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)Phyto.nls <- nls(y ~ Ymax/(1 + x/x50),data=Phytopath,start=list

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:

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),

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

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

comments on that?

read this in depth today.

Depending on you purpose, log(dose+0.015) might be

the confidence intervals? I guess I could try several methods, but

how would I go about choosing the right one? Criteria?

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

______________________________________________

R-help at stat.math.ethz.ch mailing list

https://www.stat.math.ethz.ch/mailman/listinfo/r-help

;-)

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 mycelialindependent 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?

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 theamount 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 toopercentage

increase

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

comments on that?

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 willlog(dose) is not acceptable in a model like this. You need a

model like

Peter suggested.

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) ? How would this affect the LD50 value calculated andanswer your

question. If not, perhaps this solution will help you find a better

solution.

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:

p = 14: -2.451018 0.04858572

OK, I will need to try this (later today). I don't see "dose.p" innow 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 SEp = 14: -2.451018 0.04858572

exp(LD50P100p[1,1]+c(-2,0,2)*LD50P100p[1,2])-0.015

[1] 0.06322317 0.07120579 0.08000303this?

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

______________________________________________

R-help at stat.math.ethz.ch mailing list

https://www.stat.math.ethz.ch/mailman/listinfo/r-help