Scheffé’s Method
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Scheffé’s Method:
Suppose
y = Xβ + ε,
where
ε ∼ N(0, σ 2 I).
Let
c01 β, . . . , c0q β
be q estimable functions, where
c1 , . . . , cq
are linearly independent.
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Let
 
c01
.
.
C=
.
c0q
and define
  

θ1
c01 β
.  . 
.  . 
θ=
 .  =  .  = Cβ.
θq
c0q β
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Let W = C(X0 X)− C0 with diagonal elements denoted
w11 , . . . , wqq .
Then
Var(Cβ̂) = σ 2 W
and
Var(θ̂j ) = Var(c0j β̂) = σ 2 wjj
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j = 1, . . . , q.
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For any q × 1 vector u and any k ∈ R, let L(u, k) denote the interval
√
√
[u0 θ̂ − k σ̂ 2 u0 Wu, u0 θ̂ + k σ̂ 2 u0 Wu].
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We want to find k 3
P(u0 θ ∈ L(u, k) ∀ u ∈ Rq ) = 1 − α.
Thus, we seek simultaneously coverage probability 1 − α for an infinite
set of intervals.
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Show that
u0 θ ∈ L(u, k) ∀ u ∈ Rq
⇐⇒
k2
(Cβ̂ − Cβ)0 (C(X0 X)− C0 )−1 (Cβ̂ − Cβ)
≤
.
qσ̂ 2
q
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Statistics 611
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First, note that
u0 θ ∈ L(u, k)
∀ u ∈ Rq
⇐⇒ u0 θ ∈ L(u, k)
∀ u ∈ Rq \ {0}
∵ 00 θ = 0 ∈ [0, 0] = L(0, k).
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√
√
u0 θ ∈ L(u, k) ⇐⇒ u0 θ̂ − k σ̂ 2 u0 Wu ≤ u0 θ ≤ u0 θ̂ + k σ̂ 2 u0 Wu
√
√
⇐⇒ −k σ̂ 2 u0 Wu ≤ u0 θ − u0 θ̂ ≤ k σ̂ 2 u0 Wu
u0 θ − u0 θ̂
⇐⇒ −k ≤ √
≤k
σ̂ 2 u0 Wu
0
u θ − u0 θ̂ ≤k
⇐⇒ √
σ̂ 2 u0 Wu
[u0 (θ − θ̂)]2
≤ k2 .
⇐⇒
σ̂ 2 u0 Wu
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Dan Nettleton (Iowa State University)
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∴ u0 θ ∈ L(u, k) ∀ u ∈ Rq \ {0}
[u0 (θ − θ̂)]2
≤ k2 ∀ u ∈ Rq \ {0}
σ̂ 2 u0 Wu
[u0 (θ − θ̂)]2
⇐⇒ max
≤ k2
u6=0
σ̂ 2 u0 Wu
⇐⇒
⇐⇒
(θ̂ − θ)0 W −1 (θ̂ − θ)
≤ k2
σ̂ 2
(by C-S generalization in previous notes)
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Dan Nettleton (Iowa State University)
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⇐⇒
(θ̂ − θ)0 W −1 (θ̂ − θ)
k2
≤
qσ̂ 2
q
⇐⇒
(Cβ̂ − Cβ)0 (C(X0 X)− C0 )−1 (Cβ̂ − Cβ)
k2
≤
.
qσ̂ 2
q
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Dan Nettleton (Iowa State University)
Statistics 611
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Thus, we have
P(u0 θ ∈ L(u, k) ∀ u ∈ Rq )
(Cβ̂ − Cβ)0 (C(X0 X)− C0 )−1 (Cβ̂ − Cβ)
k2
=P
≤
.
qσ̂ 2
q
What shall we choose for k to make this probability equal to 1 − α?
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(Cβ̂ − Cβ)0 (C(X0 X)− C0 )−1 (Cβ̂ − Cβ)
∼ Fq,n−r .
qσ̂ 2
Thus, if
k=
then
p
qFq,n−r,α ,
k2
= Fq,n−r,α
q
so that the simultaneous coverage probability is 1 − α.
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Example:
Suppose an experiment was conducted using a completely
randomized design with 10 subjects in each of 4 treatment groups.
The treatment groups were defined by the combinations of levels from
2 factors: diet (1 or 2) and exercise program (1 or 2).
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The model
yijk = µij + εijk
was fit the response data
yijk =
measure of overall health
for diet i, exercise program j,
subject k (i = 1, 2; j = 1, 2; k = 1, . . . , 10).
The parameter µij represents the mean response for diet i, exercise
program j (i = 1, 2; j = 1, 2).
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The εijk terms are assumed to be iid N(0, σ 2 ).
A summary of the data is as follows:
ȳ11· = 9
ȳ12· = 7
ȳ21· = 8
ȳ22· = 3
σ̂ 2 = 5.
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Suppose we want to construct a set of confidence intervals using a
method that gives simultaneous coverage probability at least 95%.
Suppose the confidence intervals will be used to address the following
questions:
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1. Diet main effect?
2. Exercise program main effect?
3. Diet-by-exercise program interaction?
4. Difference between diet 1 and diet 2 under exercise program 1?
5. Difference between exercise program 1 and 2 under diet 1?
6. Diet 1, exercise program 1 vs. mean of other treatments?
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What estimable function of
 
µ
 11 
µ 
 12 
β= 
µ21 
 
µ22
is of interest in each of the questions 1 through 6, respectively?
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1.
µ11 +µ12
2
−
µ21 +µ22
2
=
µ11 +µ12 −µ21 −µ22
2
2.
µ11 +µ21
2
−
µ12 +µ22
2
=
µ11 −µ12 +µ21 −µ22
2
3. (µ11 − µ12 ) − (µ21 − µ22 ) = µ11 − µ12 − µ21 + µ22
4. µ11 − µ21
5. µ11 − µ12
6. µ11 −
(µ12 +µ21 +µ22 )
3
=
3µ11 −µ12 −µ21 −µ22
.
3
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Statistics 611
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Compute an estimate and standard error for each estimable function of
interest.
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X0 X = 10 I
4×4
   
9
ȳ
 11·   
ȳ  7
 12·   
β̂ = (X0 X)− X0 y =   =  
ȳ21·  8
   
3
ȳ22·
X= I ⊗ 1 ,
4×4
10×1
σ2 0
Var(c0 β̂) = σ 2 c0 (X0 X)− c =
cc
10
r
r
p
σ̂ 2 0
5 0
se(c0 β̂) =
cc=
c c = c0 c/2.
10
10
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1. c01 β̂ = 2.5
se1 =
p
1/2
2. c02 β̂ = 3.5
se2 =
p
1/2
3. c03 β̂ = −3
se3 =
4. c04 β̂ = 2
se4 = 1
5. c05 β̂ = 1
se5 = 1
6. c06 β̂ = 3
se6 =
√
2
p
2/3.
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Statistics 611
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Determine appropriate intervals to address questions 1 through 6.
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Statistics 611
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Each interval is of the form
c0j β̂ ± ksej .
How shall we choose k?
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Statistics 611
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If we use the Bonferroni approach, then
k = t40−4,
0.05
(2)(6)
≈ 2.79.
This approach would be legitimate if we were interested in these 6
intervals, and only these 6 intervals, prior to observing the data.
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Alternatively, we can consider Scheffé intervals, k =
p
qFq,40−4,0.05 .
What is the value of q in our situation?
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Recall that Scheffé intervals have the property
P(u0 θ ∈ L(u, k) ∀ u ∈ Rq ) = 1 − α,
where
θ = Cβ
for someq×p
C of rank q.
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If we chooseq×p
C so that
c01 β, . . . , c0q β ∈ {u0 θ : u ∈ Rq },
then we will have Scheffé intervals that give simultaneous coverage at
least 1 − α for c01 β, . . . , c06 β.
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Because qFq,n−r,α is an increasing function of q, we would like to pick q
no larger than necessary.
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The matrix
  
1
2c0
 1 
2c0  1
 2 
 0 
 c3  1
 =
 0 
 c4  1
  
 c0  1
 5 
3
3c06
1 −1 −1
−1
−1
0
−1
−1


1 −1


−1
1
 has rank 3.

−1
0

0
0

−1 −1
To see this, note that each row is a contrast vector (sums to zero) and
is thus in N ( 10 ), which has dim 3.
4×1
Thus, rank ≤ 3. The last 3 rows are LI, so the rank is exactly 3.
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Thus,
dim(span{c1 , . . . , c6 }) = 3.
We can take q = 3 to obtain
k=
p
3F3,40−4,0.05 ≈ 2.93.
The resulting intervals
c0j β̂ j ± 2.93sej j = 1, . . . , 6
have simultaneous coverage at least 95%.
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We could also include additional intervals for other c0 β without
changing k or losing the guarantee of simultaneous coverage
probability at least 95%, as long as c ∈ C(C0 ).
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We have not specified C explicitly, but we can choose C 3 the rows of
C form a basis for the set of all contrasts; i.e., N ( 10 ).
4×1
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This Scheffé method for all possible contrasts allows us to construct as
many intervals as we wish and still have simultaneous coverage
probability at least 1 − α, provided each interval is for a c0 β 3 10 c = 0.
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We can even examine the data to decide which contrasts appear to be
of most interest.
When using the Bonferroni method, intervals of interest must be
preplanned before observing the data.
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If the contrasts in this example were preplanned, Bonferroni would be
preferred over Scheffé because the Bonferroni intervals are narrower.
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