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Case Control Study : Analysis. Odds and Probability. 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. - PowerPoint PPT Presentation
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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