Since it is small sample you can calculate exact distribution to arrive at confidence interval.
$Z= X_1+ \cdots + X_{10} \sim $ Poisson($10 \lambda$).
The mle of $\lambda = \hat{\lambda} = \bar{X}= Z/10$. From here you can find the confidence interval.
EDIT: Confidence interval
$Z= 10* \bar{X} \sim$ Poisson($10 \lambda$).
Let $\lambda_L$ and $\lambda_U$ be the confidence limits of $\lambda$.
And $\lambda_{L^{*}}$ and $\lambda_{U^{*}}$ be the confidence limits of $10\lambda$.
$ \lambda_{U^{*}}=10\lambda_U, \quad $ and $ \lambda_{L^{*}}=10\lambda_L \quad $
Then
$\exp(-\lambda_{L^{*}}) \sum\limits_{j=z}^{\infty}\frac{\lambda_{L^{*}}^j}{j!}= \frac{\alpha}{2}, \quad $ and $ \quad \exp(-\lambda_{U^{*}}) \sum\limits_{j=0}^{z}\frac{\lambda_{U^{*}}^j}{j!}= \frac{\alpha}{2}$
From the relationship between poisson and gamma (chisqure) it can write as
$10*\lambda_L =0.5 \chi^2_{2z,\alpha/2}, \quad$
$10*\lambda_L =0.5 \chi^2_{2(z+1),1-\alpha/2}$
which leads the CI of $\lambda$ as
(qchisq(0.025,2*z)/(2*10), qchisq(0.975,2*(z+1))/(2*10))
This answer is based on the clarification offered in comments:
I'd like to make a statement such as ... "I am 68% sure the mean is between $3.1−σ_−$ and $3.1+σ_+$", and I want to calculate $σ_+$ and $σ_−$.
I think that at least in the physics world this is called a confidence interval.
Let's take it as given that in response to my question "confidence interval for what?" you responded that want a confidence interval for the mean (and as made clear, not some other interval for values from the distribution).
There's one issue to clear up first - "I am 68% sure the mean is between" isn't really the usual interpretation placed on a confidence interval. Rather, it's that if you repeated the procedure that generated the interval many times, 68% of such intervals would contain the parameter.
Now to address the confidence interval for the mean.
I agree with your calculation of mean and sd of the data:
> x=c(1,2,3,5,1,2,2,3,7,2,3,4,1,5,7,6,4,1,2,2,3,9,2,1,2,2,3)
> mean(x);sd(x)
[1] 3.148148
[1] 2.106833
However, the mean doesn't have the same sd as the population the data was drawn from.
The standard error of the mean is $\sigma/\sqrt{n}$. We could estimate that from the sample sd (though if the data were truly Poisson, this isn't the most efficient method):
> sd(x)/sqrt(length(x))
[1] 0.4054603
If we assumed that the sample mean was approximately normally distributed (but did not take advantage of the possible Poisson assumption for the original data), and assumed that $\sigma=s$ (invoking Slutsky, in effect) then an approximate 68% interval for the mean would be $3.15\pm 0.41$.
However, the sample isn't really large enough for Slutsky. A better interval would take account of the uncertainty in $\hat \sigma$, which is to say, a 68% t$_{26}$-interval for the mean would be
$3.15\pm 1.013843\times 0.41$
which is just a fraction wider.
Now, as for whether the sample size is large enough to apply the normal theory CI we just used, that depends on your criteria. Simulations at similar Poisson means (in particular, ones deliberately chosen to be somewhat smaller than the observed one) at this sample size suggest that using a t-interval will work quite well for similar Poisson rates and 27 observations or more.
If we take account of the fact that the data are (supposedly) Poisson, we can get a more efficient estimate of the standard deviation and an interval for $\mu$, but if there's any risk the Poisson assumption could be wrong - a chance of overdispersion caused by some homogeneity of Poisson parameters, say - then the t-interval would probably be better.
Nevertheless, we should consider that specific question - "how to get a confidence interval for the population mean of Poisson variables" -- but this more specific question has already been answered here on CV - for example, see the fine answers here.
Best Answer
This is not quite true. Indeed, in this framework, the unknown mean is fixed (either it lies inside the interval or it lies outside, there is no probability attached to it). Instead, you should say that the probability that the interval (which is random) covers the true mean equals 95%. Or that there is a 5% chance that the interval does not cover the true mean.
$(45.7467, 48.617)$ is meant to provide information concerning the mean of the Poisson distribution, i.e. $\lambda$. Not about an observation from the Poisson distribution.
Perhaps you are interested in a prediction interval...
$\Pr(X > 330)$ is the probability that the first Poisson variable (with mean 326) returns a count larger than the mean of the second Poisson variable. It does not answer your question (which I can rephrase: are the two means equivalent?).
To convince yourself, you can compute Pr(Pois(326) > 326), i.e. just as if the second Poisson has the same mean as the first ("equivalence").
What you might want to do instead is to construct a (90%) confidence interval for the mean difference and check that it lies within some equivalence margins. This is called an equivalence test (for means) and this approach is known as TOST.