similar to: Survey proportions... Can I use population as denominator?

Hello, I want to specify a linear regression model in which the metric outcome is predicted by two factors and their interaction. glm() computes effects for each factor level and the levels of the interaction. In the case of singularities glm() displays "NA" for the corresponding coefficients. However, svyglm() aborts with an error message. Is there a possibility that svyglm()

Is there a way to use one of the functions in the survey package to get a table of estimated percentages (or proportions) and the standard error for each? For example, suppose that AGECODE AND SEX are two factors with 5 and 2 levels. The 5x2 AGECODE x SEX table would have the estimated percentage of persons in each cell, 100*(sum of weights in the cell) / (sum of all weights) and the std

I'm trying to use the survey package to calculate a risk difference with confidence interval for binge drinking between sexes. Variables are X_RFBING2 (Yes, No) and SEX. Both are factors. I can get the group prevalences easily enough with result <- svyby(~X_RFBING2, ~SEX, la04.svy, svymean, na.rm = TRUE) and then extract components from the svyby object with SE() and coef() to do the

Hi everyone, apologies if the answer to this is in an obvious place. I've been searching for about a day and haven't found anything.. I'm trying to replicate Stata's confidence intervals in R with the survey package, and the numbers are very very close but not exact. My ultimate goal is to replicate Berkeley's SDA website with R (http://sda.berkeley.edu/), which seems to

How do I retrieve the replicates estimates from a crosstab done using svyby? Here is an example from the help page for svyby in the package: > data(api) > dclus1<-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) > rclus1<-as.svrepdesign(dclus1) > > a <- svyby(~api99, ~stype, rclus1, svymean, return.replicates=TRUE) > a$replicates NULL But, compare to > b

Hi, I am trying to write a function to compute many cross-tabulations with the -svytable- command. Here is a simplified example of the structure of my code (adapted from the -svytable- help file): data(api) func.example<-function(variable){ dclus1<-svydesign(id=~1, weights=~pw,data=apiclus1, fpc=~fpc) svytable(~ variable, dclus1) } When I call this function with:

Dear WizaRds, I am struggling to compute correctly a cluster sampling design. I want to do one stage clustering with different parametric changes: Let M be the total number of clusters in the population, and m the number sampled. Let N be the total of elements in the population and n the number sampled. y are the values sampled. This is my example data: clus1 <-

svymean w/ svyby is working for me... > svyby(~visitcnt, ~agegrp3.f, svymean, design=svydes) agegrp3.f visitcnt se.visitcnt 18-44 18-44 8.755552 0.4953235 45-64 45-64 10.131555 0.5347806 65+ 65+ 9.588802 0.4323629 svyquantile is working for me... > svyquantile(~visitcnt, quantiles=c(.25, .5, .75), ties="rounded", design=svydes) 0.25

I am having trouble understanding (i.e. getting) confidence intervals from the survey package. I am using R version 2.8.1 (2008-12-22) and survey package (3.11-2) on FC7 linux. To simplify my question I use an example from that package: R> data(api) R> dclus1<-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) R> (tst <- svyby(~api99, ~stype,

I am facing a problem with a function in survey package. The function svyvar gives the estimated population variance from a given sampling scheme. I am working with a data having more than four continuous variables. In order to have have population total for all those cont. variables I have written in the following format svyvar(~var1+var2+var3+var4+var5+var6,data) ; var1,var2,...,var6 are 6

Should the svyby function be able to work with svyquantile? I get the error below ... data(api) dclus1<-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) svyby(~api00, design=dclus1, by = ~stype, quantiles=c(.25,.5,.75), FUN=svyquantile, na.rm=T ) > Error in object$coefficients : $ operator is invalid for atomic vectors A

Hi R users, I'm using the survey package to calculate summary statistics for a large health survey (the Demographic and Health Survey for Honduras, 2006), and when I try to calculate the variances for several variables, I get negative numbers. I thought it may be my data, so I ran the example on the help page: data(api) ## one-stage cluster sample dclus1<-svydesign(id=~dnum, weights=~pw,

I have survey data that I am working on. I need to make some multi-way tables and regression analyses on the data. After attaching the data, this is the code I use for tables for four variables (sweight is the weight variable): > a <- xtabs(sweight~research.area + gender + a2n2 + age) > tmp <- ftable(a) Is this correct? I don't think I need to use the strata and cluster

Good afternoon! I'm trying to use the Survey package for a stratified sample which has 4 criteria on which the stratification is based. I would like to get the corrected weights and for every element i get a weight of 1 E.g: tipping design <- svydesign (id=~1, strata= ~regiune + size_loc + age_rec_hhh + size_hh, data= tabel) and then weights(design) gives

Hello, svyvar from the survey package computes variances (with standard errors) from survey design objects. Is there any way to compute standard deviations and their standard errors in a similar manner? Thanks a lot, Sebastian

Hi all, I want to know how to perform the test Breslow-Day test for homogeneity of odds ratios (OR) stratified for svytable. This test is obtained with the following code: epi.2by2 (dat = daty, method = "case.control" conf.level = 0.95, units = 100, homogeneity = "breslow.day", verbose = TRUE) where "daty" is the object type table svytable consider it, but

hello I need to know the dfn and dfd of my Anova. But in the Anova output there is only "Df". Is this the dfn or the dfd? and how do I get both of it in R? Thanks for any answers -- View this message in context: http://r.789695.n4.nabble.com/df-of-numerator-and-denominator-tp3765526p3765526.html Sent from the R help mailing list archive at Nabble.com.

Hi friends - I'm plotting a ratio of bicarbonates i ggplot2 and ylab(expression(paste(frac("additive BIC","true BIC")))) worked OK - but now I have been asked to put the chemistry instead - so I wrote ?ylab(expression(paste(frac("additive",HCO[3]^"-","true",HCO[3]^"-")))) - and frac saw that as additive = numerator and HCO3- =

?s 10:01 de 02/02/2024, Troels Ring escreveu: > Hi friends - I'm plotting a ratio of bicarbonates i ggplot2 and > > ylab(expression(paste(frac("additive BIC","true BIC")))) worked OK - but > now I have been asked to put the chemistry instead - so I wrote > >

Every day I get a csv file containing the names of the 64 schools in our county, the number of students sent home ill, and the number of students absent (plus lots of other variables). The file is cumulative since fall of 2009. It is in "long" format: one line per school per day. Each line is also supposed to contain the total number of students enrolled in the school. That number