Discussiones Mathematicae Probability and Statistics 22 (2002 ) 37–51
EXACT DISTRIBUTION FOR THE GENERALIZED F TESTS ˜ o Tiago Mexia Miguel Fonseca, JoA Department of Mathematics, Faculty of Science and Technology New University of Lisbon Monte da Caparica 2829–516 Caparica, Portugal e-mail:
[email protected]
and Roman Zmy´ slony Institute of Mathematics, University of Zielona G´ ora Podg´ orna 50 65–246 Zielona G´ ora, Poland e-mail:
[email protected]
Abstract Generalized F statistics are the quotients of convex combinations of central chi-squares divided by their degrees of freedom. Exact expressions are obtained for the distribution of these statistics when the degrees of freedom either in the numerator or in the denominator are even. An example is given to show how these expressions may be used to check the accuracy of Monte-Carlo methods in tabling these distributions. Moreover, when carrying out adaptative tests, these expressions enable us to estimate the p-values whenever they are available. Keywords: exact distribution theory, hypothesis testing, generalized F distribution, adaptative test. 2000 Mathematics Subject Classification: 62E15, 62H10, 62H15, 62J10.
38
M. Fonseca, J.T. Mexia and R. Zmy´ slony 1. Introduction
When, under the tested hypothesis, the test statistic may be written as the quotient of convex combinations of independent central chi-squares divided by their degrees of freedom we have a generalized F test. Such tests were introduced by [3] and [4]. In the first of these papers, the nullity of variance components is tested. The statistics were the quotients of the positive part by the negative part of quadratic unbiased estimators. In the other paper, this technique was extended to mixed models containing also fixed parameters. If the degrees of freedom in the numerator [denominator] are even we say that the first [second] evenness condition holds. We are going to obtain exact expressions for the distribution of generalized F test statistics when either of these conditions hold. These expressions are useful in checking the precision of Monte-Carlo methods in tabling such distributions. Moreover, if adaptative tests, see [1], are carried out we can use the exact expressions of the distributions, if available, to estimate the p-values. 2. Exact expressions r , cs , gr , gs ) be the distribution of Let F (u|c ∼1 ∼2 ∼ 1 ∼ 2 r P
(1)
F=
j=1 s P j=1
c1j χ2g1j c2j χ2g2j
when the chi-squares are independent. Clearly, the c1j and g1j , j = 1, ...r r [cs and [c2j and g2j , j = 1, ...s] will be the components of ∼ cr1 and g ∼2 ∼1 s ], respectively. The generalized F distributions will belong to this famg ∼2 r , cs > 0s and ily, namely they correspond to the case in which ∼ cr1 > 0 ∼ ∼2 ∼ Ps Pr j=1 c2j g2j . j=1 c1j g1j = r , cs > 0s and that gr = 2mr . Our We start by assuming that ∼ cr1 > 0 ∼ ∼2 ∼ ∼ ∼1 s = 2ms , since results extend directly to the case in which g ∼ ∼2
(2)
µ ¶ µ ¶ r s r s −1 s r s r F u|c , c , g , 2m = 1 − F u |c , c , 2m ,g . ∼1 ∼2 ∼ 1 ∼ ∼2 ∼1 ∼ ∼1
Exact distribution for the generalized
tests
F
39
Thus we will cover the cases of generalized F distributions whenever one of the evenness conditions holds. To make the notation clear we put ai = c1i and gi = g1i , i = 1, ..., r, as well as ai+r = c2i and gi+r = g2i , i = 1, ..., s. Define also Xi = ai χ2gi i , i = 1, ..., r + s. We now carry out successive integrations. The density of Xi will be xmi −1 − x e 2ai , x > 0, i = 1, ..., r f (x) = i m (mi − 1)!(2ai ) i gi (3) x x 2 −1 − 2a i , x > 0, i = r + 1, ..., r + s gi e fi (x) = gi Γ( 2 )(2ai ) 2 Then r s r s F (u|c , c , 2m ,g ) ∼1 ∼2 ∼ ∼2
" r X
=P
Z (4)
Xi ≤ u
i=1
Z
0
i=r+1
0 u
r+s P i=r+1
···
r+s P
+∞ Z u
··· Z
Xi
i=r+1
+∞
=
#
r+s X
xi
···
0 xi −
r P
i=2
xi r+s Y
0
fj (xj )dxj .
j=1
We now point out that, with b a non-negative integer, through a successive integration by parts we get Z d b j X x d d (5) xb e− a dx = ab+1 b! 1 − e− a . j j! a 0 j=0
This expression can be rewritten as Z (6)
d
x
xb e− a dx = ab+1 b!
0
Using (6), with y = u
1 X k=0
r+s P i=r+1
kd
(−1)k e− a
kb X dj . aj j! j=0
xi and bi = mi − 1, i = 1, ..., r, and in the first
M. Fonseca, J.T. Mexia and R. Zmy´ slony
40 place, a = 2a1 and d = y − Z
y−
r P
i=2
Pr
i=2 xi ,
Z
xi
r P i=2
f1 (x1 ) dx1 =
0
y−
we get x1
xb11 e− a dx1 b1 !(2a1 )b1 +1
xi
0
µ
1 X
=
k
k1
− 2a1
y−
1
(−1) e
¶
r P
i=2
xi
(7) =
µ k
− 2a1
(−1)k1 e
y−
1
¶
r P
i=2
xi
X j1 P
i=1
à where
j1 P
i=1
k 1 b1 X j1 =0
k1 =0
Ã
j1 !(2a1 )j1
j1 =0
k1 =0
1 X
µ ¶j1 r P y− xi k 1 b1 X i=2
(−1)
t1,1
t r Y x 1,i
t1,1 !
t1,i !
j1 −t1,1 y
!
1 (2a1 )j1
i
i=2
t1,i =j1
! t1,i = j1
indicates summation for all sets of non negative inte-
gers t1,1 , ..., t1,r with sum j1 . Thus, we can apply again (6) to get Z
y−
r P i=3
xiZ y−
r P
i=2
0
xi
f1 (x1 )f2 (x2 ) dx1 dx2 =
0 Ã
(8)
1 X
k
(−1)k1 e
− 2a1
1
y−
!
r P y=3
xi
k 1 b1 X j1 =0
k1 =0
µ
r P
i=1
Z
y−
r P i=3
0
xi
b +t
x22 1,2 − e (2a2 )b2 +1
³
k 1 − 2a1 2a2 1
t r y t1,1 Y xi1,i 1 · × t1,i ! b2 !t1,2 ! ¶t1,1 !
X t1,i =j1
´
dx2 =
i=3
k1 b1 1 X (−1)k1 X 1 × b +1 2 (2a2 ) (2a1 )j1
k1 =0
j1 =0
Exact distribution for the generalized
µ
X µ
(−1)
j1 −t1,1
¶
r P
¶−(b2 +t1,2 +1)
1 k1 − 2a2 2a1
41
(b2 + t1,2 )! × b2 !t1,2 !
t1,i =j1
i=1
³
1 X
k2
(−1) e
−
k1 +k2 2a1
³
k 1 − 2a1 2a2 2
´´µ y−
r P
i=3
¶ xi
k2 (b2 +t1,2 ) µ
X
j2 =0
k2 =0
X µr−1 P i=1
tests
F
(−1)
t1,1 +t2,1
t +t r Y x 1,i 2,i−1
t1,1 !t2,1 !
t1,i !t2,i−1 !
j2 −t2,1 y
¶ t2,1 =j2
j−1 P
Now, with b+ j = bj +
u=1
i
i=3
k1 1 − 2a2 2a1
¶j2 ×
.
tu,j+1−u , j = 1, ..., w, we will have b+ 1 = b1 and
n = (k , ..., k )0 , we have k2 = (k , k )0 . b+ 1 2 1 n 2 = b2 + t1,2 . Besides this, taking k ∼ ∼ n ) = 0 and, for kn 6= 0n , d(kn ) = Let d(0 ∼ ∼ ∼ ∼
1 awn
with wn the largest index
n , n = 1, ..., r. It is easy to check that k1 + for non null components of k ∼ 2a1 ³ ´ k1 1 1 n n+1 2 n ) + kn+1 2an+1 − d(k ). ), and that d(k ) = d(k k2 ( 2a2 − 2a1 ) = d(k ∼ ∼ ∼ ∼
We may now rewrite (8) as
Z
y−
r P i=3
0
(9)
xiZ y−
r P
i=2
0
xi
f1 (x1 )f2 (x2 )dx1 dx2 = +
1 b1 1 kX X 1 1 b +1 2 (2a1 )j1 (2a2 )
k1 =0 j1 =0
+
2 b2 1 kX X
X µ
r P
i=1
¶ t1,i =j1
k2 =0 j2 =0
X µ
r P
i=1
¶ t2,i =j2
M. Fonseca, J.T. Mexia and R. Zmy´ slony
42
2 P
(−1)
(ki +ji −ti,1 )
i=1
−d(k r Y e ∼
2 )x i
t
b+ 2! b2 !t1,2 !
µ
k1 1 − 2a2 2a1
¶j2 −b+ 2 −1
e
2 −d(k ∼ )y t1,1 +t2,1 y
t1,1 !t2,1 !
×
+t
xi1,i 2,i−1 . t1,i !t2,i−1 !
i=3
Let us establish Lemma 1. With Z Lr,m (y, xm+1 , ..., xr ) =
y−
r P
i=m+1
Z
xi
r P
y−
i=2
···
0
0
m xi Y
fv (xv )
v=1
m Y
dxv
v=1
we have Z Lr,m (y, xm + 1, ..., xr ) = +
1 b1 1 kX X
1 m Q
(2av )bv +1
k1 =0 j1 =0
v=2
y−
r P
i=m+1
r P
i=1
r P
y−
i=2
···
0
0
(ki +ji −ti,1 )
(−1)i=1
1 kX m bm X
···
¶
km =0 jm
t1,i =j1
e
m Q i=1
m P
y i=1
ti,1 !
dxv =
v=1
µr+1−m P
¶ tm,i =jm
¶jv −b+ µ m v −1 Y 1 b+ 1 v! v−1 − d(k ) v−1 ∼ Q (2a1 )j1 2av v=2 b ! t ! v i,v+1−i m P
ti,1
m Y
X
i=1
i=1
m −d(k ∼ )y
fv (xv )
v=1
m P
m xi Y
+
X µ
Z
xi
r Y i=m+1
m d(k ∼ )xi
e
m Q v=1
xv=1 i
tv,i+1−v
tv,i+1−v !
.
Exact distribution for the generalized
F
tests
43
P roof. We use induction in m. If m = 2 the thesis holds since, then, we have (9). Let us assume that it holds for m ≤ w < r, since b+ w+1 = w P bw + tv,w+1−v , we get using (6) v=1
w P
Z
y−
r P
i=w+2
xi
e
w d(k ∼ )xw+1
0
tv,w+2−v
xv=1 w+1 fw+1 (xw+1 ) dxw+1 = w Q tv,w+2−v !
v=1
1 bw+1 ! ³
e
−
w Q v=1
tv,w+2−v !
1 · (2aw+1 )bw+1 +1
´
1 w −d(k ∼ ) 2aw+1
xw+1
dxw+1
Z
y−
r P
i=w+2
xi
0
bw+1 +
xw+1
w P
(tv,w+2−v )
v=1
b+ 1 w+1 ! = · w Q (2aw+1 )bw+1 +1 bw+1 ! tv,w+2−v ! v=1
µ
1 2aw+1
¶−b+ 1 X w+1 −1 w − d(k ) ∼
³
kw+1 b+ w+1
X
e
−kw+1
´
1 w −d(k ∼ ) 2aw+1
à y−
r P
i=w+2
! xi
kw+1 =0 jw+1 =0
µ
(−1)kw+1 jw+1 !
1 2aw+1
!jw+1 ¶jw+1 Ã r X w − d(k ) y− xi = ∼ i=w+2
1 X b+ 1 w+1 ! · w Q (2aw+1 )bw+1 +1 kw+1 =0 bw+1 ! tv,w+2−v !
kw+1 b+ w+1
X
jw+1 =0
v=1
X Ã
r+1−(w+1) P i=1
µ (−1)
! tw+1,i =jw+1
kw+1 +jw+1 −tw+1,1
1 2aw+1
¶jw+1 −b+ w+1 −1 − d(k ) × ∼ w
M. Fonseca, J.T. Mexia and R. Zmy´ slony
44
³
e
−kw+1
w) + k Since d(k w+1 ∼
³
2aw+1
µ
w Q i=1
(ki +ji −ti,1 )
¶
y i=1
ti,1
ti,1 !
r P
w Y v=2
bv !
xi
v−1 Q i=1
µ
ti,v+1−1 !
w −d(k ∼ )xi
r Y
e
w Q v=1
e
w d(k ∼ )xw+1
w Q v=1
(2av )bv +1
v=1
µr+1−w P
¶ tw,i =jw
¶jv −b+ v −1 1 v−1 − d(k ) ∼ 2av
tv,i+1−v
xi
tv,i+1−v !
tv,w+2−v
xv=1 w+1
fw+1 (xw+1 ) dxw+1
tv,w+2−v ! +
1 v=2
X
i=1
b+ v!
i=w+2
0
w+1 Q
kw =0 jw =0
t1,i =j1
w P
w P
i=w+2
=
···
w P
y−
+
1 kX w bw X
w −d(k ∼ )y
Z
r P
i=1
(−1)i=1 (2a1 )j1
e
Lw.r (y, xw+1 , ..., xr )fw+1 (xw+1 ) dxw+1
X
k1 =0 j1 =0
1 b1 1 kX X
k1 =0 j1 =0
.
xi
0
v=2
w P
r P
w+2
1 b1 1 kX X
(2av )bv +1
xi tw+1,(i+1)−(w+1) xi
tw+1,(i+1)−(w+1) !
i=w+2
y−
+
w Q
´
1 w −d(k ∼ ) 2aw+1
´ w ) = d(kw+1 ), we will have − d(k ∼ ∼
1
Lw+1.r (y, xw+2 , ..., xr ) =
1
kw+1 r Y e
tw+1,1 !
Z
=
³
´ t y y w+1,1
1 w −d(k ∼ ) 2aw+1
X µ
r P
i=1
+
···
¶ t1,i =j1
1 kX w bw X
kw =0 jw =0
X µr+1−w P i=1
¶ tw,i =jw
Exact distribution for the generalized
tests
w+1 P
kw+1 b+ w+1
1 X
F
X
X r+1−(w+1) P i=1
ki +ji −ti,1
(−1) i=1 (2a1 )j1 !
Ã
kw+1 =0 jw+1 =0
w+1 Y
e
×
tw+1,i =jw+1
v=2
45
µ
b+ v! bv !
v−1 Q i=1
w+1 )y −d(k ∼
w+1 Q i=1
ti,v+1−1 !
w+1 P
y i=1
¶jv −b+ v −1 1 v−1 − d(k ) × ∼ 2av
w+1 P
ti,1
ti,1 !
r Y
e
w+1 )x −d(k i ∼
w+1 Q
i=w+2
v=1
xiv=1
tv,i+1−v
tv,i+1−v !
which completes the proof. We can now establish Proposition 1. We have
+
r s r s F (u|c , c , 2m ,g ) ∼1 ∼2 ∼ ∼2
=
1 b1 1 kX X
1 r Q
(2av )bv +1
k1 =0 j1 =0
v=2
r P
X µ
1 P
i=1
tr,i =jr
µ
r P
i=1
+
···
¶ t1,i =j1
1 kX r br X
kr =0 jr =0
µ ¶ b+ 1 i−1 ji −b+ i ! i −1 −d(k ) × i−1 ∼ 2ai Q i=2 b ! tv,i+1−v ! i
(ki +ji −ti,1 ) r Y
(−1)i=1 (2a1 )j1 ¶
X
v=1
M. Fonseca, J.T. Mexia and R. Zmy´ slony
46
µ
r P
i=1
¶ P rtv,1 ti,1 !uv=1 r Q i=1
X Ã
ti,1 !
r+s P
i=r+1
li =
r P
v=1
! tv,1
à ¡ ¢ ¶−(li + gi ) ! µ 2 Γ li + g2i 1 1 r ¡ gi ¢ · + d(k )u . gi ∼ 2ai l !Γ 2 i (2a ) i 2 i=r+1 r+s Y
P roof. We can use Lemma 1 to write
Z r s r s F (u|c , c , 2m ,g ) ∼1 ∼2 ∼ ∼2
Z
+∞
=
Ã
+∞
···
Lr,r
0
0
1
1 b1 1 kX X
r+s X
u
!
r+s Y
xi
fi (xi )dxi
i=r+1
i=r+1
X
1 kX r br X
Z
Z
+∞
=
··· 0
0
+
+∞
Q r
+
···
µ r ¶ (2av )bv +1 k1 =0 j1 =0 P t1,i =j1 v=2 i=1
r P
X µ
1 P
i=1
µ
kv +jv −tv,1
(−1)v=1 (2a1 )j1 ¶
tr,i =jr
¶ 1 w−1 − d(k ) ∼ 2aw
jv −b+ v −1
e
r Y w=2
r −d(k ∼ )u
r Q i=1
kr =0 jr =0
b+ w! bw ! r P
ui=1
ti,1 !
w−1 Q v=1
tv,w+1−v !
ti,1
Y r+s fi (xi )dxi i=r+1
Exact distribution for the generalized
(we point out that
r Q i=r+1
upper one) thus getting
r P t d(∼ k r )xi i=1 i,1 e xi r Q ti,1 ! i=1
F
tests
47
= 1 since the lower index exceeds the
r s r s F (u|c , c , 2m ,g )= ∼1 ∼2 ∼ ∼2
+ 1 b1 1 kX X
1
X
r Q
···
+ 1 kX r br X
r P
X
µ r ¶ µ 1 kr =0 jr =0 P (2av )bv +1 k1 =0 j1 =0 P t1,i =j1 tr,i =jr v=2 i=1
(kv +jv −tv,1 )
(−1)v=1 (2a1 )j1 ¶
i=1
r Y w=2
e
b+ w! bw !
w−1 Q v=1
r+s P
r −d(k ∼ )u
i=r+1
tv,w+1−v !
xi
¶jv −b+ Z +∞ Z +∞ µ 1 v −1 w−1 − d(k ) ··· × ∼ 2aw 0 0
r µ r+s ¶ P ti,1 P i=1 u xi
i=r+1 r Q i=1
ti,1 !
r+s Y
fi (xi )dxi .
i=r+1
Now
Z
+∞
Z
··· 0
r +∞ −d(k ∼ )u
e
0
r+s P i=r+1
xi
r µ r+s ¶ P ti,1 P i=1 u xi
i=r+1 r Q i=1
ti,1 !
r+s Y i=r+1
fi (xi )dxi
×
M. Fonseca, J.T. Mexia and R. Zmy´ slony
48
µ =
r ¶ P tv,1 tv,1 !uv=1 r+s Y Z v=1
X Ã
r+s P
i=r+1
li =
r P
v=1
r P
r Q
! tv,1
tv,1 !
v=1
r+s Q v=r+1
µ li +
xi
gi −1 2
dxi =
X Ã
r+s P
i=r+1
r+s Y i=r+1
Ã
li =
r P
v=1
! tv,1
e
Γ
i=r+1 0
li !
r P
v=1 r Q v=1
³ ´ 1 r ∼ ) xi +∞ − 2ai +d(k
¡ gi ¢ 2
gi
(2ai ) 2
×
r ¶ P tv,1 tv,1 !uv=1
tv,1 !
r+s Q v=r+1
× li !
¢ µ ¡ ¶−(li + gi ) ! 2 Γ li + g2i 1 r + d(k )u . ¡ gi ¢ gi ∼ Γ 2 (2ai ) 2 2ai
To ´ last integrals it suffices to use the transformation zi = ³ compute the 1 r )u xi , i = r + 1, ..., r + s, and use the well known gamma func∼ 2ai + d(k tion. To complete the proof we need only to substitute the result we just r , cs , 2mr , gs ). obtained in the expression of F (u|c ∼ ∼1 ∼2 ∼2
3. Monte carlo methods r , cs , g r , g s ) Monte Carlo methods can be used to table distributions F (u|c ∼1 ∼2 ∼ 1 ∼ 2
in a straightforward way since it is easy to generate independent chi-squares. The exact formula derived above may be used to check the accuracy of these methods. For instance, in a balanced variance components model in which the first factor crosses with the second that nests a third, the variance component σ 2 (β) associated with the second factor is not the difference between two ANOVA mean squares, see [2] pg. 35. Thus, a usual F test cannot be derived for the nullity of σ 2 (β), but using the known results on that model it is straightforward to get a generalized F test with r = s = 2.
Exact distribution for the generalized
F
tests
49
The coefficients will be c11 = p1 , c12 = 1 − p1 , c21 = p2 and c22 = 1 − p2 with max{p2 , 1 − p2 } ≤ p1 . Moreover, if the factors have three levels, we will have g1 = 2, g2 = 12, g3 = 4 and g4 = 6, so that both evenness conditions hold. In this case the exact expression for the distribution will be
F (z|p1 , p2 ) = 1 −
i µ ¶µ X i j=0
j
³ 1+
3p2 (1−p1 ) z
´−2 ³ 1+
2(1−p2 ) (1−p1 ) z
´−3
2
2 (1 − p1 ) p2 4 (1 − p1 ) + 12p2 z
¶j
5 X
(12z)i
i=0
i! (2 (1 − p1 ))i
µ
¶i−j 2 (1 − p1 ) (1 − p2 ) (2+j −1)! × 6 (1 − p1 ) + 12 (1 − p2 ) z
µ ¶ µ ¶−2 µ ¶ (1 − p1 ) −6 p2 (1 − p2 ) −3 (3 + i − j − 1)! − 1 − 1+ z 1+ z 6p1 2p1 6p1
+
³ 1−
(1−p1 ) 6p1
´−6 ³
1+
p2 2p1 z
120
´−2 ³
1+
(1−p2 ) 6p1 z
´−3
7 X (12p1 − 2 (1 − p1 ))k z k k=0
k! (2p1 (1 − p1 ))k
k µ ¶µ X k l=0
l
¶l µ ¶k−l 2 (1 − p1 ) p2 2 (1−p1 ) (1−p2 ) l! (2+k−l)!, 4 (1 − p1 ) + 12p2 z 6 (1−p1 )+12 (1−p2) z
where for the sake of clarity of notation we omitted the degrees of freedom. In the next table we present the values given by the exact expression and by Monte Carlo methods (10000 runs for each value), the first value in each pair being given by the exact expression. For each pair we consider the values 1, 5, 10 and 20 for z.
M. Fonseca, J.T. Mexia and R. Zmy´ slony
50
p2 0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
z 1 5 10 20 1 5 10 20 1 5 10 20 1 5 10 20 1 5 10 20 1 5 10 20 1 5 10 20 1 5 10 20 1 5 10 20
p1 0.5 0.506,0.501 0.98,0.978 0.998,0.998 1,1 0.516,0.516 0.985,0.987 0.999,0.999 1,1 0.522,0.519 0.987,0.985 0.999,0.999 1,1 0.525,0.535 0.988,0.987 0.999,1 1,1 0.522,0.525 0.987,0.987 0.999,0.999 1,1 0.516,0.515 0.986,0.986 0.999,0.999 1,1 0.505,0.502 0.982,0.982 0.998,0.999 1,1 0.492,0.491 0.975,0.974 0.997,0.996 1,1 0.479,0.484 0.961,0.961 0.994,0.994 0.999,1
0.6 0.524,0.526 0.977,0.98 0.997,0.997 1,1 0.534,0.537 0.982,0.982 0.998,0.998 1,1 0.541,0.537 0.984,0.984 0.999,0.999 1,1 0.543,0.537 0.985,0.985 0.999,0.999 1,1 0.541,0.555 0.984,0.985 0.999,0.999 1,1 0.535,0.543 0.982,0.984 0.998,0.999 1,1 0.524,0.518 0.979,0.979 0.998,0.998 1,1 0.511,0.522 0.972,0.975 0.996,0.996 1,0.999 0.496,0.502 0.958,0.959 0.993,0.993 0.999,0.999
0.7 0.543,0.542 0.973,0.973 0.996,0.996 1,1 0.554,0.552 0.978,0.977 0.998,0.998 1,1 0.56,0.563 0.981,0.982 0.998,0.998 1,1 0.562,0.552 0.982,0.981 0.998,0.998 1,1 0.56,0.554 0.981,0.981 0.998,0.999 1,1 0.554,0.553 0.979,0.98 0.998,0.998 1,1 0.544,0.541 0.975,0.975 0.997,0.997 1,0.999 0.531,0.529 0.967,0.97 0.995,0.995 1,1 0.516,0.512 0.954,0.952 0.991,0.991 0.999,0.999
0.8 0.576,0.574 0.968,0.97 0.995,0.996 1,1 0.584,0.578 0.974,0.972 0.997,0.996 1,1 0.588,0.591 0.977,0.978 0.997,0.997 1,1 0.59,0.592 0.978,0.974 0.998,0.998 1,1 0.588,0.587 0.977,0.976 0.997,0.997 1,1 0.584,0.59 0.975,0.975 0.997,0.998 1,1 0.577,0.578 0.97,0.967 0.996,0.996 1,1 0.567,0.578 0.963,0.965 0.994,0.994 0.999,0.999 0.554,0.552 0.95,0.951 0.99,0.99 0.999,0.999
0.9 0.576,0.574 0.964,0.963 0.994,0.995 0.999,0.999 0.584,0.578 0.97,0.97 0.996,0.996 1,1 0.588,0.591 0.973,0.975 0.997,0.997 1,1 0.59,0.592 0.973,0.972 0.997,0.997 1,1 0.588,0.587 0.973,0.974 0.997,0.996 1,1 0.584,0.59 0.97,0.972 0.996,0.996 1,1 0.577,0.578 0.966,0.966 0.995,0.996 1,1 0.567,0.578 0.958,0.958 0.993,0.993 0.999,0.999 0.554,0.552 0.945,0.942 0.988,0.989 0.998,0.999
Exact distribution for the generalized
F
tests
51
References [1] E. G¸asiorek, A. Michalski and R. Zmy´slony, Tests of independence of normal random variables with known and unknown variance ratio, Discussiones Mathematicae - Probability and Statistics 2 (2000), 237–247. [2] A.I. Khuri, T. Matthew and B.K. Sinha, Statistical Tests for Mixed Linear Models, John Wiley & Sons New York 1998. [3] A. Michalski and R. Zmy´slony, Testing hypothesis for variance components in mixed linear models, Statistics 27 (1996), 297–310. [4] A. Michalski and R. Zmy´slony, Testing hypothesis for linear functions of parameters in mixed linear models, Tatra Mt. Math. Pub. 17 (1999), 103–110. Received 15 October 2002
