Calculating median age for a group of US census blocks?
Hi Kevin and all, Given the binned data, you could count the number of people per age class for those 10 blocks. You can then express that in a number of different ways, like percent under 25 years old, or by calculating the dependency ratio <https://www.who.int/data/gho/indicator-metadata-registry/imr-details/1119#:~:text=Definition%3A,a%20specific%20point%20in%20time.> . I do think it is feasible to calculate an estimated mean from the counts within groups representing ranges. See, for example, here: https://stackoverflow.com/questions/18887382/how-to-calculate-the-median-on-grouped-dataset Since you are working in Baltimore, you may consider looking at The Baltimore Neighborhood Indicators Alliance https://bniajfi.org/vital_signs/. They provide useful data on a range of issues (transportation, crime, education, environment etc.) including summaries from Census-derived demographics. What you are seeking may already exist. BNIA creates neighborhoods or "community statistical areas" (n=55) based on aggregates of Census data. Although not pertaining to age, Baltimore City Planning has paid Census in the past to aggregate from individual-level Census data to the more colloquially-used definitions of Baltimore shown here (n = 273): https://data.baltimorecity.gov/datasets/neighborhood-1/explore?location=39.284832%2C-76.620516%2C12.91 Best, Dexter https://dexterlocke.com/ On Mon, Aug 7, 2023 at 3:02?PM Kevin Zembower via R-sig-Geo <
r-sig-geo at r-project.org> wrote:
Josiah, thanks for your reply. Regarding my objective, I'm trying to compile census statistics for the blocks that make up the neighborhood where I live. It consists of ten census blocks, of which I selected three for simplicity in my example. The census block-group which contains these ten blocks also contains some blocks which are outside of my neighborhood and shouldn't be counted or included. Since I won't be able to calculate the median age from the age and count data, and since the individual data doesn't seem to be available, is it your thought that I can't produce a valid median age for a group of census blocks? Thanks so much for your advice. -Kevin On 8/7/23 14:38, Josiah Parry wrote:
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 at r-project.org <mailto:r-sig-geo at 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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