On 12-Jan-2015 10:32:41 Erik B Svensson wrote:
> Hello
> I've got a problem I don't know how to solve. I have got a dataset that
> contains age intervals (age groups) of people and the number of persons in
> each age group each year (y1994-y1996). The number of persons varies each
> year. I only have access to the age intervals, not the age of each person,
> which would make things easier.
>
> I want to know the median age interval (not the median number) for each
> year. Let's say that in y1994 23 corresponds to the median age interval
> "45-54", I want to "45-54" as a result. How is that done?
>
> This is the sample dataset:
>
agegrp <-
c("<1","1-4","5-14","15-24","25-34","35-44","45-54","55-64","65-74",
"75-84","84-")
y1994 <- c(0,5,7,9,25,44,23,32,40,36,8)
y1995 <- c(2,4,1,7,20,39,32,18,21,23,5)
y1996 <- c(1,3,1,4,22,37,41,24,24,26,8)
> I look forward to your response
>
> Best regards,
> Erik Svensson
In principle, this is straightforward. But in practice you may
need to be careful about how to deal with borderline cases -- and
about what you mean by "median age interval".
The underlying idea is based on:
cumsum(y1994)/sum(y1994)
# [1] 0.00000000 0.02183406 0.05240175 0.09170306 0.20087336
# [6] 0.39301310 0.49344978 0.63318777 0.80786026 0.96506550 1.00000000
Thus age intervals 1-7 ("<1" - "45-64") contain less that 50%
(0.49344978...), though "45-64" almost gets there. However,
age groups 1-8 ("<1" - 55-64" contain more than 50%. Hence
the median age is within "49-64".
Implementing the above as a procedure:
agegrp[max(which(cumsum(y1994)/sum(y1994)<0.5)+1)]
# [1] "55-64"
Note that the "obvious solution":
agegrp[max(which(cumsum(y1994)/sum(y1994) <= 0.5))]
# [1] "45-54"
gives an incorrect answer, since with these data it returns a group
whose maximum age is below the median. This is because the "<=" is
satisfied by "<" also.
Hoping this helps!
Ted.
-------------------------------------------------
E-Mail: (Ted Harding) <ted.hard...@wlandres.net>
Date: 12-Jan-2015 Time: 11:12:39
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