F73DA2 INTRODUCTORY DATA ANALYSIS ANALYSIS OF VARIANCE

1. F73DA2 INTRODUCTORY DATA ANALYSIS ANALYSIS OF VARIANCE

2. regression: x is a quantitative explanatory variable

3. type is a qualitative variable (a factor)

4. Illustration Company 1: 36 28 32 43 30 21 33 37 26 34 Company 2: 26 21 31 29 27 35 23 33 Company 3: 39 28 45 37 21 49 34 38 44

6. Explanatory variable qualitative i.e. categorical - a factor • Analysis of variance • linear models for comparative experiments

9. Using Factor Commands • The display is different if “type” is declared as a factor.

11. We could check for significant differences between two companies using t tests. • t.test(company1,company2) • This calculates a 95% Confidence Interval for difference between means

12. Includes 0 so no significant difference

13. Instead use an analysis of variance technique

14. Taking all the results together

15. Taking all the results together We calculate the total variation for the system which is the sum of squares of individual values – 32.59259

17. We can also work out the sum of squares within each company This sums to 1114.431

18. The total sum of squares of the situation must be made up of a contribution from variation WITHIN the companies and variation BETWEEN the companies. • This means that the variation between the companies equals 356.0884

19. This can all be shown in an analysis of variance table which has the format:

22. Using the R package, the command is similar to that for linear regression

23. Theory Data: yij is the jth observation using treatment i Model: where the errors ij are i.i.d. N(0,s2)

24. The response variables Yij are independent Yij ~ N(µ + τi , σ2) Constraint:

25. Derivation of least-squares estimators

26. The fitted values are the treatment means

27. Partitioning the observed total variation SSB SST SSRES SST = SSB + SSRES

28. The following results hold

29. Back to the example

30. Fitted values: Company 1: 320/10 = 32 Company 2: 225/8 = 28.125 Company 3: 335/9 = 37.222 Residuals: Company 1: 1j = y1j- 32 Company 2: 2j= y2j- 28.125 Company 3: 3j = y3j - 37.222

31. SST = 30152 – 8802/27 = 1470.52 SSB = (3202/10 + 2252/8 + 3352/9) – 8802/27 = 356.09 SSRES = 1470.52 – 356.09 = 1114.43

32. ANOVA table Source of Degrees of Sum Mean F variation freedom of squares squares Between 2 356.09 178.04 3.83 treatments Residual 24 1114.43 46.44 Total 26 1470.52

33. Testing H0 : τi= 0 , i = 1,2,3 v H1 : not H0 (i.e. τi 0 for at least one i) Under H0, F = 3.83 on 2,24 df. P-value = P(F2,24 > 3.83) = 0.036 so we can reject H0 at levels of testing down to 3.6%.

34. Conclusion Results differ among the three companies (P-value 3.6%)

35. The fit of the model can be investigated by examining the residuals: the residual for response yij is this is just the difference between the response and its fitted value (the appropriate sample mean).

36. Plotting the residuals in various ways may reveal ● a pattern (e.g. lack of randomness, suggesting that an additional, uncontrolled factor is present) ● non-normality (a transformation may help) ● heteroscedasticity (error variance differs among treatments – for example it may increase with treatment mean: again a transformation – perhaps log - may be required)

39. In this example, samples are small, but one might question the validity of the assumptions of normality (Company 2) and homoscedasticity (equality of variances, Company 2 v Companies 1/3).

42. plot(residuals(lm(company~type))~ fitted.values(lm(company~type)),pch=8)

44. plot(residuals(lm(company~type))~ fitted.values(lm(company~type)),pch=8) • abline(h=0,lty=2)

46. It is also possible to compare with an analysis using “type” as a qualitative explanatory variable • type=c(rep(1,10),rep(2,8),rep(3,9)) • No “factor” command

48. Note low R2 The equation is company = 27.666+2.510 x type