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The binomial applied: absolute
and relative risks, chi-square
Probability speak
(just shorthand!)…







P(X) = “the probability of event X”
P(D) = “the probability of disease”
P(E) = “the probability of exposure”
P(~D) = “the probability of not getting the
disease”
P(~E)= “the probability of not being exposed”
P(D/E) = “the probability of disease given
exposure” or “the probability of disease among the
exposed”
P(D/~E) = “the probability of disease given
unexposed” or “the probability of disease among
the unexposed”
Things that follow a binomial
distribution…
Cohort study (or cross-sectional):


The number of exposed individuals in your sample
that develop the disease
The number of unexposed individuals in your sample
that develop the disease
Case-control study:


The number of cases that have had the exposure
The number of controls that have had the exposure
Cohort study example:

You sample 100 smokers and 100 nonsmokers and follow them for 5 years to
see who develops heart disease.
Seeing it as a binomial…


The number of smokers that develop
heart disease in your study follows a
binomial distribution with N=100,
p=pd/e
The number of non-smokers that
develop heart disease in your study
follows a binomial distribution with
N=100, pd/~e
A possible outcome:
Smoker (E)
Heart disease (D)
21
Non-smoker
(~E)
13
No Disease (~D)
79
87
100
100
Statistics for these data



1. Risk ratio (relative risk)
2. Difference in proportions (absolute
risk)
3. Chi-square test of independence

For 2x2 tables, mathematically equivalent
to difference in proportions Z test.
1. Risk ratio (relative risk)
Exposure (E)
Disease (D)
a
No Exposure
(~E)
b
No Disease (~D)
c
d
a+c
b+d
risk to the exposed
a /( a c)
RR 
b /(b d )
risk to the unexposed
In probability terms…
Exposure (E)
Disease (D)
a
No Exposure
(~E)
b
No Disease (~D)
c
d
a+c
b+d
Risk of disease in
the exposed
P( D / E )
a /(ac)
RR 

b /(bd ) P( D / ~ E )
risk of disease in the
unexposed
Risk ratio calculation:
Smoker (E)
Heart disease (D)
21
Non-smoker
(~E)
13
No Disease (~D)
79
87
100
100
RR  21/100  21  1.61
13/100
13
Interpretation: there is a 61% increase in risk of
heart disease in smokers vs. nonsmokers

Inferences about risk ratio…



Is our observed risk ratio statistically different
from 1.0? What is the p-value?
I’m going to present statistical inference for
odds ratio; risk ratio is similar.
So, for now, just get answer from SAS:


95% confidence interval: 0.86 to 3.04
P-value>.05
2. Difference in proportions
Exposure (E)
Disease (D)
a
No Exposure
(~E)
b
No Disease (~D)
c
d
a+c
b+d
P( D / E )  P( D / ~ E )  a /( a  c)  b /(b  d )
2. Difference in proportions
Smoker (E)
Heart disease (D)
21
Non-smoker
(~E)
13
No Disease (~D)
79
87
100
100
P( D / E )  P( D / ~ E )  21%  13%  8%
Absolute, rather than relative risk difference!
Difference in proportions test

Null hypothesis: difference in proportions = 0
Under the null, the groups have the same risk
of heart disease (=overall risk in the study):


The number of smokers that develop heart
disease in your study follows a binomial
distribution with N=100, p=.17
The number of non-smokers that develop heart
disease in your study follows a binomial
distribution with N=100, p=.17
Follows a normal
because binomial can be
approximated with
normal
Difference in proportions test
Null hypothesis: The difference in proportions is 0.
Z
p
p1  p2
p * (1  p ) p * (1  p )

n1
n2
Recall, variance of a
proportion is p(1-p)/n
n1 p1  n2 p2
(just average proportion )
n1  n2
p1  proportion in group 1
p2  proportion in group 2
n1  number in group 1
n2  number in group 2
Use average (or pooled)
proportion in standard
error formula, because
under the null
hypothesis, groups have
equal proportions.
Z-test applied here…
21  13
p
 .17
200
.08
.08
Z

 1.51
.17 * .83 .17 * .83 .053

100
100
Corresponding two-sided p-value is .131.
Corresponding 95%
confidence interval…
.08  1.96 * stderror
.08  1.96 * (.053)  .02 to .18
If the 95% confidence interval crosses the null value
(here=0), then p>.05
OR, use computer simulation
to make inferences…





1. In SAS, assume infinite population of smokers
and non-smokers with equal disease risk, p=.17
(UNDER THE NULL!)
2. Use the random binomial function to randomly
select n=100 smokers and n=100 non-smokers,
each with p=.17
3. Calculate the observed difference in proportions.
4. Repeat this 1000 times (or some large number of
times).
5. Observe the distribution of differences under the
null hypothesis.
Computer Simulation Results
Empirical standard
error is about 5.3%
P-value from our simulation…
We also got 82
results as small
or smaller than
–8%.
When we ran this
study 1000 times,
by chance, we got
72 results as big
or bigger than
8%.
P-value
From our simulation, we
estimate the p-value to be:
154/1000 or .154
3. chi-square test of
independence
Smoker (E)
Heart disease (D)
21
Non-smoker
(~E)
13
No Disease (~D)
79
87
100
100
Null hypothesis: smoking and heart disease are independent
What does it mean to be
“independent” in stats?
Under independence, P(A&B)=P(A)*P(B)
In words the “joint probability” equals the product of
the “marginal probabilities.”
OR
The probability of both A and B happening is equal to the
probability of A times the probability of B.
If smoking and heart disease are independent, then
P(smoker&heart disease)=P(smoker)*P(heart disease)
Calculate expected counts
under independence…
Smoker (E)
Heart disease (D)
21
Non-smoker
(~E)
13
No Disease (~D)
79
87
100
100
IF smoking and heart disease are independent THEN:
P(HeartDisesae&Smoker)=P(HeartDisease)*P(Smoker)
P(HeartDisease)=34/100=17%
P(Smoker)=100/200=50%
IF INDEPENDENT, then P(HeartDisease&Smoker) should be 8.5%;
8.5% of 200 = 17
Fill in the expected table…
Smoker (E)
Heart disease (D)
17
No Disease (~D)
83
100
Non-smoker
(~E)
Marginals are fixed!
17
34
83
156
100
Notice that the rest of the table is
determined after you fill in 17 for cell A.
There are no degrees of freedom left! (This
table has only 1 degree of freedom).
Compare expected and
observed counts…
Smoker (E)
Heart disease (D)
No Disease (~D)
17
Non-smoker
(~E)
17

83
Smoker (E)
expected
83
Heart disease (D)
21
Non-smoker
(~E)
13
No Disease (~D)
79
87

observed
Chi-Square test
(observed - expected) 2
 
expected
2
1
2
2.25=1.5squared. The chi-square
test produces exactly the
square of the Z-test and
the same p-value.
(21  17) 2 (13  17) 2 (79  83) 2 (87  83) 2




 2.25
17
17
83
83
Degrees of freedom = (rows-1)*(columns-1)=(2-1)*(2-1)=1
Rule of thumb: if the chi-square statistic is much greater than it’s degrees of freedom,
indicates statistical significance. Here 2.25 not quite big enough—p=.131.
Bonus material:
The Chi-Square distribution:
is sum of squared normal deviates
df
 2 d f   Z 2 ; where Z ~ Normal(0,1 )
i 1
The expected value and
variance of a chi-square:
E(x)=df
Var(x)=2(df)
Case-control study example:

You sample 50 stroke patients and 50
controls without stroke and ask about
their smoking in the past.
Possible study results:
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
50
50
Statistics for these data



1. Odds ratio (relative risk)
2. Difference in proportions exposed
(absolute risk)
3. Chi-square
What’s the risk ratio here?
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
50
50
Tricky: There is no risk ratio, because we
cannot calculate the risk of disease!!
The odds ratio…

We cannot calculate a risk ratio from a case-control
study.

BUT, we can calculate a measure called the odds
ratio…
Odds vs. Risk
If the risk is…
½ (50%)
¾ (75%)
1/10 (10%)
1/100 (1%)
Then the odds
are…
1:1
3:1
1:9
1:99
Note: An odds is always higher than its corresponding probability,
unless the probability is 100%.
The Odds Ratio (OR)
Exposure (E)
Disease (D)
a
No Disease (~D)
c
No Exposure
(~E)
b
d
Odds of exposure
inThe
theproportion
cases
of cases to
controls
Pare
( E set
/ Dby
) the
investigator;
they
P (~ E therefore,
/ D)
do not represent the risk
P ( E / ~ofDdeveloping
)
(probability)
P (~ E / ~ D )
disease.
Odds of exposure
in the controls
OR 
a+b=cases
c+d=controls
a
ad
b
 
c bc
d
The Odds Ratio (OR)
Odds of exposure
in the cases
OR 
P(E / D)
This
P (~ E / D ) expression is
mathematically
P ( E / ~ D ) equivalent to:
P (~ E / ~ D )
Odds of exposure
in the controls
Backward from what we
want…
Odds of disease in
the exposed
P(D / E)
P (~ D / E )
P(D /~E )
P (~ D / ~ E )
Odds of disease in
the unexposed
The direction of interest!
Proof via Bayes’ Rule (optional)
P( E / D)
P(~ E / D)
P( E / ~ D)
P(~ E / ~ D)
Odds of exposure in the cases
Odds of exposure in the controls
Bayes’ Rule
P( D / E ) P( E )
P( D)
P( D / ~ E ) P(~ E )
P( D)
P(~ D / E ) P( E )
P(~ D)
P(~ D / ~ E ) P(~ E )
P(~ D)
P( D / E )
P(~ D / E )
P( D / ~ E )
P(~ D / ~ E )
=
Odds of disease in the exposed
What we want!
Odds of disease in the unexposed
The odds ratio
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
50
50
ad 15 * 42
OR 

 2.25
bc
35 * 8
Interpretation: there is a 2.25-fold higher odds of stroke
in smokers vs. non-smokers.

Inferences about the odds
ratio…


Does the sampling distribution follow a
normal distribution?
What is the standard error?
Simulation…





1. In SAS, assume infinite population of cases and
controls with equal proportion of smokers
(exposure), p=.23 (UNDER THE NULL!)
2. Use the random binomial function to randomly
select n=50 cases and n=50 controls each with
p=.23 chance of being a smoker.
3. Calculate the observed odds ratio for the
resulting 2x2 table.
4. Repeat this 1000 times (or some large number
of times).
5. Observe the distribution of odds ratios under
the null hypothesis.
Properties of the OR (simulation)
(50 cases/50 controls/23% exposed)
Under the null, this is the expected
variability of the sample ORnote
the right skew
Properties of the lnOR
Normal!
Properties of the lnOR
From the simulation,
can get the empirical
standard error (~0.5)
and p-valuE (~.10)
Properties of the lnOR
Or, in general, standard error
=
1 1 1 1
  
a b c d
Inferences about the ln(OR)
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
50
50
OR  2.25
ln( OR )  0.81
Z
ln( 2.25)  0
0.81

 1.64
0.494
1 1
1
1
  
8 15 35 42
p=.10
Confidence interval…
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
50
50
95% CI ln OR  0.81  1.96 * 0.494  0.16,1.78
95% CI OR  e
.16
1.78
,e
 0.85,5.92
Final answer: 2.25 (0.85,5.92)
2. Difference in proportions
exposed
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
50
50
P( E / D)  P( E / ~ D)  15 / 50  8 / 50
 30%  16%  14%
2. Difference in proportions
exposed
Z
14%  0%
.14

 1.67
.23 * .77 .23 * .77 .084

50
50
95% CI : 0.14  1.96 * .084  0.03 to .31
3. chi-square test of
independence
Smoker (E)
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42
Expected count for cell A:
proportion: 0.5*.23=.115
count: .115*100= 11.5
expected and observed
counts…
Smoker (E)
Stroke (D)
11.5
Non-smoker
(~E)
38.5

No Stroke (~D)
11.5
Smoker (E)
expected
38.5
Stroke (D)
15
Non-smoker
(~E)
35
No Stroke (~D)
8
42

observed
Chi-Square test
squared.
1

2
2.78=1.67-
(15  11.5) 2 (8  11.5) 2 (35  38.5) 2 (42  38.5) 2




 2.78
11.5
11.5
38.5
38.5
Not quite sufficient evidence to reject null…