Ph.D. course in epidemiology: Fall 2011. Analysis of...

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Ph.D. course in epidemiology: Fall 2011. Analysis of cohort studies. C & H, Ch. 6, 14-15. 29 September 2011 www.biostat.ku.dk/~pka/epiE11 Per Kragh Andersen 1

Transcript of Ph.D. course in epidemiology: Fall 2011. Analysis of...

Page 1: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Ph.D. course in epidemiology: Fall 2011.

Analysis of cohort studies.

C & H, Ch. 6, 14-15.

29 September 2011

www.biostat.ku.dk/~pka/epiE11

Per Kragh Andersen

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Page 2: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Confounding

• Epidemiology relies on observational studies or experiments of

nature

• Often these are poor experiments

— no control for confounding by extraneous influences

• Definition:

A confounder is a variable whose influence we would have

controlled if we had been able to design the natural

experiment.

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Page 3: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Example: confounding by age, Fig. 14.1

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Page 4: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

• Probability of failure for unexposed:

(0.8 × 0.1) + (0.2 × 0.3) = 0.14

• Probability of failure for exposed:

(0.4 × 0.1) + (0.6 × 0.3) = 0.22

• Difference entirely due to difference in age structure.

• When there is a true effect, its magnitude can be distorted by

such influences.

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Page 5: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Confounding when RR = 2

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Page 6: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Results.

• The true relative risk, RRT = 0.2/0.1 = 0.4/0.2 = 2

• Probability of failure for unexposed:

(0.8 × 0.1) + (0.2 × 0.2) = 0.12

• Probability of failure for exposed:

(0.4 × 0.2) + (0.6 × 0.4) = 0.32

• The apparent relative risk:

RRO = 0.32/0.12 = 2.67

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Page 7: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

ConfoundingA confounder is:

• associated with outcome:

e.g., older persons have higher disease probability,

• associated with the exposure:

e.g., older persons are more / less likely to be exposed,

• not a result of exposure, i.e. not an intermediate variable.

Not a statistical property; cannot be seen from tables; common

sense is required!

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Page 8: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Confounding: schematically.A variable C is a potential confounder for the relation:

E → O

if it is

• 1) related to the exposure:

E − C

• 2) an independent risk factor for the outcome:

C → O

• 3) not a consequence of the exposure:

E → C → O

That is:

E − C

ց ւ

O

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Page 9: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Confounding.

The problem is that we do not always get a fair comparison between

exposed and non-exposed.

Young

OldOld

Young

NON-EXPOSEDEXPOSED

A randomly selected exposed person tends to be older than a

randomly chosen non-exposed.

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Page 10: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Controlling confounding, Sect. 14.2

In controlled experiments there are two ways of controlling

confounding:

1. Randomization of subjects to experimental groups so that the

distributions of the confounder are the same.

2. Hold the confounder constant.

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Page 11: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Standardization is a classical statistical technique for controlling

for extraneous variables (in particular: age) in the analysis of an

observational study

1. Direct standardization simulates randomization by equalizing

the distribution of extraneous variables.

2. Indirect standardization simulates the second method: holding

extraneous variables constant.

We first discuss direct standardization and then later turn to the

main ways of “holding the confounder constant”:

• stratified (“Mantel-Haenszel”) analysis

• or (more importantly) regression analysis: logistic, Poisson, Cox.

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Page 12: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Direct standardization, sect. 14.3

1. Estimate age-specific rates (or risks) in each group,

2. Calculate marginal rates (risks) if the age distribution were fixed

to that of some agreed standard population.

A standard population is another term for a common

age-distribution.

3. Direct standardization is good for illustrative purposes as it

provides absolute rates.

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Page 13: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

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Marginal failure probability (with 50-50 age distribution) is

(0.5 × 0.1) + (0.5 × 0.3) = 0.2 for both groups

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Page 14: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

The Diet data

Exposed Unexposed

Current (< 2750 kcal) (≥ 2750 kcal)

age D Y Rate D Y Rate RR

40–49 2 311.9 6.41 4 607.9 6.58 0.97

50–59 12 878.1 13.67 5 1271.1 3.93 3.48

60–69 14 667.5 20.97 8 888.9 9.00 2.33

Total 28 1857.5 15.07 17 2768.9 6.14 2.46

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Page 15: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Direct standardization in the diet data.

We can standardize the age-specific rates to a population with equal

numbers of person–years in each age group.

Exposed:(

1

3× 6.41

)

+

(

1

3× 13.67

)

+

(

1

3× 20.97

)

= 13.67

Unexposed:(

1

3× 6.58

)

+

(

1

3× 3.93

)

+

(

1

3× 9.00

)

= 6.50

Estimate of rate ratio is 13.67/6.50 = 2.10.

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Page 16: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Choice of weights

• Sometimes overall age structure of the whole study is used

• Use of a standard age structure can facilitate comparison with

other work.

• In cancer epidemiology standard populations approximating the

European, US or World population age-distribution are used.

• Equal weights essentially give a comparison between cumulative

rates in the two groups

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Page 17: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Stratified (Mantel-Haenszel) analysis, Ch. 15.

• Aim is to hold age constant.

• Compare exposed and unexposed persons within age strata.

• Compute a combined estimate of effect over all strata.

• This implies a model in which there is no (systematic) variation

of effect over strata.

• If estimates are similar we combine them, by a suitable average.

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Page 18: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

If the effect of exposure is the same in all age-strata, we can

re-parameterize rates as:

Exposed Unexposed

Age Low energy High energy Rate Ratio

40–49 λ01 = θλ0

0 λ00 θ

50–59 λ11 = θλ1

0 λ10 θ

60–69 λ21 = θλ2

0 λ20 θ

This is the proportional hazards model:

For every stratum a: λa1 = θλa

0 .

θ is the effect of exposure “controlled for” age.

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Page 19: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Data

Exposed Unexposed

Age (a) Low energy (1) High energy (0)

40–49 (a = 0) D10, Y10 D00, Y00

50–59 (a = 1) D11, Y11 D01, Y01

60–69 (a = 2) D12, Y12 D02, Y02

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Page 20: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

The Mantel-Haenszel estimate

The MH-estimate for θ is (the weighted average):

θMH =

a

D1aY0a

Y0a+Y1a

a

D0aY1a

Y0a+Y1a

=

aQa

aRa

=Q

R.

This may be calculated by hand.

Note that only θ is estimated, not the λ’s.

Maximum likelihood estimation of all parameters: later.

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Page 21: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

An approximate confidence interval for θ can be obtained using a

standard error for log(θ̂) and then calculate the error factor in the

usual way:

sd(log(θMH)) =

V

QR

where

V =∑

a

Va =∑

a

(D0a + D1a)Y0aY1a

(Y0a + Y1a)2.

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Page 22: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

The Mantel-Haenszel test

The Mantel-Haenszel test for no exposure effect is:

U2/V

where

U =∑

a

Ua

and

Ua = D1a − (D0a + D1a)Y1a

Y0a + Y1a

(NB: calculations by hand). This test may also be based on the

likelihood principle.

When θ = 1, this is approximately χ21−distributed.

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Page 23: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Is it reasonable to assume constant rate ratio?

Estimate θ and compute the expected number of unexposed cases

given the total number of cases and the split of risk time between

exposed and unexposed:

E0a = (D0a + D1a)Y0a

Y0a + θMHY1a

(cases should occur in proportion Y0a : θMHY1a). Then, compute the

“Breslow-Day” test statistic for homogeneity over strata:

A∑

a=1

(D0a − E0a)2

E0a

∼ χ2A−1,

(where A is the number of age strata). If this is sufficiently small,

accept that the rate ratio is constant.

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Page 24: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

The diet data.

• θMH = 2.40,

• 90% c.i. from 1.44 to 4.01,

• MH-test statistic: 8.48 ∼ χ21, P = 0.004,

• Breslow-Day test statistic: 1.65 ∼ χ22, P = 0.44.

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Page 25: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Fixed follow-up time.

If all cohort members are followed for the same time (say, from t0 to

t1) then data from stratum a may be summarized in a (2 × 2)−table:

Group F(ailure) S(urvival) Total

Non-exp. D0a n0a − D0a n0a

Exposed. D1a n1a − D1a n1a

M-H estimate and M-H test for an assumed common risk ratio may

be obtained as for the rates replacing Y0a by n0a and Y1a by n1a.

M-H analysis of OR may also be performed.

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Page 26: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Cohorts where all are exposed: indirect standardization.C & H: Sect. 15.6.

When there is no comparison group we may ask:

Do mortality rates in cohort differ from those of an external

population, for example:

• Occupational cohorts

• Patient cohorts

compared with reference rates obtained from:

• Population statistics (mortality rates)

• Disease registers (hospital discharge registers)

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Page 27: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Accounting for age composition

• Compare rates in a study group with a standard set of

age–specific rates

• Reference rates are normally based on large numbers of cases, so

they can be assumed to be known

• If we use the Mantel-Haenszel estimator when

D0a is large, Y0a is large,D0a

Y0a

= λa

0

then θMH = SMR = D/E

• Calculate “expected” number of cases, E =∑

aλa

0Y1a, if the

standard rates had applied in our study group, and compare this

with the observed number of cases, D =∑

aD1a:

• Similarly, sd(log[SMR]) =√

1/D

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Page 28: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

Example: C & H, p.56.

974 women treated with hormone replacement therapy were followed up.

In this cohort 15 incident cases of breast cancer were observed. The

woman–years of observation and corresponding E & W rates were:

person- E & W rate

Age years per 100 000 py E

40–44 975 113 1.10

45–49 1079 162 1.75

50–54 2161 151 3.26

55–59 2793 183 5.11

60–64 3096 179 5.54P

16.77

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Page 29: Ph.D. course in epidemiology: Fall 2011. Analysis of ...publicifsv.sund.ku.dk/~pka/epiE11/epi-E11-trans4.pdf · Standardization is a classical statistical technique for controlling

• “Expected” cases at ages 40–44:

975 ×113

100 000= 1.10

• Total “expected” cases is E = 16.77

• The SMR is 15/16.77 = 0.89, or 89%.

• Error-factor: exp(1.645 ×p

1/15) = 1.53

• 90% confidence interval is:

0.89 × / ÷ 1.53 = (0.58, 1.36)

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