Re: svyby and svyratio in the Thomas Lumley's survey package
On Thu, 23 Feb 2006, Smith, Phil wrote:
> Dear R-ers:
> I'm using Thomas Lumley's "survey" package.
> I'd like to compute survey ratio estimates (numerator=~utd ,
> denominator=~one) for each of serval domains using by=~factor(domain).
> I can't quite work out the syntax for the call to the "svyby" function. I try:
> svyby( numerator=~ utd, denominator=~one , by=~ factor(domain) ,
> design=nis , svyratio )
> and I get an error that says that "domain" is not found. (nis is the
> design object)
Hmm. I don't get the same error message (perhaps a question of versions?),
but there is a problem. svyby is not really set up to work with svyratio.
I will try to fix this, but there are simple workarounds.
If you do
result <- svyby(~api.stu,by=~factor(stype), denominator=~enroll,
(ie, don't name the numerator argument) then the computations are done as
you want. The result does not print correctly, because svyby() returns
some extra attributes that print() doesn't know how to handle.
You can extract the results you need with coef() or just remove the
Error in unlist(x, recursive, use.names) :
argument not a list
E 0.8518163 4.945408e-05
H 0.8105702 0.0004193180
M 0.8356958 0.0003307829
factor(stype) statistic.ratio statistic.var SE
E E 0.8518163 4.945408e-05 0.00703236
H H 0.8105702 0.0004193180 0.02047726
M M 0.8356958 0.0003307829 0.01818744
Finally, the name of the denominator variable looks suspicious. If `one'
is a variable whose values are all 1 and you just want domain means you
can do the equivalent of
to get the mean of api00 within each value of stype. You don't need
svyratio(). In fact, part of the test suite for the survey package is to
check that svyratio, svymean, and svyglm give the same answers for domain