Hello, I also tried with ``` library(MASS) > dose.p(model,p=.95) Dose SE p = 0.95: 1.70912 96.26511 ``` which is closer to the expected 1.72 but with a very large error (I expected 1.10-2.34). Is this regression correct?
On Sat, Oct 2, 2021 at 10:14 AM Luigi Marongiu <marongiu.lu...@gmail.com> wrote: > > Hello, > I have set a glm model using probit. I would like to use it to predict > X given Y. I have followed this example: > ``` > f2<-data.frame(age=c(10,20,30),weight=c(100,200,300)) > f3<-data.frame(age=c(15,25)) > f4<-data.frame(age=18) > mod<-lm(weight~age,data=f2) > > predict(mod,f3) > 1 > 150 > > predict(mod,f4) > 1 > 180 > ``` > > I have set the following: > ``` > df <- data.frame(concentration = c(1, 10, 100, 1000, 10000), > positivity = c(0.86, 1, 1, 1, 1)) > model <- glm(positivity~concentration,family = binomial(link = > "logit"), data=df) > > e3<-data.frame(concentration=c(11, 101), positivity=c(1, 1)) > > predict(model, e3) > 1 2 > 5.645045 46.727573 > ``` > but: > ``` > > e4<-data.frame(positivity=0.95) > > e4 > positivity > 1 0.95 > > predict(model, e4) > Error in eval(predvars, data, env) : object 'concentration' not found > ``` > Why did the thing worked for f4 but not e4? How do I get X given Y? > Do I need to find the inverse function of logit (which one?) and apply > this to the regression or is there a simpler method? > Also, is it possible to plot the model to get a smooter line than > `plot(positivity ~ concentration, data = df, log = "x", type="o")`? > Thanks > -- > Best regards, > Luigi -- Best regards, Luigi ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.