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Ch. 7: Test on covariance matrix
I. Introduction
A. Before testing a mean difference as in MANOVA, we can test
the covariance matrices related to some assumptions.
B. Specific patterns of a covariance matrix can be tested for
assumptions such as sphericity and compound symmetry.
C. The equality and independence of different matrices can also
be tested.
II. Testing a specific pattern for Σ
A. Testing Ho: Σ = Σ 0
- Ho : Σ = Σ 0 H1: Σ ≠ Σ0,
( Σ 0 from either theory or practice.)
- TS a) u = (n-1)[ln| Σ 0 | - ln| S | + tr( SΣ 0 - ) – p].
u is approximately distributed as χ
2 [(1/2)p(p+1)] when
(n-1) is large.
b) When n is moderate (n<30), then
u’ = [1 - 6 ( 1 ) 1
n − −
(2p + 1 - 1
p +
)]u.
u’ is approximately distributed as χ
2 [(1/2)p(p+1)].
- If u ≥ χ
2 (crit), reject Ho. The sample covariance
matrix is not equal to the target covariance matrix.
- If we want to test Ho for independence and unit variance
for variables, then
Ho: Σ = I ( Σ 0 = I ).
B. Testing sphericity (Mauchly, 1940, for repeated measures
ANOVA)
- Ho: Σ = σ
2 I
(identical variance and zero covariance among y-
variables)
TS
LR =
/ 2
n
p trS p
S
where p: # of y-variables
n: sample size in each group.
χ
2 (obs) = -2 ln (LR) = -2 ln
/ 2
n
p tr S p
S
= -2(n/2) ln (^)
p trS p
S
= -n ln (^)
p
p
tr S
p S
( ( ))
= -n ln u.
let u = (^) p
p
trS
p S
( ( ))
p
p
i
i
p
i
i
p p
1
1
=
=
λ
λ
, and df = (1/n)(p(p+1) – 1)
Decision: If χ
2 (obs) ≥ χ
2 (crit), then reject Ho.
- Ho: CΣC’ = σ
2 I (test the difference in variances,
sphericity assumption test,
C(p-1)Xp = orthogonal contrast coefficient
matrix)
TS
χ
2 = -2 ln (^)
−
−
1
1
p
p
trCSC
p CSC with df = (1/n)p(p-1) – 1.
Decision: If χ
2 (obs) ≥ χ
2 (crit), then reject Ho.
C. Testing Ho: Σ = σ
2 [(1-ρ) I + ρ J ]
(compound symmetry: all variances are equal, and all
covariances are equal)
- Ho: Σ = σ
2 [(1-ρ) I + ρ J ] = σ
2
2. TS
χ
2 (obs) = | |
S 0
S
with df = (1/2)p(p+1) – 2
where S = sample covariance matrix,
S 0 = a (pXp) mean (centroid) matrix with s
2 on the main-
diagonals and s
2 r on the off-diagonals,
s
2
=
p
p
sii p (^) 1
, and
s
2
r = ∑
− (^) i ≠ j
sij p ( p 1 )
So =
2 1
2
2
2 23
2 2 21
2
1
2 13
2 12
2 2
sr s
sr s sr sr
s s r sr sr
p
p
p
r 12 = 1 2
12
s s
s , ∴ s 12 = rs 1 s 2. If s
2 1 = s
2 2 (under Ho), s 12 =
s
2 r.
S pl = N J
E
=
=
J
j
j
J
j
j j
n
n S
1
1
and χ
2 (obs) = -2(1-c 1 ) ln M is an approximation of
χ
2 -distribution with df = (1/2)(J-1)p(p+1).
where
c 1 = 6 ( 1 )( 1 )
2
J n p
J p p = (^)
= 6 (^1 )(^1 )
2
1 p J
p p
n
J
j j
, and
ln M = ( 1 ) ln| | 2
( 1 )ln| | 2
1 1
pl
J
j
J
j
∑ nj Sj ∑ nj S
= =
- If reject Ho (χ
2 (obs) ≥ χ
2 (crit)), then two covariance
matrices are significantly different.
- Example: J=2, n=
SM =
SF =
Spl =
Ho: ΣM = ΣF H1: ΣM ≠ ΣF
α = .05 χ
2 (crit)[df = (1/2)(J-1)p(p+1) = 10] = 18.
TS χ
2 (obs) = u = -2(1-c 1 ) ln M
ln M = ( 1 ) ln| | 2
( 1 )ln| | 2
1 1
pl
J
j
J
j
∑ nj Sj ∑ nj S
= =
= (1/2)[31 ln (7917.7)+31 ln (58958.1)
- (1/2)(31+31) ln (27325.2)
= -7.
c 1 = 6 ( 1 )( 1 )
2
J n p
J p p
6 ( 2 )( 31 )( 4 1 )
( 2 1 )[( 2 )( 4 ) 3 ( 4 ) 1 ]
2
∴χ
2 (obs) = -2(1-c 1 ) ln M
= -2(1-.06935)(-7.2803) = 13.
Decision χ
2 (obs) = 13.551 NGT 18.307 = χ
2 (crit)
Thus, we fail to reject Ho. Two covariance matrices are not significantly different.
IV. Testing the independence of sub-matrices
A. Situation
- A covariance matrix is formed from y (pX1) and x (qX1).
- Thus, Σ (p+q)X(p+q) =
xy xx
yy yx , S =
xy xx
yy yx
S S
S S
, R =
xy xx
yy yx
R R
R R
- Want to test if Σ yx = 0 x and y are independence.
B. Testing procedure
- Ho: Σ yx = 0 or Ho: Σ (p+q)X(p+q) = (^)
xx
yy
O
O
2. TS
yy xx Ryy R xx
R
S S
S
= with Λ(crit)[p, q, (n-1-q)]
- Decision
If Λ(obs) ≤ Λ(crit), then reject Ho. not independent.
C. Example: n=46, y 1 , y2, x 1 , x 2 , x 3
S =
xy xx
yy yx
S S
S S
Syy S xx
S
9
9
X
X
= .724 < .730 = Λ(crit)
We reject Ho of the independence hypothesis
