Lecture 2 The analysis of crosstabulations Crosstabulations Tables

Lecture 2 The analysis of cross-tabulations

Cross-tabulations • Tables of countable entities or frequencies • Made to analyze the association, relationship, 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 tenure Preterm Total Owner-occupier 50 849 899 Council tentant 29 258 Private tentant 11 164 175 Lives with parents 6 66 72 Other 3 36 39 Total 99 1344 1443

Delivery and housing tenure • Expected number without any association between delivery and housing tenure Housing tenure Pre Term Total Owner-occupier 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39 Total 99 1344 1443

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 tenure Pre Term Total Owner-occupier 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39 Total 99 1344 1443

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 tenure Pre Owner-occupier 61. 7 Term Total 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39 Total 99 1344 1443

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 tenure Pre Term Total Owner-occupier 61. 7 837. 3 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39 Total 99 1344 1443

Delivery and housing tenure If the null-hypothesis is true • In general row total * column total grand total Housing tenure Pre Term Total Owner-occupier 61. 7 837. 3 899 Council tenant 17. 7 240. 3 258 Private tenant 12. 0 163. 0 175 Lives with parents 4. 9 67. 1 72 Other 2. 7 36. 3 39 Total 99 1344 1443

Delivery and housing tenure If the null-hypothesis is true • In general row total * column total grand total Housing tenure Pre Term Total Owner-occupier 50(61. 7) 849(837. 3) 899 Council tenant 29(17. 7) 229(240. 3) 258 Private tenant 11(12. 0) 164(163. 0) 175 Lives with parents 6(4. 9) 66(67. 1) 72 Other 3(2. 7) 36(36. 3) 39 Total 99 1344 1443

Delivery and housing tenure If the null-hypothesis is true Housing tenure Pre Term Total Owner-occupier 50(61. 7) 849(837. 3) 899 Council tenant 29(17. 7) 229(240. 3) 258 Private tenant 11(12. 0) 164(163. 0) 175 Lives with parents 6(4. 9) 66(67. 1) 72 Other 3(2. 7) 36(36. 3) 39 Total 99 1344 1443

Delivery and housing tenure test for association • If the Null- Hyp. is true and large numbers 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 • Although we have shown that there is an association between housing and the time of delivery, it is difficult to say anything about the nature of the association. Housing tenure Pre Term Total Owner-occupier 50(61. 7) 849(837. 3) 899 Council tenant 29(17. 7) 229(240. 3) 258 Private tenant 11(12. 0) 164(163. 0) 175 Lives with parents 6(4. 9) 66(67. 1) 72 Other 3(2. 7) 36(36. 3) 39 Total 99 1344 1443

2 by 2 tables Bronchitis No bronchitis Total Cough 26 44 70 No Cough 247 1002 1249 Total 273 1046 1319

2 by 2 tables Cough No Cough Total Bronchitis No bronchitis Total 26 (14. 49) 44 (55. 51) 70 247 (258. 51) 1002 (990. 49) 1249 273 1046 1319

Chi Squared Table

Chi-squared test for small samples • Expected valued – > 80% >5 – All >1 Streptomycin Control Total Improvement 13 (8. 4) 5 (9. 6) 18 Deterioration 2 (4. 2) 7 (4. 8) 9 Death 0 (2. 3) 5 (2. 7) 5 Total 15 17 32

Chi-squared test for small samples • Expected valued – > 80% >5 – All >1 Streptomycin Control Total Improvement 13 (8. 4) 5 (9. 6) 18 Deterioration and death 2 (6. 6) 12 (7. 4) 14 15 17 32 Total

Fisher’s exact test • An example S D T A 4 0 4 A 3 1 4 B 1 3 4 B 2 2 4 5 3 8 S D T A 2 2 4 A 1 3 4 B 3 1 4 B 4 0 4 5 3 8

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 S D T A 4 0 4 A 3 1 4 B 1 3 4 B 2 2 4 5 3 8 S D T A 2 2 4 A 1 3 4 B 3 1 4 B 4 0 4 5 3 8

Fisher’s exact test • Survivers: – a, b, c, d, e • Deaths: • The properties of finding table 2 or a more extreme is: – F, g, h • • • Table 1 can be made in 5 ways Table 2: 30 Table 3: 30 Table 4: 5 70 ways in total

Fisher’s exact test S D T A 3 1 4 A f 11 f 12 r 1 B 2 2 4 B f 21 f 22 r 2 5 3 8 c 1 c 2 n S D T A 4 0 4 A f 11 f 12 r 1 B 1 3 4 B f 21 f 22 r 2 5 3 8 c 1 c 2 n

Yates’ correction for 2 x 2 • Yates correction: Streptomycin Control Total Improvement 13 (8. 4) 5 (9. 6) 18 Deterioration and death 2 (6. 6) 12 (7. 4) 14 15 17 32 Total

Chi Squared Table

Yates’ correction for 2 x 2 • Table 13. 7 – Fisher: – ‘Two-sided’ – χ2: – Yates’ p = 0. 001455384362148 p = 0. 0029 p = 0. 001121814118023 p = 0. 0037

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

Odds Bronchitis No bronchitis Total Cough 26 (a) 44 (b) 70 No Cough 247 (c) 1002 (d) 1249 273 1046 1319 Total • 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

Odds Cough No Cough Total Bronchitis No bronchitis Total 0. 105 (a) 0. 0439 (b) 70 9. 50 (c) 22. 8 (d) 1249 273 1046 1319

Odds ratio Bronchitis No bronchitis Total Cough 26; 0. 105 (a) 44; 0. 0439 (b) 70 No Cough 247; 9. 50 (c) 1002; 22. 8 (d) 1249 273 1046 1319 Total • The odds ratio; the ratio of odds for experiencing coughs in kids with and kids without a history of bronchitis.

Is the odds ratio different form 1? Bronchitis No bronchitis Total Cough 26 (a) 44 (b) 70 No Cough 247 (c) 1002 (d) 1249 273 1046 1319 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 Bronchitis No bronchitis Total Cough 26 (a) 44 (b) 70 No Cough 247 (c) 1002 (d) 1249 273 1046 1319 Total • 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