Using the srvyr package and replicate weights REPWT, I would like to calculate the standard error and margin of error of the proportion (p) of households (HHWT) in the dataframe "data" that are paying over housing tax credit level rents, this is denoted in the dataframe by a "1" in the field OVERLIHTC. I'm able to get as far as shown below, but I don't know how to finish the code and print the results. Ideas?
p <- sum(data$HHWT[data$OVERLIHTC == 1]) / sum(data$HHWT)
svy <- as_survey(data, weight = HHWT , repweights = matches("REPWT[0-9]+"), type = "JK1", scale = 4/ 80 , rscales = rep(1, 80 ), mse = TRUE)
sub_design <- subset(svy, OverLIHTC == 1 )
Dear Elizabeth,
1.96 would be a 95% confidence interval [qnorm(0.975) ]which is typical for many statistical analyses. However the Census uses a 90% confidence interval for the MoE so the "scale" factor would be approximately 1.65
See the ACS handbook chapter 8 for the calculation
https://www.census.gov/content/dam/Census/library/publications/2020/acs/acs_general_handbook_2020_ch08.pdf
formula (2) page 60 that gives the relationship between SE (standard error) and the MoE (margin of error)
Dave
hi, the instructions on this page hopefully allow you to create a survey design and run the most common analysis commands.. the latter section also includes the `srvyr` conversion in case you'd like to work with dplyr syntax. hope this helps :-)
https://asdfree.com/american-community-survey-acs.html#analysis-examples-with-srvyr