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
..
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 KAILASHKAPUR
 Normal Distribution, Variance, Chisquare distribution, yi.

Click to edit the document details