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?