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)`

)

They do give you a standard error and a $z$ value, which suggests that the SAS developers thought it reasonable to use the normal distribution to approximate the sampling distribution of the statistic involved. So, if you can confirm that your $z$ statistic is equal to the estimated log rate ratio (call it $LRR$) divided by the standard error, then I would be comfortable in using $LRR \pm 2\times SE$ as an approximate 95% CI for the true log rate ratio. Supposing that this results in an interval $(\ell,u)$, then the interval $(e^\ell, e^u)$ provides a 95% CI for the incidence rate ratio itself.

## Best Answer

Assuming I understand the situation correctly:

$\text{Var}(\hat{\lambda}) = \text{Var}(\frac{N}{T}) = \frac{1}{T^2}\text{Var}(N)=\frac{1}{T^2}\text{Var}(\sum_iY_i)=\frac{1}{T^2}(\sum_i\lambda T_i)=\lambda/T$

Hence the estimated standard error is $\sqrt{\hat{\lambda}/T} = \frac{\sqrt{N}}{T} = \frac{N\times \frac{1}{\sqrt{N}}}{T}=\frac{N}{T}\times \frac{1}{\sqrt{N}}=\frac{\text{IR}}{\sqrt{N}}$

So (if $N>0$),

eitherof these two formulas is okay:$\text{s.e.}(\text{IR}) = \sqrt{\frac{IR}{T}} = \frac{IR}{\sqrt{N}}$

and so a large sample (large-$N$) interval for the $\text{IR}$ would multiply that by a suitable $Z$ value to get the interval half-width either side of the estimate.

You seem to have perhaps muddled up the roles of $N$ and $T$ in your calculation

Actually, it looks like the link you gave explained how their formula came about pretty clearly in the couple of lines under "If $N$ is large".