1
Solution to Homework #2 Stat c151/c225 Winter 2011
3.7
(a)
All entries can be determined as below.
Source
Degrees of Freedom
Sum of Squares
Mean Squares
Block
2
520
260
Treatment
4
498
124.5
Residual
8
40
5
total
14
(b) The
t
statistics for Tukey method are calculated below (using (2.11) with
ˆ
σ
=
5
,
b
=
3
)
Group
A(45)
B(58)
C(46)
D(45)
E(56)
A(45)
7.12
0.55
0
6.02
B(58)
6.57
7.12
1.10
C(46)
0.55
5.48
D(45)
6.02
E(56)
68
.
4
414
.
1
62
.
6
2
1
2
1
01
.
0
,
8
,
5
),
1
)(
1
(
,
=
=
=
−
−
q
q
k
b
k
α
. By comparing the
t
values with 4.68, Tukey
method declares that the following pairs are significantly different: A&B, A&E, B&C, B&D,
C&E and D&E.
(c) From (b), Tukey method declares 6 pairs of treatments to be significantly different at level 0.01.
The null hypothesis of the
F
test is that all the treatments are same to each other. Based on the
conclusion of (b), we know the null hypothesis of
F
test is not true at level 0.01. Hence the
F
test
will reject the null hypothesis at level 0.01.
3.8
(a)
This is a paired comparison design because a pair of catalysts is compared in each of the six
batches. Each batch can be considered as a block of size 2. In this design, the two catalysts are
compared on two portions of the same batch, this eliminates a source of variation due to the
difference among different batches, which were known to be quite different from one another.
(b) Use paired ttest,
1
2
3
4
5
6
A
9
19
28
22
18
8
B
10
22
30
21
23
12
d
i
1
3
2
1
5
4
33
.
2
6
14
1
1
=
=
=
∑
=
N
i
i
d
N
d
,
=
−
−
=
∑
=
2
/
1
1
2
)
(
1
1
N
i
i
d
d
d
N
s
2.16.
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