Second Edition
113
11.7 Let
U
i
=
¯
Y
i
·

θ
i
. Then
k
X
i
=1
n
i
[(
¯
Y
i
·

¯
¯
Y
)

(
θ
i

¯
θ
)]
2
=
k
X
i
=1
n
i
(
U
i

¯
U
)
2
.
The
U
i
are clearly n(0
,σ
2
/n
i
). For
K
= 2 we have
S
2
2
=
n
1
(
U
1

¯
U
)
2
+
n
2
(
U
2

¯
U
)
2
=
n
1
±
U
1

n
1
¯
U
1
+
n
2
¯
U
2
n
1
+
n
2
²
2
+
n
2
±
U
2

n
1
¯
U
1
+
n
2
¯
U
2
n
1
+
n
2
²
2
=
(
U
1

U
2
)
2
"
n
1
±
n
2
n
1
+
n
2
²
2
+
n
2
±
n
1
n
1
+
n
2
²
2
#
=
(
U
1

U
2
)
2
1
n
1
+
1
n
2
.
Since
U
1

U
2
∼
n(0
,σ
2
(1
/n
1
+ 1
/n
2
)),
S
2
2
/σ
2
∼
χ
2
1
. Let
¯
U
k
be the weighted mean of
k U
i
s,
and note that
¯
U
k
+1
=
¯
U
k
+
n
k
+1
N
k
+1
(
U
k
+1

¯
U
k
)
,
where
N
k
=
∑
k
j
=1
n
j
. Then
S
2
k
+1
=
k
+1
X
i
=1
n
i
(
U
i

¯
U
k
+1
)
2
=
k
+1
X
i
=1
n
i
³
(
U
i

¯
U
k
)

n
k
+1
N
k
+1
(
U
k
+1

¯
U
k
)
´
2
=
S
2
k
+
n
k
+1
N
k
N
k
+1
(
U
k
+1

¯
U
k
)
2
,
where we have expanded the square, noted that the crossterm (summed up to
k
) is zero, and
did a boatload of algebra. Now since
U
k
+1

¯
U
k
∼
n(0
,σ
2
(1
/n
k
+1
+ 1
/N
k
)) = n(0
,σ
2
(
N
k
+1
/n
k
+1
N
k
))
,
independent of
S
2
k
, the rest of the argument is the same as in the proof of Theorem 5.3.1(c).
11.8 Under the oneway ANOVA assumptions,
Y
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 Spring '12
 Dr.Hackney
 Statistics, Trigraph, UK

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