Measures of Associations and of Causal Effects

EPI 200AOctober 13, 2009

In-Class Lecture OutlineNon-random relationships between exposure and outcome occurrenceAssociation versus causal effect (causation)Overview of measures of relationships (association versus causal effect)Measures of associationMeasures of effectAttributable, etiologic and other fractions

Analytical epidemiology is about making comparisons.

To compare exposed and not exposed, coming as close to the counterfactual ideal as possible

We compare frequencies of diseases or states of health (usually prevalence or incidence) in populations

Most measures of associations are relative or absolute for example relative risks or risk differences and the P-value is not a measure of association.

Comparison

When we compare incidence or prevalence data for specific health outcomes. We want these health outcomes to be the same.

The Kappa Coefficient

Epidemiologists often have to work with imperfect data.We get health data from existing records, health examinations or interviews. All of the data sources are subject to measurement error or other sources of bias.We often need to know something about the magnitude of bias.

The Kappa Coefficient cont.

We can get that information if we can compare our recorded health data with a gold standard.We may compare self-reported data on colic with sound recordings. We may compare self-reported data on ADHD with psychiatric diagnosing.If we have no gold standard we may just compare data from different observers.

Kappa Measures

Kappa measures the proportion of non-random agreement between two measures.

Random Agreement

Two people flipping the same coin 1000 times:

1 2 Head

Tails

500Head 250 250 Tails 250 250 500

500 500 1000

A measure of the proportion of agreement above the agreement expected by chance alone.Pathologists reading cell smears on cervical cancer.

Pathologists reading cell smears on cervical cancer

Pathologist B Pathologist A Positive

Negative

All

37Positive 25 12 Negative 15 230 245 All

40 242 282

Proportion of agreement Examples

Proportion of agreement (25+230)/282=0.904

Expected by chancePAT B 37/282 pos=0.131 x 40=5.25PAT B 245/282 neg=0.869 x 242=210.25

Cont.

All = 215.50Out of 282 =0.7640.904 agreement, but 0.764 may be due to chance

Kappa =0.904-0.764 1-0.764

= 0.58

Pathology Example

PAT B PAT A POS NEG

POS a b

NEG c d

Po proportion of observed agreement =(a+d)/(a+b+c+d)Pe proportion of expected agreement due to change = [(a+b) (a+c)] +[(b+d) (c+d)] / (a+b+c+d)2

Kappa =PO

-PE

1-PE

More levelsObserver 1 (i) Observer 2 (j)

1 2 K

1 M11 M12 M1K M1. 2 M21 K MK1 MK.

M.1 M.K

M

Po=

Pe=

K=

K∑i=j=1

Mij

/M.

K∑i=j=1

Mi.

M.j

/M2

Po-Pe1-Pe

Population Prevalence

Kappa depends not only on the level of agreement, but also on the frequency of the variable to be studiedPop A true prevalence 0.50, obs 1 always correctPop B true prevalence 0.01, obs 2 misclassifies 10% of cases and 10% non cases

Population A Example

Obs 1 Obs 2 Positive +

Negative -

500Obs + 450 50 Obs - 50 450 500 500 500 1000

PO

=0.90

PE

= 0.50K=0.80

Population B ExampleObs 1 Obs 2

Positive +

Negative -

10Obs + 9 1 Obs - 99 891 990 108 892 1000

PO

=0.90

PE

= 0.884

K=0.14

Interpreting Kappa – only guidelines0.81-0.99 almost perfect0.61-0.80 substantial agreement0.41-0.60 moderate agreement0.21-0.40 fair agreement0.01-0.20 slight agreement

Correlations (associations) can be generated by the design, the conduct of the study, the data you use, other correlated factors (confounders), chance or by causation.

We try to eliminate non-causal factors; what remains may be causes but we will never know (mostly untestable assumptions without experimentation)

Closed population, no loss to follow-up

Exposure N D Obs

time/person yearsNoYes

90,00010,000

900500

89,5509,750

RR = Rexp

= 500/10,000 = 5.0Rnot

exp

900/90,000

IRR =

IRexp

= 500/9750years

= 5.1 IRnot

exp

900/89,550years

When would the two measures be almost the same?

Closed population, no loss to follow-up, cont.

RD = Rexp

– Rnot exp

= 500/10,000 –

900/90,000 = 0.040

IRD = IRexp

– IRnot exp

= 500/9750 years –

900/89,550years

= 0.041 years-1

For statistical reasons associations are often measured as odds ratios (OR).

The odds ratio is a ratio between 2 odds

Odds ratios are frequently used because of good statistical properties

Odds =

If R = 0.01; odds = 0.01/0.99 = 0.0101

If R = 0.10; odds = 0.10/0.90 = 0.1111

If R = 0.40; odds = 0.40/0.60 = 0.6666

If R = 0.50; odds = 0.50/0.50 = 1

If R = 0.60; odds = 0.60/0.40 = 1.5000

R1-R

OR = (Rexp/1-Rexp) / (Rnot

exp/1-Rnot exp)

=

=

=

OR = 5.21

OR ≈

RR if R is small since then 1-R ≈

1

(500/10000) / (1-500/10000)

(900/90000) / (1-900/90000)

(500/10000) / (9500/10000)

(900/90000) / (89100/90000)

500 / 9500900 / 89100

A RR of 5 is equivalent to a 400% increase in risk

RR = = 5

x 100 = 100

= 400%Rexp

– Rnot exp

Rnot exp

500/10000 -

900/90000900/90000

500/10000900/90000

(RR-1) x 100 equals increase in percent and a RR of 1 equals no association (RD=0)

Measure Range DimensionalityRRRDIRRIRD

0, ∞-1, +10, ∞-∞, +∞

NoneNoneNoneTime -1

If the disease is rare (Δt is short)

Rexp

IRexp

x Δt IRexp

Rnot exp

IRnot exp

x Δt IRnot exp

When Δt goes towards 0; RR goes towards IRR

When Δt goes towards ∞; RR goes towards 1

Since the upper limit for R is 1 the R for not exposed sets the upper limit for RR

If the Rnot exp for an abortion is 0.30, RR cannot be more than 1/0.30 = 3.33

~ =

Measures of effect

Humes two definitions of causality

1.

A cause is an object followed by another E D where all objects similar to the first one are followed by objects similar to the second

2.

Or in other words, where the first object had not been the second would never exist

The second definition inspired counterfactual thinking as the ideal for causal inference, for being able to talk about causal effect.

Causal rate difference = A+/T+ - Ao/To

Causal risk difference = A+/N – Ao/NCausal difference in average disease free time = T+/N – To/NCausal rate ratio = (A+/T+) / (Ao/To)Causal risk ratio = (A+/N) / (Ao/N) = R+/Ro

Causal odds ration = (A+/(N-A+)) / (Ao/(N-Ao)) = (R+/S+) / (Ro/So)

To define an effect of an exposure is to define it in relation to noexposure under counterfactual conditions.

Cohort D N TWomen exposed to HRTCounterfactual cohort

A+

Ao

NN

T+

To

We are interested in effects but we measure associations.

OK to use the effect word when you describe aims/hypotheses, but not when you describe actual findings

Attributable fractions

To describe the potential importance of an exposure we sometimes use estimates like attributable fractions. Since this term has causal implications you should use it with care

Attributable fractions

If we have estimated an effect of the exposure, what is then the fraction of the disease among exposed related to the exposure?

Let the number of exposed be N+

N+ (R+

- Ro) R+

- RoN+ R+

R+=

Attributable fractions for exposed

R+

- Ro

RD

1

RR –

1R+

R+

RR

RR

= Attributable fractions for exposed

=

= 0.80 or 80%

= = =1-

Exp N D R RR+O

10,00090,000

500900

0.050.01 5.0

RR –

1RR

5 –

15

The attributable fraction for the population

AF =

AF =

= 400/1400 = 0.286

Closed population

No competing risk

10,000 x 0.05 x90,000 x 0.01 + 10,000 x 0.05

N+

x R+

xN+

x R+

+ No

x Ro

RR -1RR

5 -15

ExposureIR per 100,000 years

Lung cancer CHPSmokersNon-smokers

14010

669413

IRR =

= 14

IRR = = 1.62 LC

CHD

IRD = 140-10 = 130 years-1

IRD = 669-413 = 256 years-1

LC

CHD

14010

669413

Attributable fraction among exposed

AF =

= 0.929

AF = = 0.383 LC

CHD

Attributable fraction in the population will depend upon the proportion of smokers in the population

14-114

1.62-11.62

20% smoke

Lung CancerO E

CHDO E

SmokersNon-smokers

200,000800,000

280

2080

801338

8263304

3304All

Pop Attr. Fraction

360

100 4642

4130

= 0.72 = 0.11

360-100360

4642-41304642

60% smoke

Lung CancerO E

CHDO E

SmokersNon-smokers

600,000400,000

840

6040

404014

24781652

1652All

Pop Attr. Fraction

880

100 5666

4130

= 0.89 = 0.27

880-100880

5666-41305666

14-1140.6 x 140 x

0.6 x 140 + 0.4 x 10Check: = 78

88 = 0.89

Preventable fraction

Attributable fraction R1

– R0

-

let exposure be no exposureR1

then

= 1 –

RR preventable fractionR0

– R1

R0

BMJ Oct. 20, 2006

Around 40% of the fall in number of deaths from cancer among US men from 1991 to 2003 can be attributed to the decline in smoking, say researchers from the American Cancer Society.

Etiologic fraction

Attributable fractions have also been called etiologic fractions (EF), but if we define the etiologic fraction to represent those who had a particular exposure to cause their disease (the disease would not have occurred had they not been exposed) attributable fractions (AFs) need not be similar to EFs.

If the exposure is a cause for some but a preventive factor for others, the attributable fraction may be 0 and the etiologic fraction > 0.

Etiologic fractionAssume genetically susceptible will always get lung cancer if exposed to either cigarette smoke or asbestos. Assume the time sequence for a smoker and the counter factual non-smoker would always be

smoke

I+

non-smoking

I0

If the cohort is unexposed, all cases are attributable to C1

If the cohort is exposed and G is always present all cases will be attributable to E (smoking) or E and G. The etiologic fraction is 100% for the exposed, not

G C1

C1

I+

- I-I+

Deterministic model

4 causal types

1.

No effect

doomed

case

case2.

Causative

case

non case3.

Preventative

non case case

4.

No effect (immune)

non case non case

If all type 2 RR =

1/0 = ∞

RD = 1 –

0 = 1

If 10% are type 1, 10% type 2 and 80% type 4

C1If exposed

if not exposed

Deterministic model

X

X

If exposed

2 D

If not exposed 1 D

RR = 2/1 = 2

RD = 0.20 –

0.10 = 0.10

C

If exposed 5 would get the disease

5 would not get the disease

If not exposed 5 would not get the

disease5

would get the disease

C

C

C

C

C C

C

RR = 5/5 = 1

RD = 0.5 –

0.5

= 0

If 50% are type 2 and 50% type 3

Even in situations (real life) where the exposure only causes the disease (no diseases are prevented by the exposure) there will be a competition between which causal field onsets the disease.

E E

C1