Suppose $X \sim \mathrm{Bin}(n,p)$ is a binomial random variable with success probability $p$ and $n$ trials, where $p$ is known and $n$ is unknown. Now only one observation of $X$ is given and I want to estimate the value of $n$. How should I estimate $n$ and give a 95% confidence interval?
I think an intuitive estimator might be $\hat{n} = X/p$. However, I have no idea about how to give a confidence interval here.
I might be particularly interested in the case where $n$ is large. For example, how would I give an estimation and 95% confidence interval of total number of people in a company, when I was told that 300 people in this company was born in January? This would be the case of $X\sim\mathrm{Bin}(n,1/12)$ and the only observation is $X=300$.
As you say, the estimate $\hat n = X / p$ seems intuitive for $n$. There are many ways you could build a confidence interval around it, just as when estimating $p$; one simple one would be to take a normal approximation for $n$ large.
We have $\mathbb E(\hat n) = n$, $\mathbb V \mathrm{ar}(\hat n) = n(1-p)/p$, and as $n \to \infty$, the CLT says $\hat n$ approaches a normal distribution. Thus a simple $(1-\alpha)$ confidence interval for $n$ would be $\hat n \pm z\sqrt{\hat{n}(1-p)/p}$, where $z$ is the $(1 - \frac\alpha2)$-quantile of the standard normal distribution.