1. doc.Ing. Zlata Sojková,CSc.1 Analysis of Variance

2. doc.Ing. Zlata Sojková,CSc.2 §In practice it is often necessary to compare a large number of independent random selections in terms of level, we are interested in hypothesis: for at least one i (i = 1, 2,…m) for m > 2, when  i, i =1, 2, …m are mean values of normally distributed populations with equal variances  2, t.j. N( ,  2 ) §To verify this hypothesis is used important statistical method called Analysis of variance, abbreviated ANOVA ( resp. AV)

3. doc.Ing. Zlata Sojková,CSc.3 §In practice is AV used for examination of the impact of one, or more factors (treatments) on the statistical sign. §Factors are labeled A, B,…in AV they will be regarded as qualitative attributes with different variations – levels of factor §Result will be quantitative statistical sign denoted Y §AV is frequently used in the evaluation of biological experiments § The simplest case is AV with single factor called One factor analysis of variance

4. doc.Ing. Zlata Sojková,CSc.4 §Level of the factor refer to : l certain amount of quantitative factor, e.g. Amount of pure nutrients in manure, different income groups of households l Certain kind of qualitative factor, e.g. different types of the same crop, methods of products placing in stores, §AV is a generalization of Student's t-test for independent choices §AV also examines the impact of qualitative factors resulting in a quantitative character -> analyzes the relationships between attributes

5. doc.Ing. Zlata Sojková,CSc.5 Scheme of single-factor experiment “balanced attempt” A 12…j… nY i.y i. 1y 11 y 12 y 1j y 1n Y 1.y 1. 2 y 21 y 22 y 2j y 2n Y 2. y 2. ………….. i y i1 y i2 y ij y in Y i. y i. ………….. m y m1 y m2 y mj y mn Y m. y m. Y.. y.. Repetition Levels of the factor row sum row average Overall average Total sum

6. doc.Ing. Zlata Sojková,CSc.6 Row sum:Total sum: Row average: Overall average:

7. doc.Ing. Zlata Sojková,CSc.7 Model for resulting observed value:  - expected values for all levels of the factor and observed values  i - impact of i-th level of the factor A e ij - random error, every measurement is biased, resp. impact of random factors where i = 1, 2,…, m j = 1,2,…, n

8. doc.Ing. Zlata Sojková,CSc.8 Then we can formulate null hypothesis: H o :  1 =  2 =…  i =  m = 0 -> effects of all levels of factor A are zero, insignificant, against the alternative hypothesis H 1 :  i  0 for at least one i (i = 1,2…m) effect  i at least one i – level of the factor is significant, => significantly different from zero or

9. doc.Ing. Zlata Sojková,CSc.9 Estimates of parameters are sample characteristics: : What can be rewrited:

10. doc.Ing. Zlata Sojková,CSc.10 Comparison of two experiments with three levels of factor 123 312

11. doc.Ing. Zlata Sojková,CSc.11 Principle of the ANOVA Essence of the analysis of variance lies in the decomposition of the total variability of the investigated sign. Total variability Variability between levels of factor, caused by the action of factor A, “variability between groups” Random variability, residual, “variability within groups“ ScSc S1S1 SrSr

12. doc.Ing. Zlata Sojková,CSc.12 Variability between groups Variability within groups Total variability Variability 1 Sum of squares (SS) 2 Degrees of freedom m-1 m.n - m N-1= m.n-1 3 Mean square (MS) (1/2) S1S1 SrSr ScSc s12s12 sr2sr2 4 F critical

13. doc.Ing. Zlata Sojková,CSc.13 Test statistics for one factor ANOVA can be written: F value will be compared with appropriate table value for F- distribution: F , with (m-1) and (m.n - m) degrees of freedom

14. doc.Ing. Zlata Sojková,CSc.14 Decision about test result: §If F vyp  F . ((m-1,(N-m))  We reject H 0, In that case is effect of at least one level of the factor significant, thus average level of the indicator is significantly different from others. => At least one effect  i is statistically significantly different from zero. Rejection region H 0 FF Acceptance regon Ho If F vyp  F  Do not reject Ho

15. doc.Ing. Zlata Sojková,CSc.15 If null hypothesis is rejected: §We found only that effect of the factor on examined attribute is significant. §It is also necessary to identify levels of the factor, which are significantly different - for this purpose are used tests of contrasts §Test of contrast: Duncan test, Scheffe test, Tuckey test and others…..

16. doc.Ing. Zlata Sojková,CSc.16 Terms of use AV: §Samples have normal distribution, violating of this assumption has significant effect on the results of AV §statistical independence of random errors eij §Identical residual variances  1 2 =  2 2 = …. =  2, t.j. D(e ij ) =  2 for all i = 1,2…., m, j=1,2, …n this assumption is more serious and can be verified by Cochran, resp. Bartlett test.

17. doc.Ing. Zlata Sojková,CSc.17 Scheme of single-factor experiment “unbalanced attempt” A 12…j … n i Y i.y i. 1y 11 y 12 y 1j... n 1 Y 1.y 1. 2 y 21 y 22 y 2j... n 2 Y 2. y 2. ………….. i y i1 y i2 y ij... n i Y i. y i. ………….. m y m1 y m2 y mj... n m Y m. y m. Y.. y.. Different number of repetitions Levels of the factor row sum Row average Overall average Where

18. doc.Ing. Zlata Sojková,CSc.18 Variability between groups Variability within groups Total variability Variability 1 Sum of squares (SS) 2 Degrees of freedom m-1 N - m N-1 3 Mean square (MS) (1/2) S1S1 SrSr S s12s12 sr2sr2 4 F- critical

19. doc.Ing. Zlata Sojková,CSc.19 Two-factor analysis of variance with one observation in each subclass.... TAV §Consider the effect of factor A, which we investigate on the m - levels, i = 1,2,...., m §Then consider the effect of factor B, which is observed on n - levels, j = 1,2, …, n §On every i-level of factor A and j-level of factor B we have only one observation (repetition) y ij §=>We are veryfying two null hypothesis

20. doc.Ing. Zlata Sojková,CSc.20 Scheme for Two-factor experiment with one observation in each subclass TAV A 12 …j … nY i.y i. 1y 11 y 12 y 1j y 1n Y 1.Y 1. 2 y 21 y 22 y 2j y 2n Y 2. y 2. ………….. i y i1 y i2 y ij y in Y i. y i. ………….. m y m1 y m2 y mj y mn Y m. y m. Y. 1 Y. 2... Y. j... Y. 1 Y.. y. 1 y. 2... y. j... y. 1 y.. n- levels of factor B m-levels of factor A row sum Row average Overall average B Column sum Column average

21. doc.Ing. Zlata Sojková,CSc.21 We are verifying the validity of two null hypothesis Hypothesis for factor A: H o 1 :  1 =  2 =…  i =  m = 0 t.j. All effects of factor A levels are equal to zero, thus insignificant, against alternative hypothesis H 1 1 :  i  0 for at least one i (i = 1,2…m) effect  i of at least one i – level of factor A is significant, significantly different from zero We can write model for examined attribute as follows:

22. doc.Ing. Zlata Sojková,CSc.22 Hypothesis for factor B: H o 2 :  1 =  2 =…  j =  n = 0 => All effects of factor A levels are equal to zero, thus insignificant, against alternative hypothesis H 1 2 :  j  0 for at least one j (j = 1,2…m) effect  j of at least one j – level of the factor B is significant, significantly different from zero

23. doc.Ing. Zlata Sojková,CSc.23 Variability between rows Residual variability Total variability TAV Variability 1 Sum of squares (SS) 2 Degrees of freedom m-1 n-1 (m-1)(n-1) 3 Mean square (MS) (1/2) SrSr ScSc s12s12 sr2sr2 4 F - critical S1S1 S2S2 Variability between columns m.n -1 s22s22

24. doc.Ing. Zlata Sojková,CSc.24 Decomposition of the total variability S c = S 1 + S 2 + S r Variability between rows, effect of factor A Variability between columns, Effect of factor B Residual variability Total variability

25. doc.Ing. Zlata Sojková,CSc.25 Investigating the relationships between statistical attributes §Investigating the relationship between qualitative attributes, e.g. A  B, called measurement of the association §Investigating the relationship between quantitative attributes - regression and correlation analysis

26. doc.Ing. Zlata Sojková,CSc.26 Inestigating the association §Based on the association, resp. pivot tables §For testing the existence of significant relationship between qualitative signs we use  2 - test of independence Ho: two signs A and B are independent H 1 : signsA and B are dependent Attribute A has m - levels, variations Attribute B has k - levels, variation

27. doc.Ing. Zlata Sojková,CSc.27 Hypotheses formulation §Dependence of the attributes will appear in different frequency §E.g. We examine wheter the size of the package is affected by the size of the family §H o : Choice of the package size depend on the count of family members §H 1 : Choice of package size is affected by the size of the family §The procedure lies in comparing empirical and theoretical frequencies, (how should be empirical frequencies, if the attributes A and B were independent

28. doc.Ing. Zlata Sojková,CSc.28 Simultanous frequencies, frequencies of the second order (a i b j ) Package size Size of the family 1-23-45 <Total (b 1 ) (b 2 ) (b 3 ) do 100g2537870 (a 1 )(a 1 b 1 )(a 1 b 2 ) 100-150g106253125 (a 2 ) 250g < 54159105 (a 3 )(a 3 b 3 ) Total 40140 120 300 Marginal frequencies (a i ) resp.(b j ) Total count of the respondents n

29. doc.Ing. Zlata Sojková,CSc.29 Determination of theoretical frequencies Based on the sentence about independence of the random events A and B: P(A  B) = P(A). P(B), thus signs A and B are independent, then: P(a i b j ) = P(a i ).P(b j ) Estimate based on the relative frequencies: (a i b j ) o = (a i ). (b j )  (a i b j ) o = (a i ).(b j ) n n n n Theoretical frequencies

30. doc.Ing. Zlata Sojková,CSc.30 Calculation of theoretical frequencies (a 1 b 1 ) o = 70.40/300 = 9,33 Package size Family size 1-23-45 and <Total (b 1 ) (b 2 ) (b 3 ) do 100g2537870 (a 1 )9.33 32,6728.00 100-150g106253125 (a 2 )16.6758.3350 250g < 54159105 (a 3 )14.004942 Total40140120300 Total count of respondents n

31. doc.Ing. Zlata Sojková,CSc.31 Calculation of test criteria and decision: If  2 calculated   2 for significance  for degrees of freedom (m-1).(k-1)  H o is rejected => signs A and B are dependent In our case it means, that count of the family members significantly affects choice of the package size. Further, we should measure strength (power) of the dependence.