ST4241: Design and Analysis of Clinical Trials - The Department of ...

Sum up (3) over j yields
(5) X·· = grµ + k ∑ s i + gκ
(5)-(1) yields that κ = 0. From (2) and (3), we have
(6)
(7)
µ + s i = ¯X i· − 1 k
∑
j ′ ∈T i
α j
′
X·j = rα j + ∑ i∈S j
(µ + s i )
Substituting (6) into (7) yields
It yields
(ii) Show that
X·j = rα j + ∑ ( ¯X i· − 1 ∑
α
k j
′)
i∈S j j ′ ∈T i
= rα j + rM j − 1 ∑ ∑
k
α j
′
i∈S j j ′ ∈T i
= rα j + rM j − r − λ
k α j.
(1 − r − λ
kr )α j = ¯X·j − M j , i.e., α j = 1
eff ( ¯X·j − M j ).
Var(a j ) =
where eff = g(k−1)
k(g−1) .
σ2 e
reff
g − 1
, Cov(a j , a k ) = − σ2 e
g
1
reff g ,
Write
¯X·j − M j = 1 ∑
(X ij −
r
¯X i·)
i∈S j
= 1 ∑
(α j − 1 ∑
α
r k j
′ + ɛ ij − 1 ∑
ɛ
k ij
′).
i∈S j j ′ ∈T i j ′ ∈T i
2