Please, I wish to ask if any could help correct this error: NOTE: Module FUNC defined. 47   interval = {A, B}; 48   call quad(answer, "Func", interval); param must be numeric ERROR: (execution) Character argument should be numeric This is data set and code: Data sim; Input A B; cards; 1   3 2   7 3   10 4   12 ; Run; proc iml; use sim; read all  var{A B} into DM; close; A = DM[,1]; B = DM[,2]; n = nrow(DM); start Func(x);    return(  exp(2-x)#cdf("Normal", x) ); finish; interval = {A, B}; call quad(answer, "Func", interval); create kaplan1n var{A  B Answer }; append; quit; ... View more

Please help correct this macro: %macro surv(replicates=); %do p=1 %to &replicates; %macro predict_&p (outest=, out=_last_, xbeta=, time=); %predict_&p (outtest=a_&p, out=b_&p, xbeta=lp, time=30); data _pred_&p; _p_=1; set &outest point=_p_; set &out; lp=&xbeta; gamma=1/_scale_; alpha=exp(-lp*gamma); prob=0; _dist_=upcase(_dist_); if _dist_='WEIBULL' or _dist_='EXPONENTIAL' or _dist_='EXPONENT' then prob=exp(-alpha*time**gamma); if _dist_='LOGNORMAL' or _dist_='LNORMAL' then prob=1-probnorm((log(time)-lp)/_scale_); if  _dist_='LLOGISTIC' or _dist_='LLOGISTC' then prob=1/(1+alpha*time**gamma); if _dist_='GAMMA' then do; d=_shape1_; k=1/(d*d); u=(t*exp(-lp))**gamma; prob=1-probgam(k*u**d,k); if d lt 0 then prob=1-prob; end; drop lp gamma alpha _dist_ _scale_ intercept _shepe1_ _model_ _name_ _type_ _status_ _prob_ _lnlike_ d k u; run; proc print data=_pred_&p; run; %mend predict_&p; DATA test_&p; one   = 1; beta0 =  1.0; beta1 = 0.75; beta2 =  -.5; subsd =  -.5; errsd =  1; seed  = 974657731; n=1; do while (n le 100); ni=1; sun = rannor(seed); do while (ni le 10); x1 = abs(normal(6));    error = rannor(seed);    time = ni-1;   test=7*ranuni(seed)+1;  /** uniform (1,8)  */;   m1=min(time, test);   m2=max(0,time-test);   olf = (time lt test);   *grp = 0; if n le 50 then grp=1;   y = beta0*x1 + beta1*m1 +beta2*m2 + subsd*sun + errsd*error;      output test_&p;   ni+1; end;    n+1; end; proc lifereg data=test_&p outest=a_&p; class n;       model (m1, m2) = x1/D=LNORMAL;    output out=b_&p xbeta=lp;    run; %PREDICT_&P(OUTEST=a_&p, out=b_&p, xbeta=lp, time=30); %end; %mend surv; %surv(replicates=2); ods html; ... View more

How do I translate this R code to SAS: library(lme4) sim<-function(m,n,b0=1,b1=-.5,b2=-1,testlow=1,testhi=10,xtest,xolftrue){ a0hat<-0 b0hat<-0 b1hat<-0 b2hat<-0 d0hat<-0 c0hat<-0 c1hat<-0 c2hat<-0 c3hat<-0 for(i in 1:m){ olf<-runif(n,1,10) olftrue<-as.vector(t(matrix(olf,nrow=n,ncol=10,byrow=F))) test<-sample(testlow:testhi,n,replace=T) olftest<-ifelse(olf<test,test,0) olfind<-ifelse(olf<test,1,0) olft<-matrix(0,nrow=n,ncol=10,byrow=T) olft[cbind(1:n,olftest)]<-1 olft<-t(apply(olft,1,cumsum)) olft<-as.vector(t(matrix(olft,nrow=n*10,ncol=1,byrow=T))) x<-rnorm(n,3,1) mx<-matrix(x,nrow=n*10,ncol=1,byrow=T) x<-as.vector(mx) alpha<-rnorm(n,0,.5) malpha<-matrix(alpha,nrow=n*10,ncol=1,byrow=T) alpha<-as.vector(malpha) eps<-rnorm(n*10,0,1) time<-matrix(1:10,nrow=n*10,ncol=1,byrow=T) time<-as.vector(time) time1<-pmin(time,olftrue) time2<-pmax(time-olftrue,0) y<-alpha+b0*x+b1*time1+b2*time2+eps id<-as.vector(t(matrix(1:n,nrow=n,ncol=10,byrow=F))) ot<-olft*time try<-lmer(y~1+(1|id)+x+time1+time2) a0hat<-a0hat+fixef(try)[[1]] b0hat<-b0hat+fixef(try)[[2]] b1hat<-b1hat+fixef(try)[[3]] b2hat<-b2hat+fixef(try)[[4]] timenew<-time-as.vector(t(matrix(olftest,nrow=n,ncol=10,byrow=F))) olfind<-as.vector(t(matrix(olfind,nrow=n,ncol=10,byrow=F))) try2<-lmer(y~1+(1|id)+x+timenew+olfind+timenew*olfind) d0hat<-d0hat+fixef(try2)[[1]] c0hat<-c0hat+fixef(try2)[[2]] c1hat<-c1hat+fixef(try2)[[3]] c2hat<-c2hat+fixef(try2)[[4]] c3hat<-c3hat+fixef(try2)[[5]] } ... View more

Please can some one explain to me how SAS compute standard error is proc logistic ... View more

Hi I wish to ask why this logistic simulated data does give the correct estimates for t  when I ran the simulation with 1000 replicate; Any help? data test; intercept = 0.01; beta = 0.5; beta1=0.1; do i = 1 to 1000; t = rand("WEIBULL", .5, .25; ); X=ranbin(0,1,.5); linpred = intercept + t*beta + x*beta1; prob = exp(linpred)/ (1 + exp(linpred)); U=.6*(uniform(0)) + 0.4; y = U lt prob; output; end; run; proc logistic data=test; model Y(event='1') = t x; run; ... View more