HSRP 734: HSRP 734: Advanced Advanced

Statistical MethodsStatistical MethodsMay 29, 2008May 29, 2008

Finish talking about Association Measures:

Odds Ratio

• OR=2 of Disease for Exposed vs. Not exposed

• What is the interpretation?

• “Exposed patients have twice the odds of disease versus patients that were not exposed.”

Finish talking about Association Measures:

Relative Risk

• RR=2.5 of Disease for Exposed vs. Not exposed

• What is the interpretation?

• “Exposed patients are 2.5 times as likely to have the disease versus patients that were not exposed.”

Finish talking about Association Measures

• OR is not close to RR

• Unless Pr(disease) for Exposed, Not exposed low

• “Rare” disease

Finish talking about Association Measures

• Confidence intervals for Odds Ratio

• Confidence intervals for Relative Risk

Measures of Disease Association

Exposed E

Not Exposed Total

Disease D

a b n1

No Diseasec d n0

Total m1 m0 N

D

E

Confidence limits are based on the sampling distribution of

which is normal or approximately normal with

and

Confidence Interval for Odds Ratio

dcbaVariance

1111

bc

adMean ln

)ˆ1(ˆ

)ˆ1(ˆ

lnln

2

2

1

1

pp

pp

bc

ad

Confidence Interval for Odds Ratio

S1.96M CI 95% Calculate .3

1111S Calculate .2

lnORlnM Calculate .1

e

dcba

bc

ad

*Use if N>25

Confidence limits are based on the sampling distribution of

which is normal or approximately normal with

and

Confidence Interval for Risk Ratio

2

1lnp

pMean

2

1

ˆ

ˆln

)/(

)/(ln

p

p

dcc

baa

)()( dcc

d

baa

bVariance

Confidence Interval for Risk Ratio

S1.96M CI 95% Calculate .3

S Calculate .2

lnRRlnM Calculate .1

e

dcc

d

baa

b

dcc

baa

*Use if N>25

SAS Enterprise:

or_rr.sas7bdat

SAS websites

• Online help:

http://support.sas.com/onlinedoc/913/docMainpage.jsp

• UCLA: http://www.ats.ucla.edu/stat/SAS/

• SAS SUGI:

http://support.sas.com/events/sasglobalforum/previous/index.html

Categorical Data Analysis

1. Understand the Multinomial probability mass function

2. Compute Goodness-of-fit tests and chi-squared tests for association

3. Test for association in the presence of a possibly confounding third factor

(e.g., disease versus exposure from 3 sites)

Categorical Data Analysis

• Motivation – How do we estimate and test the magnitude of

posited relationship when the outcome of interest is categorical?

– e.g., An international study examines the relationship between age at first birth and the development of breast cancer

• Age = categorized into age groups

Categorical Data Analysis

<20 20-24 25-29 30-34 >=35 Total

Cancer 320 1206 1011 463 220 3220

No cancer

1422 4432 2893 1092 406 10245

Total 1742 5638 3904 1555 626 13465

Categorical Data Analysis

• Research question

– Is there a relationship between age at first birth and Cancer status?

• Better to convert the table into percentages (easier to see)

• Turns out that there is a significant relationship (p<0.001)

Categorical Data Analysis

• Statistical techniques involve

– Probability distribution for categorical data

– Tests for relationship in a RxC table

R = # of Rows in Table

C = # of Columns in Table

Probability Distribution for Categorical Outcomes

• Fun for Friday night:

– Go home and flip a quarter 10,000 times. Determine if there is evidence that one side is falling down more.

Probability Distributions for Categorical Data

– Bernoulli (1 toss of a coin, outcome=H,T)

– Binomial (10 tosses of a coin, outcome=0,1,2..,10 heads)

– Multinomial (throw 10 balls into 4 pigeon holes ABCD, outcome= (3A,2B,1C,4D))

Why use multinomial for testing?

• Relationship between 2 categorical variables

– RxC table analysis

– Based on multinomial distribution

Why use multinomial for testing?

• Example:

2 level exposure status (Exposed, Not exposed),

3 level outcome (severe, mild, no disease)

– Treat 2x3=6 outcomes as categorical or a multinomial distribution with 6 pigeon holes

– The expected probability of the pigeon holes are specified under some kind of assumptions (e.g., independence)

Level of Measurement

– Categorical response

• dichotomous

• ordinal (>2 categories, ordered)

• nominal (>2 categories, not ordered)

– Dichotomous use Binomial distribution

– Ordinal, Nominal use Multinomial distribution

Multinomial Distribution

• Multinomial experiment:1. Experiment consists of n identical and independent

trials2. Each trial results in one of K outcomes3. Let pi be the probability of outcome i

a. Each pi remains constant for each experimentb.

• The pmf for k outcomes is:

• Notes:

11

K

iip

kn

k

nn

k

k pppnnn

nnnnP ...

!!...!

!),...,,( 21

21

21

21

ii

k

ii

k

ii pnnEnnp

)(;;1

11

Example of a Multinomial Experiment

Consider an unfair die and 6 tosses:

Let

Find the probability of this outcome

1 2 3 4 5 6

Pi 0.3 0.1 0.1 0.1 0.0 0.4

ni 2 0 1 1 0 2

02592.0000144.04

720

)4.0)(0.0)(1.0)(1.0)(1.0)(3.0(!2 !0 !1 !1 !0 !2

!6

)2,0,1,1,0,2Pr(

201102

Simple Multinomial Experiments

Classical example: Mendel Sample from the second generation of seeds resulting from crossing yellow round peas and green wrinkled peas (N=556)

Yellow Green

Round Wrinkled Round Wrinkled

315 101 108 32

Mendel’s Laws of Inheritance suggest that we should expect the following ratios:

9/16, 3/16, 3/16, 1/16

For N = 556, the expected number of each outcome is:

E(YR) = 556 x 9/16 = 312.75

E(YW) = 556 x 3/16 = 104.25

E(GR) = 556 x 3/16 = 104.25

E(GW) = 556 x 1/16 = 34.75

Yellow Green

Round Wrinkled Round Wrinkled

315(312.75)

101

(104.25)

108

(104.25)

32

(34.75)

(Expected counts)

Multinomial distribution

• The observed cell counts are not identical to the expected cell counts

• Under the assumption of a multinomial model with the stated probabilities, how might we determine how unlikely it is to observe these data?

Chi-square GOF Test

• Hypothesis: observed cell counts are consistent with the multinomial probabilities

• Theoretical result

• Require that expected cell counts not too small• Expected counts > 5.

21

1

2

1

2 )()(

k

distk

i i

iik

i i

ii

Np

Npn

Expected

ExpectedObserved

Chi-square distribution

• Remarks about Chi-squared distribution:

1.Nonsymmetric

2.Strictly positive

3.Different chi-squared distribution for each df.

Chi-square GOF Test

• Applying this test to Mendel’s peas example yields

• H0: pYR = 9/16, pYW = 3/16, pGR = 3/16, pGW = 1/16

• H1: at least one pi differs from hypothesized value

Yellow Green

Round Wrinkled Round Wrinkled

Observed (ni) 315 101 108 32

Expected (Npi)

312.75 104.25 104.25 34.75

Chi-square GOF Test

47.0

75.34

75.3432

25.104

25.104108

25.104

25.104101

75.312

75.312315

)()(

2222

1

2

1

22

k

i i

iik

i i

ii

Np

Npn

Expected

ExpectedObserved

Chi-square GOF Test

• Therefore, we observed 2 = 0.47 from a multinomial experiment with k = 4. Thus, df = k-1 = 3.

For = 0.05,

• Thus, the observed chi-squared statistic is not greater than the critical value for = 0.05 and df = 3.

• We fail to find evidence that these data depart from the hypothesized probabilities. i.e., model fits well to data

81.723,95.0

214,05.01

21,1 k

Testing association in 2x2 table

• This method translates to testing cross-tabulation tables for RxC cases

• Here the cells are formed by cross-classification of 2 variables

• Null hypothesis is the 2 variables are independent

• Simplest case : 2x2 table

Testing association in 2x2 table

• Testing for independence or no association

• Similar idea to checking goodness-of-fit

– Compare what to see to what you hypothesized to be true

– You did, in fact, hypothesize “independence”

Basic Inference for 2x2 Tables

• 2x2 Contingency Table

Column Levels

Row Levels

1 2 Total

1 n11 n12 n1+

2 n21 n22 n2+

Total n+1 n+2 N

Chi-square GOF Test for 2x2 Tables

• H0: There is no association between row and columns• Under H0, the expected cell counts are the product of the

marginal probabilities and the sample size. Why?

• The classic Pearson’s chi-squared test of independence

• df = (2-1) x (2-1) = 1• Conservatively, we require EXPECTEDij ≥ 5 for all i, j

OVERALL

COLROWjiij Total

TotalTotal

N

n

N

nNEXPECTED

*

22 221

1 1

( )ij ij dist

i j ij

Observed Expected

Expected

Other Tests for 2x2 Tables

• Two alternative tests

– Yate’s continuity corrected chi-square statistic

– Mantel-Haenszel chi-square statistic

• For sufficiently large sample size, all three Chi-squared statistics are approximately equal and all have a Chi-squared distribution with 1 df

When to use Chi-square vs. Fisher’s Exact

• When the expected cell counts are less than 5, it is better to use the Fisher’s exact test.

Summary of the Use of 2 test

• Test of goodness-of-fit

Determine whether or not a sample of observed values of some random variable is compatible with the hypothesis that the sample was drawn from a population with a specified distributional form (e.g., specified probabilities of certain events)

Summary of the Use of 2 test

• Test of independenceTest the null hypothesis that two criteria of classification (variables) are independent

Summary of the Use of 2 test

• Test of homogeneity

Test the null hypothesis that the samples are drawn from populations that are homogeneous with respect to some factor (i.e., no association between group and factor)

Summary of the Use of 2 test

• Could consider this test as answering:

“Are the Row factor and Column factor associated?”

Categorical Data Analysis

• Ideas of multinomial and chi-squared test generalize to testing RxC association and RxCxK association

• Example:

– 2 exposure status, 2 disease status, 3 sites

– 2x2x3 association analysis

Test of General Association (R x C Table)

• Consider a study designed to test whether there exists an association between political party affiliation and residency within specific counties

County

Party Buncombe Transylvania Halifax

Democrat 221 160 360

Independent 200 291 160

Republican 208 106 316

• Notation for general RxC table

Response Variable Categories

Group 1 2 … c Total

1 n11 n12 … n1c n1+_

2 n21 n22 … n2c n2+

… … … … … …

r nr1 nr2 … nrc nr+

Total n+1 n+2 … n+c N

Test of General Association

• H0: There is no association between rows and columnsH1: There exists a dependence between rows and columns

• Under H0,the expected cell counts are the product of the corresponding marginal probabilities and the sample size.

• The classic Pearson’s chi squared test of independence

2)1)(1(

1 1

22

crdist

r

i

c

j ij

ijij

Expected

ExpectedObserved

OVERALL

COLROWjiij Total

TotalTotal

N

n

N

nNExpected

*

SAS Enterprise:

chisq.sas7bdat

Mantel-Haenszel test

• Often, there are other factors in a RxC test

• Mantel-Haenszel test (or Cochran Mantel Haenzsel CMH) can be used for controlling for “nuisance” factors

• Typically used for rxcx2 table– e.g., 2x2x2 cross classification– e.g., Association between disease status and

exposure controlling for age group (strata)

Stratified Analysis

• Examples of commonly used strata•Age group•Gender•Study site (hospital, country)•ethnic group

Stratified Analysis

• Myocardial infarction and anticoagulant use by Coronary Care Unit

AC use MI No MI Total

Stratum 1 No 43 56

CCU+ Yes 20 90

Total 209

Stratum 2 No 137 437

CCU- Yes 32 341

Total 947

Stratified Analysis

• Idea: test for an association while controlling for CCU effects

• Denote the counts from the first cell within the hth subtable as nh11,

• Construct the CMH test of association controlling for CCU

Stratified Analysis

• Test assumes the direction of effect within each table is the same

• The Cochran-Mantel-Haenszel approach partially removes the confounding influences of the explanatory variable (e.g., CCU)

• May improve power

Mantel-Haenszel Test

• The expected value of nh11 for h = 1,2,…,g is

and the variance of nh11

This leads to the Cochrane-Mantel-Haenszel test

h

hhhh n

nnmnE 11

1111)(

122121

11

hh

hhhhh nn

nnnnnVar

21

111

2

1 11111

distg

hh

g

h

g

hhh

nVar

mn

Direction of effects across Strata

• Note that if directions of conditional ORs are not the same, discrepancies between observed and expected from different strata may cancel out one another

• Lead to poor power and biased result

MH “Pooled” Odds Ratio

g

gg

g

gg

g

hh

hh

g

hh

hh

MH

n

nn

nnn

nnn

n

nn

nnn

nnn

n

nnn

nn

OR1221

2

212221

1

112121

2211

2

222211

1

122111

1

1221

1

2211

...

...

MH test decision list

• Z = strata of potential confounder

-> If ORc ≈ (ORZ=1 ≈ ORZ=2 ≈…) Z is not a confounder, report crude OR (ORc)

-> If ORc ≠ (ORZ=1 ≈ ORZ=2 ≈…) Z is a confounder, report adjusted OR (ORMH)

-> If ORZ=1 ≠ ORZ=2 ≠ … Z is an effect modifier, report strata specific OR’s (don’t adjust!)

Breslow Day test

• (More formal approach) Can also test for homogeneity of odds ratio across strata

• If Breslow Day test is significant => odds ratios within strata are not homogeneous. Thus, => ORMH would be inappropriate!

SAS Enterprise:

cmh.sas7bdat

Results from cmh.sas7bdat

ORcrude = 3.76 (2.01, 7.05)ORcenter1 = 4.01 (1.67, 9.66)ORcenter2 = 4.05 (1.55, 10.60)ORMH = 4.03 (2.11, 7.71)

Breslow-Day p-value = 0.99

MH Chi-square = 18.41, p-value < 0.0001

Take home messages

• Multinomial and the Chi-square test are the “workhorse” for testing of goodness-of-fit

• Idea is to compare expected counts (calculated from a pre-determined set of probabilities) and the observed counts

• The same idea can be applied to testing statistical assumptions such as no association

• CMH test is for testing association when a confounding effect (strata) may be present

For Next Class 6/5

• HW #1 key posted

• HW #2 will be due

• Read Kleinbaum Ch. 1,2