Epide chap 4.pptx epidemology assignment year one

Measurement in Epidemiology
By Mengistu Handiso (MPHE)
Prepared by Mengistu H(MPHE) 1

Outline
 Epidemiological measurement variables
 Measurement of disease occurrence in
population
 Mortality measures
 Measurement of association
 Evaluation of causation

Measurement in Epidemiology
 Epidemiology is mainly a quantitative
discipline, so we should quantify health and
health related events.
 Epidemiological measurement variables
 Absolute number
 Ratio
 Proportion
 Rate

 Measurement of disease occurrence in population
1. Incidence rate
 Cumulative Incidence
 incidence Density
2. Prevalence rate
 Point Prevalence
 Period Prevalence
 Measurement of mortality (death rate )
 Measurement of association
1. Risk ratio
2. Odd ratio,
3. Risk difference

Measuring Disease Occurrence
Absolute number
 The number of cases(Absolute number ) in a
given community can give more epidemiologic
sense if they are related to the size of the
population.
 Such tie of the number of cases with the
population size can be determined by
calculating ratios, proportions, and rates

Ratio x፡y
 Ratio: The value of x and y may be completely
independent. example (Male: Female)
Proportion x/ x+y
 Proportion: is a ratio (expressed as a percent)
in which x is included in y.
 Example Female/Both sexes
(proportion of female in a community)

Rate
 Rate is a Proportion, it measures the
occurrence of an event in a population over
time.
 The time component is important in the
definition.
Measles cases in under five children in 1995
Under five children in 1995

 Notice three important aspects of this formula.
1. The persons in the denominator must reflect
the population from which the cases in the
numerator arose.
2. The counts in the numerator and denominator
should cover the same time period.
3. In theory, the persons in the denominator
must be “at risk” for the event, that is, it should
have been possible for them to experience the
event

Table1.1 Neonatal sepsis, Hospital A, Ethiopia, 2003

The line listing in the Table 1.1 presents some of the
information collected on infants born at Hospital
A with neonatal sepsis.
1. What is the ratio of males to females?
2. What proportion of infants lived?
3. What proportion of infants were delivered in a
delivery room?
4. What is the ratio of operating room deliveries to
delivery room deliveries?

Incidence rate
 Incidence rates are the most common way of
measuring and comparing the frequency of
disease in populations.
 It measures the rate at which people without
the disease develop the disease during a
specified period of time.
 The incidence rate expresses the probability or
risk of illness in a population over a period of
time.

 Since incidence is a measure of risk,
when one population has a higher
incidence of disease than another, we
say that the first population is at a higher
risk of developing disease than the
second, all other factors being equal.

 It is used to study disease aetiology
 Required data:
 period of observation;( Time)
 number of new cases; (n)
 time of disease onset (diagnosis Time );
 population at risk /denominator(N).
 The formula for calculating an incidence rate follows:

 The numerator of an incidence rate should reflect
new cases of disease which occurred or were
diagnosed during the specified period.
 The denominator is the population at risk.
 This means that persons who are included in the
denominator should be able to develop the disease
that is being described during the time period
covered.

Risk (Cumulative Incidence)
 Measures the risk (the likelihood, probability) that an
individual will contract the disease during a certain
time period or before a given age.
= New cases occurring during a given time period
Population at risk during the same period
 Cumulative incidence relates occurrences of new cases
to the population at the beginning of the study period.

Example
 Two surveys were done of the same
community 12 months apart. Of 5,000 people
surveyed the first time, 25 had antibodies to
histoplasmosis.
 Twelve months later, 35 had antibodies,
including the original 25.
1. What is cumulative/ Incidence during the 12-month
period:
x = Number of new positives during the 12-month
period = 35 - 25 = 10
y = population at risk = 5,000 - 25 = 4,975
x/y . 10n = 10/4,975 x1,000 = 2 per 1,000
2. Prevalence at the second survey:
x = antibody positive at second survey = 35
y = population = 5,000
n

Incidence Density
 Incidence density represents rate at which new
cases are occurring.
 Does not indicate the risk for any individual in a
population.
= New cases occurring in a specified period
Total person-time at risk in the same period
 An incidence measure is a speed, usually
expressed as number of cases per 1,000 or
100,000 units of follow-up time

 The population at risk is dynamic and
each person in the population
contributes the amount of time that
they remained under observation and
free from disease (person-time)
 The numerator is still the number of new
cases, but the denominator is the sum of
the time each person is observed,
totalled for all persons.

 Person-time rates are often used in
cohort (follow-up) studies of diseases
with long incubation or latency periods,
such as occupationally related diseases,
AIDS, and chronic diseases.
 Time unit=month, year, day
 person-time=person-year, person-
month,

Example:-
 Investigators enrolled 2,100 men in a study
and followed them over 4 years to determine
the rate of heart disease.
 We assume that persons diagnosed with
disease and those lost to follow-up were
disease-free for half of the year, and thus
contribute ½ year to the denominator.

Example
 Initial enrolment: 2,100 men free of disease
 After 1 year: 2,000 disease-free, 0 with
disease, 100 lost to follow-up
 After 2 years: 1,900 disease-free, 1 with
 After 3 years: 1,100 disease-free, 7 with
 After 4 years: 700 disease-free, 8 with
1. Identify x: x = cases diagnosed = 1 + 7 + 8 = 16
2. Calculate y, the person-years of observation

 700 men x 4.0 years = 2,800 person-years
 8 + 392 = 400 menx3.5 years = 1,400 person-
years
 7 + 793 = 800 menx2.5 years = 2,000 person-
years
 1 + 99 = 100 menx1.5 years = 150 person-years
 0 + 100 = 100 menx0.5 years = 50 person-years
 Total = 6,400 person-years of observation

Prevalence rate
 Prevalence, sometimes referred to as prevalence rate,
is the proportion of persons in a population who have
a particular disease or attribute at a specified point in
time or over a specified period of time.
 Numerator is number of existing cases
 Denominator is your population of interest
(including all those in numerator)
 The formula for presence of disease is:

Point prevalence:-
Point prevalence is the amount of disease present
in a population at a single point in time.
Point prevalence =
all the cases of factor of interest at a given time
x10n
total population

Period prevalence.
The numerator in period prevalence is the
number of persons who had a particular
disease or attribute at any time during a
particular interval (week, month, year, decade,
or any other specified time period).
Period prevalence =
all cases (old and new) of the factor of interest during the time period x
10n
average population during the given period of time

Exercise 2.4
 Recall that Figure 2.1 represents ten
episodes of an illness in a population of
20 over a period of 16 months.
 Each horizontal line represents the portion
of time one person spends being ill. The
line begins on the date of onset and ends
on the date of death or recovery.
28
Prepared by Mengistu H(MPHE)

29

Calculate the following rates:
a. Point prevalence on October 1, 1990
b. Period prevalence, October 1, 1990 to September
30, 1991
C, point prevalence on September 30,1991
30

a. Point prevalence on October 1, 1990:
x = cases present on 10/1/90 = 6
y = population = 20
= x /y .10n=
6/20 × 100 = 30%
31

b. Period prevalence, October 1, 1990 to
September 30, 1991
x = cases present between 10/1/90 and
9/30/91 = 6 old case , 4 new case =10
y = population = 20
x/y × 10n =
 10 /20× 100 = 50%
32

C, point prevalence rate on September
30,1991
x= 6 old case, 4 new case , 5 death
x/y =6+4-5/20
= 5/20*100
= 25%
33

Calculation : in f/up {cohort study )
Incidence rate;-
 Subtract old case from denominator(risk
)
 Subtract dead, cure case from the
numerator=new case=point type)
Point Prevalence rate :-
 Subtract dead, cure case from the
numerator = (old &new case )=point
prevalence 34

In a study population of 1000 inhabitants there are 60
cases of tuberculosis and 20 cases of hypertension.
During the first six months there are 17 new cases of
tuberculosis and 8 new cases of hypertension. At the end
of the year the total of new cases is 27 for tuberculosis
and 13 for hypertension. During the same period, 8 new
cases of tuberculosis died and 8 were cured. In the same
period, 5 cases of hypertension died and no case of
hypertension was cured.
Calculate the following rates :-
1. -Prevalence rate at the beginning of the study
2. -Incidence rate A) tuberculosis and B) hypertension at the
end of the first six months
3. -Incidence rate of tuberculosis at the end of the year
4. -Prevalence rate of A)tuberculosis and B) hypertension at
the end of the year
35
Class work

Possible answer

Characteristics of Prevalence
Cause and effect measured simultaneously
-Impossible to infer causation
Useful for planning (e.g. beds, clinics, workforce
needs)
High prevalence  high risk
 could reflect increased survival(improved care,
behavior change - long duration)/old case
Low prevalence  low risk
 could reflect rapid fatal or cure process -
short duration)
Easy to obtain need only one measurement

Figure 3.2 Relationship between prevalence and
incidence
Migration
Care,
behavioral
change ,
-Dx

Relationship between prevalence and incidence
 Prevalence is based on both incidence
(risk) and duration of disease.
P ≈ I*D
 High prevalence of a disease within a
population may reflect high risk, or it may
reflect prolonged survival without cure.
 Conversely, low prevalence may indicate
low incidence, a rapidly fatal process, or
rapid recovery.

Comparison of prevalence and incidence
 The prevalence and incidence of disease are
frequently confused.
 They are similar, but differ in what cases are
included in the numerator.
 Numerator of Incidence = new cases occurring
during a given time period
 Numerator of Prevalence = all cases present
during a given time period
 As you can see, the numerator of an incidence rate
consists only of persons whose illness began during
a specified interval.

Prevalence rate increased by
•Longer duration of the disease
•Prolongation of life of patients
without cure.
•Increase in new cases
•Immigration of cases
•Out migration of health people
•In-migration of susceptible people
•Improved diagnostic facility

Prevalence rate decreased by
•Shorter duration of disease.
•High case-fatality rate from
disease.
•Decreases in new cases (decrease
in incidence)
•In-migration of healthy people.
•Out migration of cases.
•Improved cure rate of case

Attack Rate
 An attack rate is a variant of an incidence rate,
applied to a narrowly defined population observed
for a limited time, such as during an epidemic.
 The attack rate is usually expressed as a per cent, so
10n equals 100.
 For a defined population (the population at risk),
during a limited time period.

 Of 75 persons who attended a church picnic, 46
subsequently developed gastroenteritis.
 To calculate the attack rate of gastroenteritis we
first define the numerator and denominator:
X = Cases of gastroenteritis occurring within the
incubation period for gastroenteritis among
persons who attended the picnic =46
Y = Number of persons at the picnic = 75
Then, the attack rate for gastroenteritis is :
Example

Secondary Attack Rate
 A secondary attack rate is a measure of the
frequency of new cases of a disease among the
contacts of known cases/primary case.
 The formula is as follows:-

EXAMPLE
 To calculate the total number of household
contacts, we usually subtract the number of
primary cases from the total number of people
residing in those households.
 Seven(7) cases of hepatitis A occurred among
70 children attending a child care centre.
 Each infected child came from a different
family. The total number of persons in the 7
affected families was 32.

o One incubation period later, 5 family
members of the 7 infected children also
developed hepatitis A.
o Calculate the attack rate in the child care
centre and the secondary attack rate
among family contacts of those cases?.
o Attack rate in child care centre:
x = cases of hepatitis A among children in
child care center = 7
y = number of children enrolled in the child
care centre = 70

Figure 3.4

Mortality Measures
Mortality Rates:-
 A mortality rate is a measure of the frequency
of occurrence of death in a defined population
during a specified interval. For a defined
population, over a specified period of time.
 The following are frequently used mortality
measures,

Case-fatality rate
 The case-fatality rate is the proportion of persons
with a particular condition (cases) who die from that
condition.

 Proportionate mortality describes the proportion
of deaths in a specified population over a period of
time attributable to different causes. Each cause is
expressed as a percentage of all deaths, and the sum
of the causes must add to 100%.
Proportionate mortality Ratio

CDR = No. of deaths in a year x 1000
Total mid-year population
= Hard to use to compare different
populations, due to distortions by
differences in age-sex composition
Crude Death Rate (CDR):

Age Specific Death Rate
ASDR = total Deaths of at ages a or age group a *
1000
Mid year population at age a or age
group a
Distinguishes the mortality level of different age,
sex, or occupational groups.
Specific death rate
Cause Specific death rate (CSDR)
 CSDR = Total deaths from a given cause * 1000
population at risk
.

 Infant Mortality Rate (IMR)
IMR = No. of deaths of under 1 in a year x
1000
Live births in a year
= 52 death from 1000 live birth , 2011
EDHS
 Early Neonatal M. R
Deaths under1 week x 1000
Live births

 Late Neonatal MR:
Deaths 1-4 weeks x 1000
Live births
 Post Neonatal MR :
Deaths 4-52 weeks x 1000
Live births
IMR = ENMR+LNMR+PNMR

 Child Mortality Rate:
CMR = deaths of children in 1-4 years age
group*1000
Mid-year population of children 1-4
years

 Under five mortality rate : the
probability of dying between birth & age
five per 1000 live births.
U5MR= No. of deaths of under five children* 1000
total live births
58

Maternal Mortality
 Definition:-
A maternal death :- the death of a woman while
pregnant or Within 42 days of termination of
pregnancy, irrespective of the duration and the site
of the pregnancy, from any cause related to or
aggravated by the pregnancy (WHO 1993).
- requires specific cause of death information

 Maternal Mortality Ratio (MMR) :
MMR = No. of maternal deaths in a year *
100,000
No. of Live births
Example, Eth 2005 ;- 673/100,000 live births
 Represents the risk associated with each
pregnancy, i.e., the obstetric risk

Maternal mortality measures…
 Maternal Mortality Rate :
= No. of maternal deaths in a specified
period*1000
No. of women of reproductive age
 Represents both the obstetric risk and the
frequency with which women are exposed to
this risk

Natality Frequency Measures
 Natality measures are used in the area of maternal
and child health and less so in other areas. The
following Table shows a summary for some frequently
used measures of natality.

Class work
 A total of 2,123,323 deaths were
recorded in the United States in 1987.
The mid-year population was estimated
to be 243,401,000. HIV-related mortality
and population data by age for all
residents and for black males are shown
in Table 2.9. We will use these data to
calculate and interpret the following
four mortality rates:

 a. Crude mortality rate.
 b. HIV-(cause)-specific mortality rate for
the entire population
 c. HIV-specific mortality among 35- to
44-year-olds
 d. HIV-specific mortality among 35- to
44-year-old black males

Answer
A. 872.4 deaths per 100,000 populations
B. 5.5 HIV related death per 100, 000
C. 14.0 HIV related death per 100,000 35-
44years old
D. 72.9 HIV related death per 100,000 35-
44 years old black male

Relative Risk, Odds Ratio,
Measure of Association
Prepared Mengistu Handiso(MPHE) 67

Measurement of risk:
 A number N of subjects exposed to a factor of
risk for a period of time T. The risk of
becoming ill is the proportion of people in this
cohort of N people, becoming ill (new cases)
during the period T.
Risk (R) = number with the event
Number observed
The risk is therefore a proportion, its minimum
value is zero and its maximum value is 1.

Testing and measuring the association between
risk factor and diseases
a b a+b
c+d
c d
a+c b+d a+b+c+d
Disease
Yes No
Exposed
No
yes
Total

Relative risk
 The Relative risk estimates the magnitude of
an association between exposure and disease
and indicates the likelihood of developing the
disease in the exposed group relative to those
who are not exposed.
Risk in exposed = a/a+b
Risk in non-exposed=c/c+d
Therefore, RR = a/a+b
c/c+d
 When the risk in the exposed and the risk in the
non exposed is known, the ratio of these risks
can be calculated and represent the relative
risk. 70

Relative Risk (RR) =Risk in the
exposed
Risk in the non
exposed
 Interpretation :The value of the RR
reflects the magnitude of the
association between exposure and
disease. A relative risk of 5 means that
the probability to develop the disease
in the exposed is 5 times the
probability to develop it in the non
exposed.

Use of the Relative Risk
 The relative risk measures whether there is
an association between the risk factor and
the frequency of the disease.
 It is a first step into evaluating a causal
relationship. On its own a relative risk
does not allow to conclude about
causality. It only shows an association that
may not be causal.

a. RR=1 There is no association between exposure
and disease. That means the risk of developing
disease in the exposed and non-exposed groups
are identical.
b. RR>1 There is an association between exposure
and disease and the exposure is associated with an
increase of the frequency of the disease.
c. RR<1 There is an association between exposure
and disease and the exposure is associated with a
decrease of the frequency of the disease

Calculation of the relative risk
Example :
 A cohort study was carried to examine the effect of
early weaning on diarrhoea. 120 children aged 6
months who were already weaned (exposed) were
compared to 480 control, of the same age, still on
breast milk (non exposed, 4 controls per case).
 The two cohorts were followed up for three months
and any episode of diarrhoea was recorded. Once a
child had had one episode of diarrhoea, she/he was
withdrawn from the study and the risk was then
calculated.

Interpretation #1: risk in Weaning early is 1.12
times higher than not weaning early
interpretation #2: children who were weaned
early had 1.12 times higher risk of developing
diarrhea when compared to not weaned early .
 Risk in the exposed =
89/120 = 0.74
 Risk in the non exposed =
318/480 = 0.66
 Relative risk = 0.74/0.66 =
1.12
 This is not substantially
different from 1, there is
therefore no association
between early weaning
and diarrhoea, which can
be interpreted as: Breast
milk does not protect
from diarrhoea.
89 31 120
480
318 162
407 193 600
Wean
ed
early
Not
weaned
early
Diarrhoea no diarrhoea

Odds ratio
 The word "odds" means the chances of an event to
happen.
 The Odds of an event is the ratio of the event to
happen over the event not to happen.
 If a is the number of event happening and b the
number of event not happening, the odds are a/b.
Thus odds= Number with events
Number without events

 The risk of becoming ill cannot be
calculated in a case control study, the
odds of being ill can be. If in the cohort
study the risk of becoming ill was
a/(a+b), the odds of being ill versus not
being ill is a/b in exposed .
 The ratio of the odds in the exposed
over the odds in the non exposed is
called the Odds Ratio and can be
calculated.
 Again to the prototype two-by-two table
can be used to calculate the odds ratio.

Calculation of the Odds Ratio
 Odds of being ill in exposed=__a__
b
 Odds of being ill in non exposed = __ c__
d
 Odds ratio (OR) =_Odds in exposed
Odds in non exposed
Odds Ratio = ad/ b c

Interpretation
 OR=1 There is no association between exposure
and disease. That means the odds of developing
disease in the exposed and non-exposed groups
are identical.
 OR>1 There is an association between exposure
and disease and the exposure is associated with
an increase of the odds of the disease.
 OR<1 There is an association between exposure
and disease and the exposure is associated with a
decrease of the odds of the disease.

Example
 A case control study of lung cancer and
cigarette smoking was carried out. All the
patients diagnosed to have lung cancer in a
hospital in a certain period of time were
included and their smoking habit recorded.
 For each case (lung cancer) one control (a
patient of same age and sex admitted in
the same hospital, at the same time, for an
other complaint) was chosen and its
smoking habit recorded. The following
table shows the results:

Interpretation #1
Odds of being ill in smoker is 5.4 times higher than the
Odds of being ill in non smoker.
The odd ratio is: OR=
70x70/30 x 30
= 5.4
 This suggests a fairly
strong association
between smoking
and lung cancer (by
the way the
association is
statistically significant
which means that the
results of this sample
can be extrapolated
to the population as a
whole).
70 30 100
100
30 70
100 100 200
Smokers
Non-
smokers
Lung
cancer
No lung
cancer
Interpretation #2
• The odd of pt who were current
smoke had 5. 4 times higher risk of
developing lung cancer when
compared to non-smoker.

Impact Measures
82
1. Attributable Risk (AR)
2. Attributable Risk Percent (ARP)
3. Population Attributable Risk (PAR)
4. Population Attributable Risk Percent
(PARP)

Attributable Risk (AR) / Risk
Difference (RD)
83
 AR provides information about the absolute
effect of the exposure or the excess risk of
disease in those exposed.
AR = Incidence among exposed (Ie) _ Incidence among non-
exposed (Io)
 AR quantifies the risk of disease in the exposed
group that is attributable to the exposure by
removing the risk of disease that occurs due to
other causes.

Alternative Naming of AR
84
 Risk difference
 Rate difference
 Cumulative incidence difference
 Incidence density difference

D. Attributable Risk Percent
85
 Estimates the proportion of the disease among
the exposed that is attributable to the
exposure,
 The proportion of the disease in the exposed
group that could be prevented by eliminating
the exposure
AR % = (Ie - Io) X 100
Ie

Example 1
86
 Among 2390 women aged 16 to 49 years
who were free from bacteriuria, 482 were
OC users at the initial survey in 1973,
while 1908 were not. At a second survey
in 1976, 27 of the OC users had
developed bacteriuria, as had 77 of the
non users. Calculate the measure of
association( RR), AR, AR% and interpret
it.
 Draw epidemiological 2x2 table
 What is study design
 Calculate Measure of association

Example:
87
O
C
U
S
E
Bacteruria
Yes No Total
Yes 27 455 482
No 77 1831 1908
Total 104 2286 2390

88
Cohort study
Calculate RR
RR = Ie = 27
lo [ 482 ] = 1.4
[ 77 ]
1908
 Interpretation: women who used oral
contraceptive had 1.4 times higher risk of
developing bacteruria when compared to
non-users.

89
Example: Refer to example 1 and calculate AR
AR = 27 _ 77
482 1908
=0.0156=1566 per 100,000 OC users
Interpretation: The excess occurrence of
bacteruria among OC users attributable to
their OC use is 1566 per 100,000 OC users

Example:
90
Refer to example 1 and calculate AR%
AR % = (27/482-77/1908)/27/482x100
=1566/105 X 100 =27.96 %
27/482
AR % = Ie –Io/Ie X 100
 Interpretation: If OC use causes bacteruria,
about 28 % of bacteruria among women who
use OC can be attributed to their OC use and
can be eliminated if they did not use oral
contraceptives

POSSIBLE OUTCOMES IN STUDYING THE RELATIONSHIP
91
 No association between exposure and disease
AR=0, OR/RR=1
 Positive association between exposure and
disease (more exposure, more disease)
OR/RR>1, AR>0
 Negative association between exposure and
disease
(more exposure, less disease)
AR<0 (negative), OR/ RR <1(fraction, mean
decimal )

Evaluation of causation
Should I believe my
measurement ?

Smoking Lung cancer
Associated
OR = 9.1
Due to,
-Chance?
-Confounding?
-Bias?
True association
may be:
-causal
-non-causal
93

Judging Observed Association
Apply the criteria and
make judgment of causality
Could it be due to selection or measurement
bias?
NO
Could it be due to confounding?
NO
Could it be A result of chance?
NO probability
Could it be Causal?

Common Problems in observed
findings
 Inadequacy of the observed sample
 Inappropriate selection of study
subjects
 Inappropriate/unfair data collection
methods
 Comparing unequal

Accuracy
 Accuracy = Validity + Precision
.Validity= is the finding a reflection of the
truth?
. Precision= is the finding due to sampling
variation?

 Validity Vs Reliability

Precision
 Precision in measurement and estimation
corresponds to the reduction of random error.
 Mostly related to sampling variation or
sampling error.
Solution:
 Increase sample size
 Improving the efficiency of measurement

Validity
 Internal: are we measuring what we intend to
measure
 Do we have alternative explanations for the
observed findings:
 Chance
 Bias
 Confounding
 External (generalizability) : can we make
inferences beyond the subjects of the study

Chance/random error
››.››reduced by increasing sample size
 Chance can often be an alternative explanation
to observed findings must always be
considered.
 Evaluation of chance is a domain of statistics
involving:
1. Hypothesis Testing (Test of Statistical
Significance)
2. Estimation of Confidence Interval
 But , Statistical significance do not provide

P-value ≤ 0.05 􀃆 Chance is unlikely
explanation
 ∴ Reject the null hypothesis
 ∴ There is Statistically significant
difference
Confidence Interval
 Provide information that p-value gives.
– If null value is included in a 95%
confidence interval, by definition the
corresponding P-value is >0.05. so , it is 101

Definition of bias
Any systematic error in an epidemiological
study, that results in an incorrect estimate of the
association between exposure and risk of disease
An alternative explanation for an observed
association is the possibility that some aspect of
the design or conduct of a study has introduced
a bias into the results.

Bias
 Unlike “chance” and “confounding,” which can be
evaluated quantitatively, the effects of bias are far
more difficult to evaluate and may even be
impossible to take into account in the analysis.
General class of
Bias
Selection Observation
bias (Information)
bias

Observation/Information Bias
 Results from systematic differences in
the way data on exposure or outcome
are obtained from the various study
groups
››››››››› data collection ?

Recall Bias
 Sick individuals more likely to remember
and report exposures than healthy
individuals
 Problematic in case-control studies

Control of Bias in the Design
Phase
1. Choice of Study Population-to reduce
selection bias
2. Data Collection Methods- to reduce
Observation bias
 Use standardized questionnaires
 Train data collectors/interviewers
 Method of data collection should be
similar for all study groups

A mixing of the effect of the exposure
under study on the disease with that of a
third factor
• A factor which is associated with the
exposure variable, and independent
of the exposure, is related to the
outcome/disease (that is, it’s a risk
Confounding

Criteria of confounder variable
 It must not intermediate
 It should be risk factor/cause for outcome/
disease with other main variable
 It must be risk for the outcome/disease
independently
Eg
 Coffee and acute MI
 Low density lipoprotein and MI

Interrelationship
EXPOSURE(A) DISEASE
OR crude ≠ ORA = ORB
CONFOUNDING FACTOR(B)

Causation
 Epidemiology and statistics review
 Epidemiological concepts of
 disease incidence and prevalence
 relative risk
 Statistical concepts of
 p-values
 confidence intervals
 Epidemiological study designs
 Randomized controlled trials
 Cohort studies
 Case-control studies
 Cross-sectional studies
 Randomized studies tend to offer stronger evidence than
observational studies

Sir Austin Bradford Hill
 In 1965
 Proceedings of the Royal Society of Medicine
 Bradford Hill’s listed the following criteria in causality
in attempting to distinguish causal and non-causal
associations
1. Strength of association
2. Consistency of findings
3. Biological gradient (dose-response)
4. Temporal sequence
5. Biological plausibility
6. Coherence with established facts
7. Specificity of association

Strength of the Association
 The Stronger the association (OR
0.00 or + ∞ ), then less likely the
relationship is totally due to the effect of an
uncontrolled confounding variable
 A strong association serves only to rule out
hypothesis that association is entirely due
to weak unmeasured confounder or other
sources of bias
 But weak association does not rule out a
causal association

Biological Credibility / Plausibility
 The belief in the existence of a cause and
effect relationship is enhanced if there is a
known or postulated biologic mechanism
by which the exposure might reasonably
alter the risk of developing the disease
 Alcohol and CHD (HDL)
 OC use and circulatory disease (platelet
adhesiveness; arterial wall changes)
 Smoking and lung cancer (hundreds of
carcinogens and promoters)

 Since what is considered biologically
plausible at any given time depends on
the current state of knowledge, the lack
of a known or postulated mechanism
does not necessarily mean that a
particular association is not causal

Consistency with Other
Investigations
 Have multiple studies conducted by
multiple investigators concluded the
same thing?
 Relationships that are demonstrated in
multiple studies are more likely to be
causal, i.e., consistent results are found
 in different populations,
 in different circumstances, and
 with different study designs.

Time Sequence / Temporality
 Exposure of interest has to precede the
outcome (by a period of time that
biologically makes sense)
 Smoking and lung ca; induction/latency

Dose-Response
 Smoke more, higher CHD death rates
 Difficulty: The presence of a dose-response
relationship doesn’t mean that the association is one
of cause and effect. Could be, for example, due to
confounding.
 Smoking and hepatic cirrhosis (alcohol)
 Absence of a dose-response relationship does not
mean that a cause-effect relationship does not exist.
 Sometimes there is a convincing association but not a
dose-response relationship

Coherence
 Causal mechanism proposed must not
contradict what is known about the
natural history and biology of the
disease, but the causal relationship may
be indirect data may not be available to
directly support the proposed
mechanism

Thank you !!
119

Epide chap 4.pptx epidemology assignment year one

More Related Content

What's hot (20)

Similar to Epide chap 4.pptx epidemology assignment year one (20)

More from GetahunAlega (20)

Recently uploaded (20)

Epide chap 4.pptx epidemology assignment year one