Lecture 5: Contingency Tables

Transcription

1 Lecture 5: Contingency Tables Dipankar Bandyopadhyay, Ph.D. BMTRY 711: Analysis of Categorical Data Spring 2011 Division of Biostatistics and Epidemiology Medical University of South Carolina Lecture 5: Contingency Tables p. 1/46

2 Overview Over the next few lectures, we will examine the 2 2 contingency table Some authors refer to this as a four fold table We will consider various study designs and their impact on the summary measures of association Lecture 5: Contingency Tables p. 2/46

3 Rows Fixed: Product Binomial Case - Generally Prospective Studies Question of interest: Does treatment affect outcome? OUTCOME NO COLD COLD Total T R VITAMIN E C <-- A (fixed T by M NO design) E VITAMIN <-- N C T Total Lecture 5: Contingency Tables p. 3/46

4 Columns Fixed: Also Product Binomial - Generally Retrospective Studies Question of interest: Does alcohol exposure vary among cases and controls? STATUS CASE CONTROL Total A L 80+ C (gm/day) O H O 0-79 L (gm/day) Total ˆ ˆ (fixed by design) Lecture 5: Contingency Tables p. 4/46

5 N Fixed: Multinomial Case - Generally Cross-Sectional Studies Question of interest: Is there an association among cancer stage and smoking status? CANCER STAGE NOT SPREAD SPREAD Total S M YES O K E NO Total <---(fixed by design) Lecture 5: Contingency Tables p. 5/46

6 Rows and Columns Fixed: Hypergeometric Case Question of Interest: Is there gender bias in juror selection? SELECTED FOR JURY YES NO Total G E N FEMALE D E R MALE Total This distribution is will be used in Fisher s Exact testing. Lecture 5: Contingency Tables p. 6/46

7 Prospective Studies We are going to begin examining contingency tables first by looking at prospective studies. Number on each treatment (or experimental) arm fixed by design. Rows are independent binomials. Question of interest: Does treatment affect outcome? Usually the design for Experimental Studies, Clinical Trials. In general, the 2 2 table is written as Outcome 1 2 Treatment 1 Y 1 n 1 Y 1 n 1 2 Y 2 n 2 Y 2 n 2 Lecture 5: Contingency Tables p. 7/46

8 Facts about the distribution n 1 and n 2 are fixed by design Y 1 and Y 2 are independent with distributions: Y 1 Bin(n 1, p 1 ) Y 2 Bin(n 2, p 2 ) The distribution is the product of 2 independent binomials; often called the product binomial : P(y 1, y 2 p 1, p 2 ) = P(Y 1 = y 1 p 1 )P(Y 2 = y 2 p 2 ) = n 1 y 1! n 2 y 2! p y 1 1 (1 p 1) n 1 y 1 p y 2 2 (1 p 2) n 2 y 2 Lecture 5: Contingency Tables p. 8/46

9 Question of interest (all the same) Does treatment affect outcome? Are treatment and outcome associated? Is the probability of success the same on both treatments? How do we quantify treatment differences? Also, what test statistics can we use for H 0 :p 1 = p 2 = p and the alternative is H A :p 1 p 2 Lecture 5: Contingency Tables p. 9/46

10 MLE and estimated SEs of treatment differences To estimate these treatment differences, we must estimate the success probabilities p 1 and p 2. Intuitively, thinking of the two groups separately, the MLE s should be the proportion of successes in the two groups, i.e., bp 1 = Y 1 n 1 and bp 2 = Y 2 n 2. However, we will derive these based on the likelihood of the product binomial. Lecture 5: Contingency Tables p. 10/46

11 The Likelihood for (p 1, p 2 ) is the product binomial distribution of (y 1, y 2, p 1, p 2 ). L(p 1, p 2 ) = P(Y 1 = y 1 p 1 )P(Y 2 = y 2 p 2 ) = n 1 y 1! n 2 y 2! p y 1 1 (1 p 1) n 1 y 1 p y 2 2 (1 p 2) n 2 y 2 Lecture 5: Contingency Tables p. 11/46

12 Then the log-likelihood is the sum of the two pieces, log L(p " 1, p 2 )! = log n 1 y 1 p y 1 1 (1 p 1) n 1 y 1 # + log " n 2 y 2! p y 2 2 (1 p 2) n 2 y 2 # Similar to before, to find the MLE, we set the partial derivatives of log L(p 1, p 2 ) with respect to p 1 and p 2 to 0, and solve for bp 1 and bp 2 : Note: Agresti (and most statisticians) simply denote the natural logarithm as log instead of the ln as you would see in mathematics or physics. In this class, all references of log are consider the log to base e. Lecture 5: Contingency Tables p. 12/46

13 Now, log L(p 1, p 2 ) = log " n 1 y 1! p y 1 1 (1 p 1) n 1 y 1 # + log " n 2 y 2! p y 2 2 (1 p 2) n 2 y 2 # The derivative of the log-likelihood with respect to p 1 is d log L(p 1,p 2 ) dp 1 = = d dp 1 log d dp 1 log " d dp 1 log since the the second part is not a function of p 1. " n 1 y 1 " n 1 y 1! n 2 y 2! p y 1 1 (1 p 1) n 1 y 1! # + # p y 2 2 (1 p 2) n 2 y 2 p y 1 1 (1 p 1) n 1 y 1 # + 0 Lecture 5: Contingency Tables p. 13/46

14 Note, though, dlog L(p 1, p 2 ) dp 1 " = d log dp 1 n 1 y 1! p y 1 1 (1 p 1) n 1 y 1 # is just the derivative of a binomial log-likelihood with respect to its parameter p 1. From before, we have bp 1 = y 1 n 1 To explicitly show this, in the single binomial section, we showed that d log L(p 1, p 2 ) dp 1 " = d log dp 1 n 1 y 1! p y 1 1 (1 p 1) n 1 y 1 # = y 1 n 1 p 1 p 1 (1 p 1 ) Similarly, d log L(p 1, p 2 ) dp 2 = y 2 n 2 p 2 p 2 (1 p 2 ) Lecture 5: Contingency Tables p. 14/46

15 Then, the MLE s are found by simultaneously solving and which gives d log L(p 1, p 2 ) dp 1 = y 1 n 1 bp 1 bp 1 (1 bp 1 ) = 0 d log L(p 1, p 2 ) dp 2 = y 2 n 2 bp 2 bp 2 (1 bp 2 ) = 0 bp 1 = y 1 n 1 and bp 2 = y 2 n 2. provided that bp 1, bp 2 0,1 Lecture 5: Contingency Tables p. 15/46

16 Since Y 1 and Y 2 are independent binomials we know that and V ar(bp 1 ) = p 1(1 p 1 ) n 1 V ar(bp 2 ) = p 2(1 p 2 ) n 2 Lecture 5: Contingency Tables p. 16/46

17 Estimating the treatment differences To obtain the MLE of the log-odds ratio, we just plug bp 1 and bp 2 in to get log(dor) = log bp1 /(1 bp 1 ) bp 2 /(1 bp 2 ) = logit(bp 1 ) logit(bp 2 ) Now, suppose we want to estimate the variance of log( d OR). Since the treatment groups are independent, logit(bp 1 ) and logit(bp 2 ) are independent, so that Cov[logit(bp 1 ), logit(bp 2 )] = 0, The variance of differences of independent random variables is V ar[log( d OR)] = V ar[logit(bp 1 ) logit(bp 2 )] = V ar[logit(bp 1 )] + V ar[logit(bp 2 )] Lecture 5: Contingency Tables p. 17/46

18 Delta method approximation The V ar[log( d OR)] can be approximated by the delta method To do so we need to calculate d d p [log(p) log(1 p)] = 1 p 1 1 p = 1 p(1 p) Therefore, Var log( p 1 p ) = = = 2 1 p(1 p) p(1 p) n 1 np(1 p) 1 np + 1 n(1 p) Using these results from the Delta Method, we have and V ar[logit(bp 1 )] = 1 n 1 p 1 + V ar[logit(bp 2 )] = 1 n 2 p n 1 (1 p 1 ) 1 n 2 (1 p 2 ) Lecture 5: Contingency Tables p. 18/46

19 Then, V ar[log( d OR)] = V ar[logit(bp 1 )] + V ar[logit(bp 2 )] = 1 n 1 p n 1 (1 p 1 ) n 2 p 2 1 n 2 (1 p 2 ) which we estimate by replacing p 1 and p 2 with bp 1 and bp 2, dv ar[log( d OR)] = 1 n 1 bp n 1 (1 bp 1 ) n 2 bp 2 1 n 2 (1 bp 2 ) = 1 y n 1 y y n 2 y 2 Note: This is the same result we obtained in the previous lecture; however, in this case we assumed two independent binomial distributions. Lecture 5: Contingency Tables p. 19/46

20 General formula for variance of treatment difference The MLE of a treatment difference θ = g(p 1 ) g(p 2 ) is ˆθ = g(bp 1 ) g(bp 2 ) Also, since bp 1 and bp 2 are independent, so g(bp 1 ) and g(bp 2 ) are independent. Recall, the variance of a difference of two independent random variables is V ar[g(bp 1 ) g(bp 2 )] = V ar[g(bp 1 )] + V ar[g(bp 2 )] Then, to obtain the large sample variance, we can apply the delta method to g(bp 1 ) to get V ar[g(bp 1 )] and to g(bp 2 ) to get V ar[g(bp 2 )] and then sum the two. Lecture 5: Contingency Tables p. 20/46

21 The results are summarized in the following table: TREATMENT DIFFERENCE ESTIMATE Var(ESTIMATE) RISK DIFF bp 1 bp 2 p 1 (1 p 1 ) n 1 + p 2(1 p 2 ) n 2 log (RR) log bp1 bp2 log (OR) log 1 p 1 n 1 p p 2 n 2 p 2 h bp1 /(1 bp 1 ) 1 + bp 2 /(1 bp 2 ) n 1 p 1 i h n 1 (1 p 1 ) n 2 p 2 i 1 n 2 (1 p 2 ) Lecture 5: Contingency Tables p. 21/46

22 ESTIMATES of Standard Error, and LARGE SAMPLE CONFIDENCE INTERVALS To estimate the variances, we can replace p 1 and p 2 with bp 1 and bp 2. dv ar(bp 1 bp 2 ) = bp 1(1 bp 1 ) n 1 + bp 2(1 bp 2 ) n 2 ; dv ar[log( d RR)] = 1 bp 1 n 1 bp bp 2 n 2 bp 2 ; dv ar[log( d OR)] = 1 n 1 bp n 1 (1 bp 1 ) n 2 bp 2 1 n 2 (1 bp 2 ) = 1 y n 1 y y n 2 y 2 Lecture 5: Contingency Tables p. 22/46

23 Then Large Sample 95% confidence interval for treatment differences can be obtained via (bp 1 bp 2 ) ± 1.96 s bp 1 (1 bp 1 ) n 1 + bp 2(1 bp 2 ) n 2 log(drr) ± 1.96 s 1 bp 1 n 1 bp bp 2 n 2 bp 2 and log(dor) ± 1.96 s 1 y n 1 y y n 2 y 2 Lecture 5: Contingency Tables p. 23/46

24 Confidence Interval for OR and RR You want a confidence interval for RR or OR that is assured to be in the interval (0, ). Similar to what we did for a confidence interval for p, it is first better to get confidence intervals for log(rr) or log(or), and to exponentiate the endpoints : i.e., q exp{log( OR) d ± 1.96 dv ar[log( OR)]}, d and q exp{log( RR) d ± 1.96 dv ar[log( RR)]}, d Lecture 5: Contingency Tables p. 24/46

25 Example: MI example Suppose clinical trial participants are randomized to either Placebo or Aspirin The subjects are followed prospectively for 5 years to determine whether or not an MI (or heart attack) occurs The following table summarizes the results Myocardial Infarction Heart or No Total per Attack Attack Arm Placebo ,034 Aspirin ,037 About randomized to each treatment Overall probability of heart attack is low The disease is rare = 1.33% Lecture 5: Contingency Tables p. 25/46

26 Estimates and Test Statistics The test statistics for H 0 : p 1 = p 2 versus H A : p 1 p 2 Estimated Z Statistic Parameter Estimate Standard Error (Est/SE) RISK DIFF log(rr) (RR=1.818) log(or) (OR=1.832) In each case, we reject the null, and the Z statistic is about 5. Lecture 5: Contingency Tables p. 26/46

27 Confidence Intervals Creation The following are the 95% confidence intervals Parameter Estimate 95% C.I. RISK DIFF.0077 [.0047,.0107] RR [1.433,2.306] OR [1.440,2.331] For the OR and RR, we exponentiated the 95% confidence intervals for the log(or) and log(rr), respectively. None of the confidence intervals contain the null value for no association (0 for the RISK DIFFERENCE, 1 for the OR and RR). Lecture 5: Contingency Tables p. 27/46

28 Interpretation The risk difference has the interpretation that the Excess Risk of a heart attack on Placebo is This fraction is not very meaningful for rare diseases, but stated in terms of subjects, we can say that we would expect 77 more heart attacks in placebo subjects than in aspirin users. The relative risk has the interpretation that Individuals on Placebo have almost twice (1.8) the risk (or probability) of a heart attack than individuals on Aspirin The odds ratio has the interpretation that Individuals on Placebo have almost twice (1.8) the odds of a heart attack versus no heart attack than individuals on Aspirin Lecture 5: Contingency Tables p. 28/46

29 Relationship between OR and RR Recall, OR = p 1/(1 p 1 ) p 2 /(1 p 2 ) OR = p 1/(1 p 1 ) p 2 /(1 p 2 ) = h i p1 1 p2 p 2 1 p 1 h i = RR 1 p2 1 p 1 When the disease is rare (in the example, bp 2 < bp 1 < 2%),» 1 p2 1 1 p 1 1 = 1; and OR RR. In the example, d OR = 1.832, d RR = 1.818; i.e., they are almost identical. Lecture 5: Contingency Tables p. 29/46

30 Likelihood Ratio Test Now, we want to test the null hypothesis H 0 :p 1 = p 2 = p versus the alternative H A :p 1 p 2 with the likelihood ratio statistic (the likelihood ratio statistic generally has a two-sided alternative...i.e., it is χ 2 based). The general likelihood ratio statistic involves the estimate of p 1 = p 2 = p under the null and (p 1, p 2 ) under the alternative. Thus, unlike the simple single binomial sample we discussed earlier, in which the null was H 0 :p =.5 the parameters are not completely specified under the null. i.e., we must still estimate a common p under the null for the likelihood ratio. Lecture 5: Contingency Tables p. 30/46

31 General Likelihood Ratio Statistic The likelihood is a function of the parameter vector p = [p 1, p 2 ]. In large samples, it can be shown that 2 log n o L(bp1,bp 2 H A ) L( p 1, p 2 H 0 ) = 2[log L(bp 1, bp 2 H A ) log L( p 1, p 2 H 0 )] χ 2 df where L(bp 1, bp 2 H A ) is the likelihood after replacing [p 1, p 2 ] by its estimate, [bp 1, bp 2 ] under H A and L( p 1, p 2 H 0 ) is the likelihood after replacing [p 1, p 2 ] by its estimate, [ p 1, p 2 ], under H 0 (In our case, [ p 1, p 2 ] = [bp, bp] since p 1 = p 2 = p under the null ). and the degrees-of-freedom df is the difference in the number of parameters estimated under the alternative and null (In our example, df = 2 1 = 1). Lecture 5: Contingency Tables p. 31/46

32 MLE under the Null Thus, to use the likelihood ratio statistic, we need to estimate the common p under the null hypothesis. When H 0 : p 1 = p 2 = p, E(Y 1 ) = n 1 p and Then, E(Y 2 ) = n 2 p E(Y 1 + Y 2 ) = E(Y 1 ) + E(Y 2 ) = n 1 p + n 2 p = (n 1 + n 2 )p The pooled estimate of p is bp = «Y1 + Y 2 n 1 + n 2 = «total # successes total sample size which is unbiased and the MLE. Intuitively, when the probability of success is the same on both treatments, the best estimate (MLE) of p is obtained by pooling over the treatments. Lecture 5: Contingency Tables p. 32/46

33 Using the likelihood to obtain the MLE under the null Under the null H 0 :p 1 = p 2 = p, the MLE of p is obtained from the likelihood L(p) = n 1 y 1! n 2 y 2! p y 1(1 p) n 1 y 1 p y 2(1 p) n 2 y 2 = n 1 y 1! n 2 y 2! p y 1+y 2 (1 p) (n 1+n 2 ) (y 1 +y 2 ), Then, d log L(p) dp = " d log dp 1 n 1 y 1! n 2 y 2!# + d dp 1 log[p y 1+y 2 (1 p) (n 1+n 2 ) (y 1 +y 2 ) ] = y 1 + y 2 (n 1 + n 2 )p p(1 p) Lecture 5: Contingency Tables p. 33/46

34 This is the same first derivative as a single binomial sample, in fact, under the null, and it is easily shown that the solution is Y 1 + Y 2 Bin(n 1 + n 2, p), bp = Y 1 + Y 2 n 1 + n 2 Lecture 5: Contingency Tables p. 34/46

35 Using the Estimates to obtain the Likelihood Ratio Statistic Under the alternative, and log[l(bp 1, bp 2 H! A )] = log log bp 1 = Y 1 n 1 and bp 2 = Y 2 n 2, n 1 y 1 n 2 y 2 + y 1 log(bp 1 ) + (n 1 y 1 ) log(1 bp 1 )+! + y 2 log(bp 2 ) + (n 2 y 2 )log(1 bp 2 ) Lecture 5: Contingency Tables p. 35/46

36 Then, log[l(bp, bp H 0 )]! = log log n 1 y 1 n 2 y 2 + y 1 log(bp) + (n 1 y 1 ) log(1 bp)+! + y 2 log(bp) + (n 2 y 2 )log(1 bp) Under the alternative, we estimate 2 parameters, under the null, we estimated 1, so df = 2 1 = 1. Then, we take 2 times the differences in the log-likelihoods and compare it to a chi-square with 1 df. Lecture 5: Contingency Tables p. 36/46

37 Simplification of the Likelihood Ratio Statistic Then, the likelihood ratio statistic equals 2 times the difference in the log-likelihoods under the alternative and null, or G 2 = 2[y 1 log bp1bp + (n 1 y 1 )log +y 2 log bp2bp + (n 2 y 2 )log = 2[y 1 log y1 n 1 bp +y 2 log y2 n 2 bp χ 2 1 under the null, in large samples. (1 bp1 ) (1 bp) (1 bp2 ) (1 bp) ] + (n 1 y 1 )log n1 y 1 n 1 (1 bp) + (n 2 y 2 )log n2 y 2 n 2 (1 bp) ] + Lecture 5: Contingency Tables p. 37/46

38 OBSERVED and EXPECTED Cell Counts First, let s look at the (2 2) table of OBSERVED Cell Counts. OUTCOME 1 2 TRT 1 Y 1 (n 1 Y 1 ) n 1 2 Y 2 (n 2 Y 2 ) n 2 total Y 1 + Y 2 [(n 1 + n 2 ) (n 1 + n 2 ) (Y 1 + Y 1 )] Lecture 5: Contingency Tables p. 38/46

39 If we look at the likelihood ratio statistic, G 2 = 2[y 1 log y1 n 1 bp +y 2 log y2 n 2 bp + (n 1 y 1 )log n1 y 1 n 1 (1 bp) + (n 2 y 2 )log n2 y 2 n 2 (1 bp) In the numerator of the log s, we have the observed cell counts for the 4 cells in the table. Sometimes, statisticians let O ij denote the observed count in row i, column j, ] + O 11 = Y 1, O 12 = n 1 Y 1, O 21 = Y 2, O 22 = n 2 Y 2 Lecture 5: Contingency Tables p. 39/46

40 Then, we can rewrite the observed table as OUTCOME 1 2 TRT 1 O 11 O 12 O 11 + O 12 2 O 21 O 22 O 21 + O 22 total O 11 + O 21 O 12 + O 22 We will show that the likelihood ratio statistic is often written as G 2 = 2[y 1 log y1 n 1 bp +y 2 log y2 n 2 bp + (n 1 y 1 )log n1 y 1 n 1 (1 bp) + (n 2 y 2 )log n2 y 2 n 2 (1 bp) = 2 P 2 P 2 i=1 j=1 O Oij ij log, E ij ] + Lecture 5: Contingency Tables p. 40/46

41 Simple Form of the Estimated Expected Counts First, suppose p 1 p 2, Then, the (2 2) table of expected cell counts is OUTCOME 1 2 TRT 1 n 1 p 1 n 1 (1 p 1 ) n 1 2 n 2 p 2 n 2 (1 p 2 ) n 2 total n 1 p 1 + n 2 p 2 [(n 1 + n 2 ) (n 1 + n 2 ) (n 1 p 1 + n 2 p 2 )] If we look at the n 1 subjects in the first row, we expect n 1 p 1 subjects to have outcome 1, and n 1 (1 p 1 ) of them to have outcome 2. Similarly, if we look at the n 2 subjects in the second row, we expect n 2 p 2 subjects to have outcome 1, and n 2 (1 p 2 ) of them to have outcome 2. Lecture 5: Contingency Tables p. 41/46

42 Under the null, when the probability of success is the same on both treatments, p 1 = p 2 = p, the table of expected counts looks like OUTCOME 1 2 TRT 1 n 1 p n 1 (1 p) n 1 2 n 2 p n 2 (1 p) n 2 total (n 1 + n 2 )p [(n 1 + n 2 )(1 p)] (n 1 + n 2 ) Here, if we look at the n 1 subjects in the first row, we expect n 1 p subjects to have outcome 1, and n 1 (1 p) of them to have outcome 2. Similarly, if we look at the n 2 subjects in the second row, we expect n 2 p subjects to have outcome 1, and n 2 (1 p) of them to have outcome 2. Lecture 5: Contingency Tables p. 42/46

43 Under H 0 : p 1 = p 2 = p, the table of estimated expected counts looks like OUTCOME 1 2 TRT 1 n 1 bp n 1 (1 bp) n 1 2 n 2 bp n 2 (1 bp) n 2 total (n 1 + n 2 )bp [(n 1 + n 2 )(1 bp)] (n 1 + n 2 ) where, recall, bp is the pooled estimate of p, bp = «Y1 + Y 2 n 1 + n 2 = «total # successes. total sample size These estimated expected counts are denoted E ij, (i th row, j th column), and are found in the denominator of the likelihood ratio statistic, with E 11 = n 1 bp, E 12 = n 1 (1 bp), E 21 = n 2 bp, E 22 = n 2 (1 bp) Lecture 5: Contingency Tables p. 43/46

44 Simplification of Expected Cell Counts Substituting and bp = Y 1 + Y 2 n 1 + n 2, 1 bp = 1 Y 1 + Y 2 n 1 + n 2 = (n 1 + n 2 ) (Y 1 + Y 2 ) n 1 + n 2, in the table, we get the E ij s, TRT 1 2 OUTCOME 1 2 n 1 (Y 1 +Y 2 ) n 1 +n 2 n 1 [(n 1 +n 2 ) (Y 1 +Y 2 )] n 1 +n 2 n 1 n 2 (Y 1 +Y 2 ) n 1 +n 2 n 2 [(n 1 +n 2 ) (Y 1 +Y 2 )] n 1 +n 2 n 2 total (Y 1 + Y 2 ) [(n 1 + n 2 ) (n 1 + n 2 ) (Y 1 + Y 2 )] Lecture 5: Contingency Tables p. 44/46

45 From this table, you can see that E ij = [ith row total] [j th column total] [total sample size (n 1 + n 2 ) ] Lecture 5: Contingency Tables p. 45/46

46 Summary We did all this to show that 2X 2X G 2 = 2 O ij log i=1 j=1 Oij E ij «Note that, we can also write this as 2X 2X G 2 = 2 O ij [log(o ij ) log(e ij )] i=1 j=1 Writing it this way, we see that the likelihood ratio measures the discrepancy between the log of the observed counts, and the log of estimated expected counts under the null; if they are similar, you would expect the statistic to be small, and the null not to be rejected. Lecture 5: Contingency Tables p. 46/46