Adjusting for extraneous factors

Topics for today

• Stratified analysis of 2x2 tables

• Regression

Readings

• Jewell Chapter 9

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?

Admit Reject

Male 1198 1493

Female 557 1278

Log odds ratio =

95% conf interval:

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)

Dept Male Female

1 825 62% 108 82%

2 560 63% 25 68%

3 325 37% 593 34%

4 417 33% 375 35%

5 191 28% 393 24%

6 373 6% 341 7%

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

Diseased Not Diseased

Exposed ai bi

Unexposed ci di

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

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

Admit Reject

Male 1211 1480

Female 716 1119

Expected data

Admit Reject

Male

Female

Chi-Squared test for a 2x2 table

22 22

1 1

( )ij ij

i i ij

O E

E

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:

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 i D Not D

E ai bi

Not E ci di

2

1 12

1

2

where ( )( ) /

and ( )( )( )( ) /[ ( 1)]

I I

i ii i

CMH i i i i i iI

ii

i i i i i i i i i i i

a A

A a b a c nV

V a b c d a c b d n n

Berkeley Admissions

Male Female

stratum a b c d

1 512 313 89 19

2 353 207 17 8

3 120 205 202 391

4 138 279 131 244

5 53 138 94 299

6 22 351 24 317

Estimating a common effect

• Wolf method (averages the log odds ratios)

• Mantel-Haenszel (averages the odds ratios)

• Regression-based

Wolf’s average log-odds ratio

1 1

1

1

ˆ ˆlog( ) log( )

ˆwhere log( )) log log

1 1 1 1ˆvar(log( ))

ˆ(log( ))

I I

W i i ii i

i ii

i i

i ii i i i

I

W ii

OR w OR w

a cOR

b d

w ORa b c d

Var OR w

Can add .5 to cell entries if sample sizes are small

Applying Wolf method to Berkeley data

stratum a b c d lor v w=1/v w*lor

1 512 313 89 19 -0.457 0.069 14.489 -6.62

2 353 207 17 8 -0.096 0.1915 5.2223 -0.499

3 120 205 202 391 0.054 0.0207 48.264 2.6185

4 138 279 131 244 -0.036 0.0226 44.321 -1.578

5 53 138 94 299 0.087 0.0401 24.939 2.1682

6 22 351 24 317 -0.082 0.0931 10.738 -0.881

Wolf estimate of LOR is .03, with variance .0068. What is 95% CI?

Corresponding OR estimate is

Wolf’s average log-odds ratio

1 1

1

1

ˆ ˆlog( ) log( )

ˆwhere log( )) log log

1 1 1 1ˆvar(log( ))

ˆ(log( ))

I I

W i i ii i

i ii

i i

i ii i i i

I

W ii

OR w OR w

a cOR

b d

w ORa b c d

Var OR w

Can add .5 to cell entries if sample sizes are small

Applying Wolf method to Berkeley data

stratum a b c d lor v w=1/v w*lor

1 512 313 89 19 -0.457 0.069 14.489 -6.62

2 353 207 17 8 -0.096 0.1915 5.2223 -0.499

3 120 205 202 391 0.054 0.0207 48.264 2.6185

4 138 279 131 244 -0.036 0.0226 44.321 -1.578

5 53 138 94 299 0.087 0.0401 24.939 2.1682

6 22 351 24 317 -0.082 0.0931 10.738 -0.881

Wolf estimate of LOR is .03, with variance .0068. What is 95% CI?

Corresponding OR estimate is

Mantel-Haenszel average odds ratio

* *

1 1

*

ˆ ˆ

ˆwhere

ˆ( ) page 131 of Jewell!

i

I I

MH i i ii i

ii

i i

i ii

i

MH

OR w OR w

a dOR

c b

b cw

n

Var OR

Applying Wolf method to Berkeley data

stratum a b c d lor v w=1/v w*lor

1 512 313 89 19 -0.457 0.069 14.489 -6.62

2 353 207 17 8 -0.096 0.1915 5.2223 -0.499

3 120 205 202 391 0.054 0.0207 48.264 2.6185

4 138 279 131 244 -0.036 0.0226 44.321 -1.578

5 53 138 94 299 0.087 0.0401 24.939 2.1682

6 22 351 24 317 -0.082 0.0931 10.738 -0.881

Wolf estimate of LOR is .03, with variance .0068. What is 95% CI?

Corresponding OR estimate is

Regression-based analysis for

Berkeley data

data berkeley;input stratum male a b ;cards;1 1 512 3131 0 89 192 1 353 2072 0 17 83 1 120 2053 0 202 3914 1 138 2794 0 131 2445 1 53 1385 0 94 2996 1 22 3516 0 24 317run;

data berkeley; set berkeley;n=a+b;

Unstratified analysis;proc genmod;model

a/n=male/dist=binomial;run;

Code continued

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

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

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;

Epilepsy data in SAS

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

Why don’t drug estimates change much??

Hint – look at association between drug and smoking

proc freq ;

table one3*cig2 /chisq;

run;