search for: svycontrast

...;)) # Baja un fichero de 1.5M m <- svyglm( intuse ~ I(age-18)*sex , design = shs, family = quasibinomial()) m2 <- svyglm( intuse ~ (pmin(age, 35) + pmax(age, 35))*sex , design = shs, family = quasibinomial()) summary(m) # OK. Coincide con el libro summary(m2) # OK. Coincide con el libro svycontrast(m2, quote('pmin(age, 35)' +'pmin(age, 35):sexfemale' )) # ¡Falla! ## Error en deriv.default(expr, names(datalist)) : ## expresión inválida en 'FindSubexprs' ## sessionInfo() # Mi configuración ## R version 2.12.0 (2010-10-15) ## Platform: i486-pc-linux-gnu (32-bit) ## l...

Dear list colleagues, I'm trying to come up with a test question for undergraduates to illustrate comparison of means from a complex survey design. The data for the example looks roughly like this: mytest<-data.frame(harper=rnorm(500, mean=60, sd=1), party=sample(c("BQ", "NDP", "Conservative", "Liberal", "None", NA), size=500,

Hello all, This is something that I am sure has a really suave solution in R, but I can't quite figure out the best (or even a basic) way to do it. I have a simple linear regression that is fit with lm for which I would like to estimate the x intercept with some measure of error around it (confidence interval). In biology, there is the concept of a developmental zero - a temperature under

Using the survey package I find it is convenient and easy to get estimated proportions using svymean, and their corresponding estimated standard errors. But is there any elegant/simple way to calculate corresponding confidence intervals for those proportions? Of course +/- 1.96 s.e. is a reasonable approximation for a 95% CI, but (incorrectly) assumes symmetrical distribution for a proportion.

Hello, svyvar from the survey package computes variances (with standard errors) from survey design objects. Is there any way to compute standard deviations and their standard errors in a similar manner? Thanks a lot, Sebastian

I'm trying to use the survey package to calculate a risk difference with confidence interval for binge drinking between sexes. Variables are X_RFBING2 (Yes, No) and SEX. Both are factors. I can get the group prevalences easily enough with result <- svyby(~X_RFBING2, ~SEX, la04.svy, svymean, na.rm = TRUE) and then extract components from the svyby object with SE() and coef() to do the