1. Lecture 7 The Odds/ Log Odds Ratios
Outline of Today:
The Odds Ratio
Estimation
Asymptotic Distribution
Example
The Log Odds Ratio
11/29/2018
SA3202, Lecture 7

2. Definition
Problem of Interest It is usually useful to summarize the difference between the two groups in a single measure. Two such measurements are
Measure 1: the proportion difference p1-p2
Measure 2: the proportion ratio p1/p2
Odds Ratio: The more satisfactory measure is the “odds ratio”:
theta= Odds Ratio=(Odds of Positive Response in Group1)/ (Odds of Positive Response in Group1)
=[p1/(1-p1)]/[p2/(1-p2)]
Relationship :
p1<p means theta<1
p1=p theta=1
p1>p theta>1
11/29/2018
SA3202, Lecture 7

3. Example
The odds ratio may be regarded as a measure of degree of association (relationship, dependence) between the response variable R and the explanatory (classification) variable C. It is a more satisfactory measure than the proportion difference p1-p2.
Example Suppose
Smoker Non-smoker
Probability for Lung Cancer
Then
the odds ratio = (.020/(1-.020))/(.001/(1-.001))=20.38
the proportion difference= =.019
Thus the odds ratio provides a more satisfactory measure of the effect of smoking on lung cancer. Roughly speaking, smokers are 20 times more at risk than nonsmokers.
11/29/2018
SA3202, Lecture 7

4. Estimation
11/29/2018
SA3202, Lecture 7

5. Asymptotic Distribution
11/29/2018
SA3202, Lecture 7

6. A Simple Formula Using the usual notation for two way tables
Response Group Group 2
Positive X X12
Negative X X22
Then
The odds ratio =
The s.e. Of the odds ratio =
11/29/2018
SA3202, Lecture 7

7. Example For the Vitamin C Data Placebo Vitamin C Cold 31 17
Not Cold
Total
The odds of catching cold for the Placebo group are 31 to 109, whereas for the Vitamin C group the odds are 17 to 122. The odds ratio is
(31/109)/(17/122)=2.04
i.e., “The risk of catching cold for not taking Vitamin C, is twice the risk for taking Vitamin C”
Alternatively, the odds ratio for “not catching cold” is of course
(109/31)/(122/17)=1/2.04=.49
i.e. “the risk of catching cold is halved by taking Vitamin C”
11/29/2018
SA3202, Lecture 7

8. Example For the Vitamin C Data Placebo Vitamin C Cold 31 17
Not Cold
Total
The s.e. of the estimated odds ratio is
s.e. ( )=2.04 (1/31+1/109+1/17+1/122)^(1/2)=.672
The 95% CI for is
2.04+/ *.672=[.72, 3.36]
Therefore, H0 is not rejected. That is, the effect of Vitamin C may not be all that significant. This is inconsistent with the previous result based on the proportion difference test.
11/29/2018
SA3202, Lecture 7

9. Testing Equality of the Proportions H0: p1=p2 is equivalent to H0: =1
The latter hypothesis may be tested by a Z-test based on Under H0,
Var( |H0)=theta^2 (1/n1+1/n2) /(p(1-p))
Estimating p by the pooled estimator of p:
We get
s.e. ( |H0)=
Therefore, asymptotically,
Z=( )/s.e.( |H0)~ AN(0,1)
Thus, the Z-test is conducted via comparing the Z-value with the table value of N(0,1).
11/29/2018
SA3202, Lecture 7

10. Example For the Vitamin C Data Placebo Vitamin C Cold 31 17
Not Cold
Total
Under H0, the pooled sample proportion and the s.e. ( |H0) are
(31+17)/( )=.172
s.e. ( |H0 )=2.04 ((1/140+1/139)/(.172*(1-.172)))^(1/2)=.647
(compare with .672 not under H0)
Thus Z=(2.04-1)/.647=1.61 < , Do not reject H0: =1 against H1: >1
The 95% CI under H0 for is
2.04+/ *.647=[.77, 3.31] (compare with [.72, 3. 36] not under H0)
Therefore, H0 is NOT rejected.
11/29/2018
SA3202, Lecture 7

11. The Log Odds Ratio
Log Odds Ratio: It is sometimes more convenient to work with the logarithm of the odds ratio
Main Features: Take any value in the real line
2. The zero value corresponds to “independence” (p1=p2)
3. The magnitude reflects the degree of the association
4. The sign reflects the direction
p1<p means the log odds ratio <0
p1=p =0
p1>p >0
5. It has a simpler expression of the asymptotic variance
11/29/2018
SA3202, Lecture 7

12. Statistical Inferences
Estimation The natural estimator of the log odds ratio is the sample log odds ratio
Asymptotic Normality For large sample sizes, the estimator is asymptotically normal with mean
And variance
Var( )=(1/(n1p1(1-p1))+1/(n2p2(1-p2)))
The s.e. Of is then
s.e. ( )=
which can be simplified as
11/29/2018
SA3202, Lecture 7

13. Testing Equality of the Proportions H0: p1=p2 is equivalent to H0: =0
The latter hypothesis may be tested by a Z-test based on Under H0,
Var( |H0)=(1/n1+1/n2) /(p(1-p))
Estimating p by the pooled estimator of p:
We get
s.e. ( |H0)=
Therefore, asymptotically,
Z=( )/s.e.( |H0)~ AN(0,1)
Thus, the Z-test is conducted via comparing the Z-value with the table value of N(0,1).
11/29/2018
SA3202, Lecture 7

14. Example For the Vitamin C Data Placebo Vitamin C Cold 31 17
Not Cold
Total
The estimated log odds ratio is
log(2.04)=.7129
Under H0, the pooled sample proportion and the s.e. ( |H0) are
(31+17)/( )=.172
s.e. ( |H0 )= ((1/140+1/139)/(.172*(1-.172)))^(1/2)=.3173
Thus Z=( )/.3173= >> , Reject H0: =0 in favor H1: >0 .
The 95% CI under H0 for is
.7129+/ *.3173=[.091, 1.335]
Therefore, H0 is rejected. That is, the effect of Vitamin C is significant.
11/29/2018
SA3202, Lecture 7

15. Summary For the test of equality of the proportions
“ H0:p1=p2 “ means “H0: Independence between two groups”
we can based on the following tests:
1. Use the sample proportion difference to test H0: p1-p2=0
2. Use the sample odds ratio to test H0: =1
3. Use the sample log odds-ratio to test H0: =0
However, it can be showed that Test 2 is less powerful in the sense that it can not detect the difference between p1 and p2 while the other two can.
11/29/2018
SA3202, Lecture 7