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STAT 514 Design of Experiments
Homework 4 Solution
Problem 1
(a) SAS generates the following ANOVA table.
Sum of
Source
DF
Squares
Mean Square
F Value
Pr > F
Model
3
6026.83333
2008.94444
6.97
0.0022
Error
20
5767.00000
288.35000
Corrected Total
23
11793.83333
The
p
value for the
F
statistic is 0
.
0022, much less than 0.05. Hence we reject the null hypothesis
and conclude that there exists diﬀerence between the treatments.
(b) Since the experiment is a balanced design, two contrasts are orthogonal to each other if and
only their inner product is 0. Let Γ
1
,Γ
2
and Γ
3
be the contrasts of “Hormone I vs Hormone II”, “Low
Level vs High Level” and “Equivalence of Level”, respectively. Then
Γ
t
1
Γ
2
=1
*
1+1
*
(

1) + (

1)
*
1+(

1)
*
(

1) = 1

1

1+1=0
,
Γ
t
1
Γ
3
*
*
(

1) + (

1)
*
(

1) + (

1)
*
1=1


1=0
,
Γ
t
2
Γ
3
*

1)
*
(

1) + 1
*
(

1) + (

1)
*
1=1+1

1

.
Hence the three contrasts are orthogonal to each other.
(c) The SAS output for contrast sums of squares and contrasts testing is as follows.
Contrast
DF
Contrast SS
Mean Square
F Value
Pr > F
C1
1
864.000000
864.000000
3.00
0.0989
C2
1
5162.666667
5162.666667
17.90
0.0004
C3
1
0.166667
0.166667
0.00
0.9811
Based the table above, Contrast Γ
1
(or
C
1
) is not signiﬁcant because the
p
value 0.0989 is greater
than
α
. Therefore, the average eﬀect of hormone I and the average eﬀect of hormone II on are not
diﬀerent from each other; Contrast Γ
2
(or
C
2
) is signiﬁcant (
p
value = 0
.
0004), this suggests that the
average eﬀect of the high levels of hormones and the average eﬀect of the low levels of hormones are
diﬀerent. Contrast Γ (or
C
3
) is not signiﬁcant at all (
p
value = 0
.
9811), so the diﬀerence between
the highlevel and lowlevel of hormone I is the same as that between the highlevel and lowlevel of
hormone II.
Problem 2
(a) The ANOVA table is given below.
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 Fall '08
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