I'm working with data tables from 2020 NSDUH data. The data tables I have access to give me the following information:
I would like to use a 2 Proportion Z-test to see if there is a statistically significantly difference between people with a substance use disorder based on their type of insurance. In R, prop.test requires you have the number of "successes", but I don't have the raw data to populate that info accurately.
Is there another way to do this or different test I should use that would allow me to use the proportions that have already been calculated along with the standard error?
Here is some sample data
idd_all <- matrix(c(36280, 0.066, 0.0021,
23690, 0.053, 0.0024,
6990, 0.118, 0.0067),
byrow = T, nrow = 3, ncol = 3,
dimnames = list(c("total", "medicaid 12+", "private 12+"),
c("sample size", "% IDD", "SE")))
| sample size | est % IDD | SE | |
|---|---|---|---|
| total | 36280 | 0.066 | 0.0021 |
| medicaid 12+ | 23690 | 0.053 | 0.0024 |
| private 12+ | 6990 | 0.118 | 0.0067 |
note: There are obviously more insurance types that we are not interested in, we want primarily interested in the difference between people on Medicaid and privately insured individuals.
I believe the following answers your question on how to test the null hypothesis of equal proportions between medicaid 12+ and private 12+.
total_medicaid <- 23690 / 0.053
total_private <- 6990 / 0.118
x <- c(23690, 6990)
n <- c(total_medicaid, total_private)
prop.test(x, n)
Result
# 2-sample test for equality of proportions with continuity correction
#
# data: x out of n
# X-squared = 3880.4, df = 1, p-value < 2.2e-16
# alternative hypothesis: two.sided
# 95 percent confidence interval:
# -0.06768921 -0.06231079
# sample estimates:
# prop 1 prop 2
# 0.053 0.118