1
TwoWay Analysis of Variance  no interaction
Example:
Tests were conducted to assess the effects of two factors, engine type, and propellant type, on
propellant burn rate in fired missiles. Three engine types and four propellant types were tested.
Twentyfour missiles were selected from a large production batch. The missiles were randomly split into three
groups of size eight. The first group of eight had engine type 1 installed, the second group had engine type 2,
and the third group received engine type 3.‘
Each group of eight was randomly divided into four groups of two. The first such group was assigned propellant
type 1, the second group was assigned propellant type 2, and so on.
Data on burn rate were collected, as follows:
Engine
Propellant Type
type
1
2
3
4
1
34.0
30.1
29.8
29.0
32.7
32.8
26.7
28.9
2
32.0
30.2
28.7
27.6
33.2
29.8
28.1
27.8
3
28.4
27.3
29.7
28.8
29.3
28.9
27.3
29.1
We want to determine whether either
factor
, engine type (factor A) or propellant type (factor B), has a
significant effect on burn rate.
Let
Y
ijk
denote the k’th observation at the i’th level of factor A and the j’th level of factor B.
The two factor model (without interaction) is:
Y
ijk
=
μ
+
α
i
+
β
j
+
ijk
i
= 1
,
2
,
3,
j
= 1
,
2
,
3
,
4,
k
= 1
,
2, where
1.
μ
is the overall mean
2.
∑
i
α
i
= 0
3.
∑
j
β
j
= 0
4. we assume
ijk
are iid
N
(0
, σ
2
)
5. The mean of
Y
ijk
is
μ
ijk
=
E
[
Y
ijk
] =
μ
+
α
i
+
β
j
This model specifies that a plot of the mean against the levels of factor A consists of parallel lines for each
different level of factor B, and a plot of the mean against the levels of factor B consists of parallel lines
for each different level of factor A.
More generally, there will be
I
levels of factor A,
J
levels of factor B, and
K
replicates at each combination
of levels of factors A and B.
Y
ijk
=
μ
+
α
i
+
β
j
+
ijk
i
= 1
,
2
, . . . , I
,
j
= 1
,
2
, . . . , J
,
k
= 1
,
2
, . . . , K
.
•
Here,
I
= 3,
J
= 4,
K
= 2, and there are
n
= 24 observations in total. There are
K
= 2
replicates
at each level the factors A and B, and the experimental design is said to be
balanced
, because there
are the same number of replicates in each cell.
The following table gives the cell means:
¯
Y
ij.
Engine
Propellant Type
type
1
2
3
4
1
33.35
31.45
28.25
28.95
2
32.60
30.00
28.40
27.70
3
28.85
28.10
28.50
28.95
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2
•
Estimated grand mean: ˆ
μ
= ¯
y
...
= 29
.
5917
•
Estimated factor A level means:
¯
y
1
..
= (33
.
35 + 31
.
45 + 28
.
25 + 29
.
95)
/
4 = 30
.
50
¯
y
2
..
= (32
.
6 + 30 + 28
.
4 + 27
.
7)
/
4 = 29
.
675
¯
y
3
..
= (28
.
85 + 28
.
1 + 28
.
5 + 28
.
95)
/
4 = 28
.
60
•
Estimated factor B level means:
¯
y
.
1
.
= (33
.
35 + 32
.
6 + 28
.
85)
/
3 = 31
.
60
¯
y
.
2
.
= (31
.
45 + 30 + 28
.
1)
/
3 = 29
.
85
¯
y
.
3
.
= (28
.
25 + 28
.
4 + 28
.
5)
/
3 = 28
.
383
¯
y
.
4
.
= (28
.
95 + 27
.
7 + 28
.
95)
/
3 = 28
.
533
Estimation of Model Parameters
•
ˆ
μ
= ¯
y
...
= (30
.
5 + 29
.
675 + 28
.
6)
/
3 = (31
.
6 + 29
.
85 + 28
.
383 + 28
.
533)
/
4 = 29
.
5917
•
ˆ
α
i
= ¯
y
i..

¯
y
...
ˆ
α
1
= ¯
y
1
..

¯
y
...
= 30
.
50

29
.
5917 =
.
908
ˆ
α
2
= ¯
y
2
..

¯
y
...
= 29
.
675

29
.
5917 =
.
083
ˆ
α
3
= ¯
y
3
..

¯
y
...
= 28
.
60

29
.
5917 =

.
992
Note that ˆ
α
1
+ ˆ
α
2
+ ˆ
α
3
= 0.
•
ˆ
β
j
= ¯
y
.j.

¯
y
...
ˆ
β
1
= ¯
y
.
1
.

¯
y
...
= 31
.
60

29
.
5917 = 2
.
0083
ˆ
β
2
= ¯
y
.
2
.

¯
y
...
= 29
.
85

29
.
5917 =
.
2583
ˆ
β
3
= ¯
y
.
3
.
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 Spring '12
 daniel
 Regression Analysis, engine type, B. Yijk

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