STATS 203: HOMEWORK SOLUTION #2
THANKS TO ROHAN TANDON, RONGZHI LU
Question 1
We are given a oneway ANOVA model
Y
ij
=
μ
+
α
i
+
±
ij
,
±
ij
∼
N
(0
,σ
2
)
The mean sumoftreatmentsquares term is
MSTR
=
∑
r
i
=1
n
i
(
¯
Y
i
·

¯
Y
··
)
2
r

1
and we assume that each group
i
has
n
observations, i.e.
n
i
=
n
.
(a).
From the deﬁnition of a oneway ANOVA model, we see immediately that
¯
Y
i
·
=
1
n
n
X
j
=1
Y
ij
=
1
n
n
X
j
=1
(
μ
+
α
i
+
±
ij
) =
μ
+
α
i
+ ¯
±
i
·
¯
Y
··
=
1
r
r
X
i
=1
1
n
n
X
j
=1
Y
ij
=
1
r
r
X
i
=1
1
n
n
X
j
=1
(
μ
+
α
i
+
±
ij
)
=
1
r
r
X
i
=1
(
μ
+
α
i
+ ¯
±
i
·
)
where ¯
±
i
·
=
1
n
∑
n
j
=1
±
ij
. Then clearly, we see that
¯
Y
i
·

¯
Y
··
= (
μ
+
α
i
+ ¯
±
i
·
)

1
r
r
X
i
=1
(
μ
+
α
i
+ ¯
±
i
·
)
= (
μ
+
α
i
+ ¯
±
i
·
)

μ

1
r
r
X
i
=1
α
i
!
+
1
r
r
X
i
=1
1
n
n
X
j
=1
±
ij
= (
μ
+
α
i
+ ¯
±
i
·
)

μ

1
r
·
0 +
1
nr
r
X
i
=1
n
X
j
=1
±
ij
=
μ
+
α
i
+ ¯
±
i
·

μ
+ ¯
±
··
=
α
i
+ (¯
±
i
·

¯
±
··
)
where we have used the “identiﬁability” assumption of the oneway ANOVA model that
∑
r
i
=1
α
i
= 0.
±
Date
: Submitted on February 4, 2010.
1
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2
(b).
Using the result from part (a), we have that
¯
Y
i
·

¯
Y
··
=
α
i
+ (¯
±
i
·

¯
±
··
). Hence, it follows directly from the
quadratic expansion that
r
X
i
=1
(
¯
Y
i
·

¯
Y
··
)
2
=
r
X
i
=1
[
α
i
+ (¯
±
i
·

¯
±
··
)]
2
=
r
X
i
=1
h
α
2
i
+ 2
α
i
(¯
±
i
·

¯
±
··
) + (¯
±
i
·

¯
±
··
)
2
i
=
r
X
i
=1
α
2
i
+
r
X
i
=1
2
α
i
(¯
±
i
·

¯
±
··
) +
r
X
i
=1
(¯
±
i
·

¯
±
··
)
2
=
r
X
i
=1
α
2
i
+ 2
r
X
i
=1
α
i
(¯
±
i
·

¯
±
··
) +
r
X
i
=1
(¯
±
i
·

¯
±
··
)
2
which proves the desired result.
±
(c).
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 Null hypothesis, Oneway ANOVA model

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