1 / 35

Case Control Study : Analysis

Case Control Study : Analysis. Odds and Probability. Odds and Probability. what is odds? Let’s say that probability of success is 0 .8, thus P = .8 Then the probability of failure is = 1-p = q = .2 The odds of success are The odds of failure would be.

Télécharger la présentation

Case Control Study : Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Case Control Study : Analysis

  2. Odds and Probability

  3. Odds and Probability what is odds? Let’s say that probability of success is 0 .8, thus P = .8 Then the probability of failure is = 1-p = q = .2 The odds of success are The odds of failure would be • Odds of <1 mean the outcome occurs less than half the time • Odds of 1 mean the outcome occurs half the time • Odds of >1 mean the outcome occurs more than half the time

  4. odds ratio= The odds of success are 16 times greater for failure.

  5. ODDS RATIO: Case Control a+b a b + - Exposure c+d c d a+c b+d Odds of disease among exposed = a/ b Odds of disease among unexposed = c/d

  6. MI Controls 172 173 345 yes no Ever smokers 436 90 346 262 519 Example : The following data refer to a study that investigated the relationship between MI and smoking.The first column refers to 262 young and middle - aged women admitted to 30 coronary care units with acute MI. Cross classification of smoking status and MI

  7. The odds of having MI is 3.82 times more for smokers as compared nonsmokers.

  8. Rare disease example: MI Yes No placebo aspirin 11034 189 10845 Group 10933 104 11037 293 21778

  9. Yes No 299 374 Stroke unit 673 403 332 Control 735 702 706 Example of OR with common outcome Mortality The frequency of poor outcome (i.e., mortality in the control group) was very high (55%) The OR underestimates the RR by 19%

  10. TESTING HYPOTHESIS USING CI 0 1  1.58 3.78 2.0 .5 .8 .3

  11. ISSUES IN APPROXIMATING ‘OR’ INTO ‘RR’ • When the risks (or odds) in the two groups being compared are both small (say less than 20%) then the OR will approximate to the RR • The discrepancy between odds ratio and relative risk depends on the risk (odds) in both groups (i.e., the discrepancy depends on the initial risk and the OR itself ) • OR may be non-intuitive in interpretation, but in almost all realistic cases interpreting them as though they were RRs is unlikely to change any qualitative assessment of the study findings

  12. Since the odds ratio is difficult to interpret, • why is it so widely used? • Odds ratio is valuable in case control studies where events are usually rare • The odds ratio remains especially useful when researchers need to adjust for other variables, for which LR is usual approach • Odds ratios are a common way of presenting the results of a Meta - Analysis - a statistical analysis for combining the results of several studies, used within systematic reviews

  13. Examples of Confounding and Effect Modification ____________________________________________________ RR (exposed vs non exposed) control var is ------------------------------------------------------------ str1 str2 str3 Total(crude) Conf EM ------------------------------------------------------------------------------ Example A 1.4 1.8 1.3 2.5 Y N Example B 1.4 1.8 1.3 1.5 N N Example C 1.4 1.8 0.2 1.5 N Y Example D 1.4 1.8 0.2 2.5 Y Y ___________________________________________________ Level of Control Variable:

  14. At 24 hours Sex Total Dead Dead % OR RR (95% CI) (95% CI) Female 544 15 2.8 1.96 1.93 Male 3366 48 1.4 (1.1-3.5) (1.09-3.39) Total 3910 63 1.6 Twenty-Four-Hour Mortality After Coronary Artery Bypass Surgery, by Sex

  15. At 24 hours Total Dead Dead % OR RR (95% CI) (95% CI) Low BSA 1250 35 2.8 2.71 2.66 High BSA 2660 28 1.1 (1.68-4.37) (1.66-4.25) Total 3910 63 1.6 Twenty-Four-Hour Mortality After Coronary Artery Bypass Surgery, by Body Surface Area (BSA)

  16. Twenty-Four-Hour Mortality After Coronary Artery Bypass Surgery, by Sex within Each BSA Category At 24 hours Total Dead Dead % OR RR (95% CI) (95% CI) Low BSA Female 483 14 2.9 1.06 1.06 Male 767 21 2.7 (0.54-2.10) (0.54-2.10 High BSA Female 61 1 1.6 1.59 1.58 Male 2599 27 1.0 (0.21-11.79) (0.22-11.40

  17. Dead Sex Yes No Total ai E(ai) V(ai) Low BSA Female 14 469 483 Male 21 746 767 Total 35 1215 1250 14 13.524 8.027 High BSA Female 1 60 61 Male 27 2572 2599 Total 28 2632 2660 1 0.642 0.621 Calculation of Mantel-Haenzel Chi-Square

  18. Dead Exposure Yes No Total OR RR Low BSA Female 14 469 483 8.36/7.88=1.06 8.59/8.11=1.06 Male 21 746 767 Total 35 1215 1250 High BSA Female 1 60 61 0.97/0.61=1.59 0.98/0.62=1.58 Male 27 2572 2599 Total 28 2632 2660 Adjusted Measure 9.32/4.89=1.1 9.57/8.73=1.1 Calculation of Mantel-Haenzel Adjusted Measures

  19. Basics for matching • Study design procedure: • select referent group • comparable to index group on one or more • matching factors

  20. Matching factor = age referent group constraint to have same age structure Case-control study: referent = controls index = cases Follow-up study: referent = unexposed index = exposed

  21. Category Matching Factor A: A1, A2, A3 Factor B: B1, B2, B3 Factor C: C1, C2, C3 Example: AGE: 20-39, 30-39, 40-49, 50-59, 60-69 GENDER: Male, Female STAGE: I, II, III

  22. Category Matching Factor A: A1, A2, A3 Factor B: B1, B2, B3 Factor C: C1, C2, C3 Example: AGE: 20-39, 30-39, 40-49, 50-59, 60-69 GENDER: Male, Female STAGE: I, II, III Control has same age-gender-stage combination as Case

  23. Category Matching Factor A: A1, A2, A3 Factor B: B1, B2, B3 Factor C: C1, C2, C3 Example: AGE: 20-39, 30-39, 40-49, 50-59, 60-69 GENDER: Male, Female STAGE: I, II, III Control has same age-gender-stage combination as Case

  24. Category Matching Factor A: A1, A2, A3 Factor B: B1, B2, B3 Factor C: C1, C2, C3 Example: AGE: 20-39, 30-39, 40-49, 50-59, 60-69 GENDER: Male, Female STAGE: I, II, III

  25. Case Number of Type Controls • R R- to 1 • 1 1-1 or pair matching R may vary from case to case e.g R=3 for some cases R=2 for other cases R=1 for other cases Not always possible to find exactly R controls for each case .

  26. Usual Display of Matched Case-Control Data with Dichotomous Exposure

  27. Analysis of matched data Comparison: proportion of exposed cases vs proportion of exposed controls Information regarding differential exposure is given by the discordant pairs

  28. Odds ratio “Inference regarding the difference in proportions in matched pairs is made solely on the basis of discordant pairs” McNemar (1947) Estimate of OR conditional on the number of discordant pairs is given as Kraus (1960), Cox (1958)

  29. To match or not to match • Advantages: • Matching can be statistically efficient • i,.e may gain precision using confidence interval • Disadvantages: • Matching is costly • To find matches • Information loss due to discarding controls

  30. Match on strong risk factors expected to be confounders. Matching No matching Correct estimate ? YES YES Appropriate analysis? YES YES Matched Standard stratified (Stratified) OR1, OR2, OR3 Combine Validity is not an important reason for matching Match to gain efficiency or precession

  31. Matched Analysis using Stratification: Strata = matched sets Special case: case-control study 100 matched pairs, n=200 100 strata = 100 matched pairs 2 observations per stratum 1st pair 2nd pair 100th pair E E E E E E D D D D D D --------

  32. 1 1 1 0 W pairs 1 0 D D 1 0 1 1 X pairs 0 1 D D 0 1 1 1 Y pairs 1 0 D D 1 1 0 1 Z pairs 0 1 E E D D W + X + Y + Z = total number of pairs

  33. 2. McNemar chi-square test Ch-square = (X-Y)2/(X-Y) = (30-10)2/(30+10) = 10.0 (df=1, p<0.01) McNemar’s test = MH test for pair matching MOR = X/Y = 3 E E D D D 1 1 1 W pairs (30) 0 E E 1 0 E E W Y X Z D D D 1 0 1 1 X pairs (30) 0 1 Y pairs (10) D D 0 1 1 1 1 0 Z pairs (30) D D 1 1 0 1 0 1

  34. Example: • W = 30, X=30, Y =10, Z =30 • W + X + Y + Z = 100 • Analysis : Two equivalent ways • Mantel -Haenzel Chi-square test Stratum 1 Stratum 2 Stratum 100 -------- Compute Mantel-Haenzel chi-square and MOR

  35. Analysis for R-to-1 and mixed matching : Use stratified analysis Example: If, R = 4 E E 1 • 0 • 1 3 1 4 5 • Each stratum contains five subjects • Compute MH chi-square and MOR based on stratified data D D

More Related