FAQ
Vincent:

Here is a simple solution using Prof. Bates' non-linear least squares
algorithm:

Best,
Ravi.
Phytopath <- data.frame(x=c(0, 0.03, 0.1), y=c(28, 21, 11))
Phyto.nls <- nls(y ~ Ymax/(1 + x/x50),data=Phytopath,start=list
(Ymax .0,x50=0.01),trace=T)
404.3058 : 20.00 0.01
15.76932 : 27.96313636 0.04960484
2.043625 : 28.2145584 0.0694645
1.851401 : 28.33886844 0.07198951
1.851231 : 28.34892493 0.07185953
1.851230 : 28.34843670 0.07186804
1.851230 : 28.3484688 0.0718675
summary(Phyto.nls)
Formula: y ~ Ymax/(1 + x/x50)

Parameters:
Estimate Std. Error t value Pr(>|t|)
Ymax 28.34847 1.31522 21.554 0.0295 *
x50 0.07187 0.01348 5.332 0.1180
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

Residual standard error: 1.361 on 1 degrees of freedom

Correlation of Parameter Estimates:
Ymax
x50 -0.6001

----- Original Message -----
From: Vincent Philion <vincent.philion@irda.qc.ca>
Date: Friday, July 25, 2003 9:25 am
Subject: Re: [R] inverse prediction and Poisson regression
Hi, ... and good morning!

;-)

On 2003-07-25 08:43:35 -0400 Spencer Graves
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!!!
;-)

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 and that I should simply look at percentage... Any

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

Depending on you purpose, log(dose+0.015) might be
sufficiently close to a model like what Peter suggested to
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) ? How would this affect the LD50 value calculated and
the confidence intervals? 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,

--
Vincent Philion, M.Sc. agr.
Phytopathologiste
Institut de Recherche et de D?veloppement en Agroenvironnement (IRDA)
3300 Sicotte, St-Hyacinthe
Qu?bec
J2S 7B8

t?l?phone: 450-778-6522 poste 233
courriel: vincent.philion at irda.qc.ca
Site internet : www.irda.qc.ca

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