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.
About all I can think of is to transform onto some scale like the logit scale, use the delta method to estimate standard errors, and transform back for appropriate confidence intervals. Or perhaps use the machinery of the survey package svyglm function with logit link (family=quasibinomial) to do the calculations for me. Still, seems rather cumbersome. Is there a better solution? Thanks for any suggestions or pointing out the errors in my logic! Peter Holck ______________________________________________ R-help@r-project.org mailing list 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.