5. Tests of hypotheses and
con dence 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)
speci es the values for one or more elements of
speci es 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 (signi cance 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 E ects 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
By de nition
E (Yij ) = +
0
0
0
1
1
1
i
3
2 3
2
3
77 6 7
77 64 75 + 666 777
75
64 75
We can test
H0 : + 1 = 60 seconds
against
HA : + 1 =
6 60 seconds
(two-sided alternative)
11
12
1
21
2
31
3
33
32
Or we can test
H0 : + 1 = 60 seconds
against
HA : + 1 < 60 seconds
(one-sided alternative)
is estimable.
291
In this case
+ 1 = cT
213
6 7
where c = 664 10 775
0
Gauss-Markov Model
Note that this quantity is estimable, e.g.,
= + 1 = E (( 12 21 0 0 0 0)Y):
Then, any solution
b = (X T ;1X );X T ;1Y
to the generalized least squares estimating
equations
X T ;1X b = X T ;1Y
yields the same value for cT b and it is the
unique blue for cT .
cT
We will reject H0 : cT = 60 if
cT b = cT (X T ;1X );X T ;1Y
is too far away from 60.
292
293
If Var(Y) = 2I , then any solution
b = (X T X );X T Y
to the least squares estimating equations
XT Xb = XT Y
yields the same value for cT b, and cT b 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.
294
(ii) Di erence between the mean response for
two treatments is estimable
1 ; 3 = ( + 1) ; ( + 3)
= 12 12 0 ;31 ;31 ;31 E (Y)
ans we can test
H0 : 1 ; 3 = 0 vs. HA : 1 ; 3 6= 0
If Var(Y) = 2I , 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.
(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.
Di erent solutions to the normal equations
produce di erent values for
^1 = cT b = cT (X T X );X T Y
To make a statement about 1, an
additional restriction must be imposed on
the parameters in the model to give 1 a
precise meaning.
295
In Example 3.2 (pages 154-159) we found several solutions to the normal equations:
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:
69
3
2 1 ;1 ;1 0 3 2 Y 3 2 69 3
66 ;1 2:5 1 0 77 66 Y1::: 77 66 ;8 77
1
b = 64
3 ;1 1 4 0 75 64 Y2: 75 = 64 2 75
0
0 0 0 0 Y3:
2 2 ;1 ;1
66 61 61 6
0
6;
b = 66 61 2 1
64 ; 6 0 1
; 16 0
;1 3 2
6 7 Y::
0 777 666 Y1:
0 775 4 Y2:
0 13 Y3:
3 2 66:66 3
77 666 ;5:66 777
7
75 = 66
4 4:33 75
2:33
297
296
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.
298
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
speci ed 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
299
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
H0 : 01 10 00 ;11
"
#
"
#
= ;10
60
= ;10
70
"
#
"
#
50
H0 : 12 21 00 ;11 = 130
are all equivalent.
300
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
We will only consider null hypotheses of the
form H0 : C = d where rank(C ) = number of
rows in C .
301
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
302
Normal Theory Gauss-Markov Model
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
2 3
Y1
Y = 64 .. 75 N (X ; 2I )
Yn
A least squares estimator b for minimizes
(Y ; X b)T (Y ; X b)
For any generalized inverse of X T X ,
b = (X T X );X T Y
is a solution to the normal equations
(X T X )b = X T Y :
303
Result 5.1. (Results for the Gauss-Markov
model)
For a testable null hypothesis
H0 : C = d
the OLS estimator for C ,
C b = C (X T X );X T Y ;
has the following properties:
(i) Since C is estimable, C b is invariant to
the choice of (X T X );. (Result 3.10).
(ii) Since C is estimable, C b is the unique
b.l.u.e. for C . (Result 3.11).
305
304
(iii)
E (C b ; d) = C ; d
V ar(C b ; d) = V ar(C b)
= 2C (X T X );C T
The latter 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 [(X 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 X T AT
= 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
306
(iv) C b ; d N (C ; d, 2C (X T X );C T )
This follows from normality,
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 )
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
(vi) De ne
SSH0 = (C b ; d)T [C (X T X );C T ];1(C b ; d)
then
1
2 2
2 SSH0 m( )
where m = rank(C ) and
2 = 12 (C ; d)T [C (X T X );C T ];1(C ; d)
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 de nite
and (C (X T X );C T );1 exists. Then,
d.f. = rank(A) = rank(C (X T X );C T )
=m
Clearly, A = I is idempodent.
307
Since C (X T X );C T is positive de nite, 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.
309
308
To obtain an estimate of
V ar(C b ; d) = 2C (X T X );C T
we need an estimate of 2.
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 .
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
310
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.
(viii)
where
and
Result (viii) is obtained by applying Result 4.6
to
SSresidual = YT(I ; PX)Y
E (YT AY) = T A + tr(A)
= T X T (I ; PX )X + tr((I ; PX )2I )
E (SSresiduals) = (n ; k)2
"
this is a zero matrix
= 2tr(I ; PX )
= 2 (n ; k )
where k = rank(X).
k = rank(X) = rank(PX)
n ; k = rank(I ; PX)
and it follows that
MSresiduals = SSnresiduals
;k
This used the assumption of a Gauss-Markov
model, but does not use the normality assumption.
is an unbiased estimator of 2.
312
311
(ix) 12 SSresiduals 2n;k
To show this use the assumption that
Y N (X ; 2I ) and apply Result 4.7 to
(x) SSH0 and SSresiduals are independently
distributed.
To show this note that SSH0 is a
function of
12 SSresiduals = YT h 12 (I ; PX) i Y
%
E (Y) = X = V ar(Y) = 2I = this is A
C b = C (X T X );X T Y
Clearly A = 12 (I ; PX )2I = I ; PX is
idempotent and the noncentrality parameter is
2 = T A
= 12 T X T (I ; PX )X = 0
%
this is a matrix
of zeros
313
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 ).
314
Then, by Result 4.4, C b and e are independent
because
Cov(C b; e) = 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
(xi)
F =
=
SSH0 !
2
SS m residuals
(n;k)2
SSH0
m
SSresiduals
n;k
Fm;n;k ( 2 )
with noncentrality parameter
2 = 12 (C ; d)T [C (X T X );C T ];1(C ; d)
0
and 2 = 0 if and only if H0 : C = d is true.
315
Perform the test by rejecting H0 : C = d if
316
Type I error level:
n
F > F(m;n;k);
where is a speci ed signi cance level (Type
I error level) for the test.
= Pr frejectH0j H0 is trueg
= 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.
o
This is the probability of incorrectly rejecting
a null hypothesis that is true.
317
318
Type II error level:
0.8
= PrfType II errorg
0.6
= Prffail to reject H0 H0 is falseg
0.4
= PrfF < Fm;n;k; H0 is falseg
%
when H0 is false;
F = MSMSH0
0.0
0.2
f(x;n)
Central F Distribution with (4,20) df
0
1
2
3
4
residuals
5
has a noncentral F distribution
with (m; n ; k) d:f : and
noncentrality parameter > 0:
x
320
319
Central and Noncentral
F Distributions with (4,20) df
0.8
Power of a test:
%
this determines the value
of the noncentrality
parameter.
0.4
Noncentral
Noncentral
F(3)
F(3)
Noncentral
Noncentral
F(5)
F(5)
0.2
H0 is falseg
Noncentral
Noncentral
F(1)
F(1)
0.0
power = 1 ;
= PrfF > Fm;n;k;
f(x;v,w)
0.6
Central
Central
FF
0
1
2
3
4
5
x
321
322
0.6
0.6
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 3
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 1.5
Central F
Central F
0.4
0.0
0.0
0.2
0.4
f(x;v,w)
Noncentral F
0.2
f(x;v,w)
Noncentral F
0
1
2
3
4
5
0
1
2
x
3
4
5
x
323
324
0.6
0.6
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 20
0.8
Central and Noncentral F Densities
with (5,20) df and noncentrality parameter = 10
Central F
Central F
0.4
0.2
0.0
0.2
0.4
f(x;v,w)
Noncentral F
0.0
f(x;v,w)
Noncentral F
0
1
2
3
4
5
0
x
1
2
3
4
5
x
325
326
Example 3.2 E ects of three diets on blood
coagulation times in rats.
For a xed type I error level (signi cance level)
, the power of the test increases as the
noncentrality parameter increases.
2 = 12 (C ; d)T [C (X T X );C T ];1(C ; d)
"
"
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 a ects
degrees of
freedom.)
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
328
2 3
# 6 7 " #
1 6 17
0
H0 : C = 00 10 01 ;
;1 64 2 75 = 0
"
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 di erent 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
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
on 3 ; 1 = 2 d:f :
2
ni
3 X
X
SSresiduals = YT (I ; PX )Y =
(Yij ; Yi:)2
MSH0 =
SSresiduals on
MSresiduals = X
3
(ni ; 1)
i=1
329
"
i=1 j =1
3
X
(ni ; 1) d:f :
i=1
330
For this particular alternative
2 3
"
#6 7 "
#
1 66 1 77 = :5
C = 00 10 01 ;
;1 4 2 5
Reject H0 in favor of Ha if
F = MSMSH0 > F(2;P3i=1(ni;1));
residuals
3
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 speci c alternative
( + 1) ; ( + 3) = 0:5
( + 2) ; ( + 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
1
:
5
2
= 2 ; 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
332
In this case,
2 3n
6
X T X = 664 nn
n
333
0.0
0.2
0.4
0.6
0.8
1.0
Power versus Sample Size
F-test for equal means
Power
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
0
10
20
30
40
50
Sample Size
334
For testing
H0 : ( + 1) = ( + 2) = = ( + k )
against
HA : ( + 1) 6= ( + j )for somei 6= j
# This file is stored in power.ssc
#===========================================
use
n <- 2:50
d2 <- n/2
power <- 1 - pf(qf(.95,2,3*n-3),2,3*n-3,d2)
#
#
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
ni i
: = i=1
k
X
#============================================
i=1
ni
336
335
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
:
77
75
k
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
337
338
Con dence intervals for
estimable functions of
0.4
Central t Densities
2 df
5 df
30 df
0.2
f(x;df)
0.3
Defn 5.2: Suppose Z N (0; 1) is distributed
independently of W 2v , and then the distribution of
t = qZW
v
0.0
0.1
is called the student t-distribution with v degrees of freedom.
We will use the notation
t tv
-4
-2
0
2
4
x
340
339
# This code is stored in:
tden.ssc
#=================================================#
# 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.
}
341
342
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)
343
Note that
Cov(cT b; (I ; PX )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
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.
344
Then,
t=
=
=
Z
r SSE
2(n;k)
pc2TcTb;(XcTT X );c
r SSE
2
(n;k)
T
T
r SSEc b;c
T T ; t(n;k)
(n;k) c (X X ) c
%
SSE
n;k
345
is the MSE
346
It follows that
Central t Density
( 5 degrees of freedom )
(
0.4
1;
= Pr ;t(n;k);
p
0.3
= Pr cT b ; t(n;k);
=2
cT b ; cT
t
MSE cT(XTX);c (n;k);
q
)
=2
MSE cT(XTX);c cT
q
0.2
cT b + t(n;k); =2 MSE cT(XTX);c
and a (1 ; ) 100% con dence interval for
is
0.1
f(x;df)
=2
0.0
cT
-4
-2
0
2
cT b ; t(n;k); =2
4
x
q
MSE cT(XTX);c ;
cT b + t(n;k); =2
q
MSE cT(XTX);c
348
347
Con dence interval for 2:
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 ) we have shown that
SSE = YT (I ; PX )Y 2
(n;k)
2
2
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 .
349
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
350
0.0
0.04
f(x;df)
The resulting (1; )100% con dence interval
for 2 is
0
1
SSE
SSE
@ 2
A
(n;k); =2 ; 2(n;k);1; =2
0.08
0.12
Central Chi-Square Density
0
5
10
15
x
351
20
25
30