similar to: survey package: proportion estimate confidence intervals using svymean

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

Estimados usuarios: Estoy intentando reproducir el ejemplo 6.4 de Thomas Lumley. Complex Survey. Editorial Wiley. 2010 (ver la página en google:

Dear list colleagues, I'm trying to come up with a test question for undergraduates to illustrate comparison of means from a complex survey design. The data for the example looks roughly like this: mytest<-data.frame(harper=rnorm(500, mean=60, sd=1), party=sample(c("BQ", "NDP", "Conservative", "Liberal", "None", NA), size=500,

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

Hi, is there a way to analyze subpopulations (e.g. women over 50, those who answered "yes" to a particular question) in a survey using Survey package? Other packages (e.g. Stata, SUDAAN) do this with a subpopulation option to identify the subpopulation for which the analysis shoud be done. I did not see this option in the Survey package. Is there another way to do this?

Howdy, Referencing the below exchange: https://stat.ethz.ch/pipermail/r-help/2006-April/103862.html I am still getting the same warning ("non-integer #successes in a binomial glm!") when using svyglm:::survey. Using the API data: library(survey) data(api) #stratified sample dstrat<-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc)

I have been asked to look at options for doing relative risk regression on some survey data. I have a binary DV and several predictor / adjustment variables. In R, would this be as "simple" as using the survey package to set up an appropriate design object and then running svyglm with family=binomial(log) ? Any other suggestions for covariate adjustment of relative risk

Hello, I'm running R 2.14.1 on OS X (x86_64-apple-darwin9.8.0/x86_64 (64-bit)), with version 3.28 of Thomas Lumley's survey package. I was using predict() from svyglm(). E.g.: data(api) dstrat<-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc) out <- svyglm(sch.wide~ell+mobility, design=dstrat, family=quasibinomial()) pred.df <-

Dear WizaRds, Working through sampling theory, I tried to comprehend the concept of stratification and apply it with Survey to a small example. My question is more of theoretic nature, so I apologize if this does not fully fit this board's intention, but I have come to a complete stop in my efforts and need an expert to help me along. Please help: age<-matrix(c(rep(1,5), rep(2,3),

Hi, I have been experimenting with the new Survey package. Specifically, I was trying to use some of the functions on the public-use survey data from NHIS (2000 Sample Adult file). Error 1): The first error I get is when I try to specify the complex survey design. nhis.design<-svydesign(ids=~psu, probs=~probs, strata=~strata, data=nhis.df, check.strata=TRUE) Error in svydesign(ids =

Hi R users, I am trying to use the aggregate function with a survey design object and survey functions, but get the following error. I think I am incorrectly using the syntax somehow, and it may not be possible to access variables directly by name in a survey-design object. Am I right? How do I fix this problem? I have used aggregate with "mean" and "weighted.mean", and

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

Dear list, I got some surprising results when using the svytotal routine from the survey package with data containing missing values. Some example code demonstrating the behaviour is included below. I have a stratified sampling design where I want to estimate the total income. In some strata some of the incomes are missing. I want to ignore these missing incomes. I would have expected that

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()

Hi, I am trying to fit a model with an ordered response variable (3 levels) and 13 predictor variables. The sample has complex survey design and I've used 'svydesign' command from the survey package to specify the sampling design. After reading the manual of 'svyglm' command, I've found that you can fit a logistic regression (binary response variable) by specifying the

Good morning! I have 4 questions which trouble me: 1. I want to test the hypothesis that the 2 proportions (the mean of a binomial) which come from 2 different samples are equal. I want to use the following function z= (p1-p2)/ sqrt((p1(1-p1)/n1)+(p2(1-p2)/n2)) which is one of the standard formulas for this case. Is there such a function in R? p1=the proportion from the first sample n1=the

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 <-

I have the following code which produces the output below it clus1 <- svydesign(ids = ~schid, data = lower_dat) items <- as.formula(paste(" ~ ", paste(lset, collapse= "+"))) rr1 <- svymean(items, clus1, deff='replace', na.rm=TRUE) > rr1 mean SE DEff W525209 0.719748 0.015606 2.4932 W525223 0.508228 0.027570 6.2802 W525035 0.827202

Dear List: I am working to understand some differences between the results of the svymean() function in the survey package and from code I have written myself. The results from svymean() also agree with results I get from SAS proc surveymeans, so, this suggests I am misunderstanding something. I am never comfortable with "I did what the software" does mentality, so I am working to

Hi, I'm trying to get standard errors for some of the variables in my data frame. One of the questions on my survey is whether faculty coordinate across curriculum to include Arts Education as subject matter. All the responses are coded in zeros and ones obviously. For some of the other variables I have a 2 for those that responded with "Don't Know". I'm getting NA for