[R] Finding LD50 from an interaction Generalised Linear model

From: Greme <boa04gmd_at_shef.ac.uk>
Date: Tue, 12 Feb 2008 09:17:20 -0800 (PST)

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-Linear-model-tp15436597p15436597.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help_at_r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Received on Tue 12 Feb 2008 - 17:27:55 GMT

Archive maintained by Robert King, hosted by the discipline of statistics at the University of Newcastle, Australia.
Archive generated by hypermail 2.2.0, at Wed 13 Feb 2008 - 01:30:13 GMT.

Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help. Please read the posting guide before posting to the list.

list of date sections of archive