similar to: Surveys ans missing values

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),

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 <-

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?

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

Dear WizaRds, having met some of you in person in Vienna, I think even more fondly of this community and hope to continue on this route. It was great talking with you and learning from you. Thank you. I am trying to work through an artificial example in post stratification. This is my dataset: library(survey) age <- data.frame(id=1:8, stratum=rep(

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, Although my R code for the svymean () and svyquantile () functions works fine, I am stuck with the svyby () and make.formula () functions. I got the following error messages. - Error: object of type 'closure' is not subsettable # svyby () - Error in xx[[1]] : subscript out of bounds # make.formula () A reproducible example is appended below. I would appreciate if

R-help, Sorry for this question (I guess it has been addressed before but I could not find it in the archives) but how can I see a function code when the following comes up: > svymean function (x, design, na.rm = FALSE, ...) { .svycheck(design) UseMethod("svymean", design) } <environment: namespace:survey> Thanks in advance

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

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 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

Hello I'm trying to get the proportion "true" for dichotomous variable for various subgroups in a survey. This works fine, but obviously doesn't give proportions directly: svytable(~SurvYear+problem.vandal, seh.dsn, round=TRUE) problem.vandal SurvYear FALSE TRUE 1995 8906 786 1997 17164 2494 1998 17890 1921 1999 18322 1669 2001 17623 2122 ...

Hi everyone, my apologies in advance if I'm overlooking something simple in this question. I am trying to use R's survey package to make a direct method age-adjustment to some complex survey data. I have played with postStratify, calibrate, rake, and simply multiplying the base weights by the correct proportions - nothing seems to hit the published numbers on the nose. I am trying to

Hello, I'm using a complex survey dataset and my goal is to simply spit out a bunch of probability-weighted outcome variable means for the different levels of covariate. So I first define the structure of the study design (I'm using the CDC's NHANES data): dhanes <- svydesign(id=~PSU, strat=~STRATA, weight=~lab_weight, data=final, nest=TRUE) No problem there. Now I use 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

Dear WizaRds, dear Thomas, First of all, I want to tell you how grateful I am for all your support. I wish I will be able to help others along one day the same way you do. Thank you so much. I am struggling with a multistage sampling design: library(survey) multi3 <- data.frame(cluster=c(1,1,1,1 ,2,2,2, 3,3), id=c(1,2,3,4, 1,2,3, 1,2), nl=c(4,4,4,4, 3,3,3, 2,2), Nl=c(100,100,100,100,

Using the survey package I find it is convenient and easy to get estimated proportions using svymean, and their corresponding estimated standard errors. But is there any elegant/simple way to calculate corresponding confidence intervals for those proportions? Of course +/- 1.96 s.e. is a reasonable approximation for a 95% CI, but (incorrectly) assumes symmetrical distribution for a proportion.

Hi all, When I try the following with pkg Survey it returns the error below: ftable(svyby(~INCOME, ~AGECL+RACECL, svymean, design=q50), rownames=list(AGECL=c("<35", "35-44", "45-54", "55-64", "65-74", ">=75"), RACECL=c("white non hispanic", "non white or

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