predict.emaxalt: Mean response and SE for specified doses for a simulated object output by function emaxalt

Description

Estimated mean and standard error for specified doses computed from the output of a model fit by function emaxalt. Also returns mean difference with placebo and their standard errors.

Usage

# S3 method for emaxalt
predict(object,dose, dref=0, ...)

Value

A list containing:

fitpred: Vector with mean dose response estimate for each specified dose.
fitdif: Corresponding differences with placebo.
sepred: SEs for fitpred.
sedif: SEs for fitdif.

Arguments

object: Output of emaxalt
dose: Vector (can be a single value) of doses where dose response curve is to be evaluated.
dref: A reference dose (0 by default) for contrasts, but other values can be specified. If specified, a single reference value must be given.
...: Optional arguments are not used.

Author

Neal Thomas

Examples

Run this code


if (FALSE) {
## random number seed changed by this example

doselev<-c(0,5,25,50,100)
n<-c(78,81,81,81,77)
dose<-rep(doselev,n)

### population parameters for simulation
e0<-2.465375 
ed50<-67.481113 

dtarget<-100
diftarget<-9.032497
emax<-solveEmax(diftarget,dtarget,log(ed50),1,e0)

sdy<-7.967897
pop.parm<-c(log(ed50),e0,emax)    
meanresp<-emaxfun(dose,pop.parm)  
y<-rnorm(sum(n),meanresp,sdy)

simout<-emaxalt(y,dose)
predict(simout,c(75,150))

simout2<-emaxalt(y,dose,modType=4)
predict(simout2,c(75,150))
}
# \dontshow{
doselev<-c(0,5,25,50,100)
n<-c(78,81,81,81,77)
dose<-rep(doselev,n)

### population parameters for simulation
e0<-2.465375 
ed50<-67.481113 

dtarget<-100
diftarget<-9.032497
emax<-solveEmax(diftarget,dtarget,log(ed50),1,e0)

sdy<-7.967897
pop.parm<-c(log(ed50),e0,emax)    
meanresp<-emaxfun(dose,pop.parm)  
y<-rnorm(sum(n),meanresp,sdy)

simout<-emaxalt(y,dose)
predict(simout,c(75,150))
# }