R help - Nov 2010 - confidence interval for logistic joinpoint regression from package ljr

I?m trying to run a logistic joinpoint regression utilising the ljr package.
I?ve been using the forward selection technique to get the number of knots for
the analysis, but I?m uncertain as to my results and the interpretation. The
documentation is rather brief ( in the package and the stats in medicine article
is quite technical) and without any good examples. At the moment I?m thinking
1)find the number of knots both using forward and backward techniques and see if
they are close
2)present the annual percent change (APC) for each of the intervals, ie my
present data (1950-2010 in 5 year intervals) is giving me

       Variables          Coef
b0     Intercept -131.20404630
g0             t    0.06146463
g1 max(t-tau1,0)   -0.51582466
g2 max(t-tau2,0)    0.43429615

Joinpoints:
              
1 tau1= 1990.5
2 tau2= 1995.5

APC 1950->1990=exp(0.06)=1.06-->6%
1990-1995=exp(0.06-0.51)=exp(-0.45)=0.63--> -37%
1995-2010=exp(0.06-0.51+0.43)-->-2%

3) Preferably a confidence interval for the APC should be given. However, this I
havent figured out yet.

//M

dear M.,
I do not know how to get the SE for the joinpoint (or breakpoint) from 
your ljr fit. However you can find useful the segmented package which 
works for any GLM (including the logistic one) and it returns 
(approximate) StErr (and Conf Int) also for the joinpoint (breakpoint in 
the segmented package). For some examples, see the R news paper
http://cran.r-project.org/doc/Rnews/Rnews_2008-1.pdf

Also, as regards to the APC, you could find interesting the following note

http://onlinelibrary.wiley.com/doi/10.1002/sim.3850/abstract

hope this helps you,
vito


Il 30/11/2010 23.54, moleps ha scritto:
> I?m trying to run a logistic joinpoint regression utilising the ljr
package. I?ve been using the forward selection technique to get the number of
knots for the analysis, but I?m uncertain as to my results and the
interpretation. The documentation is rather brief ( in the package and the stats
in medicine article is quite technical) and without any good examples. At the
moment I?m thinking
> 1)find the number of knots both using forward and backward techniques and
see if they are close
> 2)present the annual percent change (APC) for each of the intervals, ie my
present data (1950-2010 in 5 year intervals) is giving me
>
>         Variables          Coef
> b0     Intercept -131.20404630
> g0             t    0.06146463
> g1 max(t-tau1,0)   -0.51582466
> g2 max(t-tau2,0)    0.43429615
>
> Joinpoints:
>
> 1 tau1= 1990.5
> 2 tau2= 1995.5
>
> APC 1950->1990=exp(0.06)=1.06-->6%
> 1990-1995=exp(0.06-0.51)=exp(-0.45)=0.63-->  -37%
> 1995-2010=exp(0.06-0.51+0.43)-->-2%
>
> 3) Preferably a confidence interval for the APC should be given. However,
this I havent figured out yet.
>
> //M
>
> ______________________________________________
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
-- 
===================================Vito M.R. Muggeo
Dip.to Sc Statist e Matem `Vianelli'
Universit? di Palermo
viale delle Scienze, edificio 13
90128 Palermo - ITALY
tel: 091 23895240
fax: 091 485726/485612
http://dssm.unipa.it/vmuggeo