Cornell, BTRY 6030

Excerpt: ... Three-way Contingency Tables Lecture 11 ILRST 212 Partial association Example: Victim's race White Black Total Death penalty Defendant's race Yes No Percentage Yes White 53 414 11.3 Black 11 37 22.9 White 0 16 0.0 Black 4 139 2.8 White 53 430 11.0 Black 15 176 7.9 Simpson's Paradox The result that a marginal association can have different direction from the conditional associations is called Simpson's Paradox. Simpson's Paradox Firstly , lets control for death penalty: Table is given below: defendants race Victim race White Black White 467 48 Black 16 143 What is Odds ratio ? Partial association Example: Victim's race White Black Total Death penalty Defendant's race Yes No Percentage Yes White 53 414 11.3 Black 11 37 22.9 White 0 16 0.0 Black 4 139 2.8 White 53 430 11.0 Black 15 176 7.9 Simpson's Paradox Secondly, consider the percentages, regardless of defendant's race, the death penalty was considerably more likely when the victims were white than when the victims were black. This sugg ...

Cornell, BTRY 6030

Excerpt: ... Three-way Contingency Tables Lecture 13 ILRST 212 Cochran-Mantel Haenszel This section shows whether sample data are consistent with homogenous associations or conditional independence. Cochran Mantel Haenszel method present a test of conditional independence and test of homogenous association for the K conditional odds ratio s in 2*2*K tables. Example: table 3.3, page 60 Cochran-Mantel Haenszel For a 2*2*K tables, the null hypothesis that X and Y are conditionally independent, given Z, means that the conditional odds ratio xy(k) between X and Y equals 1 in each partial table. Ho: xy(1) = xy(2) = xy(3) = .xy(k)= 1 Ha: Atleast one of the odds ratio is not equal to 1 Cochran-Mantel Haenszel The test statistic utilizes n11k in each partial table. When the true odds ratio xy(k) exceeds 1.0 in partial table k, we expect to observe (n11k- 11k) >0. The test statistic combines these differences across all K tables. Cochran-Mantel Haenszel Hence CMH stat ...

Washington, DPHS 568

Excerpt: ... he interpretation of the regression coefficient, is fairly intuitive. See next slide. Interpretation of coefficients in Logistic reg. p Odds for success = 1- p p log Logistic model: 1 - p = + X odds for success when X=x is exp( + x) To compare odds of success for different values of X we can look at the Odds Ratio . In particular, the odds ratio for X = x+1 compared to X = x (i.e. the effect of a one unit increase in X) is: OR = odds( X = x + 1) exp( + x + ) = = exp( ) odds( X = x) exp( + x) is the log( odds ratio ) Hypothesis testing and Confidence Intervals The hypotheses of interest are usually: H0: = 0 vs. H1: 0, Which is equivalent to testing whether or not there is association between x and p. The test statistic is Z= b , SE (b) where b is the regression estimate of , (esimated by computer program). Compare Z to a standard Normal distribution Should have a large number of events for normal approximation to be valid. 100(1- )% confidence interval for is given by b Z 1- / 2 ...

Columbia, P 8400

Excerpt: ... g) PRISON: whether the addict had prison history, yes=1, no=0 1. Fit a linear regression model to estimate number of days in treatment from prison history and maximum methadone dose. Report the intercept, partial regression coefficients and interpret them. (6 points) 2. Run a logistic regression model to predict the completion of treatment (FINISH) with DOSE and PRISON as explanatory variables. Report and interpret the odds ratio s for DOSE and PRISON. (4 points) 3. Use Kaplan-Meier method to estimate whether prison history influences the number of days in treatment. What is the median number of days in treatment for with and without prison history? Interpret the Log-Rank test. (2 points) 4. Run a Cox model, which includes both DOSE and PRISON as explanatory variables. Report and interpret the hazards ratios for DOSE and PRISON. (4 points) 5. There are problems in fitting a linear regression model for this data. What are they? (2 points) 6. Is it right to fit a ...

Texas San Antonio, CHAPTER 7853

Excerpt: ... ed as RR = 0 1 2 1 2 When RR=1, then X and Y are independent. Case 1) RR = Case 2) RR = .45 .441 .01 .001 = 1.02 whereas = 10 Interpretation: Suppose RR of lung cancer for smokers versus non-smokers is 10, this means that the risk of lung cancer for smokers is 10 times the risk of lung cancer for non-smokers. 3) Odds Ratio : Let denote the probability of success. Then = 1- 5 is called odds of success. When = 1, success and failures are equally likely. When = 3 , then the odds of success is 3 to 2. Or P (Success) = = .6, 2 and P (F ailure) = 1 - = .4. In fact = 1+ . Now consider a 2 2 table with row marginals fixed. Success Failure Row 1 Row 2 1 2 1 - 1 1 - 2 Define the odds for success for the two rows: 1 1 - 1 2 2 = 1 - 2 1 = Then, Odds Ratio is defined as 1 1 (1 - 2 ) = 2 2 (1 - 1 ) = If we have a joint distribution for X and Y with the following structure Success Failure Row 1 Row 2 11 21 12 22 6 Then 11 12 21 2 = 22 1 = and = 1 11 22 = 2 21 12 In that sense, it is also called C ...

Michigan, STAT 406

Excerpt: ... Statistics 406 Problem Set 10 Due Tuesday December 12th 1. In this problem you will use simulation to evaluate the accuracy of the approximate value for the sampling standard deviation of the log odds ratio , given at the top of page 9 in the Binary response data notes. (a) Construct three 2 2 table of the form n11 n10 n01 n00 One of them should have log odds ratio 0, one should have log odds ratio between 0.5 and 1, and the third table should have log odds ratio between -1.5 and -1. (b) Use simulation to assess the coverage probability of the 95% CI for the population log odds ratio based on the approximate value for the sampling standard deviation of the log odds ratio (from page 9 of the notes). Consider all three tables that you constructed in part (a), and vary the total sample size from 20 to 100 in increments of 10 (change the sample size by scaling the tables you found in part (a), rounding to get integer sample sizes). 2. Suppose we have n timecourses of length T : Xit , i = 1, . . . , n and ...

University of Florida , STA 4504

Excerpt: ... STA 4504 First Exam Topic List What is the difference between ordinal and nominal categorical variables? In what situations would we use a binomial model for categorial data? How would we draw inferences for the binomial parameter pi based on observed data? What does it mean for an estimator to be the maximum likelihood estimator? What type of model is used for binary data when the number of observed responses is random? When data are cross-classified in a two-way contingency table, how do we form relative risks, odds and odds ratio s? Be able to estimate these from data. Why do we express results in terms of odds and odds ratio s? How are a relative risk and an odds ratio interpreted? Be able to find a CI for an odds ratio ? When it comes to describing a study, what is the meaning of the terms: experimental, observational, retrospective, prospective, case-control, cohort, clinical trial and cross-sectional? What does independence mean in a contingency table and why is this condition important? Be able to find t ...

Michigan, STAT 406

Excerpt: ... 12 p21 p22 . 15 We can simulate the n11 , n12 , n21 , and n22 by simulating uniform random values U1 , . . . , Un. For each Ui we add 1 to one of the cells, as follows. U < c1 c1 U < c2 c2 U < c3 c3 U < c4 add add add add 1 1 1 1 to to to to n11 n12 n21 n22 . 16 Here is a simple simulation study to show that the frequencies of the four cells agree with their probabilites. # Set the cell probabilities here. p11 = 0.2 p12 = 0.3 p21 = 0.1 p22 = 0.4 # c1 c2 c3 c4 Cumulative probabilities. = p11 = c1+p12 = c2+p21 = c3+p22 # Simulate a contingency table. U = runif(1e4) N = array(0, c(2,2) N[1,1] = sum(U <= c1) N[1,2] = sum( (U > c1) & (U <= c2) ) N[2,1] = sum( (U > c2) & (U <= c3) ) 17 N[2,2] = sum(U > c3) The odds ratio and log odds ratio For a univariate Bernoulli trial with success probability p, the odds is the ratio of the success probability to the failure probability: p/(1 p). In a contingency table, if we know that X = 1, the probability that Y = 1 is p11 /(p11 + p12 ). ...

Washington, B 536

Excerpt: ... use it when the number of nuisance parameters ('s) is greater than 10 BIOST 536 Lecture 12 5 Example Oxford data with a parameter for each year of birth 1944-1964 Cases and controls were frequency matched on year of birth Ungrouped data: n = 11,852 observations . tabulate xray y | y xray | 0 1 | Total -+-+-0 | 5,324 4,994 | 10,318 1 | 602 932 | 1,534 -+-+-Total | 5,926 5,926 | 11,852 . table yob y -| y yob | 0 1 -+-1934 | 28 28 1935 | 53 53 1936 | 108 108 1937 | 175 175 1938 | 226 226 1939 | 281 281 1940 | 345 345 1941 | 429 429 1942 | 455 455 1943 | 529 529 1944 | 486 486 1945 | 471 471 1946 | 457 457 1947 | 378 378 1948 | 405 405 1949 | 358 358 1950 | 272 272 1951 | 192 192 1952 | 146 146 1953 | 100 100 1954 | 32 32 - BIOST 536 Lecture 12 6 Example Usual odds ratio and Mantel-Haenszel odds ratio adjusting for year of birth . cc y xray Proportion | Exposed Unexpos ...

Michigan State University, EPI 826

Excerpt: ... - which indicates that for low values of z the risk of developing disease remains minimal, until you reach some threshold, then risk rises rapidly as z increases, and then again reaches its asymptotic limit and remains high once z gets large enough. So what is the relationship between logistic function, logistic regression models and an estimate of odds ratio s? 1 is dependent on the value of z. To obtain 1 + e z the logistic model from logistic function we write z as a function of independent variables Xs If you notice the value of logistic function f ( z ) z = + 1 X1 + 2 X2 + k Xk then we can write our logistic model as follows: f (z) = 1 + e ( + 1 X1 + 2 X2 + + k Xk ) 1 = e( + 1 X1 + 2 X2 + k Xk ) 1 + e( + 1 X1 + 2 X2 + k Xk ) In epidemiology, logistic function was initially used in the framework of a follow-up study. Suppose we observe at time T0 , characteristics X1 , Xk on a group of n individuals and then 1 Alka Indurkhya EPI 82 ...

Columbia, P 8400

Excerpt: ... T1 T0 = I1 I0 Causal Risk Ratio N N A0 = A1 R = 1 A0 R0 Relationship between Ratio Measures Risk Ratio R1 RN A IT = 1 = 1 = 1 1 R0 R0 N A0 I 0 T0 because R = A/N and I=A/T Risk Ratio = Rate Ratio * Ratio of Persontime Odds Ratio R1 R0 S1 S0 R1 = R0 (1 - R1 ) (1 - R0 ) = A1 N A0 N 1 - A1 N A1 = A0 1 - N A0 ( N - A1 ) ( N - A0 ) Scenarios when relative measures approximate relative risk R1 R1 I 1T 1 I1 S1 = = = & & R0 R0 I 0T 0 I0 S0 Time period sufficiently small that average T for exposed population is only slightly smaller than T for unexposed means T1 and T0 are approximately equal and rate ratio approximates risk ratio Sufficiently small proportion of onsets means R1 and R0 are small, S1 and S0 are close to 1 ( odds ratio approximates risk ratio) Odds Ratio will overestimate the risk ratio Scenario 1 where factor increases risk R 1 > R0 S1 = 1 - R1 < 1 - R0 = S 0 S0 >1 S1 R1 R1 R1 S 0 S1 1< < = R0 R0 R0 S1 S0 Odds Ratio will underestimate the risk ratio Scenario 2 ...