R help - Feb 2008 - Finding LD50 from an interaction Generalised Linear model

Hi,
I have recently been attempting to find the LD50 from two predicted fits
(For male and females) in a Generalised linear model which models the effect
of both sex + logdose (and sex*logdose interaction) on proportion survival
(formula = y ~ ldose * sex, family = "binomial", data = dat (y is the
survival data)). I can obtain the LD50 for females using the dose.p()
command in the MASS library with dose.p(mod1,c(1,2)). However I cannot find
a way to determine the LD50 of males.
Any help on finding this male LD50 would be appreciated.

Pasting of R workspace below:

> rm(list=ls())
> 
> ##checking file
> dat
   ldose sex numdead
1      0   M       0
2      1   M       3
3      2   M       9
4      3   M      16
5      4   M      18
6      5   M      20
7      0   F       0
8      1   F       2
9      2   F       6
10     3   F      10
11     4   F      11
12     5   F      14
> str(dat)
'data.frame':   12 obs. of  3 variables:
 $ ldose  : int  0 1 2 3 4 5 0 1 2 3 ...
 $ sex    : Factor w/ 2 levels "F","M": 2 2 2 2 2 2 1 1 1 1
...
 $ numdead: int  0 3 9 16 18 20 0 2 6 10 ...
> ##convert numdead to propdead
> dat$propdead<-dat$numdead/20
> ##Calculate survival from dead
> dat$numsurv<-20-dat$numdead
> dat$propsurv<-dat$numsurv/20
> ##check table
> dat
   ldose sex numdead propdead numsurv propsurv
1      0   M       0     0.00      20     1.00
2      1   M       3     0.15      17     0.85
3      2   M       9     0.45      11     0.55
4      3   M      16     0.80       4     0.20
5      4   M      18     0.90       2     0.10
6      5   M      20     1.00       0     0.00
7      0   F       0     0.00      20     1.00
8      1   F       2     0.10      18     0.90
9      2   F       6     0.30      14     0.70
10     3   F      10     0.50      10     0.50
11     4   F      11     0.55       9     0.45
12     5   F      14     0.70       6     0.30
> str(dat)
'data.frame':   12 obs. of  6 variables:
 $ ldose   : int  0 1 2 3 4 5 0 1 2 3 ...
 $ sex     : Factor w/ 2 levels "F","M": 2 2 2 2 2 2 1 1 1 1
...
 $ numdead : int  0 3 9 16 18 20 0 2 6 10 ...
 $ propdead: num  0 0.15 0.45 0.8 0.9 1 0 0.1 0.3 0.5 ...
 $ numsurv : num  20 17 11 4 2 0 20 18 14 10 ...
 $ propsurv: num  1 0.85 0.55 0.2 0.1 0 1 0.9 0.7 0.5
...
> ##load the lattice library
> library(lattice)
> ##plot data with mag + line width 1.5
>
xyplot(propsurv~ldose,groups=sex,data=dat,cex=1.5,lwd=1.5,type="b")
> ##create model
> dat$n<-c(20)
> y<-cbind(dat$numsurv,dat$n-dat$numsurv)
> y
      [,1] [,2]
 [1,]   20    0
 [2,]   17    3
 [3,]   11    9
 [4,]    4   16
 [5,]    2   18
 [6,]    0   20
 [7,]   20    0
 [8,]   18    2
 [9,]   14    6
[10,]   10   10
[11,]    9   11
[12,]    6   14
> mod1<-glm(y~ldose*sex,dat,family="binomial")
> summary(mod1)
Call:
glm(formula = y ~ ldose * sex, family = "binomial", data = dat)

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-0.94787  -0.36158   0.04914   0.63592   1.56417  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)   2.7634     0.5231   5.282 1.28e-07 ***
ldose        -0.7793     0.1550  -5.028 4.96e-07 ***
sexM          0.7219     0.8477   0.852  0.39444    
ldose:sexM   -0.8085     0.3131  -2.582  0.00981 ** 
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 136.1139  on 11  degrees of freedom
Residual deviance:   6.8938  on  8  degrees of freedom
AIC: 42.794

Number of Fisher Scoring iterations: 4
> anova(mod1,test="Chisq")
Analysis of Deviance Table

Model: binomial, link: logit

Response: y

Terms added sequentially (first to last)


          Df Deviance Resid. Df Resid. Dev P(>|Chi|)
NULL                         11    136.114          
ldose      1  106.323        10     29.791 6.266e-25
sex        1   15.063         9     14.729 1.040e-04
ldose:sex  1    7.835         8      6.894     0.005
> ##plot data
> pr<-expand.grid(sex=levels(dat$sex),ldose=seq(0,5,0.2))
>
pr2<-data.frame(pr,preds=predict(mod1,type="response",newdata=pr))
> 
> mm<-dat[dat$sex=="M",]
> ff<-dat[dat$sex=="F",]
> 
> par(mfrow=c(1,1))
>
plot(mm$numsurv/mm$n~ldose,mm,cex=1.5,cex.axis=1.5,xlab="logDose",ylab="Proportion
> Survived")
> points(ff$numsurv/ff$n~ldose,ff,cex=1.5,col=2)
> 
> ##prediction lines
>
lines(pr2[pr2$sex=="M",]$preds~pr2[pr2$sex=="M",]$ldose)
>
lines(pr2[pr2$sex=="F",]$preds~pr2[pr2$sex=="F",]$ldose,col=2)
> 
> ##lethaldose50line+legend
> abline(h=0.5,lty=2)
> text(0.5,0.48,"Find LD50")
> legend(3.5,1,c("Male","Female"),col=1:2,lty=1,cex=1.5)
> library(MASS)
> dose.p(mod1,c(1,2))
             Dose        SE
p = 0.5: 3.545981 0.3025148
> 
-- 
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Sent from the R help mailing list archive at Nabble.com.

The trick is to fit the model in a form which has the two separate
intercepts and the two separate slopes as the parameters.

You do have to realise that a*b, a*b-1, a/b, a/b-1, ... all specify the
same model, they just use different parameters for the job.  (Yes,
really!)
> dat
   ldose sex numdead
1      0   M       0
2      1   M       3
3      2   M       9
4      3   M      16
5      4   M      18
6      5   M      20
7      0   F       0
8      1   F       2
9      2   F       6
10     3   F      10
11     4   F      11
12     5   F      14
> dat <- transform(dat, Tot = 20)
> fm <- glm(numdead/20 ~ sex/ldose-1, binomial, dat, weights = Tot)
> coef(fm)
      sexF       sexM sexF:ldose sexM:ldose 
-2.7634338 -3.4853625  0.7793144  1.5877754 
> dose.p(fm, c(1,3))           ## females
             Dose        SE
p = 0.5: 3.545981 0.3025148
> dose.p(fm, c(2,4))		 ## males
             Dose        SE
p = 0.5: 2.195123 0.1790317
>  
In fact if you look at the book from which dose.p comes, you will find
an example not unlike this one done this way.  At least I think that's
what I, er, read.

W.

Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessary):  +61 7 3826 7304
Mobile:                         +61 4 8819 4402
Home Phone:                     +61 7 3286 7700
mailto:Bill.Venables at csiro.au
http://www.cmis.csiro.au/bill.venables/ 

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Greme
Sent: Wednesday, 13 February 2008 3:17 AM
To: r-help at r-project.org
Subject: [R] Finding LD50 from an interaction Generalised Linear model


Hi,
I have recently been attempting to find the LD50 from two predicted fits
(For male and females) in a Generalised linear model which models the
effect
of both sex + logdose (and sex*logdose interaction) on proportion
survival
(formula = y ~ ldose * sex, family = "binomial", data = dat (y is the
survival data)). I can obtain the LD50 for females using the dose.p()
command in the MASS library with dose.p(mod1,c(1,2)). However I cannot
find
a way to determine the LD50 of males.
Any help on finding this male LD50 would be appreciated.

Pasting of R workspace below:

> rm(list=ls())
> 
> ##checking file
> dat
   ldose sex numdead
1      0   M       0
2      1   M       3
3      2   M       9
4      3   M      16
5      4   M      18
6      5   M      20
7      0   F       0
8      1   F       2
9      2   F       6
10     3   F      10
11     4   F      11
12     5   F      14
> str(dat)
'data.frame':   12 obs. of  3 variables:
 $ ldose  : int  0 1 2 3 4 5 0 1 2 3 ...
 $ sex    : Factor w/ 2 levels "F","M": 2 2 2 2 2 2 1 1 1 1
...
 $ numdead: int  0 3 9 16 18 20 0 2 6 10 ...
> ##convert numdead to propdead
> dat$propdead<-dat$numdead/20
> ##Calculate survival from dead
> dat$numsurv<-20-dat$numdead
> dat$propsurv<-dat$numsurv/20
> ##check table
> dat
   ldose sex numdead propdead numsurv propsurv
1      0   M       0     0.00      20     1.00
2      1   M       3     0.15      17     0.85
3      2   M       9     0.45      11     0.55
4      3   M      16     0.80       4     0.20
5      4   M      18     0.90       2     0.10
6      5   M      20     1.00       0     0.00
7      0   F       0     0.00      20     1.00
8      1   F       2     0.10      18     0.90
9      2   F       6     0.30      14     0.70
10     3   F      10     0.50      10     0.50
11     4   F      11     0.55       9     0.45
12     5   F      14     0.70       6     0.30
> str(dat)
'data.frame':   12 obs. of  6 variables:
 $ ldose   : int  0 1 2 3 4 5 0 1 2 3 ...
 $ sex     : Factor w/ 2 levels "F","M": 2 2 2 2 2 2 1 1 1 1
...
 $ numdead : int  0 3 9 16 18 20 0 2 6 10 ...
 $ propdead: num  0 0.15 0.45 0.8 0.9 1 0 0.1 0.3 0.5 ...
 $ numsurv : num  20 17 11 4 2 0 20 18 14 10 ...
 $ propsurv: num  1 0.85 0.55 0.2 0.1 0 1 0.9 0.7 0.5
...
> ##load the lattice library
> library(lattice)
> ##plot data with mag + line width 1.5
>
xyplot(propsurv~ldose,groups=sex,data=dat,cex=1.5,lwd=1.5,type="b")
> ##create model
> dat$n<-c(20)
> y<-cbind(dat$numsurv,dat$n-dat$numsurv)
> y
      [,1] [,2]
 [1,]   20    0
 [2,]   17    3
 [3,]   11    9
 [4,]    4   16
 [5,]    2   18
 [6,]    0   20
 [7,]   20    0
 [8,]   18    2
 [9,]   14    6
[10,]   10   10
[11,]    9   11
[12,]    6   14
> mod1<-glm(y~ldose*sex,dat,family="binomial")
> summary(mod1)
Call:
glm(formula = y ~ ldose * sex, family = "binomial", data = dat)

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-0.94787  -0.36158   0.04914   0.63592   1.56417  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)   2.7634     0.5231   5.282 1.28e-07 ***
ldose        -0.7793     0.1550  -5.028 4.96e-07 ***
sexM          0.7219     0.8477   0.852  0.39444    
ldose:sexM   -0.8085     0.3131  -2.582  0.00981 ** 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05
'.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 136.1139  on 11  degrees of freedom
Residual deviance:   6.8938  on  8  degrees of freedom
AIC: 42.794

Number of Fisher Scoring iterations: 4
> anova(mod1,test="Chisq")
Analysis of Deviance Table

Model: binomial, link: logit

Response: y

Terms added sequentially (first to last)


          Df Deviance Resid. Df Resid. Dev P(>|Chi|)
NULL                         11    136.114          
ldose      1  106.323        10     29.791 6.266e-25
sex        1   15.063         9     14.729 1.040e-04
ldose:sex  1    7.835         8      6.894     0.005
> ##plot data
> pr<-expand.grid(sex=levels(dat$sex),ldose=seq(0,5,0.2))
>
pr2<-data.frame(pr,preds=predict(mod1,type="response",newdata=pr))
> 
> mm<-dat[dat$sex=="M",]
> ff<-dat[dat$sex=="F",]
> 
> par(mfrow=c(1,1))
>
plot(mm$numsurv/mm$n~ldose,mm,cex=1.5,cex.axis=1.5,xlab="logDose",ylab="
Proportion
> Survived")
> points(ff$numsurv/ff$n~ldose,ff,cex=1.5,col=2)
> 
> ##prediction lines
>
lines(pr2[pr2$sex=="M",]$preds~pr2[pr2$sex=="M",]$ldose)
>
lines(pr2[pr2$sex=="F",]$preds~pr2[pr2$sex=="F",]$ldose,col=2)
> 
> ##lethaldose50line+legend
> abline(h=0.5,lty=2)
> text(0.5,0.48,"Find LD50")
> legend(3.5,1,c("Male","Female"),col=1:2,lty=1,cex=1.5)
> library(MASS)
> dose.p(mod1,c(1,2))
             Dose        SE
p = 0.5: 3.545981 0.3025148
> 
-- 
View this message in context:
http://www.nabble.com/Finding-LD50-from-an-interaction-Generalised-Linea
r-model-tp15436597p15436597.html
Sent from the R help mailing list archive at Nabble.com.

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