*0.
0
20
40
60
80
100
Sample Size
340
#======================================================#
# min.power.vs.sample.size()
#
# -------------------------#
# Input : k = # of groups
#
#
n = maximum sample size in the
#
#
group(n1=n2=...=nk)
#
#
alpha = type I error level
_
#
#
w =spesifies alternative |ai-a.| >= w*sigma #
# Output: line pot (Minimum Power vs. Sample Size)
#
# Note : Unix users must activate a motif() window
#
#
before using this function.
#
#======================================================#
min.power.vs.sample.size _ function(k,n,alpha,w)
{
n _ 2:n
power _ 1 - pf(qf(1-alpha,k-1,k*(n-1)),k-1,k*(n-1),n*w^2)
}
#
#
#
#
plot(c(0,max(n)), c(0,1),type="n",xlab="Sample Size",
ylab="Power", main=paste("Minimum Power vs. Sample
Size\nAt least one difference exceeds sigma*",w,sep=""))
lines(n, power, type="l",lty=1)
Condence intervals for estimable functions of //[.2in]
Defn 5.2: Suppose Z N (0; 1) is distributed
independently of W 2v , and then the distribution of
t = qZW
v
is called the student t-distribution with v degrees of freedom.
We will use the notation
t tv
The following statement can be used to
execute this function
min.power.vs.sample.size(4,100,.01,.5)
341
342
0.4
Central t Densities
0.2
0.0
0.1
f(x;df)
0.3
2 df
5 df
30 df
-4
-2
0
2
4
x
343
# This code is stored in:
tden.spl
#=================================================#
# t.density.plot()
#
# -----------------------#
# Input : degrees of freedom; it can be a vector. #
#
(e.g.) t.density.plot(c(2,5,7,30))
#
#
creates curves for df = 2,5,7,and 30 #
# Output: density plot of Student t distribution. #
# Note: Unix users must first use
motif()
#
#
to open a grpahics window before
#
#
using this function.
#
#================================================ #
# The following code creates a legend;
}
t.density.plot <- function(df)
{
x <- seq(-4,4,,100)
#
#
#
#
#
#
#
#
# draw x,y axis and title
plot(c(-4,4), c(0,.4), type="n",
xlab="x", ylab="f(x;df)",
main="Central t Densities")
legend( x = rep(0.85,length(df)) ,
y = rep(.4,length(df)) ,
legend = paste(as.character(df),"df") ,
lty = seq(1,by=2,length=length(df)) ,
bty = "n")
This function can be executed as follows.
Windows users should delete the motif( )
command.
motif( )
source("tden.spl")
par(fin=c(7,8),cex=1.2,mex=1.5,lwd=4)
t.density.plot(c(2,5,30))
for(i in 1:length(df)) {
lty.num <- 2*i-1
# specify the line types.
f.x <- dt(x,df[i])
# calculate density.
lines(x, f.x, type="l",lty=lty.num)
# draw.
}
344
345
For the normal-theory Gauss-Markov model
Y N (X ; 2I )
and from Result 5.1.(iv) we have for an estimable function, cT , that the OLS estimator
cT b = c T ( X T X ) ; X T Y
follows a normal distribution, i.e.,
cT b N (cT ; 2cT (X T X );c) :
It follows that
T
T
Z = q c2 Tb ; cT ; N (0; 1)
c (X X ) c
From Result 5.1.(ix), we have
1
1 T
2 = 2 Y (I ; PX )Y
2(n;k)
where k = rank(X ).
Using the same argument that we used to derive Result 5.1.(x), we can show that cT b is
distributed independently of 12 SSE.
First note that
"
#
cT b
(I ; PX )Y
" T T ; T#
= c ((XI ;XP) )X Y
X
has a joint normal distribution under the
normal-theory Gauss-Markov model. (From
Result 4.1)
346
Note that
Cov(cT b; (I ; P ; X )Y)
= (cT (X T X );X T )(V ar(Y))(I ; PX )T
= (CT (X T X );X T )(2)(I ; PX )
= 2CT (X T X );X T (I ; PX )
=0
"
this is a matrix of zeros
347
Then,
t = q ZSSE
2(n;k)
=
pc2TcTb;(XcTTX );c
q
SSE
2(n;k)
= r SSEcT bT;cT T
Consequently, (by Result 4.4)
cT b is distributed independently
of e = (I ; PX )Y
which imples that
cT b is distributed independently
of SSE = eT e.
;
;k) c (X X ) c
(n
t(n;k)
%
call SSE
n;k the MSE
348
349
0.2
0.0
0.1
f(x;df)
0.3
0.4
Central t Density
( 5 degrees of freedom )
-4
-2
0
2
4
x
350
It follows that
1; =
(
T
cT t
Pr ;t(n;k);=2 p c b ;
MSE cT(XTX);c (n;k);=2
q
= Pr cT b ; t(n;k);=2 MSE cT(XTX);c cT q
cT b + t(n;k);=2 MSE cT(XTX);c
and a (1 ; ) 100% condence interval for
cT is
q
cT b ; t)(n ; k); =2
cT b + t(n;k);=2
q
MSE cT(XTX);c ;
MSE cT(XTX);c
351
)
For brevity we will also write
cT b t(n;k);=2 ScT b
where
q
ScT b = MSE cT(XTX);c :
For the normal-theory Gauss-Markov model
with Y N (X ; 2I ); the interval
cT b t(n;k);=2ScT b
is the shortest random interval with probability
(1 ; ) of containing CT .
352
Condence interval for 2:
For the normal-theory Gauss-Markov model
with Y N (X ; 2I ) we have shown that
The resulting (1;)100% condence interval
for 2 is
0
1
SSE
SSE
@ 2
A
(n;k);=2 ; 2(n;k);1;=2
SSE = YT (I ; PX )Y 2
(n;k)
2
2
Then,
1 ; = Pr 2(n;k);1;=2 SSE
2 (n;k);=2
8
9
< SSE
=
SSE
2
= Pr : 2
(n;k);1;=2 ;
(n;k);=2
353
0.0
0.04
f(x;df)
0.08
0.12
Central Chi-Square Density
0
5
10
15
x
20
25
30
354