1. Repeated measures ANOVA in SPSS Cross tabulations Survival analysis

2. Repeated measures ANOVA in SPSS 2

3. The data from Chen 2004 The number of errors are recorded for the same test repeated 4 times The degree of anxiety was different between the subjects. Stuff we would like to know: Is there a difference in the performance between the anxiety groups? Is there a reduction in errors over time – a learning effect? Does one of the groups learn faster than the other? 3

4. What to do? A mediocre suggestion and a really bad one Is there a difference in the performance between the anxiety groups? We could calculate the mean (or sum, time-to-max, min, max, etc.) for each subject and do a t-test between the anxiety groups Is there a reduction in errors over time – a learning effect? We could make 6 paired t-tests to test the difference between Does one of the groups learn faster than the other? ??? 4

5. Repeated measures ANOVA Time is a within subject factor (4 levels) Anxiety is a between subjects factor (2 levels) 5

6. Repeated measures ANOVA Plotting the data 6

7. Repeated measures ANOVA – between subjects effects 7

8. Repeated measures ANOVA – within subjects effects Is there a time effect? Univariate or Multivariate? Univariate assumes sphericity of the covariance matrix: 8 Residual SSCP Matrix Trial1Trial2Trail3Trial4 Trial14,5672,4171,3831,017 Trial22,4176,2835,1335,167 Trail31,3835,1336,4176,083 Trial41,0175,1676,0837,617 Based on Type III Sum of Squares

9. Repeated measures ANOVA – within subjects effects Univariate Mauchly's Test of Sphericity b Measure:MEASURE_1 Within Subjects Effect Mauchly's W Approx. Chi- SquaredfSig. dimension1 time,3379,4965,093 Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix. b. Design: Intercept + Anxiety Within Subjects Design: time 9 Residual SSCP Matrix Trial1Trial2Trail3Trial4 Trial14,5672,4171,3831,017 Trial22,4176,2835,1335,167 Trail31,3835,1336,4176,083 Trial41,0175,1676,0837,617 Based on Type III Sum of Squares

10. Repeated measures ANOVA – within subjects effects Univariate 10 Greenhouse-Geisser is newer than Huynh-Feldt methods Lower-bound is the most konservative Alternatively use Multible ANOVA

11. Repeated measures ANOVA – within subjects effects Multivariate 11 The covariance matrices must be equal for all levels of the between groups Tested by Box’s test

12. Repeated measures ANOVA – within subjects effects Multivariate 12 If the 4 methods disagree use Wilk’s Lambda Multivariate and Univariate renders the same conclusion in this case Which method to choose?

13. Repeated measures ANOVA – within subjects effects Which method to choose? 13

14. Repeated measures ANOVA – within subjects effects Pairwise comparisons 14 Does all these comparisons make sense?

15. Repeated measures ANOVA – within subjects effects Contrasts? 15

16. Cross tabulations 16

17. Delivery and housing tenure Housing tenurePretermTerm Owner-occupier50849 Council tentant29229 Private tentant11164 Lives with parents666 Other336

18. Cross-tabulations Tables of countable entities or frequencies Made to analyze the association, relationship, or connection between two variables This association is difficult to describe statistically Null- Hypothesis: “There is no association between the two variables” can be tested Analysis of cross-tabulations with larges samples

19. Delivery and housing tenure Housing tenurePretermTermTotal Owner-occupier50849899 Council tentant29229258 Private tentant11164175 Lives with parents66672 Other33639 Total9913441443

20. Delivery and housing tenure Expected number without any association between delivery and housing tenure Housing tenurePreTermTotal Owner-occupier899 Council tenant258 Private tenant175 Lives with parents72 Other39 Total9913441443

21. Delivery and housing tenure If the null-hypothesis is true 899/1443 = 62.3% are house owners. 62.3% of the Pre-terms should be house owners: 99*899/1443 = 61.7 Housing tenurePreTermTotal Owner-occupier899 Council tenant258 Private tenant175 Lives with parents72 Other39 Total9913441443

22. Delivery and housing tenure If the null-hypothesis is true 899/1443 = 62.3% are house owners. 62.3% of the ‘Term’s should be house owners: 1344*899/1443 = 837.3 Housing tenurePreTermTotal Owner-occupier61.7899 Council tenant258 Private tenant175 Lives with parents72 Other39 Total9913441443

23. Delivery and housing tenure If the null-hypothesis is true 258/1443 = 17.9% are council tenant. 17.9% of the ‘preterm’s should be council tenant: 99*258/1443 = 17.7 Housing tenurePreTermTotal Owner-occupier61.7837.3899 Council tenant258 Private tenant175 Lives with parents72 Other39 Total9913441443

24. Delivery and housing tenure If the null-hypothesis is true In general Housing tenurePreTermTotal Owner-occupier61.7837.3899 Council tenant17.7240.3258 Private tenant12.0163.0175 Lives with parents4.967.172 Other2.736.339 Total9913441443 row total * column total grand total

25. Delivery and housing tenure If the null-hypothesis is true Housing tenurePreTermTotal Owner-occupier50(61.7)849(837.3)899 Council tenant29(17.7)229(240.3)258 Private tenant11(12.0)164(163.0)175 Lives with parents6(4.9)66(67.1)72 Other3(2.7)36(36.3)39 Total9913441443

26. Delivery and housing tenure test for association If the numbers are large this will be chi-square distributed. The degree of freedom is (r-1)(c-1) = 4 From Table 13.3 there is a 1 - 5% probability that delivery and housing tenure is not associated

27. Chi Squared Table

28. Delivery and housing tenure If the null-hypothesis is true It is difficult to say anything about the nature of the association. Housing tenurePreTermTotal Owner-occupier50(61.7)849(837.3)899 Council tenant29(17.7)229(240.3)258 Private tenant11(12.0)164(163.0)175 Lives with parents6(4.9)66(67.1)72 Other3(2.7)36(36.3)39 Total9913441443

29. Chi-squared test for small samples Expected valued > 80% >5 All >1 StreptomycinControlTotal Improvement13 (8.4)5 (9.6)18 Deterioration2 (4.2)7 (4.8)9 Death0 (2.3)5 (2.7)5 Total151732

30. Chi-squared test for small samples Expected valued > 80% >5 All >1 StreptomycinControlTotal Improvement13 (8.4)5 (9.6)18 Deterioration and death 2 (6.6)12 (7.4)14 Total151732

31. Fisher’s exact test An example SDT A314 B224 538 SDT A404 B134 538 SDT A134 B404 538 SDT A224 B314 538

32. Fisher’s exact test Survivors: a, b, c, d, e Deaths: f, g, h Table 1 can be made in 5 ways Table 2: 30 Table 3: 30 Table 4: 5 70 ways in total The properties of finding table 2 or a more extreme is: SDT A314 B224 538 SDT A404 B134 538 SDT A134 B404 538 SDT A224 B314 538

33. Fisher’s exact test SDT A f 11 f 12 r1r1 B f 21 f 22 r2r2 c1c1 c2c2 n SDT A314 B224 538 SDT A f 11 f 12 r1r1 B f 21 f 22 r2r2 c1c1 c2c2 n SDT A404 B134 538

34. Odds and odds ratios Odds, p is the probability of an event Log odds / logit

35. Odds The probability of coughs in kids with history of bronchitis. p = 26/273 = 0.095 o = 26/247 = 0.105 The probability of coughs in kids with history without bronchitis. p = 44/1046 = 0.042 o = 44/1002 = 0.044 BronchitisNo bronchitisTotal Cough26 (a)44 (b)70 No Cough247 (c)1002 (d)1249 Total27310461319

36. Odds ratio The odds ratio; the ratio of odds for experiencing coughs in kids with and kids without a history of bronchitis. BronchitisNo bronchitisTotal Cough26; 0.105 (a)44; 0.0439 (b)70 No Cough247; 9.50 (c)1002; 22.8 (d)1249 Total27310461319

37. Is the odds ratio different form 1? BronchitisNo bronchitisTotal Cough26 (a)44 (b)70 No Cough247 (c)1002 (d)1249 Total27310461319 We could take ln to the odds ratio. Is ln(or) different from zero? 95% confidence (assumuing normailty)

38. Confidence interval of the Odds ratio ln (or) ± 1.96*SE(ln(or)) = 0.37 to 1.38 Returning to the odds ratio itself: e 0.370 to e 1.379 = 1.45 to 3.97 The interval does not contain 1, indicating a statistically significant difference BronchitisNo bronchitisTotal Cough26 (a)44 (b)70 No Cough247 (c)1002 (d)1249 Total27310461319

39. McNemar’s test Colds at 14 Total YesNo Colds at 12 Yes212144356 No256707963 Total4688511319 39

40. Survival analysis 40

41. First example of the day

42. Problem Do patients survive longer after treatment 1 than after treatment 2? Possible solutions: ANOVA on mean survival time? ANOVA on median survival time?

43. Progressively censored observations Current life table Completed dataset Cohort life table Analysis “on the fly”

44. Actuarial / life table anelysis Treatment for lung cancer

45. Actuarial / life table anelysis A sub-set of 13 patients undergoing the same treatment

46. Kaplan-Meier Simple example with only 2 ”terminal-events”.

47. Confidence interval of the Kaplan-Meier method Fx after 32 months

48. Confidence interval of the Kaplan-Meier method Survival plot for all data on treatment 1 Are there differences between the treatments?

49. Comparing Two Survival Curves One could use the confidence intervals… But what if the confidence intervals are not overlapping only at some points? Logrank-stats

50. Comparing Two Survival Curves The logrank statistics Aka Mantel-logrank statistics Aka Cox-Mantel-logrank statistics

51. Comparing Two Survival Curves Five steps to the logrank statistics table 1.Divide the data into intervals (eg. 10 months) 2.Count the number of patients at risk in the groups and in total 3.Count the number of terminal events in the groups and in total 4.Calculate the expected numbers of terminal events e.g. (31-40) 44 in grp1 and 46 in grp2, 4 terminal events. expected terminal events 4x(44/90) and 4x(46/90) 5.Calculate the total

52. Comparing Two Survival Curves Smells like Chi-Square statistics

53. Comparing Two Survival Curves