Open leeper opened 4 years ago
Is it sufficiently clear that mm() returns domain estimates rather than SEs based on subsetting the data?
x <- structure(list(level = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("John", "Kate"), class = "factor"), outcome = c(0L, 0L, 1L, 1L, 0L, 0L, 1L, 1L), weight = c(1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L)), row.names = 1:8, class = "data.frame") # what people might be expecting with(subset(x, level == "John"), sqrt(sum((outcome - mean(outcome))^2)/3/4)) svymean(~outcome, svydesign(ids = ~1, weights = ~ 1, data = subset(x, level == "John"))) # what is actually returned (all are equivalent) ## mm() mm(x, outcome ~ level) ## unweighted data, subset to John svymean(~outcome, subset(svydesign(ids = ~1, weights = ~ 1, data = x), level == "John")) ## weighted data (Kate weight == 0), subset to John svymean(~outcome, subset(svydesign(ids = ~1, weights = ~ weight, data = x), level == "John")) ## weighted data (Kate weight == 0), full data frame svymean(~outcome, svydesign(ids = ~1, weights = ~ weight, data = x))
[ ] Document this better, pointing to vignette: https://cran.r-project.org/web/packages/survey/vignettes/domain.pdf [ ] Add option to not calculate variances as if subsets are random samples of population?
Is it sufficiently clear that mm() returns domain estimates rather than SEs based on subsetting the data?
[ ] Document this better, pointing to vignette: https://cran.r-project.org/web/packages/survey/vignettes/domain.pdf [ ] Add option to not calculate variances as if subsets are random samples of population?