#STATA GLM HOW TO#

Stata easily throws a ton of info in your face, without you knowing what it means or how it is derived. I would also suggest that, when learning multiple regression with R, you start with simpler examples and gradually build up your understanding of the methods involved.ДЄfter all, my opinion is that the philosophies of Stata and R differ. Your example is a simple dataset, so you don't need to use the survey package at all. Svyglm(lenses~carrot,Dd,family=quasipoisson(log))

Xi: glm lenses i.carrot, fam(poisson) link(log) nolog robust eform I'd also like to compare Stata and R estimates.I'm using Poisson as suggested here.

#STATA GLM HOW TO#

Could anyone show me how to do this in R (my data has clusters and strata but I've taken a simpler example)? I've tried the relrisk.est function but I'd rather use the survey package as this handles very complex designs. (Intercept) age factor(vituse)2 factor(vituse)3Дegrees of Freedom: 314 Total (i.e.Could anyone help me replicate these relative risk calculations (and its confidence interval) in R?ДЄ similar procedure used in Stata is described here. You should probably use a slightly more complex model since vituse is obviously a three level factor: > mod2 mod2Д¬all: glm(formula = betaplasma ~ age + factor(vituse), family = quasipoisson, > glm(betaplasma ~ age + vituse, family=gaussian(link='log'), data=data2,Д¬all: glm(formula = betaplasma ~ age + vituse, family = gaussian(link = 'log'), If you agree (or rather if the data agrees) then consider this construction: ?familyД¬all: glm(formula = betaplasma ~ age + vituse, family = quasipoisson,Дegrees of Freedom: 314 Total (i.e. I would like to suggest that it may be the errors that could be non-normal. Two other points: (1) it's a bit odd that there's a response with a value of 0 in this data set - not impossible, but odd (2) it looks like vituse should be treated as a factor rather than as numeric ('1=Yes, fairly often, 2=Yes, not often, 3=No') - possibly ordinal.

That said, fitting this particular dataset doesn't seem too hard. 2131874 25.65 0.000 5.049738 5.885417ДЄs I said in my comment, it's probably true that Stata has more robust (in the numerical, not the statistical sense) GLM fitting than R. With this dataset loaded, I run the following model: Could anyone tell me why this is the case, or what I do wrong, or why the suggested models from the book might not be appropriate? I'd appreciate any help, thanks!Д®dit: As an example which reproduces the error consider the dataset which can be downloaded here. I tried many different models, and different datasets, but the problem is always the same (unless I only include one single independent variable). Picking other values the message is sometimes the same, but more often I get: Error in glm.fit(x = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, :ДЄs I said, in STATA I can run these models without setting starting values and without errors. When I set them all equal to zero, I always get the message that the algorithm did not converge. However, when calling the same model using R's glm-function, but specifying family=gaussian(link='log') I am asked to provide starting values. Indeed, I have no problems in STATA with the log link. I am trying to replicate in R what they do in the book in STATA. Of course they also suggest residual plots to check whether a 'normal' linear model using an identity link can still be used. The authors suggest that instead of OLS models, 'the log link is generally used for response data that take only positive values on the continuous scale'.

I am using the book 'Generalized Linear Models and Extension' by Hardin and Hilbe (second edition, 2007) at the moment.

Stata glm