Confidence interval for N, based on binomial

Question

I've searched the web and these forums but couldn't find a subject that resembled my problem. If I missed something then I'd like to apologize upfront.

My problem is this. I have a given probability p, given by an outside agency and the number of successes n from an owned datasource. I'm interested in the population total N and the confidence interval of N.

Background: I know how many passengers there are for a specific bus agency using 2013 data (as in: I have the (micro)data from that specific company). Assumption: this is my n. I also know from transportation figures (from an outside agency) that this specific bus company had a market share (as measured in passengers) of 30% in 2013. Assumption: this is my p. This assumption brings with it certain other assumptions, that is fine.

So now the question is: how many passengers were there in total (assuming this can not be taken from the transportation figures)? ie. what is N and what is the 95% confidence interval surrounding N?

So my questions are:

1) Can I just use the relation: p=n/N, therefore the point estimat for N is: N=n/p? I'd assume so.

2) Can I just use the Clopper-Pearson interval to calculate an interval for p, then using that to get to an interval for N?

3) Suppose the 95% lowerbound for p=plb. Is the 95% lower bound for N then n/plb? Same logic for the 95% upper bound.

4) When calculating the Clopper-Pearson interval, can I just use: n=n, p=p, N=n/p?

5) If any of the answers to my questions 2-4 is 'No', then what is the correct approach?

I realize that using this approach, N is not necessarily an integer. Is that problematic?

Answer 1

If we treat the $p$ as a fixed, known quantity (rather than something that was an estimate, as it seems it actually was) and we treat the $n$ as having come from a $\text{binomial}(N,p)$, then I believe this is actually a negative binomial problem.

We can view each passenger as a Bernoulli trial, where 'success' is 'travelled with the particular bus company'. As such, the number of trials (passengers) until the $n$th success is negative binomial $\text{NB}(n,p)$ (in the 'number of trials' form; if we're using the 'number of failures' form we need to add $n$ to the result).

We can use the quantiles of that conditional distribution for $N$ to get a confidence interval.

So for example, with $n=1000$ and $p=0.3$ the interval $(3163,3509)$ contains just over 95% of the probability, and has very close to 2.5% in each tail.

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If we treat $p$ as an estimate measured with error, then this becomes a bit more complex, and we'd need some information about the estimation of $p$, or to make some assumptions about it. It might be necessary to use simulation in that case. Dealing with this would tend to make the interval a good deal wider.

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However, I'd actually be somewhat more inclined to consider a Bayesian approach for this problem, I think it's a pretty natural fit; incorporating prior information about $N$ and uncertainty in $p$ can both be brought in in fairly natural ways.