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

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<p2 means theta<1 p1=p2 theta=1 p1>p2 theta>1 11/29/2018 SA3202, Lecture 7

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 .020 .001 Then the odds ratio = (.020/(1-.020))/(.001/(1-.001))=20.38 the proportion difference=.020-.001=.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

Estimation 11/29/2018 SA3202, Lecture 7

Asymptotic Distribution 11/29/2018 SA3202, Lecture 7

A Simple Formula Using the usual notation for two way tables Response Group 1 Group 2 Positive X11 X12 Negative X21 X22 Then The odds ratio = The s.e. Of the odds ratio = 11/29/2018 SA3202, Lecture 7

Example For the Vitamin C Data Placebo Vitamin C Cold 31 17 Not Cold 109 122 Total 140 139 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

Example For the Vitamin C Data Placebo Vitamin C Cold 31 17 Not Cold 109 122 Total 140 139 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+/- 1.96*.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

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=( -1)/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

Example For the Vitamin C Data Placebo Vitamin C Cold 31 17 Not Cold 109 122 Total 140 139 Under H0, the pooled sample proportion and the s.e. ( |H0) are (31+17)/(140+139)=.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 < 1.645, Do not reject H0: =1 against H1: >1 The 95% CI under H0 for is 2.04+/- 1.96*.647=[.77, 3.31] (compare with [.72, 3. 36] not under H0) Therefore, H0 is NOT rejected. 11/29/2018 SA3202, Lecture 7

The Log Odds Ratio Log Odds Ratio: It is sometimes more convenient to work with the logarithm of the odds ratio Main Features: 1. 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<p2 means the log odds ratio <0 p1=p2 =0 p1>p2 >0 5. It has a simpler expression of the asymptotic variance 11/29/2018 SA3202, Lecture 7

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

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=( -1)/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

Example For the Vitamin C Data Placebo Vitamin C Cold 31 17 Not Cold 109 122 Total 140 139 The estimated log odds ratio is log(2.04)=.7129 Under H0, the pooled sample proportion and the s.e. ( |H0) are (31+17)/(140+139)=.172 s.e. ( |H0 )= ((1/140+1/139)/(.172*(1-.172)))^(1/2)=.3173 Thus Z=(.7129-0)/.3173=2.247 >> 1.645, Reject H0: =0 in favor H1: >0 . The 95% CI under H0 for is .7129+/- 1.96*.3173=[.091, 1.335] Therefore, H0 is rejected. That is, the effect of Vitamin C is significant. 11/29/2018 SA3202, Lecture 7

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