1. Design and Analysis of Clinical Study 7. Analysis of Case-control Study Dr. Tuan V. Nguyen Garvan Institute of Medical Research Sydney, Australia

2. Overview Risk Odds ratio (OR) Standard erros of OR 95% confidence interval of OR Interpretation

3. Results of a CC study Population CasesControls Risk+Risk-Risk+Risk- abcd Risk factor CasesControls Presenceac AbsenceBd

4. Odds Let p be a probability (p is between 0 and 1) Odds is defined as: Odds is the likelihood of the occurrence of an event to the non- occurrence of the event Obviously, when p = 0.5, odds = 1.

5. Odds Ratio (OR) Risk factorCasesControls Presenceac Absencebd Odds of risk factor presence to absence in cases: Odds of risk factor presence to absence in controls: Odds ratio:

6. Interpretation of OR OR = 1: the risk of disease is the same for exposed (presence of the risk factor) and unexposed (absence of the risk factor) OR > 1: the risk of disease higher in exposed than in unexposed. In other words, the risk factor increases the probability of getting disease. OR < 1: the risk of disease lower in exposed than in unexposed.

7. Sampling Distribution of OR OR is an estimate of a true OR ( ) Estimate is subject to sampling variation Sampling variation is characterized by –Mean –Variance OR is an unbiased mean of. Variance of OR ? –Can not be determined directly –Have to determined indirectly – log(OR)

8. Variance and 95% of OR OR = (ad) / (bc) L = log(OR) var(L) = s 2 = 1/a + 1/b + 1/c + 1/d Standard error of L: s = sqrt(s 2 ) 95% confidene interval (CI) of L: L–1.96*s to L+1.96*s 95% CI of OR: e L–1.96*s to e L–1.96*s Risk factor CasesControls Presenceac Absencebd

9. An Example: Prostate cancer and Agent Orange Giri et al (2004) conducted a pilot case control study to examine the association between prostate cancer risk and Agent Orange exposure among American soldiers who served in Vietnam during the war time. The investigators identified 47 Vietnam era veterans with a diagnosis of prostate cancer. They further randomly selected 142 veterans who were hospitalized for a reason other than cancer from a hospital data base. In each group, the investigators assessed whether an individual was exposed to Agent Orange (AO) during the war. The results are as follows Exposure statusProstate CancerControls Exposed to AO1117 Not exposed to AO29106

10. An Example: Prostate Cancer and Agent Orange Exposure statusProstate CancerControls Exposed to AO1117 Not exposed to AO29106 OR = (11 x 106) / (17 x 29) = 2.37 L = log(2.37) = 0.861 var(L) = s 2 = 1/11 + 1/17 + 1/29 + 1/106 = 0.1936 s = sqrt(0.1936) = 0.44 95% confidene interval (CI) of L: = 0.861–1.96*0.44 to 0.861+1.96*0.44 = –0.00151 to 1.724 95% CI of OR = e -0.00151 to e 1.724 = 1.00 to 5.63

11. Distribution of OR Distribution of OR = 2.37 with 95% confidence interval = 1.0 to 5.63

12. Matched Case-Control Study Confounding is a potential problem of CC study Matched CC study: cases and controls are matched for a confounding factor. –Step 1: Select a case –Step 2: Record the confounding factor –Step 3: Randomly select a control who has exactly the same confounding factor as the case –Repeat steps 1-3 until the sample size is adequate. –2x2 Table Cases Controls ExposedUnexposured Exposedab Unexposedcd

13. An Example: Hip fracure and Osteoporosis A case control study was under taken to examine the association between osteoporosis and risk of hip fracture. 51 cases of hip fracture were identified, and were individually age-matched with 51 controls (who had no fracture). The results are shown in the following table (“0” is for “non-osteoporosis” and 1 is for “osteoporosis”). pair age Hip fx Control 1 61 0 0 2 62 1 0 3 64 0 0 4 64 1 0 5 64 1 1... 49 85 1 1 50 86 0 0 51 87 1 1

14. Analysis of Matched CC by the z-test Summary of results Controls Hip fracture OsteoporosisNon-osteoporosis Osteoporosis112 Non-osteoporosis2810 Proportion of hip fx who had osteoporosis: P 1 = (11+28)/51 = 0.765 Proportion of controls who had osteoporosis: P 2 = (11+2)/51 = 0.255 Difference: D = P 1 – P 2 = 0.765 – 0.255 = 0.51 Variance of D: s 2 = (28+2) / (51) 2 = 0.0115 Standard error of D: s = sqrt(0.0115) = 0.107 z = 0.51 / 0.107 = 4.77 Highly significant!

15. Analysis of Matched CC by OR Summary of results Controls Hip fracture OsteoporosisNon-osteoporosis Osteoporosis112 Non-osteoporosis2810 OR = 28/2 = 14 L = Log(OR) = log(14) = 2.64 Variance of L: s 2 = 1/28 + 1/2 = 0.536 Standard error of L: s = sqrt(0.536) = 0.732 95% CI of L: 2.64 ± 1.96*0.732 = 1.20 to 4.07 95% CI of OR: e 1.20 = 2.33 to e 4.07 = 58.8

16. Statistical Methods The relationship between [risk factor] and [outcome, disease] was assessed by the odds ratio (OR) for case-control study. In this analysis, the study’s result was summarized into a 2x2 table with two categories of risk factor and outcome. The exposure OR was then computed from the table. The statistical significance of OR was assessed by the 95% confidence interval (CI) of OR. A CI does not include unity was considered statistically signigicant at the 5% significance level.

17. Exercises Preston-Martin et al (1989) conducted a case-control study to investigate the etiology of acoustic neuromas. Eighty-six men were diagnosed to have acoustic neuromas. The researchers selected another 86 men of the same race and within 5 years of age compared to the cases. The researchers were interested in the effect of exposure to loud noise. They found 58 cases and 46 controls had had some exposure to loud noise at work. Carry out an appropriate analysis and interpret the result of analysis.

18. Exercises R Doll and B Hill. BMJ 1950; ii:739-748 Lung CancerControls Smokers647622 Non-smokers227