252anovaex4
10/25/07 (Open this document in 'page layout' view.)
Example of 3-way ANOVA
We have measurements that describe the mpg gotten by 8 drivers (Factor 1), using 6 cars (Factor
2) during 4 seasons (Factor C). For each combination of factors there are 10 measurements. The SS1 is 700,
SS2 is 400, and SS3 is 300. For the interactions SS12 = 50, SS13 = 60 and SS23=70. The within (error or
residual) sum of squares is 100 and the total sums of squares is 1700. Set up an ANOVA table showing all
interactions. Assume a 5% significance level.
Solution: The sums of squares have been copied into the table below. If we multiply the levels of the
factors together and then multiply by the number of measurements per cell, we find a total of
8  6  4 10  1920 measurements. There are a total of 1920 – 1 = 1919 degrees of freedom. Since drivers
has 8 levels there are 7 degrees of freedom. Similarly for cars there are 6 – 1 = 5 degrees of freedom and
for seasons there are 3 degrees of freedom. These are also shown below.
Source
SS
DF
MS
F.05
F
Driver (1)
700
7
Car (2)
400
5
Season (3)
300
3
Interaction (12)
50
Interaction (13)
60
Interaction (23)
70
Interaction (123)
Within
100
.
Total
1700
1919
Since the sums of squares must add up, to get the missing sum of squares, subtract the sum of the numbers
above the line (1680) from the total sum of squares (1700) and get 20. To get the degrees of freedom for
interactions multiply the degrees of freedom for the factors. For example DF12  DF1 DF 2  7  5  35.
This is shown below.
Source
Driver (1)
Car (2)
Season (3)
Interaction (12)
Interaction (13)
Interaction (23)
Interaction (123)
Within
Total
SS
DF
700
400
300
50
60
70
20
100
1700
7
5
3
35
21
15
105
.
1919
MS
F.05
F
Since the degrees of freedom must add up, add the degrees of freedom above the line to get 191 and
subtract from the total degrees of freedom to get 1919 – 191 = 1728. To get the mean squared column,
divide the items in the SS column by the corresponding items in the MS column.
Source
Driver (1)
Car (2)
Season (3)
Interaction (12)
Interaction (13)
Interaction (23)
Interaction (123)
Within
Total
SS
700
400
300
50
60
70
20
100
1700
DF
MS
7
5
3
35
21
15
105
1728
1919
100.00
80.00
100.00
1.4286
2.8571
4.6667
0.1905
0.05787
F
F.05
To get the items in the column for computed Fs, divide the mean squares by the within mean square.
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252anovaex4
10/25/07 (Open this document in 'page layout' view.)
Source
Driver (1)
Car (2)
Season (3)
Interaction (12)
Interaction (13)
Interaction (23)
Interaction (123)
Within
Total
SS
700
400
300
50
60
70
20
100
1700
DF
MS
F
7
5
3
35
21
15
105
1728
1919
100.00
80.00
100.00
1.4286
2.8571
4.6667
0.1905
0.05787
1728
1382
1728
24.68
49.37
80.64
3.29
F.05
Now look up values of F.05 on the F table. The degrees of freedom for the numerator are given in the DF
column. The degrees of freedom for the denominator are 1728.
MS
F
F.05
7
100.00
1728

F 7,1728
.05
400
5
80.00
1382
300
3
100.00
1728
Interaction (12)
50
35
1.4286
24.68
Interaction (13)
60
21
2.8571
49.37
Interaction (23)
70
15
4.6667
80.64
Interaction (123)
20
105
0.1905
3.29
Source
SS
Driver (1)
700
Car (2)
Season (3)
DF

F 5,1728
.05

F 3,1728
.05

F 35,1728
.05

F 21,1728
.05

F 15,1728
.05

F 105,1728
.05
Within
100
1728 0.05787
Total
1700
1919
Look up the table values of F. I have used the parts of the table where the denominator degrees of freedom
are 1000 and have come as close as possible for the numerator. We have tested seven null hypotheses. The
first three of these are that there are no significant differences between driver, car and season means, and
the remaining four are that there is no interaction of the type described. In every case, the null hypothesis
was rejected because the computed F exceeded the table F, so the computed F was labeled with an ‘s.’
MS
F
F.05
7
100.00
1728s

F 7,1728
.05 =2.02
400
5
80.00
1382s
300
3
100.00
1728s
Interaction (12)
50
35
1.4286
24.68s
Interaction (13)
60
21
2.8571
49.37s
Interaction (23)
70
15
4.6667
80.64s
Interaction (123)
20
105
0.1905
3.29s
100
1700
1728
1919
0.05787
Source
SS
Driver (1)
700
Car (2)
Season (3)
Within
Total
DF

F 5,1728
.05 = 2.22

F 3,1728
.05 = 2.61

F 35,1728
.05 =1.44

F 21,1728
.05 = 1.58

F 15,1728
.05 = 1.70

F 105,1728
.05 = 1.26
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