病例对照研究2

1 病例对照研究 Case-Control Study 吕筠 北京大学公共卫生学院 流行病与卫生统计学系 2011-09-15 预防医学本科07级 2 Case-Control Study Rationale Selection of Cases and Controls Matching Data Analysis Sample Size & Power Calculation Bias 2011-09-15 预防医学本科07级 3 Rationale Design of a case-control study Exposed NotExposed Exposed Not Exposed Disease NoDisease CASES CONTROLS 4 Database Format (Independent) ID Disease Exposure Case Group 1 1 1 2 1 0 3 1 1 4 1 1 …… 1 1 100 1 0 Control Group 101 0 0 102 0 0 103 0 1 104 0 0 …… 0 1 200 0 0 2011-09-15 预防医学本科07级 Disease 1=Yes 0=No Exposure 1=Yes 0=No 2011-09-15 预防医学本科07级 5 Proportions exposed 2×2 Table (1) First Select Case Controls (2) Then Measure Past Exposure Exposed a b Not exposed c d Totals a+c b+d Calculation of Proportion Exposed in a Case-Control Study 2011-09-15 预防医学本科07级 6 Recall: Cohort Study Defined Population Exposed Non-Exposed Disease No Disease NON-RANDOMIZED Disease No Disease Design of a cohort study 2 2011-09-15 预防医学本科07级 7 Incidence Proportion of Disease Recall: Cohort Study Analysis of Cohort Study (2) Then Follow to See Whether TotalsDisease Develops Disease Does Not Develop (1) First Select Exposed a b a+b Not exposed c d c+d 2011-09-15 预防医学本科07级 8 2×2 Table (1) First Select Case Controls (2) Then Measure Past Exposure Exposed a b Not exposed c d Totals a+c b+d Proportions exposed Calculation of Proportion Exposed in a Case-Control Study COMPARE Sampling Error Statistical Significance Testing 2011-09-15 预防医学本科07级 9 Chi-Square Test Question 1: Is there a relationship (association) between exposure and disease? Or: Is there statistical significant difference between cases and controls regarding the proportion of exposed people? 2011-09-15 预防医学本科07级 10 Chi-Square Test Result No Association: STOP HERE Have Association: GO ON …… 2011-09-15 预防医学本科07级 11 Strength of Association Question 2: How strong is the relationship (association) between exposure and disease? 2011-09-15 12 Recall: Cohort Study Risk Calculation in a Cohort Study (2) Then Follow to See Whether Totals Incidence Proportion of DiseaseDiseaseDevelops Disease Does Not Develop (1) First Select Exposed a b a+b Not exposed c d c+d 预防医学本科07级 3 2011-09-15 预防医学本科07级 13 Recall: Cohort Study The meaning of RR Exposed people is ?? times more likely to develop a disease than non-exposed people 2011-09-15 预防医学本科07级 14 Traditional Case-Control Design Cannot calculate the RR directly Can obtain a very good approximation of the RR from a case-control study using the odds ratio 2011-09-15 预防医学本科07级 15 Odds Probability that the event will occur divided by the probability that the event will not occur 2011-09-15 预防医学本科07级 16 Odds (1) First Select Case Controls (2) Then Measure Past Exposure Exposed a b Not exposed c d Calculation of Proportion Exposed in a Case-Control Study What are the odds that a case was exposed? What are the odds that a control was exposed? 2011-09-15 预防医学本科07级 Odds (1) First Select Case Controls (2) Then Measure Past Exposure Exposed a b Not exposed c d Calculation of Proportion Exposed in a Case-Control Study What are the odds that a case was exposed? What are the odds that a control was exposed? 17 2011-09-15 预防医学本科07级 18 Odds Ratio, OR The ratio of the odds that the cases were exposed to the odds that the controls were exposed 4 19 Odds Ratio, OR (1) First Select Case Controls (2) Then Measure Past Exposure Exposed a b Not exposed c d Analysis of Case-Control Study Cross-products ratio OR= Support hypothesis of an association: a, d – diseased people who were exposed and non-diseased people who were not exposed Negate hypothesis of an association: b, c – non-diseased people who were exposed and diseased people who were not exposed 2011-09-15 预防医学本科07级 2011-09-15 预防医学本科07级 20 Confidence Interval, C.I. Woolf’s method 2011-09-15 预防医学本科07级 21 Odds Ratio, OR Question 3: What is the direction of the association and how to interpret it? OR = 1 OR > 1 OR < 1 2011-09-15 预防医学本科07级 22 OR vs. RR Question 4: How well the OR approximates the RR? When is the OR a good estimate of the RR? Answer: When the disease being studied does not occur frequently 2011-09-15 预防医学本科07级 23 Database Format (Matched) Pair No ExposureCase Control 1 1 0 2 1 1 3 1 1 4 0 0 5 0 1 6 1 0 7 1 0 8 1 0 …… Disease 1=Yes 0=No Exposure 1=Yes 0=No Pair No Disease Exposure 1 1 1 1 0 0 2 1 1 2 0 1 3 1 1 3 0 1 4 1 0 4 0 0 …… 2011-09-15 24 Chi-Square Test Control Exposed Not Exposed Case Exposed a bNot Exposed c d Analysis of Matched Case-Control Study Step 1: Chi-Square Test 预防医学本科07级 5 2011-09-15 预防医学本科07级 25 Pairs Contribute to our knowledge 4 possible combination of exposure status for each pair + - 0 1 0 1 Exposure Case Control + - 1 0 1 0 + - 0 1 1 0 + - 1 0 0 1 26 Pairs Contribute to our knowledge Control Exposed Not Exposed Case Exposed a bNot Exposed c d Analysis of Matched Case-Control Study Concordant pairs 1. Pairs in which both the case and the control were exposed a 2. Pairs in which neither the case nor the control was exposed d Discordant pairs 3. Pairs in which the case was exposed but the control was not b 4. Pairs in which the control was exposed and the case was not c 2011-09-15 预防医学本科07级 2011-09-15 预防医学本科07级 27 Odds Ratio, OR Control Exposed Not Exposed Case Exposed a bNot Exposed c d Analysis of Matched Case-Control Study Step 2: Calculating OR 2011-09-15 预防医学本科07级 28 Odds Ratio, OR Control Exposed Not Exposed Case Exposed a bNot Exposed c d Analysis of Matched Case-Control Study OR= Support hypothesis of an association: b – the number of pairs in which the case was exposed and the control was not Negate hypothesis of an association: c – the number of pairs in which the control was exposed and the case was not 2011-09-15 预防医学本科07级 29 Confidence Interval, C.I. Control Exposed Not Exposed Case Exposed a bNot Exposed c d Analysis of Matched Case-Control Study Step 3: 95% C.I. 2011-09-15 预防医学本科07级 30 Dose Response: Test for Trend When exposure is categorized into multiple ordinal categories, it may be of interest to assess whether the observed relation between increasing (or decreasing) levels of exposure and the risk (or odds) of disease follows a linear dose-response pattern. 6 2011-09-15 预防医学本科07级 31 Dose Response: Test for Trend Stratum (i) No. of Cases (ai) No. of controls (bi) Total (ni) ORi 1 a1 b1 n1 1.0 2 a2 b2 n2 . . . . . . . . k ak bk nk Total A B N 2011-09-15 预防医学本科07级 32 Dose Response: Test for Trend  Step 1: 2 by k chi-square independence test  Step 2: calculating stratum-specific OR and their 95% C.I.  Step 3: chi-square test for linear trend 2011-09-15 33 Recall: Cohort Study ARexp Unexposed Exposed Attributable risk in the exposed Inc ide nce (p er 10 00 ) 预防医学本科07级 2011-09-15 预防医学本科07级 34 Recall: Cohort Study Population attributable risk and its dependence on the population prevalence of the exposure ARexp Unexposed Exposed Inc ide nce (p er 10 00 ) Population A Exposure is rare Pop AR 2011-09-15 35 Recall: Cohort Study Exposure is common ARexp Unexposed Exposed Inc ide nce (p er 10 00 ) Population B Pop AR 预防医学本科07级 2011-09-15 预防医学本科07级 36 AR% in Case-Control Studies When the odds ratio is a reasonable estimate of the RR, then 7 2011-09-15 预防医学本科07级 37 Case-Control Study Rationale Selection of Cases and Controls Matching Data Analysis Sample Size & Power Calculation Bias 2011-09-15 预防医学本科07级 38 Sample Size Proportion of controls with exposure 0.0 – 1.0 Expected RR / OR Power (% chance of detecting) Usually 80% Significance level (alpha) Usually 0.05 Ratio of sample size Controls / Cases 2011-09-15 39 2011-09-15 40 2011-09-15 41 2011-09-15 预防医学本科07级 42 Power or Beta Power The probability of rejecting a false null hypothesis = 1 – Beta Beta The probability of a type-II error, which occurs when a false null hypothesis is not rejected In other words, a type-II error occurs when you fail to reject the null hypothesis of equal proportions when in fact they are different 8 2011-09-15 43 OR=2.0 p0=0.30 alpha=0.05 Ncase=85 2011-09-15 预防医学本科07级 44 Case-Control Study Rationale Selection of Cases and Controls Matching Data Analysis Sample Size & Power Calculation Bias 2011-09-15 预防医学本科07级 45 Bias, Systematic Error Why considering bias? That results in a mistaken estimate of an exposure’s effect on the risk of disease When considering bias? Before a study: avoid and control After a study: control and explain 2011-09-15 预防医学本科07级 46 Bias, Systematic Error Selection bias Information bias Confounding bias 2011-09-15 预防医学本科07级 47 Selection Bias Selection bias in case-control study is caused when individuals have different probabilities of being included in the study according to the exposure of interest. 2011-09-15 48 Unbiased Biased Selection Bias 9 2011-09-15 预防医学本科07级 49 Berkson Bias E.g.,  Exposure: Uses of tobacco, alcohol, tea, and coffee  Disease: Cancer of the pancreas 2011-09-15 预防医学本科07级 50 Berkson Bias Case-control study Cases: Patients with histologically proved cancer of the pancreas Controls: Other patients under the care of the same physician of the cases with pancreatic cancer Exclusion: Patients with diseases known to be associated with smoking or alcohol consumption 2011-09-15 预防医学本科07级 51 Berkson Bias Cases Controls Coffee  1 cup/day 207 275 No coffee 9 32 OR= (207/9) / (275/32) = 2.7 (95% C.I., 1.2-6.5) 2011-09-15 预防医学本科07级 52 Berkson Bias There are many reasons to believe that this oddsratio is biased Controls who were examined were: Other patients under the care of the samephysician at the time of an interview with apatient with pancreatic cancer Most of the MDs were gastroenterologistswhose other patients were likely advised tostop using coffee 2011-09-15 预防医学本科07级 53 Berkson Bias Controls who were examined were: Patients with diseases known to be associatedwith smoking or alcohol consumption wereexcluded Smoking and alcohol use are correlated withcoffee use; therefore, sample is relativelydepleted of coffee users 2011-09-15 预防医学本科07级 54 Berkson Bias 10 2011-09-15 预防医学本科07级 55 Berkson Bias When hospitalization rates differ for different exposure groups, the relation between exposure and disease in the hospital will not reflect the relation in the population that serves as the source of hospitalized cases. Hospital controls 2011-09-15 预防医学本科07级 56 Neyman Bias 2011-09-15 预防医学本科07级 57 Neyman Bias 2011-09-15 预防医学本科07级 58 Information Bias Information bias, also known as observation, classification, or measurement bias, results from incorrect determination of exposure or outcome, or both 2011-09-15 预防医学本科07级 59 Information Bias Exposure suspicion bias Knowledge of disease status may influence the intensity and outcome of a search for exposure to the putative cause Cases may be questioned more intensively than controls Recall bias Recall of cases and controls may differ both in amount and in accuracy (selective recall) 2011-09-15 预防医学本科07级 60 Information Bias E.g., Studying the possible relationship of congenital malformations to prenatal infections Cases (with congenital malformations) Controls (without congenital malformations) Assume that: True incidence of infection (%) 15.0 15.0 Infections recalled (%) 60 10 Result will be: Infection rate as ascertained by interview (%) 9.0 1.5 11 2011-09-15 61 Confounding Bias E.g., Smoking, Matches, and Lung Cancer  A tobacco company researcher believes that exposure to matches is the cause of lung cancer  He conducts a large case-control study to test this hypothesis 预防医学本科07级 2011-09-15 预防医学本科07级 62 Confounding Bias Lung Cancer No Lung Cancer Matches 820 340 No Matches 180 660 2=472.91, P<0.01 OR = (820/180) / (340/660) = 8.8 95% C.I. (7.2, 10.9) Confounding Bias LungCancer No LungCancer Matches 820 340 No Matches 180 660 Lung Cancer No Lung Cancer Matches 810 270 No Matches 90 30 Lung Cancer No Lung Cancer Matches 10 70 No Matches 90 630 Crude Stratified Smoker Non-Smoker 2=0.00, P=1.00 ORs = (810/90) / (270/30) = 1.0 95% C.I. (0.65, 1.55) 2=0.00, P=1.00 ORns = (10/90) / (70/630) = 1.0 95% C.I. (0.50, 2.01) 2=472.91, P<0.01 OR = (820/180) / (340/660) = 8.8 95% C.I. (7.2, 10.9) 2011-09-15 63预防医学本科07级 2011-09-15 预防医学本科07级 64 Confounding Bias Test of Homogeneity Null hypothesis: The individual stratum-specific estimates of the measure of association differ only by random variation  i.e., the strength of association is homogeneous across all strata 2011-09-15 预防医学本科07级 65 Confounding Bias Assuming homogeneity is present Form a summary of the unconfounded stratum- specific estimates Mantel-Haenszel hypothesis testing and confidence interval 2011-09-15 预防医学本科07级 66 Confounding Bias Compare “adjusted” estimate to crude estimate  If “adjusted” measure differs meaningfully from crude estimate, then confounding is present 12 2011-09-15 67 Confounding Bias This example illustrates: How confounding can create an apparent effect even when there is no actual true effect  In the relationship between matches and lung cancer, smoking is a confounding factor or a confounder  Smoking confounds the relationship between matches and lung cancer 预防医学本科07级 2011-09-15 预防医学本科07级 68 Confounding Bias How to examine the relationship between smoking and lung cancer independent from the use of matches? Stratified Analysis LungCancer No LungCancer Smoking 900 300 No Smoking 100 700 Lung Cancer No Lung Cancer Smoking 810 270 No Smoking 10 70 Lung Cancer No Lung Cancer Smoking 90 30 No Smoking 90 630 Crude Stratified Matches Present 2=140.42, P<0.01 ORm = (810/10) / (270/70) = 21.0 95% C.I. (10.67, 41.32) 2=238.64, P<0.01 ORnm = (90/90) / (30/630) = 21.0 95% C.I. (13.14, 33.55) 2=750.00, P<0.01 OR = (900/100) / (300/700) = 21.0 95% C.I. (16.4, 26.9) Matches Absent 2011-09-15 69预防医学本科07级 2011-09-15 预防医学本科07级 70 Confounding Bias What is the effect of matches on the relationship between smoking and lung cancer? Smoking Lung Cancer Matches Matches Lung Cancer ?Smoking 2011-09-15 预防医学本科07级 71 Confounder A true confounder (C) must be associated with: the exposure (E) in question the disease (D) under study A variable that is an intermediate step in the causal path between the exposure in question and disease under study is not a confounding variable. E D Factor 2011-09-15 预防医学本科07级 72 Before a study Randomization Restriction Matching (+ stratification analysis) After a study Stratification analysis Multivariate techniques Control for Confounding 13 Summary 以现在确诊的患有某特定疾病的病人作为病例， 以不患有该病但具有可比性的个体作为对照，通过 询问、检查或复查病史，搜集各种可能的危 险因素的暴露史，测量并比较病例组与对照组中各 因素的暴露比例，经统计学检验，若两组差别有意 义，则可认为暴露因素与疾病之间存在着统计学上 的关联。在评估了各种偏倚对研究结果的影响后， 再借助病因推断技术，推断出某个或某些暴露因素 是疾病的危险因素，从而达到探索和检验病因假说 的目的。 2011-09-15 预防医学本科07级 73 2011-09-15 预防医学本科07级 74