Lecture 2 The analysis of cross-tabulations

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

Delivery and housing tenure Housing tenurePretermTermTotal Owner-occupier Council tentant Private tentant Lives with parents66672 Other33639 Total

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 Total

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 Total

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 = Housing tenurePreTermTotal Owner-occupier Council tenant258 Private tenant175 Lives with parents72 Other39 Total

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-occupier Council tenant258 Private tenant175 Lives with parents72 Other39 Total

Delivery and housing tenure If the null-hypothesis is true In general Housing tenurePreTermTotal Owner-occupier Council tenant Private tenant Lives with parents Other Total row total * column total grand total

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 Total row total * column total grand total

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 Total

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

Chi Squared Table

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 Total

2 by 2 tables BronchitisNo bronchitisTotal Cough No Cough Total

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

Chi Squared Table

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

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

Fisher’s exact test An example SDT A314 B SDT A404 B SDT A134 B SDT A224 B

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 B SDT A404 B SDT A134 B SDT A224 B

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:

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

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

Chi Squared Table

Yates’ correction for 2x2 Table 13.7 –Fisher: p = –‘Two-sided’p = –χ 2: p = –Yates’p =

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

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

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; (a)44; (b)70 No Cough247; 9.50 (c)1002; 22.8 (d)1249 Total

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

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 to e = 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 Total

Chi-square for goodness of fit df = = 2