Stat 372
R.J. MacKay, University of Waterloo, 2005
Chapter 7 Solutions1
Chapter 7
Exercise Solutions
1. Practice with the F distribution.
a)
Suppose
U
F
~
,
6 24
. Find c so that Pr(
)
.
U
c
≥
=
005.
From the tables, we have
c
=
251
.
b)
Estimate Pr(
)
U
≥
3 .
From the tables,
interpolating we have Pr(
)
.
U
≥
≈
3
003
c)
What is the distribution of
1/
U
?
Since
U
K
K
=
6
2
24
2
, we have 1
24 6
/
~
,
U
F
d)
Find d so that
Pr(
)
.
U
d
≤
=
005.
Using the result from c),
Pr(
)
Pr( /
/
)
U
d
U
d
≤
=
≥
1
1
and from the tables we get
1
385
/
.
d
≈
so
d
≈
026
.
e)
Show that if
W
t
k
~
then
W
2
has an F distribution. What are the degrees of freedom?
W
G
K
K
K
F
k
k
k
2
2
2
1
2
2
1
01
~
( , )
,
=
=
. Recall that
K
Z
Z
k
k
=
+ +
1
2
2
...
where
Z
G
i
~
( , )
0 1 .
2. In a small investigation, three treatments A,B,C were compared by assigning 6 units at random to
each treatment. The data are shown below.
A
B
C
11.82
15.46
15.43
13.03
12.90
14.28
10.78
14.88
14.76
14.31
13.75
12.07
14.21
18.59
13.80
8.56
13.80
13.46
average
12.12
14.90
13.97
st. dev.
2.22
2.02
1.16
a) Calculate the ANOVA table.
The treatment sum of squares
6
2
(
)
y
y
i
i
+
++

∑
is 12 times the square of the sample standard deviation
of the treatment averages. Hence the treatment sum of squares is 24.00.
The total sum of squares is 17
times the square of the sample standard deviation of the 18 response variate values or 75.77. We
calculate the residual sum of squares by subtraction. The completed table is shown below.
b) Is there any evidence of a difference among the treatments?
To test the hypothesis
t
t
t
1
2
3
0
=
=
=
, the discrepancy is 3.48 and the pvalue is
Pr(
.
)
.
,
F
2 15
348
0 057
≥
=
so there is weak evidence of a difference among the treatments.
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View Full DocumentStat 372
R.J. MacKay, University of Waterloo, 2005
Chapter 7 Solutions2
Source
Sum of
squares
Degrees of
freedom
Mean square
Ratio to
residual ms
Treatments
24.00
2
12.00
3.48
Residual
51.76
15
3.45
Total
75.765
17
c) Treatment A was a control. Is there any evidence that the average effect of treatments B and C
exceeds the effect of the control? [Use a onesided test here  why?]
From the ANOVA table, we have
$
.
s
=
186.
Consider the contrast
q
t
t
t
=
+

2
3
1
2
to compare the
average of treatments 2 and 3 to the control. We want a onesided test to see if there is evidence that
q
0 . The estimate of
q
is
$
.
q
=
+

=
+
+
+
y
y
y
2
2
1
2
2 315 and the corresponding estimator has standard
deviation
s
s
1
24
1
24
1
6
2
+
+
=
/
. To test the hypothesis, suppose that
q
=
0 . The (onesided)
discrepancy is
d
=

=
$
$
/
.49
q
s
0
2
2
and the pvalue is Pr(
.49)
.
t
15
2
0 013
≥
=
so there is strong evidence
against the hypothesis. There is strong evidence that the average treatment effect exceeds the effect of
the control.
d) Find a 95% confidence interval for the difference between the effects of B and C.
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 Spring '11
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 Regression Analysis, Standard Deviation, R.J. MacKay

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