How to compute a confidence interval for the mean of count data? The confidence interval (CI) of the mean of data is probably the first topic on CIs on most text books.
The CI for count data can be computed with poisson test (in R). As far as I understand it, this should be performed for each count performed.
I would like to compute the CI for the mean of several counts. For example:


*

*counts: 6 7 4 5

*mean = 5.5

*sd $\approx$ 1.29


Is it correct to compute the CI for the mean, in this case?
 A: It's certainly asymptotically correct, which is as much as you can say for the usual formula for CIs, which assumes that the sampling mean of a continuous distribution is normal.
You are essentially doing a fit to a Poisson distribution here. $\hat{\lambda} = \left< k \right>$ is the maximum likelihood estimator of the Poisson parameter. The second derivative of the log-likelihood is $-n / \lambda = -n / \sigma^2$. Arguably you should use $\sqrt{\lambda}$ in place of the measured $\sigma$ for consistency, but asymptotically they are equivalent.
To compute exact confidence intervals at finite n, you need the exact distribution of your estimator. It turns out we can derive this, because the sum of $n$ Poisson deviates with parameter $\lambda$ is Poisson-distributed with parameter $n \lambda$. Applying the formulas for the expectation and variance of the Poisson distribution, we find that (1) our MLE estimator $\hat{\lambda}$ is unbiased, even at finite $n$, and (2) our MLE formula for the variance of $\hat{\lambda}$ is exact, even at finite $n$. So we have got the estimate and its variance right, even at finite $n$.
To actually get an exact confidence interval at finite $n$, you need even more that this, because the usual way to compute a CI given an estimator's mean and variance assumes it is normally distributed, and in our case that last assumption only becomes true in the limit of large $n \lambda$. To get an exact $p$-percentile CI, you need to find boundaries on a Poisson distribution with parameter $n \lambda$ that integrate up to $p$. Up to factors suppressed by $n^{-1}$, these will be the same boundaries given by the usual CI formula. 
A: Just compute a CI for the sum (which is Poisson), and then back out a CI for the mean from that. If you can get an interval for $Y=\sum_i X_i$, you can figure out a CI for $Y/4$.
You simply use properties of algebra on the probability statements in the CI. 
If $\mu_Y=4\mu_X$ and the probability that the random interval includes the (fixed but unknown) $\mu_Y$ is $1-\alpha$ then you manipulate the statement as follows: 
$P(c_l \leq \mu_Y \leq c_u)\geq 1-\alpha$ 
$P(c_l \leq 4\mu_X \leq c_u)\geq  1-\alpha$ 
$P(c_l/4 \leq \mu_X \leq c_u/4)\geq 1-\alpha$ 
Note that at each step the relationships between the limits and the thing between them is simply implied by things that are known/stated.
Hence the $1-\alpha$ interval for $\mu_X$ is $(c_l/4,c_u/4)$, where $(c_l,c_u)$ was the interval for $\mu_Y$.
Note further that if it were the case that the actual achieved coverage in the first inequality was 
$1-\alpha^*$ (due to discreteness in $c$ values):
$P(c_l \leq \mu_Y \leq c_u)= 1-\alpha^*$ 
then
$P(c_l/4 \leq \mu_X \leq c_u/4)= 1-\alpha^*$ 
