Hey Kevin, I don't think you're going to be able to get individual level
data from the US Census Bureau. The closest you may be able to get is the
current population survey (CPS) which I believe is also available via
tidycensus. Regarding your first question, I'm not sure I follow what your
objective is with it. I would use a geography of census block groups as the
measure of median for census block groups. Otherwise it is unclear how you
are defining what a "group of blocks" is.

On Mon, Aug 7, 2023 at 2:34 PM Kevin Zembower via R-sig-Geo <
r-sig-geo@r-project.org> wrote:

> Hello, all,
>
> I'd like to obtain the median age for a population in a specific group
> of US Decennial census blocks. Here's an example of the problem:
>
> ## Example of calculating median age of population in census blocks.
> library(tidyverse)
> library(tidycensus)
>
> counts <- get_decennial(
>      geography = "block",
>      state = "MD",
>      county = "Baltimore city",
>      table = "P1",
>      year = 2020,
>      sumfile = "dhc") %>%
>      mutate(NAME = NULL) %>%
>      filter(substr(GEOID, 6, 11) == "271101" &
>             substr(GEOID, 12, 15) %in% c(3000, 3001, 3002)
>             )
>
> ages <- get_decennial(
>      geography = "block",
>      state = "MD",
>      county = "Baltimore city",
>      table = "P13",
>      year = 2020,
>      sumfile = "dhc") %>%
>      mutate(NAME = NULL) %>%
>      filter(substr(GEOID, 6, 11) == "271101" &
>             substr(GEOID, 12, 15) %in% c(3000, 3001, 3002)
>             )
>
> I have two questions:
>
> 1. Is it mathematically valid to multiply the population of a block by
> the median age of that block (in other words, assign the median age to
> each member of a block), then calculate the median of those numbers for
> a group of blocks?
>
> 2. Is raw data on the ages of individuals available anywhere else in the
> census data? I can find tables such as P12, that breaks down the
> population by age ranges or bins, but can't find specific data of counts
> per age in years.
>
> Thanks for your advice and help.
>
> -Kevin
>
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