Chapter 17
Comparing Two Proportions
March 16
1
In Chapter 17:
17.1 Data
[17.2 Risk Difference]
[17.3 Hypothesis Test]
17.4 Risk Ratio
[17.5 Systematic Sources of Error]
[17.6 Power and Sample Size]
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Data conditions
• Binary response variables (“success/failure”)
• Binary explanatory variable
• Notation:
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Sample Proportions
Incidence proportion, exposed
group:
a1
pˆ 1 
n1
Incidence proportion, nonexposed group:
a2
pˆ 2 
n2
Incidence proportion
≡ average risk
4
Example: WHI Estrogen Trial
Group 1
n1 = 8506
Estrogen
Treatment
Compare risks
of index disease
Random
Assignment
Group 2
n2 = 8102
Placebo
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2-by-2 Table
Exposure +
Exposure −
Total
Disease+
a1
a2
m1
Risk, exposed
a1
pˆ 1 
n1
Disease−
b1
b2
m2
Total
n1
n2
N
Risk, non-exposed
a2
pˆ 2 
n2
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WHI Data
E+
ETotal
D+
751
623
1374
D−
7755
7479
15234
Total
8506
8102
16608
751
pˆ 1 
 0.08829
8506
Compare these risks
623
pˆ 2 
 0.07689
8102
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§17.4 Proportion Ratio
(Relative Risk)
• Compare incidences from the two groups in form
of a RATIO
• Quantifies effect of the exposure in relative terms
Relative Risk
Parameter
p1
RR 
p2
Relative Risk Estimator
(“RR hat”)
pˆ 1
ˆ
RR 
pˆ 2
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Example: RR (WHI Data)
+
−
Total
Estrogen +
751
7755
8506
Estrogen −
623
7479
8102
751
pˆ 1 
 0.08829
8506
pˆ 1
ˆ
RR 
pˆ 2
0.08829

0.07689
623
pˆ 2 
 0.07689
8102
 1.1483  1.15
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Interpretation
• When p1 = p2, RR = 1  indicating “no
association”
– RR > 1  positive association
– RR < 1  negative association
• The RR indicates how much the exposure
multiplies the risk over the baseline risk of the
non-exposed group
– RR of 1.15 suggests risk in exposed group is “1.15
times” that of non-exposed group
• Baseline RR is 1!
– Thus, an RR of 1.15 is 0.15 (15%) above the baseline
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Confidence Interval for the RR
To derive information about the precision of the estimate,
calculate a (1– α)100% CI for the RR with this formula:
ln Rˆ R  z
e
where SEln RˆR 
1 
2
1
a1
SEln Rˆ R

1
n1

1
a2

1
n2
ln ≡ natural log, base e
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90% CI for RR, WHI
D+
751
623
E+
E−
D−
7755
7479
Total
8506
8102
ln Rˆ R  ln( 1.1483)  0.1382
SEln Rˆ R 
1
751
1
1
1
 8506
 623
 8102
 0.051920
For 90% confidence , z  1.645
e 0.1382(1.645)( 0.051920)  e0.13820.0854  e 0.0528,0.2236
 (1.05, 1.25)
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