S tatistik fo r M P H : 5 F ra d en 4 . u g es statistik u n d erv isn in g : F ...

Confounding.Do we always get a fair comparison between exposed and non-exposed?Not necessarily - a randomly selected exposed person will (in the diagram below)typically be older than a randomly chosen non-exposed.This is a problem if age is a risk factor for the outcome.EXPOSEDYoungNON-EXPOSEDYoungOldOld3Confounding.A variable C is a potential confounder for the association:E → Oif it is1. related to the exposure:E − C2. an independent risk factor for the outcome:C → O3. not a consequence of the exposure (not an “intermediate” variable):E → C → OThat is:E − Cց ւO4

Risk of death sentence:White murderer:722257 = 3.2%Black murderer:592606 = 2.3%⎫RR = 3.2%2.3% = 1.41 ⎬OR = 72·254759·2185 = 1.42 white vs. black⎭ln(OR) = 0.352q1L 1 = 0.352 − 1.96 + 1 + 1 + 1 = 0.003559 2547 72 2185q1L 2 = 0.352 + 1.96 + 1 + 1 + 1 = 0.70159 2547 72 218595% conf. limits: from exp(L 1 ) = 1.00 to exp(L 2 ) = 2.027Possible confounder: race of victimSentenceVictim Murderer Death Other TotalBlack White 0 111 111Black 11 2309 2320Total 11 2420 2431White White 72 2074 2146Black 48 238 286Total 120 2312 24328

How do we see if “race of victim” is a confounder?(2) SentenceVictim death otherwhite 120 2312black 11 2420(1) MurdererVictim white blackwhite 2146 286black 111 2320OR = 11.41OR = 156.8Separate analyses in strata defined by confounder:Black victim OR = 0·230911·111= 0White victim OR =(white murderer vs. black murderer)72·23848·2074= 0.1729Combined analysis over strata(= stratified analysis)(= the Mantel-Haenszel method)We have a series (here two!) of two by two tables: one from eachstratum.stratum 1stratum ka b · · · a bc d · · · c dnnIn each stratum, we can estimate odds ratio bya · db · c = a · d/nb · c/n10

The Mantel-Haenszel estimator.A common odds ratio for all strata may be estimated by theMantel-Haenszel estimator∑ a·d= OR MHn∑ b·cn(weighted average of separate OR-estimates)In the example:OR MH =0·23092431 + 72·238243211·1112431 + 48·20742432= 0.17011The Mantel-Haenszel test.In each stratum, we can calculate:OBServed = aEXPected = (a+b)(a+c)SD == E(a)√ n(a+b)(c+d)(a+c)(b+d)n 2 (n−1)= SD(a)The combined Mantel-Haenszel test statistic is( ∑ a − ∑ E(a)) 2∑ (SD(a))2= X 2 MH ∼ χ 2 1 under H 0In the example: X 2 MH =((0 + 72) −( 111·11)) 22431 + 2146·1202432111·11·2320·2420(2431) 2·2430 + 120·2146·286·2312(2432) 2·2431= (−34.39)212.32= 96.0, P < 0.001.12

Interpretation:OR MH is an estimate of the association between exposure (colour ofmurderer) and outcome (use of death sentence), adjusted for theconfounder (colour of victim).XMH 2 is a test statistic for no association between exposure andoutcome, adjusted for the confounder.Examination of confoundingNote that 1. and 2., but NOT 3. can be checked (statistically) on thedata at hand.Degree of confounding is often examined using the change in estimateprinciple: how different are adjusted and unadjusted (“marginal”)estimates.Selection of confounders is often based on prior knowledge.13Confidence limits for common odds ratio1) calculate ln(OR MH ), here: ln(0.170) = −1.7722)calculate L 1 = ln(OR MH ) − 1.96 · SDwhere SD = |ln(OR √ MH)|, X2MHL 2 = ln(OR MH ) + 1.96 · SDhere SD = 1.772 √96.0= 0.181 that is:L 1 = −1.772 − 1.96 · 0.181 = −2.126,L 2 = −1.772 + 1.96 · 0.181 = −1.4183) The desired 95% confidence limits are from exp(L 1 ) = 0.119 toexp(L 2 ) = 0.242.Alternative (more complicated) formula for SD on Silva, p.327.14

Exercise:Analyse (using the Mantel-Haenszel method) the association betweenlength of prenatal care and neonatal mortality based on the followingdata.Is clinic a confounder for this association?Mortality among newborns and amount of prenatal care (McNeil, 1996).Clinic A prenatal care

The “marginal” 2 by 2 table ignoring clinic isprenatal care

When is the stratified analysis sensible?In the stratified analysis, we average the individual OR’s from theseparate strata.This makes sense if the individual OR’s are roughly the same in allstrata (taking the random variation into account).= if there is no interaction between exposure and stratificationvariable on the outcome= if there is no effect-modification of the stratification variable onthe relation between exposure and outcome19ExampleTests for no interaction existbut go beyond what we can cover in this course.Prevalence of myocardial infarction by systolic blood pressure and age (IsraeliIschemic Heart Disease Study, Kahn & Sempos, 1989).Age≥60 MI cases MI negative TotalSBP≥ 140 9 115 124SBP< 140 6 73 79Total 15 188 203 OR=0.95Age

Adjustment for confounding in cohort studies:Stratified analysis of rates (and risks)Male stomach cancer cases in Cali and Birmingham:CaliBirminghamAge Popu- Person- No.of Rate per Popu- Person- No.of Rate perlation years cancers 100000 ys. lation years cancers 100000 ys.0-44 524220 2621100 39 1.5 1683600 6734400 79 1.245-64 76304 381520 266 69.7 581500 2326000 1037 44.665+ 22398 111990 315 281.3 291100 1164400 2352 202.0Total 622922 3114610 620 19.9 2556200 10224800 3468 33.9Crude RR = 19.933.9Age-adjustment?Previously: standardisation.= 0.59 not relevant due to age-confounding.Now: stratified (Mantel-Haenszel) analysis.21Mantel-Haenszel analysis.We have a series (here 3) of tables of the form:Events Person Stratum no. 3years“Risk factor +” a 1 y 1 a 1 = 315 y 1 = 111990“Risk factor −” a 2 y 2 a 2 = 2352 y 2 = 1164400Total a y a = 2667 y = 1276390One table from each stratum.22

The Mantel-Haenszel estimator.In each table, we may estimate the rate ratio bya 1 /y 1a 2 /y 2= a 1y 2a 2 y 1= a 1y 2 /ya 2 y 1 /yA common rate ratio for all strata may be estimated by theMantel-Haenszel estimator (Weighted average of separateRR-estimates):P a1 y 2yP a2 y 1y= RR MH .In example: RR MH =39·67344009355500+ 266·23260002707520+ 315·1164400127639079·26211009355500+ 1037·3815202707520+ 2352·1119901276390= 1.45.23The Mantel-Haenszel test.In each stratum, we can calculate:OBServed = a 1EXPected = a · y1y= (a 1 + a 2 )y 1 +y 2= E(a 1 )√ √Standard Deviation = a y 1 y 2yy y= (a 1 + a 2 ) 1 y 2(y 1 +y 2 )= SD(a 2 1 )The combined Mantel-Haenszel test statistic is:( ∑ a 1 − ∑ E(a 1 )) 2∑ (SD(a1 )) 2 = X 2 MH ∼ χ 2 1 under H 0y 1X 2 MH =((39 + 266 + 315) −(11826211009355500+ 13033815202707520))111990 2+ 26671276390118 2621100·6734400(9355500)+ 1303 381520·23260002 (2707520)+ 2667 111990·11644002 (1276390) 2= (169.33) 2 /395.00 = 72.59 ∼ χ 2 1, P < 0.001.24

95% confidence limits for common rate ratio:1) Calculate ln(RR MH ) = ln(1.45) = 0.3732) Calculate L 1 = ln(RR MH ) − 1.96 · SDand L 2 = ln(RR MH ) + 1.96 · SDHere: SD = |ln(RR √ MH)|= √ 0.373X2MH72.59= 0.044.That is: L 1 = 0.373 − 1.96 · 0.044 = 0.287,L 2 = 0.373 + 1.96 · 0.044 = 0.459.3)The desired 95% confidence limits are from exp(L 1 ) = 1.33 toexp(L 2 ) = 1.58.Note: alternative (and more complicated) formula for SD in Silva,p.330.25Example:British Doctors cohort study, comparing coronary deaths amongsmokers and non-smokers adjusted for age (McNeil, 1996)DeathsPerson-yearsAge Group SM Non-SM SM Non-SM RR35-44 32 2 52407 18790 5.7445-54 104 12 43248 10673 2.1455-64 206 28 28612 5710 1.4765-74 186 28 12663 2585 1.3675-84 102 31 5317 1462 0.90Total 630 101 142247 39220 1.72Crude RR = 630/142247101/39220 = 1.72Age-adjustment not obvious, RR varies among strata:Effect-modification/interaction !?26

27If risks (“person-denominators”) are used instead of rates(“person-years-denominators”):Events No-events TotalRisk factor + a c a + cRisk factor − b d b + dExactly the same calculations may be used:Replacing y 1 by a + cand y 2 by b + dSee: Silva, p.329.28

Exercise:Consider the British Doctors Study restricting attention to the twoage groups 45-54 and 55-64 and compare smokers and non-smokersadjusting for age.Data:45-54 Deaths PYRS 55-64 Deaths PYRSSM 104 43248 SM 206 28612Non-SM 12 10673 Non-SM 28 5710116 53921 234 3432229Solution.X 2 MH =RR MH =((104 + 206) −(116 · 4324853921104·1067353921+ 206·57103432212·4324853921+ 28·2861234322+ 234 · 2861234322= 1.66116 · 43248·10673(53921) 2 + 234 · 28612·5710(34322) 2 = 9.42 ∼ χ 2 1, P ∼ 0.002)) 2ln(RR MH ) = 0.509 , SD(ln(RR MH )) = 0.509 √9.42= 0.166L 1 = 0.509 − 1.96 × 0.166 = 0.184, L 2 = 0.509 + 1.96 × 0.166 = 0.834.95% confidence interval from exp(L 1 ) = 1.20 to exp(L 2 ) = 2.3030