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Adjusting for extraneous factors Topics for today Stratified analysis of 2x2 tables Regression Readings Jewell Chapter 9.

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Presentation on theme: "Adjusting for extraneous factors Topics for today Stratified analysis of 2x2 tables Regression Readings Jewell Chapter 9."— Presentation transcript:

1 Adjusting for extraneous factors Topics for today Stratified analysis of 2x2 tables Regression Readings Jewell Chapter 9

2 Berkeley Admissions Data 1973 study showed that 45% of 2691 male applicants were admitted, compared with only 30% of 1835 female applicants. The odds ratio is 1.84 with 95% confidence interval (1.62, 2.08). Is this evidence of sex bias? AdmitReject Male11981493 Female5571278 Log odds ratio = 95% conf interval:

3 Berkeley Admissions Data The picture changes completely once we look at admissions by department! Bickel, P.J., J.W. Hammel and J.W. O'Connell (1975) "Sex bias in graduate admissions: Data from Berkeley" in Science, 187:398-403) # applicants (% admit) DeptMaleFemale 182562%10882% 256063%2568% 332537%59334% 441733%37535% 519128%39324% 63736%3417%

4 Stratified analysis Consider relationship between a disease outcome (D in Jewell, often Y in practice) and an exposure (E in Jewell, often X in practice), but we also want to adjust for an additional factor such as age or sex that can be divided up into I distinct strata. Suppose that the data from the ith stratum can be represented as follows: Jewell Tables 9.2 & 9.3 give two examples DiseasedNot Diseased Exposedaiai bibi Unexposedcici didi

5 What do we want to do? 1.Ask whether there is a significant association between disease (D) and exposure (E), after adjusting for the additional stratification factor 2.Estimate an adjusted odds ratio, that appropriately takes into account the stratification factor. Lets start with 1. but first, we need to quickly go over another way to assess whether there is a significant association for a 2x2 table

6 Assessing association - Berkeley Admissions again We already determined that there is a significant association in this 2x2 table, based on the 95% confidence interval for the odds ratio. An alternative approach is a chi-squared test There are several variations. But basic idea is to compare observed data to what would be expected if there were no association (see J p 69) Observed data AdmitReject Male12111480 Female7161119 Expected data AdmitReject Male Female

7 Chi-Squared test for a 2x2 table The test statistic is And its “significance” can be determined by looking up the chi-squared tables with 1 degree of freedom. For the Berkeley data, we get:

8 Back to the stratified analysis Cochran-Mantel-Haenszel test combines the differences between observed and expected values over all the strata. It focuses only on the “a” element of each 2x2 table Stratum iDNot D Eaiai bibi Not Ecici didi

9 Berkeley Admissions MaleFemale stratum a b c d 15123138919 2353207178 3120205202391 4138279131244 55313894299 62235124317

10 Estimating a common effect Wolf method (averages the log odds ratios) Mantel-Haenszel (averages the odds ratios) Regression-based

11 Wolf’s average log-odds ratio Can add.5 to cell entries if sample sizes are small

12 Applying Wolf method to Berkeley data stratumabcdlorvw=1/vw*lor 15123138919-0.4570.06914.489-6.62 2353207178-0.0960.19155.2223-0.499 31202052023910.0540.020748.2642.6185 4138279131244-0.0360.022644.321-1.578 553138942990.0870.040124.9392.1682 62235124317-0.0820.093110.738-0.881 Wolf estimate of LOR is.03, with variance.0068. What is 95% CI? Corresponding OR estimate is

13 Wolf’s average log-odds ratio Can add.5 to cell entries if sample sizes are small

14 Applying Wolf method to Berkeley data stratumabcdlorvw=1/vw*lor 15123138919-0.4570.06914.489-6.62 2353207178-0.0960.19155.2223-0.499 31202052023910.0540.020748.2642.6185 4138279131244-0.0360.022644.321-1.578 553138942990.0870.040124.9392.1682 62235124317-0.0820.093110.738-0.881 Wolf estimate of LOR is.03, with variance.0068. What is 95% CI? Corresponding OR estimate is

15 Mantel-Haenszel average odds ratio

16 Applying Wolf method to Berkeley data stratumabcdlorvw=1/vw*lor 15123138919-0.4570.06914.489-6.62 2353207178-0.0960.19155.2223-0.499 31202052023910.0540.020748.2642.6185 4138279131244-0.0360.022644.321-1.578 553138942990.0870.040124.9392.1682 62235124317-0.0820.093110.738-0.881 Wolf estimate of LOR is.03, with variance.0068. What is 95% CI? Corresponding OR estimate is

17 Regression-based analysis for Berkeley data data berkeley; input stratum male a b ; cards; 11 512 313 1 0 89 19 21 353 207 20 17 8 31 120 205 30 202 391 41 138 279 40 131 244 51 53 138 50 94 299 61 22 351 60 24 317 run; data berkeley; set berkeley; n=a+b; Unstratified analysis; proc genmod; model a/n=male/dist=binomial; run; Code continued

18 Results of unstratified analysis Standard 95% Confidence Chi- Parameter DF Estimate Error Limits Square P Intercept 1 -0.8305 0.0508 -0.9300 -0.7310 267.56 <.0001 male 1 0.6104 0.0639 0.4851 0.7356 91.25 <.0001 Scale 0 1.0000 0.0000 1.0000 1.0000 Compare with our initial analysis

19 Stratified analysis proc genmod; class stratum; model a/n=male stratum/dist=binomial; run; Standard 95% Conf Chi- Parameter DF Estimate Error Limits Square Pr > ChiSq Intercept 1 -2.6246 0.1577 -2.9337 -2.3154 276.88 <.0001 male 1 -0.0999 0.0808 -0.2583 0.0586 1.53 0.2167 stratum 1 1 3.3065 0.1700 2.9733 3.6396 378.38 <.0001 stratum 2 1 3.2631 0.1788 2.9127 3.6135 333.12 <.0001 stratum 3 1 2.0439 0.1679 1.7149 2.3729 148.24 <.0001 stratum 4 1 2.0119 0.1699 1.6788 2.3449 140.18 <.0001 stratum 5 1 1.5672 0.1804 1.2135 1.9208 75.44 <.0001 stratum 6 0 0.0000 0.0000 0.0000 0.0000.. Scale 0 1.0000 0.0000 1.0000 1.0000

20 More general modeling We can add additional factors into the logistic regression model so as to obtain an estimate of the log-odds ratio, adjusting for all these additional factors. Example, smoking in the Epilepsy study. Lets look in SAS: proc freq ; table one3*cig2 /chisq; run;

21 Epilepsy data in SAS

22 Standard Wald 95% Confidence Chi- Parameter DF Estimate Error Limits Square Pr > ChiSq Intercept 1 -3.1396 0.2229 -3.5765 -2.7028 198.41 <.0001 DRUG 1 1 1.0384 0.2876 0.4748 1.6020 13.04 0.0003 DRUG 2 1 -0.2944 0.6275 -1.5243 0.9355 0.22 0.6390 DRUG 3 0 0.0000 0.0000 0.0000 0.0000.. Scale 0 1.0000 0.0000 1.0000 1.0000 Standard Wald 95% Confidence Chi- Parameter DF Estimate Error Limits Square Pr > ChiSq Intercept 1 -3.3872 0.2435 -3.8644 -2.9100 193.55 <.0001 DRUG 1 1 1.0712 0.2939 0.4952 1.6472 13.29 0.0003 DRUG 2 1 -0.3596 0.6337 -1.6016 0.8824 0.32 0.5704 DRUG 3 0 0.0000 0.0000 0.0000 0.0000.. CIG2 1 1.0721 0.3131 0.4585 1.6857 11.73 0.0006 Scale 0 1.0000 0.0000 1.0000 1.0000

23 Why don’t drug estimates change much?? Hint – look at association between drug and smoking

24 proc freq ; table one3*cig2 /chisq; run;


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