Displaying 6 results from an estimated 6 matches for "numdead".
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2008 Feb 12
1
Finding LD50 from an interaction Generalised Linear model
...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. o...
2006 Aug 21
2
Finney's fiducial confidence intervals of LD50
I am working with Probit regression (I cannot switch to logit) can anybody help me in finding out how to obtain with R Finney's fiducial confidence intervals for the levels of the predictor (Dose) needed to produce a proportion of 50% of responses(LD50, ED50 etc.)?
If the Pearson chi-square goodness-of-fit test is significant (by default), a heterogeneity factor should be used to calculate
2005 Apr 14
1
predict.glm(..., type="response") loses names (was RE: [R] A sugg estion for predict function(s))
...> >>>Ross Darnell
> >>>--
> >>>Email: <r.darnell@uq.edu.au>
> >>>
> Hi Andy
>
> Where?
>
> Try predict.glm example
> ## example from Venables and Ripley (2002, pp. 190-2.)
> ldose <- rep(0:5, 2)
> numdead <- c(1, 4, 9, 13, 18, 20, 0, 2, 6, 10, 12, 16)
> sex <- factor(rep(c("M", "F"), c(6, 6)))
> SF <- cbind(numdead, numalive=20-numdead)
> budworm.lg <- glm(SF ~ sex*ldose, family=binomial)
> ld <- seq(0, 5, 0.1)
> row.nam...
2005 Jan 07
4
glm fit with no intercept
Dear R-help list members,
I am currently trying to fit a generalized linear model using a binomial
with the canonical link. The usual solution is to use the R function glm()
in the package "stats". However, I run into problem when I want to fit a
glm without an intercept. It is indicated that the solution is in changing
the function glm.fit (also in "stats"), by specifying
2006 Oct 18
3
creating bins for a plot
Hi. I'm trying to plot the ratio of used versus unused bird houses
(coded 1 or 0) versus a continuous environmental gradient (proportion of
urban cover [purban2]) that I would like to convert into bins (0 -
0.25, 0.26 - 0.5, 0.51 - 0.75, 0.76 - 1.0) and I'm not having much luck
figuring this out. I ran a logistic regression and purban2 ends up
driving the probability of a box being
2007 Dec 07
1
paradox about the degree of freedom in a logistic regression model
...?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 124.8756 on 11 degrees of freedom
Residual deviance: 4.9937 on 8 degrees of freedom
AIC: 43.104
Number of Fisher Scoring iterations: 4
This is the data set used in regression:
numdead numalive sex ldose
1 1 19 M 0
2 4 16 M 1
3 9 11 M 2
4 13 7 M 3
5 18 2 M 4
6 20 0 M 5
7 0 20 F 0
8 2 18 F 1
9 6 14 F 2
10 10...