1. Confounding And Interaction Dr. L. Jeyaseelan Department Of Biostatistics CMC, Vellore

2. Case Study

3. Is goiter related to high altitudes? A group of researchers presented data showing rate of goiters between two areas that were different in altitudes. There was a higher rate of goiter among people who lived in counties located at high altitudes. Hence the researchers concluded that living at high altitudes was a factor associated with presence of goiter. PDQ Epidemiology

4. Is goiter related to high altitudes? Distance to reach high altitudes Iodine evaporates before reaching high altitudes….

5. Definition What looks like a causal relationship between a supposed hazard and a disease may be due to another factor not taken into consideration. This additional factor is called a confounder, something that confuses the correct interpretation of data. GAMBLINGCANCER SMOKING ALCOHOL OTHER FACTORS Unobserved association True causation

6. Hypothetical Example Male DrugPlacebo Cure No Cure 120 (60%) 80 60 (60%) 40 Total200100  2 = 0.00 Female DrugPlacebo Cure No Cure 30 (30%) 70 60 (30%) 140 Total100200  2 = 0.00

7. Hypothetical Example (Cont.) Pooled DrugPlacebo Cure No Cure 150 (50%) 150 100 (33.3%) 200 Total300  2 = 17.14 (p < 0.0001)

8. BASIC CONCEPTS IN ASSESSMENT OF RISK Situations in which F is a confounder for a disease- exposure association. ( ) non- causal association; ( ) causal association. The letters E  Exposure F  Potential matching factor (confounder) D  Disease Fig A. Indirect association between exposure and disease that is due to the factor F. Example: Association between drinking alcoholic beverages (E) and Lung cancer (D) would likely be explained in terms of an association between alcohol intake and cigarette smoking (F). F E D Fig A James. J. Schlesselman, 1982

9. Situation in which matching on a factor F is proper Fig B. E and F individually alter the risk of disease and are also associated. Failure to match or otherwise control for F in this instance would result in a biased assessment of the individual effect of E. F E D Fig B Example: Use of oral contraceptives and cigarette smoking are both risk factors for myocardial infarction. Note: OC use and smoking are positively associated, so that failure to adjust for the effect of smoking (F) results in an overestimate of the effect of the OC use (E) on the risk of a myocardial infarction. James. J. Schlesselman, 1982

10. Situations in which F is not a confounder for a disease- exposure association. E F D Fig C E F D Fig D Fig C Example: A case control study of venous thromboembolism and blood group O provides an example of avoiding unnecessary matching. Although age and sex are characteristics that bear a strong relationship to disease, they are practically unrelated to the factors is necessary Fig D Example: Hospital based case control study on Myocardial infarction (MI) and oral contraceptives. James. J. Schlesselman, 1982

11. Situations in which F is not a confounder for a disease- exposure association. E F D E D F James. J. Schlesselman, 1982

12. Confounding:  Apparent association is due to another variables - Apparent lack of association could result from failure to control for the effect of some other factor. Example: The following table shows the recent oral Contraceptive (OC) use (last use within the month before admission) among 234 cases of MI and 1742 controls. OCMIControl Yes29135 No2051607 Odds ratio = 1.68 (Shapiro et al 1979)

13. Table: age-specific Relation of MI to Recent oral Contraceptive (OC) use

14. Table : Summary of Examples Showing Confounding and/or Interaction in Randomly Sampled Data Page (246); David G. Kleinbaum 1982

15. MANTEL-HAENSZEL METHOD OF COMBINING 2 * 2 TABLES The null hypothesis of interest is: H 0 : P LESS = P MORE VsH a : P LESS  P MORE 95% CI (1.11 to 6.71 )  ² = 5.26 > 3.84 Reject H 0

16. However, these data were collected in two clinics and then combined. The data for the individual clinics are shown below together with some summary statistics. Conclusion is s that there is no association between amount of prenatal care and one-month infant survival. This contradicts our previous conclusion. Why?

17. Suitable methods have been suggested by Mantel and Haenszel 1. To test the null hypothesis that on the average there is no association. 2. To measure the average strength of the association. The formulas for the individual tables is Where X MH 2 approximately has the chi-square distribution with 1 d.f. With  indicating summation over all strata or tables.

18. With the continuity correction, The pooled estimate of the odds ratio is given by: With  indicating summation over all strata or tables.

19. Example: For the prenatal care data: Clinic 1 Clinic 2 

20. The pooled estimate of the odds ratio is given by:

21. Case Study

22. Characteristic3 day treatment (n=1095) 5 day treatment (n=1093) Mean (SD) Age (months)17.0(13.3)16.9(13.0) Mean (SD) height (cm)74.8(10.98)74.8(10.75) Mean (SD) weight (kg)8.7(2.49)8.7(2.4) Mean (SD)duration of illness days)4.7(3.43)4.5(3.12) Mean (SD) temperature ( o C)37.1(0.66)37.2(0.67) Mean (SD) respiratory rate (breath / minute): 2 – 11 months old 12 – 59 months old 56.4 47.3 (5.02) (5.58) 56.0 47.9 (4.54) (6.1) Male685(62.6)676(61.8) Age (months): 2 – 11 12 – 59 479 616 (43.7) (56.3) 475 618 (43.5) (56.5) Weight for height z score*: -2 to -1 -3 - 2 300 188 (27.4) (17.2) 303 183 (27.7) (16.7) Table1: Baseline characteristics of 2188 children with non-severe pneumonia randomised to 3 days or 5 days of treatment with amoxicillin. Values are numbers (percentages) of patients unless stated otherwise

23. Characteristic3 day treatment (n=1095) 5 day treatment (n=1093) Duration of illness (days):  3  3 538 557 (49.1) (50.9) 540 553 (49.4) (50.6) Fever833(76.1)850(77.8) Cough1081(98.7)1078(98.6) Difficulty in breathing417(38.1)387(35.4) Vomiting135(12.3)141(12.9) Diahorrea71(6.5)55(5.0) Excess respiratory rate (breaths / minute)  10  10 903 192 (82.5) (17.5) 881 212 (80.6) (19.4) Wheeze present140(12.8)147(13.4) Adherence to treatment: At day 3 At day 5 1031 937 (94.2) (85.6) 1026 928 (93.9) (84.9) RSV Positive252(23.0)261(23.9) Table1 (Cont….) *Z score given as number of standard deviations from normal value. †Rate above the age specific cut off RSV=respiratory syncytial virus.

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