similar to: LD50

Hola, saludos desde México. Me estoy iniciando en el uso de R, que me parece un muy buen programa. Yo doy la clase de bioestadística en la facultad de Medicina de Mérida, Yucatán en México. Me gustaría hacer contacto con personas que estén usando R en el área biomédica, especialmente en docencia, para compartir experiencias, dudas, ejemplos, etc. Saludos José Arturo [[alternative HTML version

Dear members in the group. I am trying to start using R. The unix systme we have in my university is solaris 7. Is there a version of R for Solaris 7?. If afirmative where could I get it? Thanks in advance. Jose A. Farfan -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

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Hi everyone: when I try to run dovecot after modifying the dovecot.conf file I have the following error messages: Mar 23 09:56:52 mailer dovecot: [ID 583609 mail.info] Dovecot v1.2.10 starting up Mar 23 09:56:52 mailer dovecot: [ID 583609 mail.info] Generating Diffie-Hellman parameters for the first time. This may take a while.. Mar 23 09:56:52 mailer dovecot: [ID 583609 mail.error]

Hi everybody any one knows, how i could edit dovecot to assign user quotas ? I have now configured my dovecot.conf on this way: protocol imap { listen = *:143 mail_plugins = quota imap_quota } protocol pop3 { listen = *:110 mail_plugins = quota } plugin { quota = fs:INBOX:mount=/

Hello, upgrade Dovecot from v1.2.10 to v2.0.8, after build a new configuration file following instructions of http://wiki2.dovecot.org/Upgrading/2.0 run dovecot I get this: auth: Error: Fatal: Dovecot version mismatch: Master is v2.0.8, dovecot-auth is v1.2.10 (if you don't care, set version_ignore=yes) log: Error: service(auth): child 3389 returned error 89 (Fatal failure) master: Error:

Dear R users, I'm a little lost on how to define pmodels for the DRC package. My goals are to produce isoboles of binary toxicity data. any tips? I really just need to know what pmodels refers to. Cheers, Pat -- View this message in context: http://r.789695.n4.nabble.com/pmodels-in-DRC-tp4190567p4190567.html Sent from the R help mailing list archive at Nabble.com.

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

Hello! If I have a matrix as 1 2 2 3 and I want to change the value 2 in 0, what can I do? Thank you -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !)

thanks a lot Renaud. but i was interested in Finney's fiducial confidence intervals of LD50 so to obtain comparable results with SPSS. But your reply leads me to the next question: does anybody know what is the best method (asymptotic, bootstrap etc.) for calculating confidence intervals of LD50? i could "get rid" of Finney's fiducial confidence intervals but

At a snail's pace I keep on translating an introduction to R into italian; I have reached the section describing the glm() function, in which some example code is presented. The very last line of code, before the beginning of the section on Poisson models is: ldp <- ld50(coef(fmp)); ldl <- ld50(coef(fmp)); c(ldp, ldl) which of course gives results 43.663 and 43.663; the correct code

At a snail's pace I keep on translating an introduction to R into italian; I have reached the section describing the glm() function, in which some example code is presented. The very last line of code, before the beginning of the section on Poisson models is: ldp <- ld50(coef(fmp)); ldl <- ld50(coef(fmp)); c(ldp, ldl) which of course gives results 43.663 and 43.663; the correct code

Classification: UNCLASSIFIED Caveats: NONE A similar question has been posted in the past but never answered. My question is this: for probit analysis, how do you program a 95% confidence interval for the LD50 (or LC50, ec50, etc.), including a heterogeneity factor as written about in "Probit Analysis" by Finney(1971)? The heterogeneity factor comes into play through the chi-squared

Hi All! I am desperately needing some help figuring out how to calculate LD50 with a GLMM (probit link) or, more importantly, the standard error of the LD50. I conducted a cold temperature experiment and am trying to assess after how long 50% of the insects had died (I had 3 different instars (non significant fixed effect) and several different blocks (I did 4 replicates at a time)=

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

Respected Sir/Madam, I have a question regarding calculation of LD50 (Lethal Dose) and IC50 (50% inhibitory concentration) of an antimicrobial experiment. I have used a compound isolated from a plant and observed its effect on the fungus *Fusarium oxysporum* by the food poisoning method. Solutions of the compound at concentrations of 0, 50, 100, 150, 200 and 250µg/ ml were added to

Estimados En ocasionas hay informaciones epidemiológicas en reportes pdf semanales como el que adjunto que quisiéramos llevar a csv o txt USANDO R para poder analizarlas estadísticamente. Apreciaríamos su ayuda si nos diesen un script, el paquete pdftable no me resultó. Saludos José -- Este mensaje le ha llegado mediante el servicio de correo electronico que ofrece Infomed para respaldar

Dear R-list, I have a data set from 20 pigs, each of which is tested at crossed 9 doses (logdose -4:4) and 3 skin treatment substances when exposed to a standard polluted environment. So there are 27 patches on each pig. The response is irritation=yes/no. I want to determine "equally effective 50% doses" (similar to old LD50), and to test the treatments against each other. I am looking

Dear R experts. I might be missing something obvious. I have been trying to fix this problem for some weeks. Please help. #data ped <- c(rep(1, 4), rep(2, 3), rep(3, 3)) y <- rnorm(10, 8, 2) # variable set 1 M1a <- sample (c(1, 2,3), 10, replace= T) M1b <- sample (c(1, 2,3), 10, replace= T) M1aP1 <- sample (c(1, 2,3), 10, replace= T) M1bP2 <- sample (c(1, 2,3), 10, replace= T)

Dear Members, I have two estimated transition matrices and I want to compare them. In fact I want to check the hypothesis if they come from the same process. I tried to look for some test but all I found was independence test of contingency tables. The following code shows that the usual chi-squared test statistic does not follow chisq distribution. MCRepl <- 5000 khi12 <- rep(0,MCRepl)