Warm Up

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 identifiedas 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 χ = (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 / χ 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 χ = (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 = = bc 30 ×100

=2.0

2.2 OR 95%C.I.(confidence interval) (1±1.96 χ 2 )

OR95%C.I . = OR =2

(1±1.96 / 6.82)

= (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

χ 2

O R

(b − c) 2 = (b + c) c = 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  Concern is in the correct exposure 

follow-up determination of a 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 classifiction  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 .

Hypertension

case

control

1000

9000

No hypertension

4000

sum 5000 χ2 =0

36000 45000

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

sum

hypertension 200 (1000) 200 (2000) 400

No hypertension 800 (4000) 400 (8000) 1200

sum

1000 (5000) 600 (10000)

1600

case

control sum

hypertention 40

100

No hypertention 160 sum

200

140

200 300

χ2 =10.58 P<0.01 OR=(40×200)/ (100×160)=0.5

500

360

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