1. Analysis variance (ANOVA).
ANOVA compares 2 or more normally distributed random variables. Their variances are the same, therefore ANOVA compares 2 or more their (theoretical) means.
If we want to compare 2 means, we can use t-test or ANOVA. The results
(probability levels) are the same.
We want to compare weights of the clumps for 4 potato varieties:
1-st variety: 0.9, 0.8, 0.6, 0.9
2-nd variety: 1.3, 1, 1.3,1.2
3-th variety: 1.3, 1.5, 1.6, 1.1
4-th variety: 1.1, 1.2, 1.1,1
Are there differences among means of varieties?
The first calculations:
The sample mean for weights of varieties: 𝑋 𝑗 = 1 𝑛 𝑖=1 𝑛 𝑋 𝑖𝑗 , 𝑗=1, …, 𝑎
(a = 4 varieties, n = 4 measurements)
𝑋 1 =0.8, 𝑋 2 =1.2, 𝑋 3 =1.375, 𝑋 4 =1.1. the grandmean 𝑋 = 1 𝑎 𝑋 𝑖 = 𝑋 =

2. 𝛼 𝑖 − 𝑒𝑟𝑟𝑜𝑟 𝑖𝑗 = α 𝑖 − 𝑒𝑟𝑟𝑜𝑟 𝑖𝑗 = α 𝑖 .
The model of analysis of variance: Xij = m + ai + error ij , where errorij is of N(0,  2 ).
𝑋 is estimation of the common value 𝜇 and we will calculate 𝑋 𝑖𝑗 − 𝑋 = ai + error ij
𝛼 𝑖 − 𝑒𝑟𝑟𝑜𝑟 𝑖𝑗 = α 𝑖 − 𝑒𝑟𝑟𝑜𝑟 𝑖𝑗 = α 𝑖 .
ANOVA is able to detected the shift of columns by α 𝑖 in positive or negative direction from common value μ
Tests in ANOVA:
H0: 𝜇 1 = 𝜇 2 = 𝜇 3 = 𝜇 4 (H0: 𝜇+a1 = 𝜇+ α 2 = 𝜇+ α 3 = 𝜇+ α 4 )
H1: at least one equality in H0 is not valid.
Equivalently
H0: 𝛼 1 = 𝛼 2 = 𝛼 3 = 𝛼 4 H1: at least one equality in H0 is not valid.

3. Assumptions of ANOVA:
normal distribution – is not tested
homogenity of variances – the base assumption – is tested
2 estimations of variances 𝜎 2 :
Variability between columns (derived from S.E.): 𝑆 1 2 =𝑛 𝑆.𝐸. 2 = 𝑛 𝑎 𝑖=1 𝑎 ( 𝑋 𝑖 − 𝑋 ) 2 , 𝑆 1 2 ≈ 𝜎 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 ,
Variability within the dataset:
𝑆 2 2 = 1 𝑎𝑛 𝑖=1 𝑎 𝑗=1 𝑛 ( 𝑋 𝑖𝑗 − 𝑋 𝑖 ) 2 , 𝑆 2 2 ≈ 𝜎 𝑤𝑖𝑡ℎ𝑖𝑛 2 .
It is possible to derive that 𝜎 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 𝜎 𝑤𝑖𝑡ℎ𝑖𝑛 2 = 𝜎 2 + 𝑛 𝑎−1 𝑖=1 𝑎 𝛼 𝑖 2 𝜎 2 ≡ 𝜎 2 + 𝜑 2 𝜎 2
If H0 is valid, then α 𝑖 =0 for all i and then 𝜎 2 + 𝜑 2 𝜎 2 = 𝜎 2 𝜎 2 = 1.
Therefore, the method is called analysis of variance (ANOVA).
Estimation of 𝜎 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 is 𝑆 1 2 , estimation of 𝜎 𝑤𝑖𝑡ℎ𝑖𝑛 2 is 𝑆 𝑆 1 2 and 𝑆 2 2 are random variables and
𝑆 𝑆 ≈𝐹(𝑎−1, 𝑎 𝑛−1 )

4. Note.
If we have different numbers of measurements in columns (in our example we have
𝑛 1 = 𝑛 2 = 𝑛 3 = 𝑛 4 =4), the the degrees of freedom are (𝑎−1, 𝑛 𝑖 −𝑎 ).
If a = 2, we can use t- test or ANOVA. The probability levels are the same, although the test’s characteristics (t, F) are different.
It is not possible to use the series of t-tests for a > 2  the error of the first type (α) is increasing.

5. In our example
P < 0.05, thereore we reject H0, therefore some of equalities in H0 are not valid.
Means 95% conf. interval
The varieties 1 a 2, 1 a 3 can be different.

6. Post – hoc tests:
H0: mean of the column i = mean of the column j, i ≠ j
H1: inequality between means of columns i and j.
Post-hoc test is answer to the question, WHERE is the difference that was detected by ANOVA.
We can see the differences between varieties 1and 2
(averages 0.8 and 1.2, P = < 0.05) and 1 and 3
(averages 0.8 a 1.375, P = < 0.05).
In our example was 1 factor “variety“  single - factor ANOVA

7. Multifactor analysis of variance.
A. Factorial design.
Example:
The rats were fed with fresh or rancid fat during 73 days.
We are interested in the consumption of the fat of various quality depending on the
sex.
fresh
rancid
sum
males
709
592
679
638
699
476
2087
1706
3793
females
657
508
594
505
677
539
1928
1552
3480
4015
3258
Males eat more than females
Fresh fat is more attractive than rancid fat.
We have two criteria:
sex: we have 2 normally distributed random variables (males, females) fat: we have 2 normally distributed random variables (rancid, fresh).
We assume homogenity of all variances
We will test the difference between two means for sex and We will test the difference between two means for fat

8. Therefore we have 2 null hypothesis:
H01: there is no difference between means of males and females
H02: there is no difference between means of consumption of fresh and rancid fat.
Model:
𝑋 𝑖𝑗 = μ+ 𝑠𝑒𝑥 𝑖 + 𝑓𝑎𝑡 𝑗 + 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑗 + 𝑒𝑟𝑟𝑜𝑟 𝑖𝑗
The lines are approximately parallel,
There are no relations between factors.
The factors are independent  we can detected shift of factor levels.
Therefore 3th null hypothesis:
H03: there are no significant interactions.
Here the interactions are not significant.
fresh
rancid
females
males
males

9. H01 we don’t reject  there is no significant difference between males and females.
H02 we reject  there is difference in consumption of the fresh and rancid fat.
H03 we don’t reject  there are no significant differences.
Nonsignificant interactions  model is OK, it is possible to use Anova.
Significant interactions  Something else (another factor) can influence
the measurements  must be justified.
Post-hoc test
For fat only: the fresh fat is consumed more than rancid.

10. The lines are not parallel, There are some relations between factors.
Modification.
fresh
rancid
sum
males
592
709
538
679
476
699
1606
2087
3693
females
657
508
594
505
677
539
1928
1552
3480
3534
3639
The lines are not parallel,
There are some relations between factors.
The factors are dependent  we cannot detected shift of factor levels.
ANOVA can not be used
rancid
fresh
H03 is rejected  interactions are significant
females
males

11. B. Hierarchic Anova (Nested design).
We have 2 containers:
To the first one is placed fresh fat, to the second container is placed rancid fat.
To the first container we placed 6 rats randomly chosen, to the second container are
placed 6 rats randomly chosen.
1-st container
2-nd container
6 rats
6 rats
2 factors:
rat - random factor
fat - fixed factor
We are in the level “individuals“ – not in the level “sex“, we talk about the differences
among individuals
The factor “rat“ is nested in the factor “fat“
This arrangement can be viewed as one way ANOVA and the rats are taken as
repetition.

12. It was measured the water temperature in depths 0, 2 a 5 metres.
ANOVA – random blocks.
Example:
It was measured the water temperature in depths 0, 2 a 5 metres.
Are there differences among temperatures in different depths?
1-st place
2-nd place
3-th place
10-th place
0 m
0 m
0 m
0 m ….
2 m
2 m
2 m
2 m
5 m
5 m
5 m
5 m
random blocks

13. Each depth in given place is measured only ones.
the temperature in 10 places
blok
0 m
2 m
5 m 1
20.8
18.5
11.1 2
21.3 20 12 3
19.9 18
9.5 4
10.2 5
20.1 19
10.1 6
19.8
9.9 7
19.5
8.5 8
19.3
8.8 9 10
This is actually a 1-factor Anova with
the factor “depth“.
1-factor ANOVA:
F(2, 27) = , P = 0
2-factors ANOVA:
F(2, 18) = , P = 0

14. Violation of assumptions
Homogenity of variances.
If this assumption is not valid, something else enters into model, some factor that was not be included into design of experiment.
Violation of homogenity of variances means “false“ conclusions in the sense rejection
or non rejection of null hypotheses.
In practice it means the new experiment.
Aditivity of the model.
It means the significant interactions. It can be non-homogenity of conditions of experiment, or it can be not including
all substantial factors of experiments.
It follows the low sensitivity of method.
In practice the interactions are often significant. If significance of interactions is
high (P0), the new experiment is necessary.

15. Normality.
Like the t-tests, we can take the same data transformation to normality, or we can use nonparametric equivalent of Anova.
Sensitivity of nonparametric tests is low and we often use the transformation to
N (0,1).
nonparametric equivalent of one faktor Anova is Kruskal-Wallis test,
nonparametric equivalent of two factors Anova is Friedmann test (block design).
Design of experiment should be as simple as possible. The number of factors should not be greater than two, as the explanation of possible
significant interactions is difficult.