If we adopt Ripley's notation, this is mu(dose) = ymax/(1+dose/dose50), where ymax and dose50 are to be estimated. We can get this, sort of, using "glm", as follows: This model is the same as log(mu(dose)) = ymax - log(1+dose/dose50). I have not tried this, but I believe that for fixed dose50 = 0.3, this model can be written for glm with the toy example I used before as follows:
Phytopath <- data.frame(x=c(0, 0.03, 0.1),
y=c(28, 21, 11))
fitP.3 <- glm(y~offset(-log(1+x/0.3)), data=Phytopath[rep(1:3, 100),],
family="poisson")
fitP.3$deviance
[1] 359.5893Now, replace 0.3 by other possible values for dose50 to find the minimum "deviance". Then get a 95% confidence interval using "profile likelihood" to find the points above and below the minimium where the "deviance" exceeds the minimum by qchisq(0.95, 1) = 3.841459. There are better ways to do this, but none that I can make work in the next 2 minutes. When I did this, I get the following:
> fitP.06 <- glm(y~offset(-log(1+x/0.06)), data=Phytopath[rep(1:3, 100),],
+ family="poisson")
> fitP.06$deviance
[1] 16.54977
> fitP.065 <- glm(y~offset(-log(1+x/0.065)), data=Phytopath[rep(1:3, 100),],
+ family="poisson")
> fitP.065$deviance
[1] 11.59525
> fitP.07 <- glm(y~offset(-log(1+x/0.07)), data=Phytopath[rep(1:3, 100),],
+ family="poisson")
> fitP.07$deviance
[1] 10.96356
> fitP.075 <- glm(y~offset(-log(1+x/0.075)), data=Phytopath[rep(1:3, 100),],
+ family="poisson")
> fitP.075$deviance
[1] 13.49366
> fitP.08 <- glm(y~offset(-log(1+x/0.08)), data=Phytopath[rep(1:3, 100),],
+ family="poisson")
> fitP.08$deviance
[1] 18.34463
From this, I get an approximate maximum likelihood estimate for dose50 of 0.07 with a 95% confidence interval by rough interpolation of 0.061 to 0.076.
hope this helps. spencer graves
p.s. Have you plotted y vs. x with appropriate transformations? If not, I suggest you do that before anything else. Doug Bates says that students in his classes who don't plot the data get instant F's. When I don't bother plotting the data several different ways, I often do stupid things.
Vincent Philion wrote:
Hi, ... and good morning!
;-)
On 2003-07-25 08:43:35 -0400 Spencer Graves <[EMAIL PROTECTED]> wrote:
The Poisson assumption means that Y is a number of independent events from a theoretically infinite population occurring in a specific time or place. The function "glm" with 'family="poisson"' with the default link = "log" assumes that the logarithm of the mean of Y is a linear model in the explanatory variable.
OK, I think my data can fit that description.
How is Y measured?
Y is the number of line intercepts which encounters mycelial growth.
i/e if mycelia intercepts the line twice, 2 is reported. This follows poisson.
If it the number out N, with N approximately 500 (and you know N),
then you have a logistic regression situation.
No, 500 spores can grow, but there is no "real" limit on the amount of
growth possible, and so no limit on the number of intercepts. So this is why I adopted Poisson, not knowing how complicated my life would become!!!
;-)and that I should simply look at percentage... Any comments on that?
In that case, section 7.2 in
Venables and Ripley (2002) should do what you want. If Y is a percentage increase
... But you may be right, that I'm making this just too complicated
How would this affect the LD50 value calculated and the confidence intervals?
When dose = 0, log(dose) = (-Inf). Since 0 is a legitimate dose, log(dose) is not acceptable in a model like this. You need a model like Peter suggested.
OK, I see I will need stronger coffee to tackle this, but I will read this in depth today.
Depending on you purpose, log(dose+0.015) might be
sufficiently close to a model like what Peter suggested to answer your question. If not, perhaps this solution will help you find a better solution.
In other words, "cheat" and model Y_0 with a "small" value = log(0.015) ?
I guess I could try several methods, but how would I go about choosing the
right one? Criteria?
I previously was able to get dose.p to work in R, and I just now was able to compute from its output. The following worked in both S-Plus 6.1 and R 1.7.1:
LD50P100p <- print(LD50P100)
Dose SE p = 14: -2.451018 0.04858572
exp(LD50P100p[1,1]+c(-2,0,2)*LD50P100p[1,2])-0.015
[1] 0.06322317 0.07120579 0.08000303
OK, I will need to try this (later today). I don't see "dose.p" in this?
again, many thanks,
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