Lecture 25. Ingo Ruczinski. November 24, Department of Biostatistics Johns Hopkins Bloomberg School of Public Health Johns Hopkins University

Transcription

1 Lecture 25 Department of Biostatistics Johns Hopkins Bloomberg School of Public Health Johns Hopkins University November 24, 2015

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3 1 Hypothesis s of homgeneity 2 Estimating risk differences 3 Estimating odds 4 A brief note on the distinction between conditional and odds

4 1 binary First Second Survey survey Approve Disapprove Total Approve Disapprove Total Cases Controls Exposed Unexposed Total Exposed Unexposed Total Both sets from Agresti, Categorical Data Analysis, second edition

5 Matched binary can arise from Measuring a response at two occasions Matching on case status in a retrospective study Matching on exposure status in a prospective or cross-sectional study The pairs on binary observations are dependent, so our existing methods do not apply We will discuss the process of making conclusions about the probabilities and odds

6 time 2 time 1 Yes No Total Yes n 11 n 12 n 1+ no n 21 n 22 n 2+ Total n +1 n +2 n time 2 time 1 Yes No Total Yes π 11 π 12 π 1+ no π 21 π 22 π 2+ Total π +1 π +2 1 Notation We assume that the (n 11, n 12, n 21, n 22 ) are multinomial with n trials and probabilities (π 11, π 12, π 21, π 22 ) π 1+ and π +1 are the probabilities of a yes response at the two occasions π 1+ = P(Yes Time 1) π +1 = P(Yes Time 2)

7 is the hypothesis H 0 : π 1+ = π +1 is equivalent to symmetry H 0 : π 12 = π 21 The obvious estimate of π 12 π 21 is n 12 /n n 21 /n Under H 0 a consistent estimate of the variance is (n 12 + n 21 )/n 2 Therefore (n 12 n 21 ) 2 n 12 + n 21 follows an asymptotic χ 2 distribution with 1 degree of freedom

8 The from the previous page is called Notice that only the discordant cells enter into the n 12 and n 21 carry the relevant information about whether or not π 1+ and π +1 differ n 11 and n 22 contribute information to estimating the magnitude of this difference

9 Example Test statistic (80 150) = P-value = Hence we reject the null hypothesis and conclude that there is evidence to suggest a change in opinion between the two polls In R mcnemar.(matrix(c(794, 86, 150, 570), 2), correct = FALSE) The correct option applies a continuity correction

10 Let ˆπ ij = n ij /n be the sample proportions d = ˆπ 1+ ˆπ +1 = (n 12 n 21 )/n estimates the difference in the proportions The variance of d is σ 2 d = {π 1+(1 π 1+ )+π +1 (1 π +1 ) 2(π 11 π 22 π 12 π 21 )}/n d (π 1+ π +1 ) ˆσ d follows an asymptotic normal distribution Compare σd 2 with what we would use if the proportions were independent

11 Example d = 944/ /1600 = =.04 ˆπ 11 =.50, ˆπ 12 =.09, ˆπ 21 =.05, ˆπ 22 =.36 ˆσ 2 d = {.59(1.59)+.55(1.55) 2( )}/1600 ˆσ d = % CI -.04 ± = [.06,.02] Note ignoring the dependence yields ˆσ d =.0175

12 Each subject s (or matched pair s) responses can be represented as one of four tables. Response Time Yes No First 1 0 Second 1 0 Response Time Yes No First 0 1 Second 1 0 Response Time Yes No First 1 0 Second 0 1 Response Time Yes No First 0 1 Second 0 1

13 Result is equivalent to the CMH where subject is the stratifying variable and each 2 2 table is the observed zero-one table for that subject This representation is only useful for conceptual purposes

14 Exact version Consider the cells n 12 and n 21 Under H 0, π 12 /(π 12 + π 21 ) =.5 Therefore, under H 0, n 21 n 21 + n 12 is binomial with success probability.5 and n 21 + n 12 trials We can use this result to come up with an exact P-value for matched pairs

15 Consider the approval rating H 0 : π 21 = π 12 H a : π 21 < π 12 (π +1 < π 1+ ) P(X ) =.000 where X is binomial with 236 trials and success probability p =.5 For two sided s, double the smaller of the two one-sided s

16 Estimating the odds ratio The odds ratio is π 1+ /π 2+ π +1 /π +2 = π 1+π +2 π +1 π 2+ The maximum likelihood estimate of the margina log odds ratio is ˆθ = log{ˆπ 1+ˆπ +2 /ˆπ +1ˆπ 2+ } The asymptotic variance of this estimator is {(π 1+ π 2+ ) 1 + (π +1 π +2 ) 1 2(π 11 π 22 π 12 π 21 )/(π 1+ π 2+ π +1 π +2 )}/n

17 Example In the approval rating example the OR compares the odds of approval at time 1 to that at time 2 ˆθ = log( / ) =.16 Estimated standard error =.039 CI for the log odds ratio =.16 ± = [.084,.236]

18 odds First Second Survey survey Approve Disapprove Total Approve Disapprove Total

19 odds n ij cell counts n total sample size π ij the multinomial probabilities The estimate of the log odds ratio is ˆθ = log{ˆπ 1+ˆπ +2 /ˆπ +1ˆπ 2+ } The asymptotic variance of this estimator is {(π 1+ π 2+ ) 1 + (π +1 π +2 ) 1 2(π 11 π 22 π 12 π 21 )/(π 1+ π 2+ π +1 π +2 )}/n

20 Consider the following model logit{p(person i says Yes at Time 1)} = α + U i logit{p(person i says Yes at Time 2)} = α + γ + U i Each U i contains person-specific effects. A person with a large U i is likely to answer Yes at both occasions. γ is the log odds ratio comparing a response of Yes at Time 1 to a response of Yes at Time 2. γ is subject specific effect. If you subtract the log odds of a yes response for two different people, the U i terms would not cancel

21 cont d One way to eliminate the U i and get a good estimate of γ is to condition on the total number of Yes responses for each person If they answered Yes or No on both occasions then you know both responses Therefore, only discordant pairs have any relevant information after conditioning The conditional estimate for γ and its SE turn out to be log{n 21 /n 12 } 1/n21 + 1/n 12

22 Distinctions in interpretations The has a interpretation. The effect is averaged over all of the values of U i. The conditional estimate has a subject specific interpretation. interpretations are more useful for policy type statements. Policy makers tend to be interested in how factors influence populations. Subject specific interpretations are more useful in clinical applications. Physicians are interested in how factors influence individuals.