Notes on RR, Odds, Or, AR, AR% - Epidemiology

8/10/2019 Notes on RR, Odds, Or, AR, AR% - Epidemiology

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Author:Dr. Raghupathy Anchala, MD MPH PDCR, IIPH Hyderabad

Epidemiology made easy for beginners

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Authors note : Succinct and lucid style of writing coupled with real-life examples, which health

professionals commonly encounter, have been illustrated to help learners understand and

appreciate the concepts behind risk, relative risk, attributable risk, attributable risk difference,

odds, odds ratios, sensitivity, specificity, true positivity, true negativity, likelihood ratios,

posterior and prior probability (and odds). The endeavor has been to negate the difficulties that

one commonly faces with measures of association and effect and to help ease the process of

recall ability. Epidemiology rests on understanding these foundational pillars of association

between an exposure and an outcome. The fundamental concepts such as, what epidemiology

aims to cover, differences between descriptive and analytical epidemiology, epidemiological

triad, causal factors, natural history, steps in epidemiology have been listed and the measures

have been explained in detail.

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Concept one - Distribution (in terms of person, place and time) and determinants (in terms of

agent, host and environment) are the two fundamental aspects that an epidemiologist uses

frequently to arrive at a hypothesis in conducting an investigational study that deals with

diseases and epidemics.

Concept two -Examining, identifying, and reporting on the frequency and distribution of disease

in a population constitutes descriptive epidemiology. Analytic Epidemiology looks at testing a

hypothesis about the cause of disease by studying how exposures relate to the disease.

Concept three Agent, host and the environment together determine the susceptibility of a

person to develop a disease. The severity of an infection depends on the host (the sufferer).

Likewise, the probability of the disease depends on the immune constitution, personal traits,

behaviors and genetic predisposition of the human body (host). Agent (biological, physical and

chemical) has been defined as the necessary factor for disease to occur. Environment (external

conditions, physical or biologic or social) contributes to the disease process. Epidemics arise

when host, agent, and environmental factors are not in homeostasis (balance).

A new agent, a change in existing agent (infectivity, pathogenicity, virulence), change in number

of susceptible population, environmental changes that affect transmission of the agent or

growth of the agent lead to the occurrence of a disease (or an epidemic excess than the

normal expected)


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Concept four The causal factors can be necessary or sufficient. Necessary factors are those

that when removed, the disease does not occur. Sufficient factors are those that contribute tosome part of the disease process. Even in the absence of sufficient factors, a disease may

develop. A combination of sufficient and necessary factors causes disease.

Examples - Without HIV infection, AIDS does not developNecessary factor

Development of tuberculosis requires M. tuberculosis and other factors, such as

immunosuppression, to cause disease. Bacteria still necessary, but not sufficient to

cause the disease

Concept five:Public Health is an integrated discipline. Health protection, disease control, riskybehavioral change, community development, primary health care and surveillance are the

notable fields in which the study of determinants and distribution (epidemiology) comes into

play.

Concept six:The natural history of disease is the history of a particular disease in the absence of

intervention, prevention or treatment. Epidemiology deals in primary, secondary and tertiary

prevention (both at an individual and population level) based on the natural history of the

disease.

Concept seven: The steps of an epidemic investigation or any causal study can be summarized

as below:

1.

Begin with a general broad problem

2.

Collect information about the problem,

3.

Study the specific information collected,

4.

Reassess the results and draw conclusions,

5.

Re-evaluate the problem,

6.

Re-formulate the question and


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7.

Collect additional information which will show the relationship between the exposure

and the outcome or the event.

Applications of epidemiology

Establish patterns of endemic and epidemic diseases

Determine origin of diseases with unknown etiology

Investigate/control diseases whose cause is unknown or poorly understood

Describe the ecology/natural history of disease

Plan and monitor control programs

Assess economic impact of disease

Development of prevention programs

Determine cost and benefits of alternate treatment, prevention, control programs

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Measures - The following examples have been devised to introduce the importance and

emphasize the measures of association and effect that we commonly encounter.

For 2 x 2 cells

Heart attack (Disease) No heart attack (No

disease)

Totals

Smoking (Exposed) (a ) 80 (b) 20 a+b (all persons

exposed to smoking)

100No smoking (Not

exposed)

(c) 10 (d) 90 c+d (all persons who

are not smokers)

10+90 = 100

Totals a+c (all persons with

heart attack)

b+d (all persons with no

heart attack)

200


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90 (80+10) 20+90 = 110

Number of persons who are exposed with disease= a = 80

Number of persons who are exposedbut NOT having disease= b = 20

Number of persons who are NOT exposed but having disease = c = 10

Number of persons whoare NOT exposed and NOT having disease = d =90

1.) Risk of exposed to have disease = Risk of smokers to have heart attack

= (Disease and exposed)/ (all exposed persons)

= a/(a+b) = 80/(80+20) = 80/100 = 0.8 = 0.8 x 100% = 80%

This means that 80% of smokers will have the risk of having a heart attack.

This also means that out of 100 people exposed to smoking (smokers), 80 will have heart attack.

2) Risk of not exposed to have disease = Risk of people who are not smoking to have heart attack

= (no smoking with disease) / all non smokers

= c/c+d = 10/100 = 10/(10+90) = 10/100 = 0.1 = 10%

This means that 10% of non smokers will have heart attack. This also means that out of 100 non

smokers, 10 will have disease.

3) Relative risk (RR) for smokers to have heart attack

Relative means compared to ..here we compare smoking to non -smokers


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So, the question is asking us how much the risk for smokers to develop heart attack is more as compared

to non smokers

= Risk of exposed to have disease / Risk of not exposed to have disease

= Risk of smokers to have heart attack /risk of non smokers to have heart attack

= 0.8 / 0.1 = 8

This means that the relative risk for smokers to develop heart attack is 8 times more (or higher) as

compared to non smokers.

This also means that the relative risk for smokers to develop heart attack is 700% more(or higher) as

compared to non smokers.

The null value for RR is 1. To reject null hypothesis, we should get values of RR to be higher or lower

than that of null value of RR

So, 8 minus 1 = 7 (7 X 100 = 700%)

4. Attributable risk (AR) = Excess risk

Is the risk difference (RD) between two groups, or the excess risk that smokers have as compared to non

smokers to develop a heart attack

So if the question is asking you - what is the AR for smokers to develop heart attackthen you subtract

the risk of smokers to have heart attack from that of the risk of non smokers to have heart attack.

AR = (risk of smokers to have heart attackrisk of non smokers to have heart attack)

= 0.80.1 = 0.7 or 70%

Therefore this 70% means that 70% is the risk difference between smokers and non smokers to have a

heart attack. This also means that smokers have an excess risk of 70% as compared to non smokers to

have heart attack


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The difference between AR and RR should be very clear

RR is telling you how many times the risk is high in smokers as compared to non smokers

AR is telling you how much the risk is bigger in smokers as compared to non smokers to develop heart

attack

5) AR % - also called as attributable proportion

Ratio of (risk difference among both groups) / (risk in the exposed group) x 100

= (risk in exposedrisk in unexposed) / (risk in exposed population ) x 100

=

(risk in smokers to have heart attackrisk in non smokers to have heart attack )

x 100

Risk in smokers to have heart attack

= (0.80.1) / 0.7 = 1 x 100 = 100%

This means that 100% of risk among smokers to develop heart attack can be attributed to smoking

6. ODDS

Odds is the chance of an event to occur divided by the chance of the event not to occur.

So, if I say that odds of the horse winning the race is 4/7; it means that 4 times the horse will win and 7

times, it will lose.


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In other words, if the horse runs in 11 ( 4+7) races, the chances of winning are in 4 races and chances of

losing are in 7 races

7. Odds Ratio (OR) of exposure

= odds of exposed to have a disease / odds of not exposed to have a disease

=

Odds of exposed to have disease

Odds of not exposed to have disease

Odds of exposed to have disease = Chance of exposed to have disease / chance of exposed to have not

have disease

Odds of not exposed to have disease = Chance of not exposed to have disease / chance of not exposed

to have no disease

Therefore,

Chance of exposed to have disease / chance of exposed to have not have disease

Chance of not exposed to have disease / chance of not exposed to have no disease


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=

Chance of smokers to have heart attack / chance of smokers to have no heart attack

Chance of non smokers to heart attack / chance of non smokers to not have heart attack

So, in this 2 x 2 cellsHeart attack (Disease) No heart attack (No

disease)

Totals


exposed to smoking)

100

No smoking (Not

exposed)

(c) 10 (d) 90 c+d (all persons who

are not smokers)

10+90 = 100


heart attack)

90 (80+10)


heart attack)

20+90 = 110

200

Odds of exposed to have disease = odds of smokers to have heart attack = Chance of smokers to have

heart attack / chance of smokers to not have heart attack = 80/20 (because{ (80/100) / (20/100)} =80/20

Odds of not exposed to have disease = odds of non smokers to have heart attack = Chance of non

smokers to have heart attack / chance of non smokers to have no heart attack = { (10/100) / (90/100)} =

10/90


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Therefore, Odds ratio for exposure =

Chance of smokers to have heart attack / chance of smokers to have no heart attack

Chance of non smokers to heart attack / chance of non smokers to not have heart attack

(80/20)

(10/90)

Therefore the odds ratio for smokers to have heart attack as compared to non smokers is

OR of exposed people to have disease = OR for smokers to have heart attack = ( 80x 90) / (20 x 10) =

36. This means that odds for smokers to have heart attack is 36 times more as compared to non

smokers. Also , note that the OR = (axd) / (bxc) this is called as cross product ratio as shown below

Heart attack (Disease) No heart attack (No

disease)

Totals


exposed to smoking)

100

No smoking (Notexposed)

(c) 10 (d) 90 c+d (all persons whoare not smokers)

10+90 = 100


heart attack)

90 (80+10)


heart attack)

20+90 = 110

200


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8. Similarly, we can also calculate the odds ratio of disease

=

Odds of disease to have exposure

Odds of no exposed to have exposure

Therefore,

Chance of diseased to have exposure / chance of diseased to have no exposure

Chance of not diseased to have exposure / chance of not diseased to have no exposure

Therefore, Odds ratio for diseased =

Chance of heart attack to have exposure to smoking / chance of heart attack to have no exposure to

smoking

Chance of no heart attack among smokers / chance of no heart attack among non smokers


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= {(80/10) / (20/90)} = (80/10) x (90/20) = 36

OD of disease = OR of heart attack = This means that people with heart attack are 36 times more

exposed to smoking as compared to people with no heart attack

Therefore, OR of disease = OR of exposure = cross product ratio

9 . Validity

Sensitivity

New Test Old test (gold standard)

New test positive

Gold standard positive) Gold standard negative Totals

(a ) 80 (b) 220 a+b (all persons

positive on new test)

300

New test negative (c) 10 (d) 90 c+d (all persons

negative on new test)

10+90 = 100

Totals a+c (all persons positive

on gold standard test)

90 (80+10)

b+d (all persons

negative on gold

standard)

20+90 = 310

400

Sensitivity of new test = (new test positive) / (all persons positive on gold standard test)

= a/(a+c) = 80/90 = 0.88 = 88 %

This means that the new test has a sensitivity to detect (catch) 88% of the people who are actually

positive

Specificity of the new test = (new test negative) / (all persons negative on gold standard)

= d/(d+b) = 90/310 = 0.29 = 29%

This means that the new test has a specificity to detect ( catch) 29% of the people who are actually

negative.


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In other words,

New Test Old test (gold standard)

New test positive

Gold standard positive) Gold standard negative Totals

(a ) 80

(True positives)

(b) 220

(False positives)

a+b (all persons

positive on new test)

300

All positive on new

test

New test negative (c) 10

(False negatives)

(d) 90

(True negatives)

c+d (all persons

negative on new test)

10+90 = 100

All negative on new

testTotals a+c (all persons positive

on gold standard test)

90 (80+10)

b+d (all persons

negative on gold

standard)

20+90 = 310

400

Therefore,

Sensitivity = True positives / (True positives + False negatives)

Specificity = True negatives / (True negatives + false negatives)

Disease

Screening

Test

Present Absent

Positive

True

positives

Negative

False

positives

False

negatives

True

negatives


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a

dc

b

True Disease Status

+ -

+

-

Sensitivity: The probability of testing

positive if the disease is truly present

Sensitivity = a / (a + c)

Validity of Screening Tests


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a

dc

b

True Disease Status

+ -

+

-

Specificity: The probability of screening

negative if the disease is truly absent

Specificity = d / (b + d)



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Screening Principles

Sensitivity

the ability of a test to correctly identify thosewho have a disease a test with high sensitivity will have few false

negatives

Specificity

the ability of a test to correctly identify those

who do not have the disease a test that has high specificity will have few false

positives


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132

6365045

983

Breast Cancer

+ -

Physical Exam

and Mammo-

graphy

+

-

Sensitivity: a / (a + c)Sensitivity = 132 / (132 + 45) = 74.6%

Specificity: d / (b + d)Specificity = 63650 / (983 + 63650) = 98.5%


Sensitivity and specificity are not able to predict the performance of the screening test in the

population

Thus, the indices of positive and negative predictive value are needed

Predictive Value Positive (PV+): People with positive screening test results will also test positive

on the diagnostic test:

Predictive Value Negative (PV-) :People with negative screening test results are actually free of

disease:


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Performance Yield

a

dc

b

True Disease Status+ -

Results of

Screening

Test

+

-

Predictive value positive (PV+): The probability that a person

actually has the disease given that he or she tests positive.

PV+ = a / (a + b)

Performance Yield

a

dc

b

True Disease Status

+ -

Results of

Screening

Test

+

-

Predictive value negative (PV-): The probability

that a person is truly disease free given that heor she tests negative.

PV- = d / (c + d)


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Performance Yield

400

98905100

995

True Disease Status

+ -

Results of

Screening

Test

+

-

Sensitivity: a / (a + c) = 400 / (400 + 100) = 80%

Specificity: d / (b + d) = 98905 / (995 + 98905) = 99%

PV+: a / (a + b) = 400 / (400 + 995) = 29%

PV-: d / (c + d) = 98905 / (100 + 98905) = 99%


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Performance Yield

400

98905100

995

True Disease Status

+ -

Results of

Screening

Test

+

-

PV+: a / (a + b) = 400 / (400 + 995) = 29%

Among persons who screen positive, 29% are found

to have the disease.


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Performance Yield

400

98905100

995

True Disease Status

+ -

Results of

Screening

Test

+

-

PV-: d / (c + d) = 98905 / (100 + 98905) = 99.9%

Among persons who screen negative, 99.9% are found

to be disease free.

Factors that influence PV+ and PV-

1. The more specific the test, the higher the PV+

2. The higher the prevalence of preclinical disease in the screened population, the higher the PV+

3. The more sensitive the test, the higher the PV-

Therefore, in revision:


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Present Absent

Positive a b

Negative c d

a + b

c + d

a + c b + d

Disease

Screening

Test

N

Sensitivity

Specificity

Quantify a screening tests

accuracy given the known

disease status of subjects

Present Absent

Positive a b

Negative c d

a + b

c + d

a + c b + d

Disease

Screening

Test

N

PPV

NPV

Quantify a screening tests

accuracy given only the

test results of subjects

Concept of Likelihood ratios

Disease + Disease -

Test + a b

Test - c d

Likelihood ratio( +) = LR + = (T+/all D+) / (T+/all D-) = (a/a+c) / (b / b+d) = sensitivity/ 1-specificity

Likelihood ratio( -) = LR- = (T-/ all D+) / ( T-/all D-) = c/(a+c) / (d/b+d) = 1-sensitivity/ specificity

Post odds + = ( LR+ ) * Pre odds

Pre also called as prior odds = Prob of disease/(1-prob of disease)

Prob of disease = {(a+c) / (a+b+c+d)}

Prior odds = {(a+c) / (a+b+c+d)} / 1-{(a+c) / (a+b+c+d)}


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Post odds (+) = (LR+) *prior odds

Post odds (+) = {(a/a+c) / (b / b+d)} * {(a+c) / (a+b+c+d)} / 1-{(a+c) / (a+b+c+d)}

Post odds ( -) = (LR - ) *prior odds

Post odds (-) ={ c/(a+c) / (d/b+d)} * {(a+c) / (a+b+c+d)} / 1-{(a+c) / (a+b+c+d)}

Receiver Operating Curve

ROC is a curve that plots false positive rate on X axis versus True positive rate on Y axis

ROC is a curve that plots (1- specificity) on X axis versus sensitivity on Y axis

Some examples of ROC curves are mentioned below


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ROC analysis provides a useful means to assess the diagnostic accuracy of a test and to compare the

performance of more than one test for the same outcome. However, the usefulness of the test must beconsidered in the light of the clinical circumstances.

Say for example,

In this curve, The ability of two continuous variables to diagnose an outcome can be compared using

ROC curves and their Area under ROC curve (AUROCs).

For example, Fig. 3 (above figure )and Table 6 (mentioned as below) show the ROC curve and AUROC for

urea in addition to those for lactate.


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Reference for figures and table is: Critical Care December 2004 Vol 8 No 6 Bewick et al.

The AUROC for urea is greater than that for lactate (0.730 for urea as compared to 0.64 for lactate),

suggesting that urea may provide a better predictive test for mortality.

Tests for Reliability

Standard approach / test to diagnose depressionclinical exam

Self reporting depression test (

new test)

Depressed Not depressed

Self reported depressed 25 5

Self reported not depressed 10 60

1. Percent Agreement: Divide the number of paired observation in the agreement cells by the total

number of paired observations

Using the data from our example:

(25+60)/100*100%=85%

Advantage

Simple to use

Can be extended to discrete score with more than two levels

Not depressed, mild depression, severe depression

Disadvantage

Values tend to be high whenever the absent-absent cell is high.


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2. Percent Positive Agreement: Divide the number of positive paired observation by the average

number of positives by both ratings.

Using the data from our example:

25/[((25+10) + (25+5))/2]*100%=76.9%

This represents the number of times both ratings provide a positive results out of the

average number of positives by either rating.

3. Kappa Statistic: The fraction of the observed agreement not due to chance in relation to the

maximum non-chance agreement.

K=(P0-Pe)/(1-Pe)

P0=the proportion of observed agreement

Pe=the proportion of agreement expected to occur by chance alone.

From our example

P0=(25+60)/100=.85

Pe

The sum of chance agreement for each cell on the diagonal

The expected for each cell is calculated by the product of the

corresponding marginals divided by the total

(25+5)*(25*10)/100=10.5

(60+5)*(60+10)/100=45.5


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Therefore, Pe=(10.5+45.5)/100=.56

K=(P0-Pe)/(1-Pe) = (.85-.56)/(1-.56)=.66

Range of Kappa: -1 to 1

-1: Complete disagreement

0: Random agreement

1: Complete agreement

Suggestedguidelines

< .4 Poor agreement beyond chance

.4-.75 Fair to good agreement beyond chance

> .75 Excellent agreement beyond chances

Reference : Landis and Koch (1977). The Measurement of observer agreement for categorical data.

Biometrics, 33:159-174.

4. Intraclass Correlation Coefficient: estimates the fraction of the total measurement variability caused

by variation among individuals.

This is an extension of the kappa; Same range of scores (-1 to 1)

ICC=Vb/(Vb+Ve)

Vb=Variance between individuals

Ve=Error variance


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Can calculate ICC from ANOVA Table

More complex approach to estimating the ICC also exist, which take into account

random effect of subjects and raters

5. Coefficient of Variability (CV): the standard deviation expressed as a percentage of the mean value

of two sets of paired observations

For each paired set of observation, calculate the variance

If have an pair of scores of 25 and 35, the mean of the two observation would be 30 and the variance

would be (25-30)2+(35-30)2=50

The CV for the pair would be the standard deviation of the paired observations divided by the mean of

the pair

SQRT(50)/30=.24

This is then repeated for each pair

The overall CV is the average of the pairwise CVs

The lower the CV, the less variation there is between the repeated measurements

If not differences between pairs, the CV would be zero

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For validity,we could use sensitivity, specificity, PV + and PV-

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Notes on RR, Odds, Or, AR, AR% - Epidemiology