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Re: st: Survey standard errors of grouped data proportions


From   Stas Kolenikov <skolenik@gmail.com>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   Re: st: Survey standard errors of grouped data proportions
Date   Mon, 16 Dec 2013 09:06:54 -0500

After -svy: total-, you can use -lincom- and -nlcom- to get the right
statistics and the right standard errors:

svy : total  weight , over(region, nolab)
local total (_b[weight:1] + _b[weight:2] + _b[weight:3] + _b[weight:4])
lincom `total'
nlcom _b[weight:1]/ `total'


-- Stas Kolenikov, PhD, PStat (ASA, SSC)
-- Senior Survey Statistician, Abt SRBI
-- Opinions stated in this email are mine only, and do not reflect the
position of my employer
-- http://stas.kolenikov.name



On Mon, Dec 16, 2013 at 1:57 AM, Philip Burgess
<filip.liljeholm@gmail.com> wrote:
> I am using Stata 13.1 SE on a Windows 7 platform.
>
>
> I have complex survey design data set, with replicate weights, etc and have
> successfully svyset using the statement
>
>
> svyset   _n [pweight=3Dwgt], vce(linearized) singleunit(missing)
>
>
> I have two variables of interest, patient visits (continuous) and patient
> severity group (5 mutually exclusive categories).
>
>
> To estimate the average number of patient visits, I simply run:
>
> svy: mean visits, over(group)
>
> This gives weighted means and appropriate standard errors (and thus the CI)
>
>
>
> What I next want to estimate in the relative proportion of all patient
> visits in each of the five groups.
>
> For example, the most severe patient group accounts for approximately 60%
> of all patient visits.
>
> I can estimate the overall total of patient visits per group using the
> command:
>
> svy: total visits, over(group)
>
> and from that can work out the point estimate of the relative proportion
> for each of the 5 groups.
>
> But I cannot figure out how to get the SE of the relative proportion.
>
>
> This is probably incredibly simple but I have had no luck searching the
> FAQ, the manual or the list archives.
>
> Thanks;
>
> Philip
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