|
See ANOVA
A statistical technique that partitions the variation in a dataset into the variation caused by the each of the factors, and the common cause variation. The purpose is to find out how much of the variation can be explained by each factor, and the statistical significance of each component of variation. See One Way ANOVA and Two Way ANOVA for examples.
The ANOVA method requires the process conforms to a normal distribution. For nonparametric tests:
Kruskal Wallace Test
Mood's Median Test
Friedman's Test (equivalent to two factor ANOVA)
The F Test is used to compare the variances of two samples to test the hypothesis that they may come from the same distribution. The hypothesis is:
H0 the samples are drawn from populations with equal variances
H1 the samples are drawn from populations with different variances
The F test is based on the ratio of the variances of two samples. It depends on the number of degrees of freedom of both samples:
| |
1
|
2
|
3
|
4
|
5
|
Variance |
|
A
|
8 .7
|
9.8
|
12.2
|
12.9
|
9.5
|
3.327
|
|
B
|
23.8
|
21.5
|
18.3
|
24.3
|
22.9
|
5.788
|
This gives an F statistic:
This can be compared to the critical F value from tables, or the p-value can be obtained using the Excel function:
=FDIST(Fo, DOFn, DOFd)
| Hypothesis Test for Variance |
|
See F Test
One Way ANOVA is used to determine whether varying levels or values of a single factor affect the process. The following data shows the strength of a fibre with varying percentages of a synthetic material:
| |
1
|
2
|
3
|
4
|
5
|
|
15%
|
8 .7
|
9.8
|
12.2
|
12.9
|
9.5
|
|
20%
|
14.5
|
17.3
|
14.9
|
15.5
|
12.7
|
|
25%
|
16.0
|
13.8
|
18.3
|
17.2
|
18.0
|
|
30%
|
23.8
|
21.5
|
18.3
|
24.3
|
22.9
|
|
35%
|
9.0
|
14.3
|
10.0
|
14.3
|
7.1
|
An ANOVA analysis tests the hypothesis:
H0 The fibres are all of equal strength
H1 At least one of the fibres is of different strength
Use a level of significance of 0.05.
The analysis is:
|
Source of Variation
|
Sum of Squares
|
Degrees of Freedom
|
Mean Square
|
F0
|
p-value |
|
Factor
|
448.32
|
4
|
112.08
|
21.78
|
0
|
|
Error
|
102.91
|
20
|
5.15
|
|
|
|
Total
|
551.23
|
24
|
|
|
|
The alternative hypothesis is accepted.
See One Way ANOVA
In the ANOVA example the five replications within each row were all taken under the same test conditions. Consider the example where three treatments are evaluated on four different patients:
|
Therapy
|
Andrew
|
Belinda
|
Chris
|
Dave
|
| Relaxed |
110
|
140
|
100
|
130
|
| Normal |
115
|
150
|
105
|
135
|
| High Intensity |
117
|
155
|
100
|
135
|
The hypothesis is:
H0 The therapies all give the same result
H1 At least one of the therapies gives a different response
Use a level of significance of 0.05.
The patients are all different, and a 'One Way ANOVA' would not cause the null hypothesis to be rejected. However a two way ANOVA separates the variation due to the therapy from that due to different patient characteristics:
|
Source of Variation
|
Sum of Squares
|
Degrees of Freedom
|
Mean Square
|
F0
|
p-value |
|
Therapy
|
113.17
|
2
|
56.58
|
5.40
|
0.046
|
|
Patient
|
3832.67
|
3
|
1277.56
|
121.99
|
0
|
|
Error
|
62.83
|
6
|
10.47
|
|
|
|
Total
|
4008.67
|
11
|
|
|
|
The null hypothesis is rejected, the therapies give different results at the 0.05 level of significance.
|
