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March 28, 30

March 28, 30. Return exam Analyses of covariance 2-way ANOVA Analyses of binary outcomes. Exam 1 Scores. N = 23 75% Q3 95 50% Median 83 25% Q1 74 Mean = 83.7 SD = 11.1. 90-100 A 89-89 B 70-79 C 30% of Grade 30% Exam 2 30% Assignments

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March 28, 30

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  1. March 28, 30 • Return exam • Analyses of covariance • 2-way ANOVA • Analyses of binary outcomes

  2. Exam 1 Scores N = 23 75% Q3 95 50% Median 83 25% Q1 74 Mean = 83.7 SD = 11.1 90-100 A 89-89 B 70-79 C 30% of Grade 30% Exam 2 30% Assignments 10% Project

  3. Analysis of Variance ANOVA simultaneously tests for difference in k means • Y - continuous • k samples from k normal distributions • each size ni, not necessarily equal • each with possibly different mean • each with constant variance 2

  4. Analyses of Covariance • Comparing k means adjusting for 1 or more other variables (covariates) • Uses • Randomized students to adjust for an imbalance among treatments in a baseline factor. • In observational studies controlling for a confounding factor • To throw light on the nature of treatment effects in a randomized study • Improve precision of estimated differences among treatments

  5. Analyses of Covariance • Comparing k means adjusting for 1 or more other variables (covariates) • Compute adjusted (or least square) means. • Sometimes called ANCOVA

  6. Original Use • Fisher (1941) the Y were yields of tea bushes in an experiment. • But the luck of the draw, some treatments will have been allotted to a more productive set of bushes than other treatments • Use as an adjustment variable the yields of the bushes in the previous year.

  7. Adjusted Means Computation YBARi = Mean of Y for group I XBARi = Mean of X for group I XBAR= Mean of X for all groups combined b = Regression slope of X with Y YBAR(A)i = Adjusted mean for group I YBAR(A)i = YBARi – b (XBARi – XBAR) Adjustment

  8. Adjusted Means ComputationObservations • YBAR(A)i = YBARi – b (XBARi – XBAR) • If b = 0 then adjusted mean equals unadjusted mean • If mean of X is same for all group then adjusted mean equals unadjusted mean

  9. Adjusted Mean Interpretation The mean of Y for the group if the mean of X for the group was at the overall mean. Uses a model to make the mean of X the same for all groups What would the means have been if all groups had the same mean of X?

  10. Example from TOMHS Compare 12-month visit mean serum cholesterol between diuretic group and placebo group. 12-mo Avg. Baseline Avg. Diuretic 231.7 230.7 Placebo 219.7 224.9 Diff: 12.0 5.8 Note: Diuretic group started out with higher cholesterols so may want to adjust for this difference.

  11. Computing the Adjusted Means • 12-mo Avg. Baseline Avg. • Diuretic 231.7 230.7 • Placebo 219.7 224.9 • Total 227.0 • =0.894 Regression slope of 12-month cholesterol on baseline cholesterol • YBAR(A) (Diur) = 231.7 – 0.894 (230.7 – 227.0) • = 231.7 – 0.894 (3.7) = 228.4 • YBAR(A) (Plac) = 219.7 – 0.894 (224.9 – 227.0) • = 219.7 – 0.894 (-3.7) = 221.6 6.8

  12. SAS Code PROCGLM; CLASS group; MODEL chol12 = group cholbl/SS3SOLUTION; MEANS group; LSMEANS group; ESTIMATE‘Adjusted Mean Dif' group 1 -1; RUN;

  13. SAS GLM Output Source DF Type III SS Mean Square F Value Pr > F GROUP 1 3666.8314 3666.8314 6.29 0.0126 cholbl 1 393014.2008 393014.2008 674.26 <.0001 Standard Parameter Estimate Error t Value Pr > |t| Intercept 18.77666226 B 7.90780076 2.37 0.0181 GROUP 3 6.79524641 B 2.70925949 2.51 0.0126 GROUP 6 0.00000000 B . . . cholbl 0.89351891 0.03441046 25.97 <.0001 SOLUTION Regression slope

  14. SAS GLM Output • Level of ------------CHOL12----------- ------------cholbl----------- • GROUP N Mean Std Dev Mean Std Dev • 3 125 231.696000 46.2561633 230.688000 38.9694703 • 6 221 219.737557 38.5904356 224.909502 37.1702588 • Least Squares Means • CHOL12 • GROUP LSMEAN • 3 228.398120 • 221.602873 • Standard • Parameter Estimate Error t Value Pr > |t| • dif 6.79524641 2.70925949 2.51 0.0126 LSMEANS group; dif ESTIMATE'dif' group 1 -1;

  15. Two-Way ANOVA • Two categorical factors related to a continuous outcome (Factor A and factor B). If factors are allocated randomly to all combinations of A and B then design called factorial design • Questions asked • Overall is A related to Y • Overall is B related to Y • Does the effect of A on Y depend on level of B • Example • A = Race; B = BP drug; Y = BP response • A = Vitamin E (y/n); aspirin use (y/n)

  16. Aspirin + Vitamin E Aspirin + Placebo for Vitamin E Placebo for Aspirin + Vitamin E Placebo for Aspirin + Placebo for Vitamin E Factorial Design Example A = Aspirin use (yes or no) B = Vitamin E use (yes or no) Placebo for aspirin and placebo for Vitamin E

  17. TOMHS Example Question: Do certain BP medications differ in lowering blood pressure in blacks compared to whites? Change in SBP (mm Hg) Diuretic a-Blocker Blacks -23.6 -8.7 Whites -21.4 -17.8 Difference -2.2 +9.1 Is the difference –2.2 significantly different from +9.1

  18. SAS Code LIBNAME tomhs 'C:\my documents\ph5415\'; DATA temp; SET tomhs.bpstudy; * Choose diuretic and alpha blocker groups; * Variable black = 1 or 2; if group in(3,4); sbpdif = sbp12 - sbpbl; RUN; PROCGLM DATA=temp; CLASS black group; MODEL sbpdif = black group black*group; MEANS black*group; RUN; Tests for interaction

  19. SAS OUTPUT The GLM Procedure Level of Level of ------------sbpdif----------- GROUP BLACK N Mean Std Dev 3 1 27 -23.6296296 14.6442379 3 2 97 -21.3505155 14.4939220 4 1 24 -8.7291667 17.5802379 4 2 104 -17.8125000 12.5978091

  20. SAS OUTPUT The GLM Procedure Dependent Variable: sbpdif Sum of Source DF Squares Mean Square F Value Pr > F Model 3 3791.89107 1263.96369 6.37 0.0004 Error 248 49197.96210 198.37888 Corrected Total 251 52989.85317 Source DF Type III SS Mean Square F Value Pr > F GROUP 1 3447.055834 3447.055834 17.38 <.0001 BLACK 1 469.412605 469.412605 2.37 0.1253 GROUP*BLACK 1 1309.006907 1309.006907 6.60 0.0108

  21. Your Turn • Using TOMHS data test

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