E509A: Principle of Biostatistics. GY Zou

Transcription

1 E509A: Principle of Biostatistics (Effect measures ) GY Zou gzou@robarts.ca

2 We have discussed inference procedures for 2 2 tables in the context of comparing two groups. Yes No Group 1 a b n 1 Group 2 c d n 2 m 1 m 2 n For hypothesis testing, we use Pearson chi-square test; For interval estimation, we use methods for p 1 p 2 (of course, NNT).

3 However, Pearson chi-square test will work only if expected value for every cell is greater than 5. For data with small cells, we can use Fisher s exact test. The idea of this test is to fix the row and column totals as the observed table, and compute the probabilities of observing as or more extreme tables in their departure from the null hypothesis (recall the definition of P -value). Fisher (1935, The logic of inductive inference JRSS A 98: 39-54) presented his test at the annual Christmas meeting of the Royal Statistical Society. The title is very good, because I ve heard that the most important contribution of statistics to science is not the formula, but logic. Still, right after his talk, a speaker compared Fisher s talk to the braying of the Golden Ass.

4 The probability of observing the table is given by Pr(a, b, c, d marginals = n 1,n 2,m 1,m 2 )= n 1!n 2!m 1!m 2! n!a!b!c!d! Fisher s procedure requires the probability of all more extreme tables to be computed, using Eq (1) repeatedly. (1) The p-value of the test is obtained by definition: Sum of all those probabilities. Thus, Fisher s exact test is essentially one-sided. If two-sided is called for, the simplest way to do it is to double the p value. This is exactly SAS proc freq gives you when n 1 = n 2.

5 Example of Fisher s exact test (p. 375). Yes No Group Group Pr(a, b, c, d 48, 9, 24) = 24!24!9!39! 48!8!16!1!23! = Amoreextremetableis: Yes No Group Group Pr(a, b, c, d 48, 9, 24) = 24!24!9!39! 48!9!15!0!24! = The p-value is then =

6 SAS function for hypergeometric probability * pdf( HYPER, a, n, n 1,m 1 ); data; bb=pdf( HYPER, 8, 48, 24, 9); cc=pdf( HYPER, 9, 48, 24, 9); dd=bb+cc; proc print; run; Obs bb cc dd Two-sided p-value is then =

7 Yes No Group Group data fisher; do i = 0 to 9; bb=pdf( HYPER, i, 44, 9, 24); output; end; proc print; run;

8 The SAS System Obs i bb One-sided p-value is = Another way called mid-p-value (Lancaster 1961 JASA 56: ): One sided mid-p value is 1/ = Two-sided p-value is 2 mid p = = Two-sided p-values is given by = This is how SAS obtains two-sided p-value.

9 The way to present the results is: Rate in group I was?, in Group II was?; difference? (95% confidence interval? to? ), P =? (Fisher s Exact test two-sided mid P).

10 McKinney et al (1989 The inexact use of Fisher s exact test in six major medical journals JAMA 261: ). Half of 70 articles reviewed either had used a one-tailed test when a two-tailed test was called for, or the authors simply had not bothered to state which test they had used.

11 If only hypothesis testing, an epidemiologist s life would be too easy. Effect estimation makes it hard, also interesting.

12 Besides randomized studies, there are more ways of generating 2 2 table: cross-sectional (naturalistic, multinomial) sampling: select a total of N subjects, followed by the determination for each subject of presence or absence of characteristics of A and B; retrospective sampling: predetermine n 1 of subjects who possess A and n 2 who do not possess A, followed by the determination of B in each group, where A is usually a disease of interest and B is a risk factor. Case-control study prospective sampling: similar to case-control, except A and B is switched. Cohort study.

13 Cross-sectional sample to estimate risk ratio (relative risk, RR) Outcome (D) Exposure (E) Yes (+) No (-) 1(Yes, +) a b n 1 2(No, -) c d n 2 Risk ratio is defined by m 1 m 2 n RR = Pr(D+ E + ) Pr(D + E ) The estimated RR is RR = a/n 1 c/n 2

14 The estimated variance for ln RR estimated by var[ln( RR)] = 1 a 1 n c 1 n 2. 95% CI for RR is obtained by obtaining CI for ln(rr) because the sampling distribution of ln RR is closer to Normal than that of RR l, u =ln( RR) ± 1.96 var(ln RR) The CI for RR is then given by exp(l), exp(u)

15 Example. Data for 200 mothers and their baby birthweight are as Outcome Maternal Age 2500 > 2500 < 20 a =10 b =40 n 1 =50 20 c =15 b = 135 n 2 = 150 m 1 =25 m 2 = 175 n = 200 RR = 10/50 15/150 =2 with variance estimate for ln RR given by var[ln( RR)] = 1 a 1 n c 1 n 2 = = % CI for RR is exp[ln(2) ± ] = (0.96, 4.16)

16 Levin s attributable risk fraction: How much risk would be reduced if the exposure is eliminated? e.g., force all the smoker in London leave town. Since people with disease include two exclusive types: those who were exposed, and those who were not exposed, we have Pr(D + ) = Pr(D + E + )+Pr(D + E ) =Pr(D + E + )Pr(E + )+Pr(D + E )Pr(E ) If E + cannot cause disease, we would expected people with exposure (E + ) have the same disease rate as those who were not exposed, i.e., Pr(D + E ). Thus, the proportion of exposed people will have disease, if the exposure could not cause disease, isgivenby Pr(D + E ) Pr(E + )

17 Levin (1953, Acta Unio Int contra Cancrum 19: ) defined Attributable Fraction as R A = actual counterfactual actual = Pr(D+ E + )Pr(E + ) Pr(D + E )Pr(E + ) = Pr(D + ) Pr(E + )[Pr(D + E + ) Pr(D + E )] Pr(D + E + )Pr(E + )+Pr(D + E )Pr(E ) = Pr(E+ )[Pr(D + E + )/ Pr(D + E ) 1] Pr(E + )Pr(D + E + )/ Pr(D + E )+Pr(E ) Pr(E = + )[RR 1] Pr(E + )RR+1 Pr(E ) R A = Pr(E+ )[RR 1] 1+Pr(E + )(RR 1)

18 R A = Pr(E+ )(RR 1) 1+Pr(E + )(RR 1) Estimated by R A = ] Pr(E + ) [ RR 1 1+ Pr(E + )( RR 1) with estimated variance for ln(1 R A ) given by [ ] var ln(1 R A ) = 1 [ ] b + R A (a + d) nc see Fleiss (1979 Am J Epidemiol 110: ).

19 Example: Infant mortality by birthweight for n = live births in New York City in What % of deaths could have been prevented if low birthweight had been eliminated? 1 yr BW Dead Alive Total 2500g a/n = b/n = (a + b)/n = > 2500g c/n = d/n = (c + d)/n = Total n/n =1 RR = / / = P (E + )=0.0717

20 R A = ] Pr(E + ) [ RR 1 1+ Pr(E + )( RR 1) =.0717( ) ( ) = Variance for ln(1 R A ) is var i.e., [ ] ln(1 R A ) ŝ.e. = ( ) [ ] ln(1 R A ) = var = = % CI for ln(1 R A ) is ln(1 R A ) ± 1.96ŝe[ln(1 R A )] = ln( ) ± =( 0.900, 0.755)

21 CI for R A is then given by [1 exp( 0.755), 1 exp( 0.900)] = (0.530, 0.593) With 95% confidence, between 53% and 59% of all infant death in New York City in 1974 could have been prevented if low birth weight had been eliminated.

22 Attributable risk among the exposed R E =1 1 RR which is widely used in the law to describe the excess risk as a fraction of the risk among those exposed to the antecedent factor. The estimator is R E =1 1 RR Estimation can be conducted through RR.

23 .

24 Cohort study may be used to estimate all of the above effect measures.

25 Case-control study and odds ratio Rothman, Modern Epidemiology, 1986, p.62 The sophisticated use and understanding of case-control studies is the most outstanding methodological development of modern epidemiology My understanding of case-control study is from Breslow Statistics in epidemiology: The case-control study. J Am Stat Assoc 91:14 28.

26 Recall case-control design involves selecting n 1 of subjects who have disease D + and n 2 who do not possess D, followed by the determination of exposure X + or X in each group. Case-control design in general can only provides the ratio of exposure odds of case group to that of the control group, i.e. OR e = exposure odds case exposure odds control Since odds is defined as P 1 P, OR e = Pr(X + D + ) 1 Pr(X + D + ) Pr(X + D ) 1 Pr(X + D ) = Pr(X+ D + )Pr(X D ) Pr(X + D )Pr(X D + )

27 For OR e to be useful, it must have some relationship with risk ratio Pr(D + X + ) Pr(D + X ). Entered Cornfield (1951 J. Natl Cancer Inst 11: ) who showed that 1) Pr(D+ X+)Pr(X+) Pr(X + D )Pr(X D + ) = Pr(D + ) OR e = Pr(X+ D + )Pr(X D ) = Pr(D+ X + )Pr(D X ) Pr(D X + )Pr(D + X ) = OR d Pr(D X )Pr(X ) Pr(D ) Pr(D X + )Pr(X + ) Pr(D Pr(D+ X )Pr(X ) ) Pr(D + ) 2) Pr(D X ) Pr(D + X ) 1, whenpr(d+ ) 0. which implies that OR d RR d. To see 2), observe Pr(D X ) = Pr(X D )P (D ) = P (X ) = Pr(X D )P (D ) Pr(X D )+Pr(X D + ) Pr(X D )P (D ) Pr(X D )Pr(D )+Pr(X D + )Pr(D + ) 1, when Pr(D+ ) 0 Similarly Pr(D X + ) 1 when Pr(D + ) 0.

28 Thus, case-control studies are indeed useful. In fact, Mantel & Haenszel (1959, J Natl Cancer Inst 22: ) stated Among the desirable attributes of the retrospective study is the ability to yield results from presently collectible data... The retrospective approach is also adapted to the limited resources of an individual investigator... For especially rare disease a retrospective study may be the only feasible approach... In the absence of important biases in the study setting, the retrospective method could be regarded, according to sound statistical theory, as the study method of choice (p. 720). This was almost 50 years ago. Recent epidemiologic literature has seen more and more prospective studies, except genetic epidemiologic literature in which case-control design is almost universal. More detailed discussion of statistical issues may be found in Zou (2006 Annals of Human Genetics 70: ).

29 Status Exposure Case Control Yes a b n 1 No c d n 2 Odds ratio in a case control study is estimated by ÔR = ad bc with variance of ln(ôr) estimated by var[ln(ôr)] = 1 a + 1 b + 1 c + 1 d

30 Thus, (1 α) 100% CI for OR is given by [ ] exp ln(ôr) ± Z 1 α/2 var[ln(ôr)]

31 Example. Sun protection during childhood by case-control status for cutaneous melanoma in Belgium, France and Germany. Exposure Sun protection Case Control Yes No ÔR = =0.72 var[ln(ôr)] = = % CI for OR is exp[ln(0.72) ± ] = exp( , ) =(0.53, 0.98)

32 Status Exposure Case Control Yes a b n 1 No c d n 2 Odds ratio in a case control study is estimated by ÔR = ad bc Status trt Yes No 1 a =0 b =14 n 1 =14 2 c =0 d =11 n 2 =11 This is a data set discussed by Parzen M, Lipsitz S, Ibrahim J, Klar N An estimate of the odds ratio that always exists. J Comput Graph Stat 11:

33 Exact confidence interval for OR It is available in SAS proc freq, which computes exact confidence limits for the odds ratio with an algorithm by Thomas (1971, Applied Statistics 20: ), ie., the limits L and U are iterative solutions for the following two equations: m1 ( n1 ) L i i=a m1 i=0 a i=0 m1 i=0 )( n2 i m 1 i ( n1 )( n2 i m 1 i ( n1 )( n2 i m 1 i ( n1 )( n2 i m 1 i ) L i = α/2 ) U i ) = α/2 U i

34 Exact may not be the best in categorical data analysis because the results may be too conservative Agresti A Dealing with discreteness: making exact confidence intervals for proportions, differences of proportions, and odds ratios more exact Stat Meth Med Res 12 (1): the inversion of the asymptotic score test seems to be a good choice. This tends to have actual level fluctuating around the nominal level. If one prefers that level to be a bit more conservative, mid-p adaptations of exact methods work well. For situations that require a bound on the error, it appears that basing conservative intervals on inverting the exact score test has reasonable performance. For teaching, the Wald-type interval of point estimate plus and minus a normal-score multiple of a standard error is simplest. Unfortunately, this can perform poorly, but simple adjustments sometimes provide much improved performance.

35 Odds ratio versus risk ratio. Most traditional statistical methods in epidemiology were developed in the case of case-control design. Specifically, OR was the effect measure of choice. Unfortunately, when it comes to prospective or cross-sectional design, the rare disease assumption may not be satisfied. In such cases, OR becomes very difficulty to interpret, sometimes misleading (see NEJM 1999;341:279 83). For a RR =2, p p OR = p 1/(1 p 1 ) p 2 /(1 p 2 )

36 My view is that we must always remember why Cornfield (1951) proposed OR. Excellent discussion on the choice of effect measures may be found in Greenland (Interpretation and choice of effect measures in epidemiologic analyses. Am J Epidemiol 1987;125:761 8).

37 Converting OR to RR (Zhang & Yu, 1998 JAMA 280: ): RR = RR = OR 1 p 2 + p 2 OR ÔR 1 p 2 + p 2 ÔR (2) Eq. (2) results in correct point estimate only if they are no confounder; Substituting confidence limits for OR to obtain CI for RR yields invalid interval for RR. More discussion can be found in McNutt et al (2003 Am J Epidemiol 157:940 3) and Zou (2004 Am J Epidemiol 159: 702 6).

38 A little trick to check calculations when a confidence interval is constructed through log-transformation (Lee PN Stat Med 18: ): Such a interval should satisfies: square of the point estimate should equal to the product of lower and upper limits. l = exp[ln point Z var(ln point)] u = exp[ln point + Z var(ln point)] thus l u =exp(2 ln point) =(point) 2

39 Sample size estimation with SAS proc power proc power; twosamplefreq test=pchi relativerisk = 1.5 refproportion = 0.2 power=0.8 ntotal=.; run;

40 The POWER Procedure Pearson Chi-square Test for Two Proportions Fixed Scenario Elements Distribution Asymptotic normal Method Normal approximation Reference (Group 1) Proportion 0.2 Relative Risk 1.5 Nominal Power 0.8 Number of Sides 2 Null Relative Risk 1 Alpha 0.05 Group 1 Weight 1 Group 2 Weight 1 Computed N Total Actual N Power Total

41 proc power; twosamplefreq test=pchi oddsratio = 2.5 refproportion = 0.3 groupweights = (1 2) ntotal =. power = 0.8; run;

42 The POWER Procedure Pearson Chi-square Test for Two Proportions Fixed Scenario Elements Distribution Asymptotic normal Method Normal approximation Reference (Group 1) Proportion 0.3 Odds Ratio 2.5 Group 1 Weight 1 Group 2 Weight 2 Nominal Power 0.8 Number of Sides 2 Null Odds Ratio 1 Alpha 0.05 Computed N Total Actual N Power Total

43 proc power; twosamplefreq test=fisher groupproportions = (.35.15) power=0.80 npergroup =.; run;

44 The POWER Procedure Fisher s Exact Conditional Test for Two Proportions Fixed Scenario Elements Distribution Exact conditional Method Walters normal approximation Group 1 Proportion 0.35 Group 2 Proportion 0.15 Nominal Power 0.8 Number of Sides 2 Alpha 0.05 Computed N Per Group Actual N Per Power Group

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46 Combine information from multiple 2 2 tables (Mantel-Haenszel methods)

47 Outcome Exposure Yes No Yes a k b k n 1k No c k d k n 2k m 1k m 2k n k MH test for no association between exposure and outcome (Mantel & Haenszel, 1959 J Natl Cancer Inst 22: ) [ ( )] 2 χ 2 k a k m 1kn 1k n k MH = k m 1k m 2k n 1k n 2k n 2 k (n k 1) which is distributed as chi-square with one degree-of-freedom, under H 0. 5 years earlier, Cochran (1954 Biometrics 10: ) proposed a test that is virtually identical to χ 2 MH (difference?) When k =1, χ 2 MH reduce to Pearson chi-square test for 2 2 table.

48 Mantel-Haenszel odds ratio estimator (1959) OR MH = k a kd k /n k k b kc k /n k For 20 years, nobody knew what was the standard error for OR MH. Hauck (1979, Biometrics 35: ) provided a formula that is valid when each table are large. Outcome Exposure Yes No Yes a k b k n 1k No c k d k n 2k m 1k m 2k n k

49 The popular variance formula for OR MH is the one derived by Robins, Breslow, & Greenland (1986, Biometrics 42: ). Recall OR MH = k a kd k /n k k b kc k /n k = k R k k S k Define two more terms P k =(a k + d k )/n k and Q k =(b k + c k )/n k var[ln( OR k MH )] = P kr k 2 ( k k R ) 2 + P ks k + k Q kr k k 2 ( k R )( k k S ) k + Q ks k k 2 ( k S k Outcome Exposure Yes No Yes a k b k n 1k No c k d k n 2k m 1k m 2k n k ) 2

50 (1 α) 100% CI for OR [ ( ) exp ln OR MH ] ± Z 1 α/2 var[ln( OR MH )

51 Example (case-control study): OR MH. Case-control studies on the role of high voltage power lines in the etiology of leukemia in children (Hanley & Thriault, 2000 Epidemiology 11(5): 613) Study 1 Study 2 Case Control Case Control < 100m a k =18 b k =25 n 1k = > 100m c k = 162 d k = 252 n 2k = m 1k = 180 m 2k = 277 n k = ÔR 1 =1.12 ÔR 2 =1.62

52 If not stratify, we have Status Case Control < 100m a =30 b = 148 > 100m c = 188 d = 683 ÔR = from leukemia. =0.74, living closer to powerlines protects children However, OR MH = = =

53 Mantel-Haenszel technique has also been used to derive RR estimator (Tarone, 1981 J Chronic Dis 34: ): RR MH = k a kn 2k /n k k c kn 1k /n k with variance given by var[ln( RR MH )] = ( k [n 1kn 2k m 1k a k c k n k ] /n 2 k k a )( kn 2k /n k k c ) kn 1k /n k Outcome Exposure Yes No Yes a k b k n 1k No c k d k n 2k m 1k m 2k n k

54 (1 α) 100% CI for RR [ ( ) exp ln RR MH ] ± Z 1 α/2 var[ln( RR MH )

55 Ex 8.5. Example (Clinical trial): RR MH Age 65+ Age 65- Drug Yes No Yes No B a k =32 b k =8 n 1k = A c k =24 d k =36 n 2k = m 1k =56 m 2k =44 n k = χ 2 MH [ = k k ( a k n 1k m 1k )] 2 n k n 1k n 2k m 1k m 2k n 2 k (n k 1) ) 2 = ( (100 1) (100 1) =18.435

56 RR MH = k a kn 2k /n k k c kn 1k /n k = = =2.0 var[ln( RR)] = 95% CI is then k (n 1kn 2k m 1k a k c k n k )/n 2 k ( k a kn 2k /n k)( k c kn 1k /n k) = = exp(ln 2 ± ) = (1.44, 2.77) =2 2 (checked)

57 Summary: Application of Mantel-Haenszel methods Adjust for confounding (the original purpose): combat Simpson s paradox; Meta-analysis: considering each study as a stratum. The method of meta-analysis is commonly referred to as fix-effect model with intention of summarizing available evidence, but not to predict future study results. For that, random-effect model must have to be adopted.

58 A note: Like fire, the chi-square test is an excellent servant and a bad master (Hill, 1965 Proc R Soc Med 58: ). Study 1 Study 2 Exposure D + D Risk D + D Risk E E RR OR p-value