Fundamentals of Data Analysis Lecture 8 ANOVA pt.2

1. Fundamentals of Data AnalysisLecture 8ANOVApt.2

2. Multifactors design On theexample of twofactors 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 rpgroups.

3. Multifactors design On theexample of twofactors 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 miof studied features treated of A withrvariants, excluding the impact of factor B: H0: m1. =... = mr. dla i = 1, ..., r. 3. of equality of all the average values miof studied features treated ofB with pvariants, excluding the impact of factor A: H0: m.1 =... = m.p dla j = 1, ..., p. 4. thatthedeviation of themeanvaluemijfrom the total value of the averagem isequal to the sum of effects of factorA and factorB: H0: mij - m = (mi. - m) + (m.j - m).

4. Twofactors 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. Twofactors design Assuming that the average grades obtained by thestudents have a normal distributions with the same varianceat the confidence level α= 95% verify the following hypotheses: • the average values of average grades for students of different departments arethe same; • the average value of average grades fordiffererntyears of study are the same; • the average value of the average grades for the first two years are the same. Example

6. Twofactors design In thatcase we haver= 3 (Departments) and p = 5 (number of years of study). After calculations we get the results shown in the table : Example

7. Twofactors design Thencomputethe sum of squareddeviations: qA = 0.5365,for df= 2, thenqA /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-statisticscalculated 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 mustcalculatenewmeanvalue: x= 3.4 and qC = 0.24 (df = 1) qR= 7.95 (df = 22) FC= 0.2400/0.3614 = 0.6641 Example

8. Twofactors design Criticalvalues: 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 Example

9. Twofactors design 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 Atthe confidence level 95% verify the following hypotheses: The values ​​of the average yield for the different varieties of potatoes are dependent on the applied fertilizer The values ​​of the average yields for the different fertilizers do not differ regardless of potato variety Exercise

10. Latinsquare design DuringLatin sQuaredesign (LQ)experimentalitemsareclassifiedin terms of the classification of three directions: rows, columns, and objects. ExperimentalfactorA presentedat p levels (containsp objects), each of which occurs exactly once in the corresponding row and column.

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

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

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

14. Latinsquare design and the mean square for factors 5. calculate the totalsum of squares

15. Latinsquare design 6. and the residual sum of squares SSE = SS - SSR - SSC - SST 7. meansquareresidual

16. Latinsquare design 8. and calculatestatistics

17. Latinsquare design Example In the experiment, on the fertilization of fieldsused 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 square6x6 (since we have 6 factors) and theresultisshownintable:

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

19. Latinsquare design Example • Harvestsfor different fertilizersare presented in Table.

20. Latinsquare design Example Degree of freedom: totaldftot = pr - 1 = 35 for rowsdfrow = r - 1 = 5 for columnsdfcol= p - 1 = 5 for factorsdftr = n - 1 = 5 for errordferror= (r-1)(p-1) - (n-1) = 35 - 5 - 5 - 5 = 20.

21. Latinsquare design Example

22. Latinsquare 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 caneffect on crop growth (excreted factor F)? • 2) Is organic fertilizer better than inorganic? • 3) IsNH4-N better thanNO3-N ? • 4) Is(NH4)2SO4better than NH4NO3 ? • 5) IsCa(NO3) better than NaNO3 ? • Suchquestionsmay of course be more, depending on the factors or groups of factors we want to compare.

23. Latinsquare design The results after the separation of values Example

24. Latinsquare design The results after the separation of values Example

25. Latinsquare 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.