R help - Jul 2012 - significant difference between Gompertz hazard parameters?

Hello, all.

I have co-opted a number of functions that can be used to plot the
hazard/survival functions and associated density distribution for a Gompertz
mortality model, given known parameters.  The Gompertz hazard model has been
shown to fit relatively well to the human adult lifespan.  For example, if I
wanted to plot the hazard (i.e., mortality) functions:

pop1 <- function (t)
	{
		x=c(0.03286343, 0.04271132)
		a3<-x[1]
		b3<-x[2]
		shift<-15     # only considering mortality after 15 years
		
		h.t<-a3*exp(b3*(t-shift))
		return<-h.t
}

pop2 <- function (t)
	{
		x=c(0.02207778, 0.04580059)
		a3<-x[1]
		b3<-x[2]
		shift<-15     # only considering mortality after 15 years
		
		h.t<-a3*exp(b3*(t-shift))
		return<-h.t
}

ylab.name <-
expression(paste(italic(h),"(",italic(a),")"))
plot(seq(15,80,1),pop1(seq(15,80,1)),type='l',ylab=ylab.name,xlab='Age
(years)',ylim=c(0,0.8))
lines(seq(15,80,1),pop2(seq(15,80,1)),lty=2)

How may I test for a significant difference in the hazard parameters that
define the mortality experience for these two populations?  Thanks in
advance.

Regards,
Trey

-----
Trey Batey---Anthropology Instructor
Division of Social Sciences
Mt. Hood Community College
Gresham, OR  97030
Alt. Email:  trey.batey[at]mhcc[dot]edu
--
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On Jul 1, 2012, at 12:04 AM, piltdownpunk wrote:
> Hello, all.
>
> I have co-opted a number of functions that can be used to plot the
> hazard/survival functions and associated density distribution for a  
> Gompertz
> mortality model, given known parameters.  The Gompertz hazard model  
> has been
> shown to fit relatively well to the human adult lifespan.  For  
> example, if I
> wanted to plot the hazard (i.e., mortality) functions:
>
> pop1 <- function (t)
> 	{
> 		x=c(0.03286343, 0.04271132)
> 		a3<-x[1]
> 		b3<-x[2]
> 		shift<-15     # only considering mortality after 15 years
> 		
> 		h.t<-a3*exp(b3*(t-shift))
> 		return<-h.t
> }
>
> pop2 <- function (t)
> 	{
> 		x=c(0.02207778, 0.04580059)
> 		a3<-x[1]
> 		b3<-x[2]
> 		shift<-15     # only considering mortality after 15 years
> 		
> 		h.t<-a3*exp(b3*(t-shift))
> 		return<-h.t
> }
>
> ylab.name <-
expression(paste(italic(h),"(",italic(a),")"))
>
plot(seq(15,80,1),pop1(seq(15,80,1)),type='l',ylab=ylab.name,xlab='Age
> (years)',ylim=c(0,0.8))
> lines(seq(15,80,1),pop2(seq(15,80,1)),lty=2)
>
> How may I test for a significant difference in the hazard parameters  
> that
> define the mortality experience for these two populations?  Thanks in
> advance.
>
You cannot test for differences in pre-specified parameters. These are  
by definition, "different". If you supply some data, possibly  
generated through simulations, you can test for differences in fit  
using parametric fits. The survival package offers facilities for  
fitting,  and in the Archives you can find several responses from  
Terry Therneau to questions about fitting data to Gompertz or Gompertz- 
Makeham distributions.
>
-- 

David Winsemius, MD
West Hartford, CT