Download CEHS 734 Lecture 2 — Review of Basic Probability

HSRP 734:
Advanced
Statistical Methods
May 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
Disease
D
No Disease
D
Total
Exposed
E
Not Exposed
E
Total
a
b
n1
c
d
n0
m1
m0
N
Confidence Interval for Odds Ratio
Confidence limits are based on the sampling
distribution of
 pˆ 1

 (1  pˆ ) 
 ad 
1
ln    ln 

ˆ
p
 bc 
 2

ˆ
(
1

p
)
2 

which is normal or approximately normal with
 ad 
Mean  ln  
 bc 
and
1 1 1 1
Variance    
a b c d
Confidence Interval for Odds Ratio
 ad 
1. Calculate M  ln OR   ln  
 bc 
1 1 1 1
2. Calculate S      
a b c d 
3. Calculate 95% CI  e
 M 1.96S 
*Use if
N>25
Confidence Interval for Risk Ratio
Confidence limits are based on the sampling
distribution of
 pˆ1 
 a /( a  b) 
  ln  
ln 
 c /( c  d ) 
 pˆ 2 
which is normal or approximately normal with
and
 p1 
Mean  ln  
 p2 
b
d
Variance 

a ( a  b ) c (c  d )
Confidence Interval for Risk Ratio
 a a  b  

1. Calculate M  ln RR   ln 
 c c  d  
 b
d 

2. Calculate S  

 aa  b  cc  d  
3. Calculate 95% CI  e
 M 1.96S 
*Use if
N>25
SAS Enterprise:
or_rr.sas7bdat
SAS websites
• Online help:
http://support.sas.com/onlinedoc/913/docM
ainpage.jsp
• UCLA:
http://www.ats.ucla.edu/stat/SAS/
• SAS SUGI:
http://support.sas.com/events/sasglobalforu
m/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
trials
2. Each trial results in one of K outcomes
3. Let pi be the probability of outcome i
a. Each pi remains constant for each experiment
K
b.  pi  1
i 1
•
The pmf for k outcomes is:
n!
P ( n1 , n2 ,..., nk ) 
p1n1  p2n2  ...  pknk
n1! n2 !...nk !
•
Notes:
k
k
i 1
i 1
 pi  1;  ni  n; E (ni )  n  pi
Example of a Multinomial Experiment
Consider an unfair die and 6 tosses:
Let
1
2
3
4
5
Pi
ni
6
0.3 0.1 0.1 0.1 0.0 0.4
2
0
1
1
0
2
Find the probability of this outcome
Pr( 2,0,1,1,0,2)
6!


2
0
1
1
0
2

(0.3 )(0.1 )(0.1 )(0.1 )(0.0 )(0.4 )
 2! 0!1!1! 0! 2! 
 720 

0.000144  0.02592
 4 
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
k
(Observedi  Expectedi ) 2
(ni  Npi ) 2 dist
2






k 1
Expected
Np
i 1
i 1
i
i
k
•
•
Require that expected cell counts not too small
Expected counts > 5.
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
Yellow
Green
Round
Wrinkled
Round
Wrinkled
Observed (ni)
315
101
108
32
Expected
(Npi)
312.75
104.25
104.25
34.75
• H0: pYR = 9/16, pYW = 3/16, pGR = 3/16, pGW = 1/16
• H1: at least one pi differs from hypothesized value
Chi-square GOF Test
2
2
k
(
Observed

Expected
)
(
n

Np
)
i
i
i
2  
 i
Expectedi
Npi
i 1
i 1
k
2
2
2
2

315  312.75 101  104.25 108  104.25 32  34.75




312.75
104.25
 0.47
104.25
34.75
Chi-square GOF Test
• Therefore, we observed 2 = 0.47 from a multinomial
experiment with k = 4. Thus, df = k-1 = 3.
For a = 0.05,
12a ,k 1  120.05, 41  02.95,3  7.81
• Thus, the observed chi-squared statistic is not greater
than the critical value for a = 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
Testing association in 2x2 table
• This method translates to testing crosstabulation tables for RxC cases
• Here the cells are formed by crossclassification 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?
ni  n j TotalROW * TotalCOL
EXPECTEDij  N 


N
N
TotalOVERALL
• The classic Pearson’s chi-squared test of independence
2
2

i 1 j 1
(Observedij  Expectedij )2
Expectedij
dist

12
• df = (2-1) x (2-1) = 1
• Conservatively, we require EXPECTEDij ≥ 5 for all i, j
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 independence
Test 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
Group
1
2
…
r
Total
Response Variable
Categories
1
2
…
c
n11
n12
…
n1c
n21
n22
…
n2c
…
…
…
…
nr1
nr2
…
nrc
n+1
n+2
…
n+c
Total
n1+_
n2+
…
nr+
N
Test of General Association
• H0: There is no association between rows and columns
H1: 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.
ni  n j TotalROW * TotalCOL
Expectedij  N 


N
N
TotalOVERALL
• The classic Pearson’s chi squared test of independence
r
c
 2  
i 1 j 1
Observedij  Expectedij 2 dist

  2
Expectedij
( r 1)(c 1)
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
Stratum 1
No
CCU+
Yes
MI
43
No MI
56
20
90
Total
Stratum 2
CCU-
No
Yes
Total
Total
209
137
32
437
341
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
E ( nh11 )  mh11
nh1  nh 1

nh
and the variance of nh11
nh1 nh 2  nh 1nh  2
Var nh11  
nh2 nh  1
This leads to the Cochrane-Mantel-Haenszel test
2


  nh11   mh11 
h 1
 h 1
 dist
2


1
g
Varnh11 
g
g
h 1
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
ng 11ng 22
n111n122 n211n222
nh11nh 22

 ... 

n1
n2
ng
h 1
nh
 g

nh 21nh12
ng 21ng 12
n121n112 n221n212

 ... 

h 1
nh
n1
n2
ng
g
ORMH
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
ORcenter1
ORcenter2
ORMH
= 3.76 (2.01, 7.05)
= 4.01 (1.67, 9.66)
= 4.05 (1.55, 10.60)
= 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