Case-control studies C&H 16 Bendix Carstensen Steno Diabetes - - PowerPoint PPT Presentation

Case-control studies

C&H 16

Bendix Carstensen

Steno Diabetes Center

& Department of Biostatistics, University of Copenhagen bxc@steno.dk http://BendixCarstensen.com

PhD-course in Epidemiology, Department of Biostatistics, Tuesday 31 January 2017

Relationship between follow–up studies and case–control studies

In a cohort study, the relationship between exposure and disease incidence is investigated by following the entire cohort and measuring the rate

• f occurrence of new cases in the different exposure

groups. The follow–up allows the investigator to register those subjects who develop the disease during the study period and to identify those who remain free

• f the disease.

Case-control studies (C&H 16) 2/ 59

Case-control study

In a case-control study the subjects who develop the disease (the cases) are registered by some other mechanism than follow-up, and a group of healthy subjects (the controls) is used to represent the subjects who do not develop the disease.

Case-control studies (C&H 16) 3/ 59

Rationale behind case-control studies

◮ In a follow-up study, rates among exposed and

non-exposed are estimated by: D1 Y1 D0 Y0

◮ and hence the rate ratio by:

D1 Y1 D0 Y0 = D1 D0 Y1 Y0

Case-control studies (C&H 16) 4/ 59

◮ In a case-control study we use the same cases,

but select controls to represent the distribution

• f risk time between exposed and unexposed:

H1 H0 ≈ Y1 Y0

◮ Therefore the rate ratio is estimated by:

D1 D0 H1 H0

◮ Controls represent risk time, not disease-free

persons.

Case-control studies (C&H 16) 5/ 59

Choice of controls (I)

s

Failures Healthy study period The period over which failures are registered as cases is called the study period. A group of subjects who remain healthy over the study period is chosen to represent the healthy part

• f the source population.

— but this is an oversimplification. . .

Case-control studies (C&H 16) 6/ 59

What about censoring and late entry?

s

Failures Healthy Censored Late entry study period Choosing controls which remains healthy throughout takes no account of censoring or late entry. Instead, choose controls who are in the study and healthy, at the times the cases are registered.

Case-control studies (C&H 16) 7/ 59

Choice of controls (II)

s

Failures Healthy Censored Late entry study period This is called incidence density sampling. Subjects can be chosen as controls more than once, and a subject who is chosen as a control can later become a case. Equivalent to sampling observation time from vertical bands drawn to enclose each case.

Case-control studies (C&H 16) 8/ 59

Case-control probability tree

Exposure

• ❅

❅ ❅ ❅ ❅ p 1 − p

Failure E1 E0

✑✑✑ ◗◗◗ π1 1 − π1 ✑✑✑ ◗◗◗ π0 1 − π0

Selection F S F S

✟✟✟ ❍❍❍ 0.97 0.03 ✟✟✟ ❍❍❍ 0.01 0.99 ✟✟✟ ❍❍❍ 0.97 0.03 ✟✟✟ ❍❍❍ 0.01 0.99

Case (D1) Control (H1) Case (D0) Control (H0) pπ1 × 0.97 p(1 − π1) × 0.01 (1 − p)π0 × 0.97 (1 − p)(1 − π0) × 0.01 Probability

Case-control studies (C&H 16) 9/ 59

Retrospective analysis of case-control studies

Retrospective: Compare the distribution of exposure between cases and controls.

◮ The proportion of cases who smoke compared

to controls

◮ The mean age of cases compared to controls

Looks at the study backwards. Only works properly for binary explanatory variables.

Case-control studies (C&H 16) 10/ 59

The retrospective argument

Selection Failure Exposure Probability

• ❅

❅ ❅ ❅ ❅

Not in study

• ❅

❅ ❅ ❅

F S

✟✟✟✟ ❍❍❍❍ ✟✟✟✟ ❍❍❍❍

E1 (Cases) E0 E1 (Controls) E0 p × π1 × 0.97 (1 − p) × π0 × 0.97 p × (1 − π1) × 0.01 (1 − p) × (1 − π0) × 0.01 Note: Parameters in the previous tree not on these branches.

Case-control studies (C&H 16) 11/ 59

Odds of exposure for cases and controls: Ωcas = p × π1 × 0.97 (1 − p) × π0 × 0.97 = p 1 − p × π1 π0 Ωctr = p × (1 − π1) × 0.01 (1 − p) × (1 − π0) × 0.01 = p 1 − p × 1 − π1 1 − π0 Odds-ratio for exposure between cases and controls: Ωcas Ωctr = π1 π0 1 − π1 1 − π0 = OR(disease)population

Case-control studies (C&H 16) 12/ 59

Prospective analysis of case-control studies

Compare the case/control ratio between exposed and non-exposed subjects — or more general: How does case-control ratio vary with exposure ? The point is that in the study it varies in the same way as in the population.

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The prospective argument

Selection Exposure Failure Probability

• ❅

❅ ❅ ❅ ❅

Not in study

• ❅

❅ ❅ ❅ p 1 − p

E1 E0

✟✟✟✟ ❍❍❍❍ π1 1 − π1 ✟✟✟✟ ❍❍❍❍ π0 1 − π0

F S F S p × π1 × 0.97 p × (1 − π1) × 0.01 (1 − p) × π0 × 0.97 (1 − p) × (1 − π0) × 0.01

Case-control studies (C&H 16) 14/ 59

Odds of disease = P {Case given inclusion} P {Control given inclusion} ω1 = p × π1 × 0.97 p × (1 − π1) × 0.01 = 0.97 0.01 × π1 1 − π1 ω0 = (1 − p) × π0 × 0.97 (1 − p) × (1 − π0) × 0.01 = 0.97 0.01 × π0 1 − π0 OR = ω1 ω0 = π1 1 − π1

• π0

1 − π0 = OR(disease)population

Case-control studies (C&H 16) 15/ 59

What is the case-control ratio?

D1 H1 = 0.97 0.01 × π1 1 − π1 = s1,cas s1,ctr × π1 1 − π1

• D0

H0 = 0.97 0.01 × π0 1 − π0 = s0,cas s0,ctr × π0 1 − π0

• D1/H1

D0/H0 = π1/(1 − π1) π0/(1 − π0) = ORpopulation — but only if the sampling fractions are identical: s1,cas = s0,cas and s1,ctr = s0,ctr.

Case-control studies (C&H 16) 16/ 59

Log-likelihood for case-control studies

Log-Likelihood (conditional on being included) is a binomial likelihood with odds-parameters ω0 and ω1 D0log(ω0)−N0log(1+ω0)+D1log(ω1)−N1log(1+ω1) where N0 = D0 + H0 and N1 = D1 + H1. Exposed: D1 cases, H1 controls Unexposed: D0 cases, H0 controls

Case-control studies (C&H 16) 17/ 59

Odds-ratio (θ) is the ratio of the odds ω1 to ω0, so: log(θ) = log ω1 ω0

• = log(ω1) − log(ω0)

Estimates of log(ω1) and log(ω0) are just the empirical odds: log D1 H1

• and

log D0 H0

• Case-control studies (C&H 16)

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The standard errors of the odds are estimated by:

• 1

D1 + 1 H1 and

• 1

D0 + 1 H0 Exposed and unexposed form two independent bodies of data (they are sampled independently), so the estimate of log(θ) [= log(OR)] is: log D1 H1

• − log

D0 H0

• ,

with s.e.

• log(OR)
• =
• 1

D1 + 1 H1 + 1 D0 + 1 H0

Case-control studies (C&H 16) 19/ 59

Confidence interval for OR

First a confidence interval for log(OR): log(OR) ± 1.96 ×

• 1

D1 + 1 H1 + 1 D0 + 1 H0 Take the exponential: OR

× ÷ exp

• 1.96 ×
• 1

D1 + 1 H1 + 1 D0 + 1 H0

• error factor

Case-control studies (C&H 16) 20/ 59

BCG vaccination and leprosy

Does BCG vaccination in early childhood protect against leprosy? New cases of leprosy were examined for presence or absence of the BCG scar. During the same period, a 100% survey of the population of this area, which included examination for BCG scar, had been carried out. The tabulated data refer only to subjects under 35, because vaccination was not widely available when

• lder persons were children.

Case-control studies (C&H 16) 21/ 59

Exercise I

BCG scar Leprosy cases Population survey Present 101 46 028 Absent 159 34 594 Estimate the odds of BCG vaccination for leprosy cases and for the controls. Estimate the odds ratio and hence the extent of protection against leprosy afforded by vaccination. Give a 95% c.i. for the OR. Use SAS for this: Exercise from the notes.

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Solution to I

OR = D1/H1 D0/H0 = 101/46028 159/34594 = 0.002194 0.004596 = 0.48 s.e.(log[OR]) =

• 1

D1 + 1 H1 + 1 D0 + 1 H0

=

• 1

101 + 1 46028 + 1 159 + 1 34594 = 0.127

The 95% limits for the odds-ratio are: OR

× ÷ exp(1.96×0.127) = 0.48 × ÷ 1.28 = (0.37, 0.61)

Case-control studies (C&H 16) 23/ 59

Exercise II

BCG scar Leprosy cases Population controls Present 101 554 Absent 159 446 The table shows the results of a computer-simulated study which picked 1000 controls at random. What is the odds ratio estimate in this study? Give a 95% c.i. for the OR. Use SAS for this: Exercise from the notes.

Case-control studies (C&H 16) 24/ 59

Solution to II

OR = D1/H1 D0/H0 = 101/554 159/446 = 0.1823 0.3565 = 0.51 s.e.(log[OR]) =

• 1

D1 + 1 H1 + 1 D0 + 1 H0 =

• 1

101 + 1 554 + 1 159 + 1 446 = 0.142 The 95% limits for the odds-ratio are: OR

× ÷ exp(1.96×0.142) = 0.51 × ÷ 1.32 = (0.39, 0.68)

Case-control studies (C&H 16) 25/ 59

More levels of exposure (William Guy)

Physical exertion at work of 1659 outpatients: 341 pulmonary consumption, 1318 other diseases.

Level of Pulmonary Other Case/ OR exertion in consumption diseases control relative

• ccupation

(Cases) (Controls) ratio to (3) Little (0) 125 385 0.325 1.643 Varied (1) 41 136 0.301 1.526 More (2) 142 630 0.225 1.141 Great (3) 33 167 0.198 1.000

The relationship of case-control ratios is what matters.

Case-control studies (C&H 16) 26/ 59

The retro/prospective argument

◮ Retrospective: Four possible outcomes

(little/varied/more/great),

◮ Prospective: Two possible outcomes

(case/control), but a large number of comparisons (between any two exposure levels).

◮ But the probability model is still a binary

model, and the argument for the analysis is still the same as before.

◮ Prospective argument applicable in deriving a

logistic regression model for case-control studies.

Case-control studies (C&H 16) 27/ 59

Odds-ratio and rate ratio

◮ If the disease probability, π, in the study period

is small: π = cumulative risik ≈ cumulative rate = λT

◮ For small π, 1 − π ≈ 1, so:

OR = π1/(1 − π1) π0/(1 − π0) ≈ π1 π0 ≈ λ1 λ0 = RR π small ⇒ OR estimate of RR.

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Important assumption behind rate ratio interpretation

The entire “study base” must have been available throughout:

◮ no censorings. ◮ no delayed entries.

This will clearly not always be the case, but it may be achieved in carefully designed studies.

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Avoiding censoring and delayed entry

◮ Can be achieved simultaneously with small π

by incidence density sampling:

◮ Subdivide calendar time in small time bands. ◮ New case-control study in each time band. ◮ Only one case in each time band. ◮ No delayed entry or censoring.

◮ If the fraction of exposed does not vary much

• ver time, all the small studies can be analysed

together as one.

◮ This is effectively matching on calendar time.

Case-control studies (C&H 16) 30/ 59

The rare disease assumption

Necessary to make the approximation: π1/(1 − π1) π0/(1 − π0) ≈ π1 π0 This is more appropriately termed: “The short study duration assumption” — each of the small studies we imagine as components of the entire study should be sufficiently short in relation to disease occurrence, so that the π (disease probability) if small.

Case-control studies (C&H 16) 31/ 59

Nested case-control studies

◮ Study base = “large” cohort ◮ Expensive to get covariate information for all

persons. (expensive analyses, tracing of histories,. . . )

◮ Covariate information only for cases and time

matched controls:

◮ To each case, choose one or more (usually

≤ 5) controls from the risk set.

Case-control studies (C&H 16) 32/ 59

How many controls per case?

The standard deviation of log(OR): Equal number of cases and controls:

• 1

D1 + 1 H1 + 1 D0 + 1 H0 =

• 1

D1 + 1 D1 + 1 D0 + 1 D0 = 1 D1 + 1 D0

• × (1 + 1)

Case-control studies (C&H 16) 33/ 59

Twice as many controls as cases:

• 1

D1 + 1 H1 + 1 D0 + 1 H0 =

• 1

D1 + 1 2D1 + 1 D0 + 1 2D0 = 1 D1 + 1 D0

• × (1 + 1/2)

m times as many cases as controls:

• 1

D1 + 1 H1 + 1 D0 + 1 H0 = 1 D1 + 1 D0

• × (1 + 1/m)

Case-control studies (C&H 16) 34/ 59

How many controls per case?

◮ The standard deviation of the log[OR] is

• 1 + 1

m times larger in a case-control study, compared to the corresponding cohort-study.

◮ Therefore, 5 controls per case is normally

• sufficient. (Only relevant if controls are

“cheap” compared to cases).

◮ But if cases and controls cost the same — and

are available — the most efficient is to have the same number of cases and controls.

Case-control studies (C&H 16) 35/ 59

SAS-intro

Bendix Carstensen

Steno Diabetes Center

& Department of Biostatistics, University of Copenhagen bxc@steno.dk http://BendixCarstensen.com

PhD-course in Epidemiology, Department of Biostatistics, Tuesday 31 January, 2017

SAS

◮ Display manager (programming):

◮ program, log, output windows ◮ reproducible ◮ easy to document

◮ SAS ANALYST

◮ menu-oriented interface ◮ writes and runs programs for you ◮ no learning by heart, no syntax errors ◮ not every thing is included ◮ it is heavy to use in the long run SAS-intro () 37/ 59

Data set example:

Blood pressure and obesity OBESE: weight/ideal weight BP: systolic blood pressure

OBS SEX OBESE BP 1 male 1.31 130 2 male 1.31 148 3 male 1.19 146 4 male 1.11 122 . . . . . . . . . . . . 101 female 1.64 136 102 female 1.73 208

SAS-intro () 38/ 59

Data

Data are in the text file BP.TXT located at www.biostat.ku.dk/~pka/epidata and contains the following variables:

◮ SEX: Character variable ($) ◮ OBESE: weight/ideal weight ◮ BP: systolic blood pressure

3 variables and 102 observations

SAS-intro () 39/ 59

Printing in SAS

We read the file bp.txt directly from www and skip the first line containing variable names (firstobs=2).

data bp; filename bpfile url ’’http://www.biostat.ku.dk/~pka/epidata/bp.txt’’; infile bpfile firstobs=2; input sex $ obese bp; run; proc print data=bp; var sex obese bp; run;

A temporary data set bp which only exists within the current program. (Permanent data sets may be saved but we will not use this feature in this course.)

SAS-intro () 40/ 59

SAS programming

◮ data-step:

data bp; ( reading ) ; ( data manipulations ) ; run;

◮ proc-step:

proc xx data=bp ; ( procedure statments ) ; run;

◮ NB: No data manipulations after run;

— only if we make a new data-step. — better to revise the first data-step.

SAS-intro () 41/ 59

Example

data bp; filename bpfile url "http://www.biostat.ku.dk/~pka/epidata/bp.txt"; infile bpfile firstobs=2; input sex obese bp; run; data bp; set bp; if bp<125 then highbp=0; if bp>=125 then highbp=1; /* an alternative way of creating the new variable highbp is: highbp = (bp>=125); */ run; proc freq data=bp; tables sex * highbp ; run;

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Example, simplfied

data bp; filename bpfile url ’’http://www.biostat.ku.dk/~pka/epidata/bp.txt’’; infile bpfile firstobs=2; input sex obese bp; if bp < 125 then highbp=0; if bp >= 125 then highbp=1; /* an alternative way of creating the new variable highbp is: highbp = (bp>=125); */ run; proc freq data=bp; tables sex * highbp ; run;

SAS-intro () 43/ 59

Typing of programs is done in the

◮ Program Editor window:

◮ Works like all other text editors: arrow keys,

backspace, delete etc.

◮ When the program is submitted (click on Submit

• r press F3), the results are in the

◮ Log-window:

◮ Here you can see how things went: ◮ how many observations you have, ◮ how many variables you have ◮ if there were any errors ◮ which pages were written by which procedures SAS-intro () 44/ 59

◮ Output-window (perhaps):

◮ In this window you will find the results (if there are

any)

◮ Graph-window (which we won’t use on this

course)

◮ Here plots are stored in order SAS-intro () 45/ 59

Making life simpler

◮ You can move between the windows by clicking

Windows in the command bar, or use that:

◮ F5 is editor window, ◮ F6 is log window, ◮ F7 is output window. SAS-intro () 46/ 59

Modifications in the program

When the program has been executed and you want to make changes:

◮ Go back to the Program-window ◮ The Log- Output- and Graph-windows

cumulate, that is output is stored consecutively

◮ Clear by choosing Clear under Edit (or press

Ctrl-E - for “erase”)

◮ Don’t print! ◮ Remember to save the the program from time

to time before SAS crashes!

SAS-intro () 47/ 59

Simple statistical models

Proportions and rates

Bendix Carstensen

Steno Diabetes Center

& Department of Biostatistics, University of Copenhagen bxc@steno.dk http://BendixCarstensen.com

PhD-course in Epidemiology, Department of Biostatistics, Tuesday 31 January, 2017

A single proportion

The log-likelihood for π, the proportion dead, if we

• bserve 4 deaths out of 10:

ℓ(π) = 4log(π) + 6log(1 − π) The log-likelihood for ω, the odds of dying, if we

• bserve 4 deaths and 6 non-deaths:

ℓ(π) = 4log(ω) − 10log(1 + ω)

Simple statistical models (Proportions and rates) 49/ 59

Programs

General purpose programs for estimating in the binomial and Poisson distribution:

◮ SAS: proc genmod ◮ R: glm ◮ Stata: glm

Here we primarily look at SAS.

Simple statistical models (Proportions and rates) 50/ 59

Estimating odds: genmod

data p ; input x n ; datalines ; 4 10 ; run ; proc genmod data= p ; model x/n = / dist=bin link=logit ; estimate "4 versus 6" intercept 1 / exp ; run ; Standard Wald 95% Confidence Parameter DF Estimate Error Limits Intercept 1

• 0.4055

0.6455

• 1.6706

0.8597 Scale 1.0000 0.0000 1.0000 1.0000 Contrast Estimate Results L’Beta Standard L’Beta Chi- Label Estimate Error Confidence Limits Square 4 versus 6

• 0.4055

0.6455

• 1.6706

0.8597 0.39 Exp(4 versus 6) 0.6667 0.4303 0.1881 2.3624

Simple statistical models (Proportions and rates) 51/ 59

Estimating a proportion: genmod

The only difference from estimation of odds is the link= argument, which is changed to log (instead

• f logit):

proc genmod data= p ; model x/n = / dist=bin link=log ; estimate "4 out of 10" intercept 1 / exp ; run ; Standard Wald 95% Confidence Parameter DF Estimate Error Limits Intercept 1

• 0.9163

0.3873

• 1.6754
• 0.1572

Scale 1.0000 0.0000 1.0000 1.0000 Contrast Estimate Results L’Beta Standard L’Beta Chi- Label Estimate Error Confidence Limits Square 4 out of 10

• 0.9163

0.3873

• 1.6754
• 0.1572

5.60 Exp(4 out of 10) 0.4000 0.1549 0.1872 0.8545

Simple statistical models (Proportions and rates) 52/ 59

A single proportion: individual records

data bissau; filename bisfile url "http://www.biostat.ku.dk/~pka/epidata/bissau.txt"; infile bisfile firstobs=2; input id fuptime dead bcg dtp age agem; run; title "Estimate odds - Bissau" ; proc genmod data=bissau descending ; model dead = / dist=bin link=logit ; estimate "odds of dying" intercept 1 / exp ; run ; Contrast Estimate Results L’Beta Standard L’Beta Label Estimate Error Confidence Limits Square

• dds of dying
• 3.1249

0.0686

• 3.2593
• 2.9905

2076.5 Exp(odds of dying) 0.0439 0.0030 0.0384 0.0503

Simple statistical models (Proportions and rates) 53/ 59

A single proportion: individual records

title "Estimate proportion - Bissau" ; proc genmod data=bissau descending ; model dead = / dist=bin link=log ; estimate "prob of dying" intercept 1 / exp ; run ; Contrast Estimate Results L’Beta Standard L’Beta Label Estimate Error Confidence Limits Square prob of dying

• 3.1679

0.0657

• 3.2966
• 3.0391

2325.8 Exp(prob of dying) 0.0421 0.0028 0.0370 0.0479

Simple statistical models (Proportions and rates) 54/ 59

Likelihood for a single rate

Recall the log-likelihood for a single rate, λ based

• n D events during Y person years:

Dlog(λ) − λY This is also the log-likelihood for a Poisson variate D with mean µ = λY . Therefor we can use a program for the Posson distribution to estimate rates, except we must “remove” the Y from the mean. Poisson distribution usually use the log-mean: log(µ) = log(λ) + log(Y ) log(Y ) extracted via the offset argument.

Simple statistical models (Proportions and rates) 55/ 59

A single rate

data r ; input d y ; ly = log(y) ; my = log(y/1000) ; datalines ; 30 261.9 ; run ; title "Estimate a rate per 1 year" ; proc genmod data= r ; model d = / dist=poisson link=log offset=ly ; estimate "30 during 261.9 - per 1 year" intercept 1 / exp ; run ; Contrast Estimate Results L’Beta Standard L’Beta Label Estimate Error Confidence 30 during 261.9 - per 1 year

• 2.1668

0.1826

• 2.5246

Exp(30 during 261.9 - per 1 year) 0.1145 0.0209 0.0801

Simple statistical models (Proportions and rates) 56/ 59

A single rate: Scaling

Remember the data step statement: my = log(y/1000) ;

title "Estimate a rate per 1000 year" ; proc genmod data= r ; model d = / dist=poisson link=log offset=my ; estimate "30 during 261.9 - per 1000 years" intercept 1 / exp run ; Contrast Estimate Results L’Beta Standard Label Estimate Error Alpha 30 during 261.9 - per 1000 years 4.7410 0.1826 0.05 Exp(30 during 261.9 - per 1000 years) 114.5475 20.9134 0.05

Simple statistical models (Proportions and rates) 57/ 59

A single rate: individual records

data bissau ; set bissau ; ld = log(fuptime) ; ly = log(fuptime/36525) ; run ; title "Estimate a rate per 1 day" ; proc genmod data=bissau ; model dead = / dist=poisson link=log offset=ld ; estimate "mortality rate - per 1 day" intercept 1 / exp ; run ; Contrast Estimate L’Beta Standard Label Estimate Error Alpha Confidence mortality rate - per 1 day

• 8.2852

0.0671 0.05

• 8.4168

Exp(mortality rate - per 1 day) 0.0003 0.0000 0.05

Simple statistical models (Proportions and rates) 58/ 59

Single rate individual records, scaling

Remember the data step statement: ly = log(fuptime/36525) ;

title "Estimate a rate per 1 year" ; proc genmod data=bissau ; model dead = / dist=poisson link=log offset=ly ; estimate "mortality rate - per 100 years" intercept 1 / exp ; run ; Contrast Estimate Results L’Beta Standard Label Estimate Error Alpha mortality rate - per 100 years 2.2205 0.0671 0.05 Exp(mortality rate - per 100 years) 9.2123 0.6183 0.05

Simple statistical models (Proportions and rates) 59/ 59