Homework 2 April 5 Refer to the Mental health example of CDA (example 7.2.4 on P. 279) the data is at the bottom of this file. mental health was coded ( Well=1, Mild=2, Moderate=3, Impaired=4 ) (1) Fit a cumulative logit model to the data, treating the SES and Life Events as explainatory variables and the mental health as responses. (basically re-produce the table 7.6). Use the fitted model to estimate the P( mental health = Mild) when SES = 0, Life Event = 4.275 Also, find the 95% confidence interval for this probability. (2) Even though clearly the responses are ordinal here, let us treat the 4 responses for the mental health as categorical, fit a baseline-category logit model to the same data. Obtain the same probability (confidence interval) as in (1) but using model obtained here. (3) In (1) we replaced the mental health responses by 1,2,3,4 and do the analysis. what if we re-labling the response as 1,2, 4,5 ? Will the outcome change? Why or WHy not? (4) Try the SIR, and compare with the results obtained above. Some general format of SAS programs: For (ordinal responses) cumulative logistic regression models: proc logistic data = CDA6; model resp = pred1 pred2 pred3; run; =============================================== For (categorical responses) baseline-Categorical logistic regression models: proc logistic data = CDA6; class resp (ref = "1"); model resp = pred1 pred2 pred3 / link = glogit; run; =========================================================== The following are the data for the problem (1) and (2) data impair; input mental ses life; datalines; 1 1 1 1 1 9 1 1 4 1 1 3 1 0 2 1 1 0 1 0 1 1 1 3 1 1 3 1 1 7 1 0 1 1 0 2 2 1 5 2 0 6 2 1 3 2 0 1 2 1 8 2 1 2 2 0 5 2 1 5 2 1 9 2 0 3 2 1 3 2 1 1 3 0 0 3 1 4 3 0 3 3 0 9 3 1 6 3 0 4 3 0 3 4 1 8 4 1 2 4 1 7 4 0 5 4 0 4 4 0 4 4 1 8 4 0 8 4 0 9 ; /** Courtesy of Feng Zhou, we provide the following SAS code example for problem (1) above **/ data pred; input mental ses life; datalines; . 0 4.275 ; proc logistic data=impair; model mental = ses life; score data = pred out= result1 CLM; run; proc print data=result1; run;