1 / 21

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

meli
Download Presentation

March 28, 30

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related