1. Fundamentals of Data Analysis Lecture 8 ANOVA pt.2

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

3. Multifactors design On the example of two factors design For this model, we verify the hypothesis : 1. of equality of mean values ​​ for all rp populations: H 0 : m ij = m dla i = 1,..., r; j = 1,..., p. 2.of equality of all the average values m i of studied features treated of A with r variants, excluding the impact of factor B: H 0 : m 1. =... = m r. dla i = 1,..., r. 3. of equality of all the average values m i of studied features treated of B with p variants, excluding the impact of factor A: H 0 : m.1 =... = m.p dla j = 1,..., p. 4. that the deviation of the mean value m ij from the total value of the average m is equal to the sum of effects of factor A and factor B : H 0 : m ij - m = (m i. - m) + (m.j - m).

4. 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

5. Two factors design Example 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.

6. 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

7. Two factors design Example Then compute the sum of squared deviations: q A = 0.5365, for df = 2, then q A /df = 0.26815 q B = 2.3797 df = 4 q B /df = 0.59492 q AB = 0.06980 df = 8 q AB /df = 0.00872 q R = 18.4050 df = 45 q R /df = 0.4090 q = 21.3908 df = 59 F-statistics calculated on this basis, have the following values : F A = 0.26815 / 0.4090 = 0.6556 F B = 0.59492 / 0.4090 = 1.4546 For the third hipothesis we must calculate new mean value: x = 3.4 and q C = 0.24 (df = 1) q R = 7.95 (df = 22) F C = 0.2400/0.3614 = 0.6641

8. Two factors design Example Critical values: F Acr = 3.20 > F A There is no reason to reject the null hypothesis F Bcr = 2.58 > F B There is no reason to reject the null hypothesis F Ccr = 4.30 > F C There is no reason to reject the null hypothesis

9. 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 1234 123123 5.6, 6.1, 5,9 5.7, 4.9, 5.1 6.3, 6.1, 6.3 6.6, 6.7, 6.6 6.5, 6.7, 6.6 6.5, 6.4, 6.2 7.7, 7.3, 7.4 6.9, 7.1, 6.5 6.6, 6.6, 6.8 6.3, 6.4, 6.3 6.6, 6.7, 6.7 6.1, 6.1, 6.0

10. 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.

11. Latin square design 1. We calculate the correction factor: 2. We calculate the sum of squares for rows

12. Latin square design and the mean square for rows 3. We calculate the sum of squares for columns

13. Latin square design and the mean square for columns 4. calculate the sum of squares for the factors

14. Latin square design and the mean square for factors 5. calculate the total sum of squares

15. Latin square design 6. and the residual sum of squares SSE = SS - SSR - SSC - SST 7. mean square residual

16. Latin square design 8. and calculate statistics

17. Latin square design Example In the experiment, on the fertilization of fields used are the following factors (fertilizers): A - (NH 4 ) 2 SO 4, B - NH 4 NO 3, C - CO(NH 2 ) 2, D - Ca(NO 3 ) 2, E - NaNO 3, F - NoN (non-fertilized). Fertilizers are used in equal doses (in g/m 2 ). 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:

18. Latin square design Example The results of the experiment as planned (achieved yields of sugar beet) are presented in Table

19. Latin square design Example Harvests for different fertilizers are presented in Table.

20. Latin square design Example Degree of freedom: totaldf tot = pr - 1 = 35 for rowsdf row = r - 1 = 5 for columnsdf col = p - 1 = 5 for factorsdf tr = n - 1 = 5 for error df error = (r-1)(p-1) - (n-1) = 35 - 5 - 5 - 5 = 20.

21. Latin square design Example

22. 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 NH 4 -N better than NO 3 -N ? 4) Is (NH 4 ) 2 SO 4 better than NH 4 NO 3 ? 5) Is Ca(NO 3 ) better than NaNO 3 ? Such questions may of course be more, depending on the factors or groups of factors we want to compare.

23. Latin square design Example The results after the separation of values

24. Latin square design Example The results after the separation of values

25. Latin square design Example Only 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.

26. 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.