Mind Expanding Exercises
1347.
)
1
(
)
(
11
2
−
−
=
∑∑
==
n
a
y
y
MS
a
i
n
j
i
ij
E
and
ij
i
ij
a
y
ε
μ
+
+
=
.
Then
.
i
ij
i
ij
y
y
−
=
−
and
1
)
(
1
.
−
−
∑
=
n
n
j
i
ij
εε
is recognized to be the sample variance of the independent random variables
in
i
i
,
,
,
2
1
…
.
Therefore,
2
1
2
.
1
)
(
σ
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
∑
=
n
E
n
j
i
ij
and
2
1
2
)
(
=
=
∑
=
a
i
E
a
MS
E
.
The development would not change if the random effects model had been specified because
.
i
ij
i
ij
y
y
−
=
−
for this model also.
1348.
The two sample ttest rejects equality of means if the statistic
t
y
y
s
y
y
s
p
nn
p
n
=
−
+
=
−

is too large. The ANOVA Ftest rejects equality of means if
12
2
E
2
..
.
i
2
1
i
MS
)
y
y
(
n
F
−
=
∑
=
is too large.
Now,
n
2
E
2
.
2
.
1
E
2
.
2
.
1
2
n
MS
)
y
y
(
MS
)
y
y
(
F
−
=
−
=
and
.
2
p
E
s
MS
=
Consequently,
. Also, the distribution of the square of a t random variable with a(n  1) degrees of
freedom is an F distribution with 1 and a(n  1) degrees of freedom. Therefore, if the tabulated t value for a
twosided ttest of size
, then the tabulated F value for the F test above is
t
Therefore, t >
whenever
F =
and the two tests are identical.
2
t
F
=
0
t
is
α
0
2
.
0
t
2
0
2
t
t
>
1349.
)
1
(
2
)
(
2
.
1
2
1
−
−
=
∑
∑
=
=
n
y
y
MS
i
ij
n
j
i
E
and
1
)
(
2
.
1
−
−
∑
=
n
y
y
i
ij
n
j
is recognized as the sample standard deviation
calculated from the data from population i. Then,
2
2
2
2
1
s
s
MS
E
+
=
which is the pooled variance estimate used
in the ttest.
1350.
from the independence of
YY
)
(
)
(
.
2
1
.
1
i
i
a
i
i
i
a
i
Y
V
c
Y
c
V
∑
∑
=
=
=
Y
a
..
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 Spring '09
 KAILASHKAPUR
 Normal Distribution, Variance, Chisquare distribution, yi.

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