HSRP 734: Advanced Statistical Methods May 29, 2008

1. HSRP 734: Advanced Statistical MethodsMay 29, 2008

2. Finish talking about Association Measures: Odds Ratio OR=2 of Disease for Exposed vs. Not exposed What is the interpretation? “Exposed patients have twice the odds of disease versus patients that were not exposed.”

3. Finish talking about Association Measures:Relative Risk RR=2.5 of Disease for Exposed vs. Not exposed What is the interpretation? “Exposed patients are 2.5 times as likely to have the disease versus patients that were not exposed.”

4. Finish talking about Association Measures OR is not close to RR Unless Pr(disease) for Exposed, Not exposed low “Rare” disease

5. Finish talking about Association Measures Confidence intervals for Odds Ratio Confidence intervals for Relative Risk

6. Measures of Disease Association

7. Confidence Interval for Odds Ratio

8. Confidence Interval for Odds Ratio

9. Confidence Interval for Risk Ratio

10. Confidence Interval for Risk Ratio

11. SAS Enterprise:or_rr.sas7bdat

12. SAS websites Online help: http://support.sas.com/onlinedoc/913/docMainpage.jsp UCLA: http://www.ats.ucla.edu/stat/SAS/ SAS SUGI: http://support.sas.com/events/sasglobalforum/previous/index.html

13. Categorical Data Analysis Understand the Multinomial probability mass function Compute Goodness-of-fit tests and chi-squared tests for association Test for association in the presence of a possibly confounding third factor (e.g., disease versus exposure from 3 sites)

14. Categorical Data Analysis Motivation How do we estimate and test the magnitude of posited relationship when the outcome of interest is categorical? e.g., An international study examines the relationship between age at first birth and the development of breast cancer Age = categorized into age groups

15. Categorical Data Analysis

16. Categorical Data Analysis Research question Is there a relationship between age at first birth and Cancer status? Better to convert the table into percentages (easier to see) Turns out that there is a significant relationship (p<0.001)

17. Categorical Data Analysis Statistical techniques involve Probability distribution for categorical data Tests for relationship in a RxC table R = # of Rows in Table C = # of Columns in Table

18. Probability Distribution for Categorical Outcomes Fun for Friday night: Go home and flip a quarter 10,000 times. Determine if there is evidence that one side is falling down more.

19. Probability Distributions for Categorical Data Bernoulli (1 toss of a coin, outcome=H,T) Binomial (10 tosses of a coin, outcome=0,1,2..,10 heads) Multinomial (throw 10 balls into 4 pigeon holes ABCD, outcome= (3A,2B,1C,4D))

20. Why use multinomial for testing? Relationship between 2 categorical variables RxC table analysis Based on multinomial distribution

21. Why use multinomial for testing? Example: 2 level exposure status (Exposed, Not exposed), 3 level outcome (severe, mild, no disease) Treat 2x3=6 outcomes as categorical or a multinomial distribution with 6 pigeon holes The expected probability of the pigeon holes are specified under some kind of assumptions (e.g., independence)

22. Level of Measurement Categorical response dichotomous ordinal (>2 categories, ordered) nominal (>2 categories, not ordered) Dichotomous use Binomial distribution Ordinal, Nominal use Multinomial distribution

23. Multinomial Distribution Multinomial experiment: Experiment consists of n identical and independent trials Each trial results in one of K outcomes Let pi be the probability of outcome i a. Each pi remains constant for each experiment b. The pmf for k outcomes is: Notes:

24. Example of a Multinomial Experiment Consider an unfair die and 6 tosses: Let Find the probability of this outcome

25. Simple Multinomial Experiments Classical example: Mendel Sample from the second generation of seeds resulting from crossing yellow round peas and green wrinkled peas (N=556) 

26. Mendel’s Laws of Inheritance suggest that we should expect the following ratios:  9/16, 3/16, 3/16, 1/16 For N = 556, the expected number of each outcome is: E(YR) = 556 x 9/16 = 312.75 E(YW) = 556 x 3/16 = 104.25 E(GR) = 556 x 3/16 = 104.25 E(GW) = 556 x 1/16 = 34.75

28. Multinomial distribution The observed cell counts are not identical to the expected cell counts Under the assumption of a multinomial model with the stated probabilities, how might we determine how unlikely it is to observe these data?

29. Chi-square GOF Test Hypothesis: observed cell counts are consistent with the multinomial probabilities Theoretical result Require that expected cell counts not too small Expected counts > 5.

30. Chi-square distribution Remarks about Chi-squared distribution: Nonsymmetric Strictly positive Different chi-squared distribution for each df.

32. Chi-square GOF Test Applying this test to Mendel’s peas example yields H0: pYR = 9/16, pYW = 3/16, pGR = 3/16, pGW = 1/16 H1: at least one pi differs from hypothesized value

33. Chi-square GOF Test

34. Chi-square GOF Test Therefore, we observed c2 = 0.47 from a multinomial experiment with k = 4. Thus, df = k-1 = 3. For a = 0.05, Thus, the observed chi-squared statistic is not greater than the critical value for a = 0.05 and df = 3. We fail to find evidence that these data depart from the hypothesized probabilities. i.e., model fits well to data

35. Testing association in 2x2 table This method translates to testing cross-tabulation tables for RxC cases Here the cells are formed by cross-classification of 2 variables Null hypothesis is the 2 variables are independent Simplest case : 2x2 table

36. Testing association in 2x2 table Testing for independence or no association Similar idea to checking goodness-of-fit Compare what to see to what you hypothesized to be true You did, in fact, hypothesize “independence”

37. Basic Inference for 2x2 Tables 2x2 Contingency Table

38. Chi-square GOF Test for 2x2 Tables H0: There is no association between row and columns Under H0, the expected cell counts are the product of the marginal probabilities and the sample size. Why? The classic Pearson’s chi-squared test of independence df = (2-1) x (2-1) = 1 Conservatively, we require EXPECTEDij = 5 for all i, j

39. Other Tests for 2x2 Tables Two alternative tests Yate’s continuity corrected chi-square statistic Mantel-Haenszel chi-square statistic For sufficiently large sample size, all three Chi-squared statistics are approximately equal and all have a Chi-squared distribution with 1 df

40. When to use Chi-square vs. Fisher’s Exact When the expected cell counts are less than 5, it is better to use the Fisher’s exact test.

41. Summary of the Use of ?2 test Test of goodness-of-fit Determine whether or not a sample of observed values of some random variable is compatible with the hypothesis that the sample was drawn from a population with a specified distributional form (e.g., specified probabilities of certain events)

42. Summary of the Use of ?2 test Test of independence Test the null hypothesis that two criteria of classification (variables) are independent

43. Summary of the Use of ?2 test Test of homogeneity Test the null hypothesis that the samples are drawn from populations that are homogeneous with respect to some factor (i.e., no association between group and factor)

44. Summary of the Use of ?2 test Could consider this test as answering: “Are the Row factor and Column factor associated?”

45. Categorical Data Analysis Ideas of multinomial and chi-squared test generalize to testing RxC association and RxCxK association Example: 2 exposure status, 2 disease status, 3 sites 2x2x3 association analysis

46. Test of General Association (R x C Table) Consider a study designed to test whether there exists an association between political party affiliation and residency within specific counties

47. Notation for general RxC table

48. Test of General Association H0: There is no association between rows and columns H1: There exists a dependence between rows and columns Under H0,the expected cell counts are the product of the corresponding marginal probabilities and the sample size. The classic Pearson’s chi squared test of independence

49. SAS Enterprise:chisq.sas7bdat

50. Mantel-Haenszel test Often, there are other factors in a RxC test Mantel-Haenszel test (or Cochran Mantel Haenzsel CMH) can be used for controlling for “nuisance” factors Typically used for rxcx2 table e.g., 2x2x2 cross classification e.g., Association between disease status and exposure controlling for age group (strata)

51. Stratified Analysis Examples of commonly used strata Age group Gender Study site (hospital, country) ethnic group

52. Stratified Analysis Myocardial infarction and anticoagulant use by Coronary Care Unit

53. Stratified Analysis Idea: test for an association while controlling for CCU effects Denote the counts from the first cell within the hth subtable as nh11, Construct the CMH test of association controlling for CCU

54. Stratified Analysis Test assumes the direction of effect within each table is the same The Cochran-Mantel-Haenszel approach partially removes the confounding influences of the explanatory variable (e.g., CCU) May improve power

55. Mantel-Haenszel Test The expected value of nh11 for h = 1,2,…,g is and the variance of nh11 This leads to the Cochrane-Mantel-Haenszel test

56. Direction of effects across Strata Note that if directions of conditional ORs are not the same, discrepancies between observed and expected from different strata may cancel out one another Lead to poor power and biased result

57. MH “Pooled” Odds Ratio

58. MH test decision list Z = strata of potential confounder -> If ORc ˜ (ORZ=1 ˜ ORZ=2 ˜…) Z is not a confounder, report crude OR (ORc) -> If ORc ? (ORZ=1 ˜ ORZ=2 ˜…) Z is a confounder, report adjusted OR (ORMH) -> If ORZ=1 ? ORZ=2 ? … Z is an effect modifier, report strata specific OR’s (don’t adjust!)

59. Breslow Day test (More formal approach) Can also test for homogeneity of odds ratio across strata If Breslow Day test is significant => odds ratios within strata are not homogeneous. Thus, => ORMH would be inappropriate!

60. SAS Enterprise:cmh.sas7bdat

61. Results from cmh.sas7bdat ORcrude = 3.76 (2.01, 7.05) ORcenter1 = 4.01 (1.67, 9.66) ORcenter2 = 4.05 (1.55, 10.60) ORMH = 4.03 (2.11, 7.71)

62. Take home messages Multinomial and the Chi-square test are the “workhorse” for testing of goodness-of-fit Idea is to compare expected counts (calculated from a pre-determined set of probabilities) and the observed counts The same idea can be applied to testing statistical assumptions such as no association CMH test is for testing association when a confounding effect (strata) may be present

63. For Next Class 6/5 HW #1 key posted HW #2 will be due Read Kleinbaum Ch. 1,2