Fundamentals of Data Analysis

Lecture 8ANOVA

pt.2

Multifactors design

With that issue we are dealing eg. in the case of alloy hardness test, which consists of two metals A and B, and their contents in the alloy determines the hardness.

We therefore divide our observations into r classes due to the characteristics of the value of A and p Classes due to the characteristics of the value of B. All observations are therefore divided into rp groups.

On the example of two factors design

Multifactors designOn the example of two factors design

For this model, we verify the hypothesis :

1. of equality of mean values for all rp populations:

H0: mij = m dla i = 1, ..., r; j = 1, ..., p.

2.of equality of all the average values mi of studied features

treated of A with r variants, excluding the impact of factor B:

H0: m1. =... = mr. dla i = 1, ..., r.

3. of equality of all the average values mi of studied features

treated of B with p variants, excluding the impact of factor A:

H0: m.1 =... = m.p dla j = 1, ..., p.

4. that the deviation of the mean value mij from the total value

of the average m is equal to the sum of effects of factor A and

factor B:

H0: mij - m = (mi. - m) + (m.j - m).

Two factors design

From three different departments of a university were drawn at l = 4 students from each year of study, and calculated the mean ratings obtained by each student in the last semester. The obtained results are shown in Table :

Example

Year Department

A B C

I 3.6, 4.1, 3.1, 2.4 3.1, 2.5, 3.3, 3.8 2.7, 4.2, 2.9, 3.7

II 2.8, 4.3, 3.8, 3.0 3.9, 2.6, 3.2, 3.3 3.0, 4.4, 3.9, 3.1

III 3.2, 4.1, 4.8, 4.0 3.4, 2.9, 4.1, 2.8 4.0, 3.3, 3.4, 3.0

IV 3.2, 3.9, 4.2, 3.6 3.6, 4.4, 2.8, 2.9 3.7, 5.0, 2.6, 3.4

V 4.0, 4.0, 3.5, 3.8 4.0, 3.0, 4.5, 3.7 3.0, 3.8, 4.8, 3.5

Two factors design

Example

Year Department

A B C

I 3.6, 4.1, 3.1, 2.4 3.1, 2.5, 3.3, 3.8 2.7, 4.2, 2.9, 3.7

II 2.8, 4.3, 3.8, 3.0 3.9, 2.6, 3.2, 3.3 3.0, 4.4, 3.9, 3.1

III 3.2, 4.1, 4.8, 4.0 3.4, 2.9, 4.1, 2.8 4.0, 3.3, 3.4, 3.0

IV 3.2, 3.9, 4.2, 3.6 3.6, 4.4, 2.8, 2.9 3.7, 5.0, 2.6, 3.4

V 4.0, 4.0, 3.5, 3.8 4.0, 3.0, 4.5, 3.7 3.0, 3.8, 4.8, 3.5

Assuming that the average grades obtained by the students have a normal distributions with the same variance at the confidence level α= 95% verify the following hypotheses:

a) the average values of average grades for students of different departments are the same;

b) the average value of average grades for differernt years of study are the same;

c) the average value of the average grades for the first two years are the same.

Two factors design

In that case we have r = 3 (Departments) and p = 5 (number of years of study). After calculations we get the results shown in the table :

Example

p r Sum

1 2 3

1 3.300 3.175 3.375 3.283

2 3.475 3.259 3.600 3.342

3 4.025 3.300 3.425 3.583

4 3.725 3.675 3.674 3.692

5 3.825 3.800 3.775 3.800

Sum 3.670 3.440 3.568 3.550

Two factors design

Example

Then compute the sum of squared deviations:

qA = 0.5365, for df = 2, then qA /df = 0.26815 qB = 2.3797 df = 4 qB /df = 0.59492 qAB = 0.06980 df = 8 qAB /df = 0.00872 qR = 18.4050 df = 45 qR /df = 0.4090 q = 21.3908 df = 59

F-statistics calculated on this basis, have the following values: FA = 0.26815 / 0.4090 = 0.6556 FB = 0.59492 / 0.4090 = 1.4546

For the third hipothesis we must calculate new mean value:x = 3.4 and qC = 0.24 (df = 1)qR = 7.95 (df = 22)FC = 0.2400/0.3614 = 0.6641

Two factors design

Example

Critical values:

FAcr = 3.20 > FA

There is no reason to reject the null hypothesis

FBcr = 2.58 > FB

There is no reason to reject the null hypothesis

FCcr = 4.30 > FC

There is no reason to reject the null hypothesis

Two factors design

Exercise

Each of the three varieties of potatoes was cultivated on 12 parcels of the same size and type. Parcels were divided into four groups of three parcels, and for each group a different type of fertilizer was used. Yields for these plots are shown in Table

At the confidence level 95% verify the following hypotheses:A) The values of the average yield for the different

varieties of potatoes are dependent on the applied fertilizer

B) The values of the average yields for the different fertilizers do not differ regardless of potato variety

Potato variety

Type of fertilizer

1 2 3 4

123

5.6, 6.1, 5,95.7, 4.9, 5.16.3, 6.1, 6.3

6.6, 6.7, 6.66.5, 6.7, 6.66.5, 6.4, 6.2

7.7, 7.3, 7.46.9, 7.1, 6.56.6, 6.6, 6.8

6.3, 6.4, 6.36.6, 6.7, 6.76.1, 6.1, 6.0

Latin square design

During Latin sQuare design (LQ) experimental items are classified in terms of the classification of three directions: rows, columns, and objects. Experimental factor A presented at p levels (contains p objects), each of which occurs exactly once in the corresponding row and column.

Latin square design

1. We calculate the correction factor:

2. We calculate the sum of squares for rows

x

rp

ijkijk

2

x

rp

ijkijk

2

rp

x

Cijk

ijk

2

Cp

xSSR i

i

2

..

Latin square design

and the mean square for rows

3. We calculate the sum of squares for columns

rowdf

SSRMSR

Cr

x

SSC jj

2

..

Latin square design

and the mean square for columns

4. calculate the sum of squares for the factors

coldf

SSCMSC

Cr

xSST k

k

2

..

Latin square design

and the mean square for factors

5. calculate the total sum of squares

tdf

SSTMST

CxSSijk

ijk 2

Latin square design

6. and the residual sum of squares

SSE = SS - SSR - SSC - SST

7. mean square residual

errordf

SSEMSE

Latin square design

8. and calculate statistics

FMSR

MSErow

FMSC

MSEcol

FMST

MSEtr

Latin square design

Example In the experiment, on the fertilization of fields used are the following factors (fertilizers): A - (NH4 )2 SO4 , B - NH4NO3, C - CO(NH2)2, D - Ca(NO3)2, E - NaNO3, F - NoN (non-fertilized). Fertilizers are used in equal doses (in g/m2).

In the first stage the draw was performed of suitable Latin square 6x6 (since we have 6 factors) and the result is shown in table:

4 2 5 1 3 6

1 B D E F A C

3 C E A D F B

5 A F C B E D

4 D A F C B E

2 F B D E C A

6 E C B A D F

Latin square design

ExampleThe results of the experiment as planned (achieved yields of sugar beet) are presented in Table

I II III IV V VI xijj

I F

28.2

D

29.1

A

32.1

B

33.1

E

31.1

C

32.4

186.0

II E

31.0

B

29.5

C

29.4

F

24.8

D

33.0

A

30.6

178.3

III D

30.6

E

28.8

F

21.7

C

30.8

A

31.9

B

30.1

173.9

IV C

33.1

A

30.4

B

28.8

D

31.4

F

26.7

E

31.9

182.3

V B

29.9

F

25.8

E

30.3

A

30.3

C

33.5

D

32.3

182.1

VI A

30.8

C

29.7

D

27.4

E

29.1

B

30.7

F

21.4

169.1

xiji

183.6 173.3 169.7 179.5 186.9 178.7 x

ij

1071 7.

Latin square design

Example

Harvests for different fertilizers are presented in Table.

Factor A B C D E F

x k

k

186.1 182.1 188.9 183.8 182.2 148.6

x k 31.0 30.4 31.5 30.6 30.4 24.8

Latin square design

Example

Degree of freedom:

total dftot = pr - 1 = 35for rows dfrow = r - 1 = 5for columns dfcol = p - 1 = 5for factors dftr = n - 1 = 5for error dferror = (r-1)(p-1) - (n-1) = 35 - 5 - 5 - 5 = 20.

Latin square design

Example

91,31903

2

rp

x

Cijk

ijk

19,32

2..

Cp

x

SSR ii

438,6rowdf

SSRMSR

67,33

2..

Cr

x

SSC jj

734,6coldf

SSCMSC

Latin square design

Example

A further step in the analysis of our data is the separation of variables (averages). Based on the result of our experiment, we can answer a number of questions:

1) Whether fertilization can effect on crop growth (excreted factor F)?2) Is organic fertilizer better than inorganic?3) Is NH4-N better than NO3-N ?4) Is (NH4)2SO4 better than NH4NO3 ?5) Is Ca(NO3) better than NaNO3 ?

Such questions may of course be more, depending on the factors or groups of factors we want to compare.

Latin square design

Example

Factor NoN (NH4)2SO

4

NH4NO3 CO(NH2)2 CA(NO3)2 NaNO3

Comparison x..k 148.6 186.1 182.1 188.9 183.8 182.2

NoN - N +5 -1 -1 -1 -1 -1

Organic-non-organic. 0 -1 -1 +4 -1 -1

NH4-N - NO3N 0 +1 +1 0 -1 -1

(NH4)2SO4 - NH4NO3 0 +1 -1 0 0 0

Ca(NO3)2 - NaNO3 0 0 0 0 +1 -1

The results after the separation of values

Latin square design

ExampleThe results after the separation of values

Źródło odchyleń df SSobs MSobs F Fkr

czynniki 5 185.77 37.15 24.56 2.71

NoN-N 1 180.20 180.20 119.10 4.35

Organ. - Nieorgan. 1 3.816 3.816 2.52 4.35

NH4-N - NO3N 1 0.202 0.202 0.13 4.35

(NH4)2SO4 - NH4NO3 1 1.334 1.334 0.88 4.35

Ca(NO3)2 - NaNO3 1 0.213 0.213 0.14 4.35

reszta 20 30.25 1.513

Latin square design

ExampleOnly when comparing the results for the test "without fertilizer - with fertilizer" calculated value is greater than the critical value, so there is no reason to reject this hypothesis. In other cases, the choice of the source of deviation is negligible.

Thanks for attention!

Books:• W. Wagner, P. Błażczak, Statystyka

matematyczna z elementami doświadczalnictwa, cz. 2, AR, Poznań 1992.

• T. M. Little, F.J. Hills, Agricultural experimentation. Design and analysis, Wiley and Sons, New York, 1987.