Case Control Study : Analysis

Odds and Probability

#with DiseaseProbability of Disease =

# with and without Disease

# with DiseaseOdds of disease =

# without Disease

Odds and Probability

what is odds?

Let’s say that probability of success is 0 .8, thus P = .8

Then the probability of failure is = 1-p = q = .2

The odds of success are

The odds of failure would be q 0.2= = = 0.25

p 0.8

p 0.8= = = 4

q 0.2

• Odds of <1 mean the outcome occurs less than half the time• Odds of 1 mean the outcome occurs half the time• Odds of >1 mean the outcome occurs more than half the time

odds ratio=

The odds of success are 16 times greater for failure.

odds (success) 4= = 16

odds (failure) 0.25

Odds of disease among exposed = a/ b Odds of disease among unexposed = c/d

a b adOdds ratio = =

c d bc

95% CI = ln OR ± 1.96(SE (ln OR))

SE (ln OR) = 1/a + 1/b + 1/c + 1/d

Exposure

Case Control

+

-

a+c b+d

dc

ba

ODDS RATIO:

a+b

c+d

Example :

The following data refer to a study that investigated the relationship between MI and smoking.The first column refers to 262 young and middle - aged women admitted to 30 coronary care units with acute MI.

Cross classification of smoking status and MI MI Controls

Ever smokers

yes

no

262 519

34690

173172

345

436

The odds of having MI is 3.82 times more for smokers as compared nonsmokers.

95% CI in lnOR 1.02

172 * 346OR = = 3.82

173 * 90 ln OR = In (3.82) = 1.343

SE(lnOR) = 1/172 + 1/346 + 1/173 +1/90

= 0.16

lnOR ± 1.96 * SE(lnOR) = 1.343 ± 0.3136

1.0294 1.6566

94 and 1.6566

95% CI on OR e and e

2.79 and 5.24

1.83 10845 * 104

10933 * 189 OR

1.820.0094

0.0171

11037104

11034189RR

MI

Yes No

Group placebo

aspirin

10845189

104 10933

11034

11037

293 21778

Rare disease example:

Example of OR with common outcome

Stroke unit

Yes No

702 706

332403

374299 673

735Control

299* 332OR = = 0.66

403* 374299/673 0.4442

RR = = = 0.81403/735 0.5483

The frequency of poor outcome (i.e., mortality in the control group) was very high (55%)

The OR underestimates the RR by 19%

Mortality

TESTING HYPOTHESIS USING CI

0 1

1.58 3.78

.5 2.0

.8.3

ISSUES IN APPROXIMATING ‘OR’ INTO ‘RR’

• When the risks (or odds) in the two groups being compared are both small (say less than 20%) then the OR will approximate to the RR

• The discrepancy between odds ratio and relative risk depends on the risk (odds) in both groups (i.e., the discrepancy depends on the initial risk and the OR itself )

• OR may be non-intuitive in interpretation, but in almost all realistic cases interpreting them as though they were RRs is unlikely to change any qualitative assessment of the study findings

Since the odds ratio is difficult to interpret, why is it so widely used?

• Odds ratio is valuable in case control studies where events are usually rare

• The odds ratio remains especially useful when researchers need to adjust for other variables, for which LR is usual approach

• Odds ratios are a common way of presenting the results of a Meta - Analysis - a statistical analysis for combining the results of several studies, used within systematic reviews

Examples of Confounding and Effect Modification

____________________________________________________

RR (exposed vs non exposed) control var is

------------------------------------------------------------

str1 str2 str3 Total(crude) Conf EM

------------------------------------------------------------------------------

Example A 1.4 1.8 1.3 2.5 Y N

Example B 1.4 1.8 1.3 1.5 N N

Example C 1.4 1.8 0.2 1.5 N Y

Example D 1.4 1.8 0.2 2.5 Y Y

___________________________________________________

Level ofControlVariable:

Twenty-Four-Hour Mortality After Coronary Artery Bypass Surgery, by Sex

At 24 hoursSex Total Dead Dead % OR RR

(95% CI) (95% CI)

Female 544 15 2.8 1.96 1.93Male 3366 48 1.4 (1.1-3.5) (1.09-3.39)

Total 3910 63 1.6

Twenty-Four-Hour Mortality After Coronary Artery Bypass Surgery, by Body Surface Area (BSA)

At 24 hoursTotal Dead Dead % OR RR

(95% CI) (95% CI)

Low BSA 1250 35 2.8 2.71 2.66High BSA 2660 28 1.1 (1.68-4.37) (1.66-4.25)

Total 3910 63 1.6

Twenty-Four-Hour Mortality After Coronary Artery Bypass Surgery, by Sex within Each BSA Category

At 24 hoursTotal Dead Dead % OR RR

(95% CI) (95% CI)

Low BSA Female 483 14 2.9 1.06 1.06 Male 767 21 2.7 (0.54-2.10) (0.54-2.10

High BSA Female 61 1 1.6 1.59 1.58 Male 2599 27 1.0 (0.21-11.79) (0.22-11.40

Calculation of Mantel-Haenzel Chi-Square

DeadSex Yes No Total ai E(ai) V(ai)

Low BSA Female 14 469 483 Male 21 746 767

Total 35 1215 1250 14 13.524 8.027

High BSA Female 1 60 61 Male 27 2572 2599

Total 28 2632 2660 1 0.642 0.621

2

2MH

15 -14.166χ = = 0.0800

8.693

Calculation of Mantel-Haenzel Adjusted Measures

DeadExposure Yes No Total OR RR

Low BSA Female 14 469 483 8.36/7.88=1.06 8.59/8.11=1.06 Male 21 746 767

Total 35 1215 1250

High BSA Female 1 60 61 0.97/0.61=1.59 0.98/0.62=1.58Male 27 2572 2599

Total 28 2632 2660

Adjusted Measure 9.32/4.89=1.1 9.57/8.73=1.1

Study design procedure:

• select referent group• comparable to index group on one or more matching factors

Basics for matching

Matching factor = age referent group constraint to have same age structure

Case-control study: referent = controls

index = cases

Follow-up study:referent = unexposed

index = exposed

Category Matching

Factor A: A1, A2, A3Factor B: B1, B2, B3Factor C: C1, C2, C3

Example:

AGE: 20-39, 30-39, 40-49, 50-59, 60-69GENDER: Male, Female STAGE: I, II, III

Category Matching

Factor A: A1, A2, A3Factor B: B1, B2, B3Factor C: C1, C2, C3

Example:

AGE: 20-39, 30-39, 40-49, 50-59, 60-69GENDER: Male, Female STAGE: I, II, III

Control has same age-gender-stage combination as Case

Category Matching

Factor A: A1, A2, A3Factor B: B1, B2, B3Factor C: C1, C2, C3

Example:

AGE: 20-39, 30-39, 40-49, 50-59, 60-69GENDER: Male, Female STAGE: I, II, III

Control has same age-gender-stage combination as Case

Category Matching

Factor A: A1, A2, A3Factor B: B1, B2, B3Factor C: C1, C2, C3

Example:

AGE: 20-39, 30-39, 40-49, 50-59, 60-69GENDER: Male, Female STAGE: I, II, III

Case Number of TypeControls

• R R- to 1• 1 1-1 or pair matching

R may vary from case to case

e.g R=3 for some casesR=2 for other casesR=1 for other cases

Not always possible to find exactly R controls for each case

.

Usual Display of Matched Case-Control Data with Dichotomous Exposure

Controls

Exposure

Cases

Exposure Present Absent Total

Present f g f+g

Absent h j h+j

Total f+h g+j n

Analysis of matched data

Control

+ - Total

Case + a b a+b

- c d c+d

Total a+c b+d n

Comparison: proportion of exposed cases vs proportion of exposed controls

a + b a + c b - c - =

n n n

Information regarding differential exposure is given by the discordant pairs

Odds ratio“Inference regarding the difference in proportions in matched pairs is made solely on the basis of discordant pairs”

McNemar (1947)

Estimate of OR conditional on the number of discordant pairs is given as

c

bOR Kraus (1960), Cox (1958)

To match or not to match

Advantages:

•Matching can be statistically efficient

i,.e may gain precision using confidence interval

Disadvantages:

Matching is costly

•To find matches

•Information loss due to discarding controls

Match on strong risk factors expected to be confounders.

Matching No matchingCorrect estimate ?

YES YESAppropriate analysis?

YES YESMatched Standard stratified(Stratified) OR1, OR2, OR3

CombineValidity is not an important reason for matching

Match to gain efficiency or precession

Matched Analysis using Stratification:

Strata = matched sets

Special case: case-control study

100 matched pairs, n=200

100 strata = 100 matched pairs

2 observations per stratum 1st pair 2nd pair 100th pair E E E E E E

D

D

D

D

D

D --------

E E D

D

W + X + Y + Z = total number of pairs

1 0

1 0

D

D

D

D

D

D

1

0

0

0

0

0

0

1

1

1

1

1

1

1

1

1

1

1

1

1

W pairs

Z pairs

Y pairs

X pairs

E E D

D

2. McNemar chi-square test

Ch-square = (X-Y)2/(X-Y)

= (30-10)2/(30+10)

= 10.0 (df=1, p<0.01)

McNemar’s test = MH test for pair matching

MOR = X/Y

= 3

1 0

1 0

D

D

D

D

D

D

1

0

0

0

0

0

0

1

1

1

1

1

1

1

1

1

1

1

1

1

W pairs

(30)

Z pairs

(30)

Y pairs

(10)

X pairs

(30)

W Y

X Z

E E D

E

ED

Example:

W = 30, X=30, Y =10, Z =30

W + X + Y + Z = 100

Analysis : Two equivalent ways

1. Mantel -Haenzel Chi-square test

--------

Stratum 1 Stratum 2 Stratum 100

Compute Mantel-Haenzel chi-square and MOR

Analysis for R-to-1 and mixed matching : Use stratified analysis

Example:

If, R = 4 E E

D

D

0

1

3

1 1

4

5

•Each stratum contains five subjects

•Compute MH chi-square and MOR based on stratified data