1. Lecture 2 The analysis of cross-tabulations

2. 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

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

4. 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

5. 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

6. 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

7. 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

8. 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

9. Delivery and housing tenure If the null-hypothesis is true In general 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 row total * column total grand total

10. 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

11. 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

12. Chi Squared Table

13. 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

14. 2 by 2 tables BronchitisNo bronchitisTotal Cough264470 No Cough24710021249 Total27310461319

15. 2 by 2 tables BronchitisNo bronchitisTotal Cough26 (14.49)44 (55.51)70 No Cough247 (258.51)1002 (990.49)1249 Total27310461319

16. Chi Squared Table

17. 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

18. 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

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

20. Fisher’s exact test Survivers: –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 SDT A314 B224 538 SDT A404 B134 538 SDT A134 B404 538 SDT A224 B314 538

21. Fisher’s exact test Survivers: –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:

22. 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

23. Yates’ correction for 2x2 Yates correction: StreptomycinControlTotal Improvement13 (8.4)5 (9.6)18 Deterioration and death 2 (6.6)12 (7.4)14 Total151732

24. Chi Squared Table

25. Yates’ correction for 2x2 Table 13.7 –Fisher: p = 0.001455384362148 –‘Two-sided’p = 0.0029 –χ 2: p = 0.001121814118023 –Yates’p = 0.0037

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

27. 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

28. 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

29. 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)

30. 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

31. Chi-square for goodness of fit df = 4-1-1 = 2