Suppose you have a population of count data, i.e., $1,2,3, \dots, k$, you have a sample of the population of size $n$, and you have a confidence interval for the proportion of $1$'s , $2$'s,\dots$n$'s in the population. Is it possible to construct a confidence interval for the sum of the counts in your sample.
For example,
Suppose you have a sample of size 1000 from a population of count data that can take any value from $1$ to $5$ and the
95% CI for $1$'s is $[.75,.76]$
95% CI for $2$'s is $[.12,.13]$
95% CI for $3$'s $[0.04, .045]$
95% CI for $4$'s is $[.048,.058]$
95% CI for $5$'s is $[.03,.035]$
Can I construct a 95 percent confidence interval for $n\mu$, where $n=1000$.
Roughly, yes. For example, take the CI for the proportion of $4$'s. It is centered at 0.053. If these are the 'usual' kind of CIs used for a sample of size $n=1000,$ then the number of $4$'s in the sample must have been $0.053(1000) = 53$.
In some cases, because the endpoints are rounded, you will not get an integer answer, but you can come very close to finding the frequency $f_i$ with which each of the digits $i$ for $i = 1,\dots,5$ appeared among 1000. From that information you can construct a CI based on the whole sample nearly as well as if you had the original data.
Just what it means to get a CI for the "sum" is unclear. Ordinarily, in such a situation one would want a CI for the population mean. Maybe your final sentence should have been "construct a CI for the population mean $\mu$ based on the sum of the counts." (For samples of size $n$, the expected total would be $n\mu.$ You could get a CI for that.)