Transcript Case Control Study

Case-Control Study
Chunhua Song
2007.10
Warm up
a
exposed
b
c
unexposed
d
Analysis of Cohort Study

CI(Cumulative Incidence)

ID(Incidence Density)

RR(Relative Risk,Risk Ratio)

AR ARP(Attributable Risk Percent )

PAR PARP(Population Attributable Risk
Percent)
In fixed cohort (the status of participants
is changeless)
CI=
number of new cases of a disease
during the follow-up period
number of participants at the
initiation of follow-up
in dynamic population (the status of
participants is protean)
number of new cases of a disease
during the follow-up period
ID=
total person-time of observation
RR (Relative Risk)
ratio of the risk (i.e., incidence
rate) in an exposed population to the
risk in an unexposed, but otherwise
similar, population.
Incidence（exposed）
RR=
Incidence（unexposed）
indicator of the strength (biological
significance) of an association between
an expose and disease.
RR>1
Research factor is a risk factor
(Positive association)
RR<1
Research factor is a protective
factor
(Negative association )
RR=1
No association between the factor
and the disease.
Incidence（exposed）
RR=
Incidence（unexposed）
AR (Attributable risk)
Numbers of cases among the
exposed that could be eliminated if
the exposure were removed.
AR is an estimate of the amount of risk that
is attributable to the risk factor after all other
known causes of the disease have been taken
into account
AR=Ie-I0
ARP (AR%) (attributable risk percent)
Proportion of disease in the exposed
population that could be eliminated if
exposure were removed.
Among the E group ,what percentage
of the total risk for disease is due to
the exposure
Ie  Io
ARP 
100%
Ie
PAR (Population Attributable Risk)
Numbers of cases among the general
population that could be eliminated if
the exposure were removed.
PAR=It-I0
PARP (PAR%) (population attributable risk
percent)
Proportion of disease in the study
population that could be eliminated
if exposure were removed.
It  Io
PARP 
100%
It
In a study of oral contraceptive use and
bacteriauria, a total of 2400 women aged
from 16 to 49 years were identified as free
from bacteriauria. Of these, 400 were OC
users at the initiation. 3 years later,20 of
the OC users had developed bacteriauria, 50
of the non-OC users had developed
bacteriauria. Based on data above, try to
evaluate the association between OC and
bacteriauria.

Ie( 3-year period CI )=(20/400) ×100%=5.0%
Io( 3-year period CI )=(50/2000) ×100%=2.5%
RR=Ie/Io=2
contraceptive use is a risk factor to
bacteriauria
AR=Ie-Io=2.5%
Ie-Io
ARP=
×100%=
Ie
5%-2.5%
5%
=50%
(20+50)
It =
×100%=2.9 %
2400
PAR=It-Io=2.9%-2.5%=0.4%
It-Io
PARP =
×100%=13.8%
It
Case-Control Study
Postulate of Case-Control Study
Types of Case-Control Study
Design of Case-Control Study
Analysis of Case-Control Study
Postulate of Case-Control Study
Case Control Study:
Subjects are selected on the basis of
whether they have a particular disease or
not .The association between the exposure
and the disease is evaluated by comparing
the two groups with respect to the
proportion having a history of an exposure
of interest.
exposure
ca
Case
Case
Non-exposure
b
c
Control
d
Types of Case-Control Study
Non-matching Case-Control Study
Case
patient
Sampling
Non patient
Control
Matching Case-Control Study
Matching:
Definition: In order to exclude the
effect of other factors, control
group are required to keep consistent
with case group in some aspects.
Age Sex Behavior
Smoke -------Lung cancer
Smoking
Method:
Frequency matching
Individual matching
Pitman efficiency increase by degrees
formula：1：R
X=2R / (R+1)
R=1, X=1
R=2, X=1.33
R=4, X=1.6
R=5, X=1.67
R=3, X=1.50
Over Matching
But if we let risk factors as matching
factor, that is ,control group are
required to keep consistent with case
group in risk factors. We call it as
over matching
Smoke -------Lung cancer
Smoking
Smoke -------Lung cancer
Drinking
Sex matching and drinking matching
If the persons who smoke
are all drinking
Smoke -------Lung cancer
The exposure factor (smoking) we study is
the same between two groups
Design of Case-Control Study
1 Ascertain the research intent
2 Define the disease and fix on the
measure method
3 A proper sample size
4 Selection of research subject
4.1 Selection of the case:
(1)Hospital-based
(2)Community-based
representational
4.2 Selection of the control
The same source with case
If control group is come from hospital, the
patient in control group should not suffer
from disease that have common causal with
the disease of interest.
Case
lung cancer
Control
bronchitis
5 Institute questionnaire
(1)Selection of variable
The factors are relate to the disease
and perhaps are the causes of disease
(2)Definition of variable
(3)Measurement of variable
6 Collection of the research-related
information
① Utilize varied routine record
② family interview
③ telephone or correspond inquiry
7 Clean-up ,enter & analyze data
Analysis of Case-Control Study
Non-Matching case control study:
Case
Control
Total
exposed
a
b
a+b
unexposed
c
d
c+d
Total
a+c
b+d
a+b+c+d=n
1 Test whether difference of exposure
proportion in 2 groups.
(ad  bc)  n
 
(a  c)(b  d )(a  b)(c  d )
2
2
Statistics: p-value
p<0.05
indicates the likelihood that a study’s
findings are due to chance in data
analysis
2 Estimate relative risk
2.1 OR(Odds Ratio)
Ratio of odds in favor of exposure among
cases to odds in favor of exposure among
controls.
ad
OR 
bc
2.2 OR 95%C.I.(confidence interval)
(11.96 /  2 )
OR95%C.I .  OR
A case-control study is conducted to reveal
the association between oral contraception
and MI. There are 150 MI patients and 150
Non-MI patients enrolled in this research.
Result is as bellows: Among the 150 MI
patients, 50 once used OC, and among the
150 Non-MI patients, 30 once used OC. Try
to evaluate the association between OC and
MI.
MI
Non-MI
Total
Exposed
to OC
50
30
80
Unexposed
100
120
220
To OC
Total
150
150
300
1 Test the difference of exposure
proportion in 2 groups.
(ad  bc)  n
 
(a  c)(b  d )(a  b)(c  d )
2
2
(50120 30100)  300

 6.82
150150 80 220
2
P<0.05 . So there is a significant
difference of the exposure proportion
in 2 groups.
2 Estimate relative risk
2.1 OR(Odds Ratio)
ad
50  120
OR 

 2.0
bc
30  100
2.2 OR 95%C.I.(confidence interval)
OR95%C.I .  OR
2
(11.96 / 6.82 )
(11.96  2 )
 (2
0.25
1.75
,2
=(1.19, 3.36)
Taking OC is a risk factor to MI
)
1:1 Matching Case Control Study
exposure
history
case
＋
＋
－
－
control
＋
－
＋
－
a
c
b
d
1:1 Matching Case Control Study
Case
Total
Control
Exposed
Non- exposed
Exposed
a
b
a+b
Non-exposed
c
d
c+d
a+c
b+d
Total
n
2
(b  c ) 2

(b  c )
c
OR 
b
Comparison of cohort and
case-control studies


provide information about a range of
effects related to a single exposure
Provide information about one effect
that afflicts the cases selected(studies
including multiple series of cases are an
exception)
Comparison of cohort and
case-control studies
Typically follow-up studies focus on
one exposure
 Provide information about a wide
range of potentially relevant
exposures

Comparison of cohort and
case-control studies
Evaluation of effects on rare
disease is problematic in follow-up
studies.
 Evaluation of effects on rare
disease are well suited to casecontrol studies.

Comparison of cohort and
case-control studies
Concern is in the follow-up
 Concern is in the determination of a
correct exposure classification

Comparison of cohort and
case-control studies
Exposure status is determined before
the presence of disease. No possibility
for the disease outcome to influence
exposure classification
 Exposure information comes from the
subject (or proxies) after disease
onset. Knowledge of disease could
affect exposure data. Greater
possibility of bias

Comparison of cohort and
case-control studies
Large, expensive, take time
 Smaller, less expensive, quick







Section 3: Calculate and answer questions:
1. In order to evaluate the hypothesized association
between the infection of HBV and liver cancer, a
case–control study was conducted. 100 patients with
liver cancer and 100 patients with diabetes were
enrolled into the research. The results of the study
were following: 80 patients with liver cancer have
had the infection of HBV, and only 10 patients with
diabetes have had the infection of HBV.
Questions:
(1) Fill in the data into a 2×2 table.
(2) Is there a significant difference of the proportion
of the infections of HBV between this two groups?
(3) Try to calculate the strength of the association
between the infection of HBV and liver cancer.
What can be wrong in the study?
Random error
Results in low precision
of the epidemiological
measure  measure is
not precise, but true
1 Imprecise measuring
2 Too small groups
Systematic errors
(= bias)
Results in low validity of
the epidemiological
measure  measure is
not true
1 Selection bias
2 Information bias
3 Confounding
Random errors
target practice
Systematic errors
Errors in epidemiological studies
Random error

Low precision because of
 Imprecise measuring
 Too small groups

Decreases with increasing group size

Can be quantified by confidence interval
Bias in epidemiology
1 Concept of bias
2 Classification and controlling of bias
2.1 selective bias
2.2 information bias
2.3 confounding bias
Overestimate?
Underestimate?
Random error：
Definition
Deviation of results and inferences
from the truth, occurring only as a
result of the operation of chance.
Bias:
Definition: Systematic, non-random
deviation of results and inferences
from the truth.
2 Classification and controlling of
bias
Time
Assembling
subjects
Selection
bias
collecting
data
Information
bias
analyzing
data
Confounding
bias
2.1 Selection bias
2.1.1 definition
Due to improper assembling method or
limitation, research population can not
represent the situation of target
population, and deviation arise from it.
2.1.2 several common Selection
biases
（1）Admission bias （Berkson’s bias)
There are 50,000 male citizen aged
30-50 years old in a community. The
prevalence of hypertension and skin cancer
are considerably high. Researcher A want
to know whether hypertension is a risk
factor of lung cancer and conduct a casecontrol study in the community .
case
control
Hypertension
1000
9000
No hypertension
4000
36000
5000
45000
sum
χ2 =0
sum
10000
40000
50000
OR=(1000×36000)/(9000 ×4000)=1
No association between
hypertension and chronic gastritis
Researcher B conduct another
case-control study in hospital of
the community.(chronic gastritis
patients as control) .
admission rate
Lung cancer & hypertension
20%
Lung cancer without hypertension
20%
chronic gastritis & hypertension 20%
chronic gastritis without hypertension 20%
case
control
hypertension 200 (1000) 200 (2000)
No hypertension 800 (4000) 400 (8000)
sum
1000 (5000) 600 (10000)
sum
400
1200
1600
case
hypertention
control
40
No hypertention 160
sum
200
χ2 =10.58
sum
100
140
200
360
300
500
P<0.01
OR=(40×200)/(100×160)=0.5
2.2 Information Bias
recalling bias
2）report bias
3）diagnostic/exposure suspicion bias
Measurement bias
2.3 Confounding bias
Definition:
The apparent effect of the exposure of
interest is distorted because the effect
of an extraneous factor is mistaken for
or mixed with the actual exposure
effect.
Properties of a Confounder:
• A confounding factor must be a risk factor for the
disease.
• The confounding factor must be associated with
the exposure under study in the source population.
• A confounding factor must not be affected by the
exposure or the disease.
The confounder cannot be an intermediate step in the
causal path between the exposure and the disease.
2.3.2 Control of confounding bias
1 In designing phase
1 ) restriction
2) randomization
3) matching
2 In analysis phase
1) Stratified analysis
(Mantal-Hazenszel’s method)
2) Standardized
3) logistic analysis