It is a well known result that for the counting process $\{N(t), t \geq 0 \}$ associated with a renewal process $\{S_n\}_{n \in \mathbb{N}}$ we have the central limit theorem
\begin{equation*} \frac{N(t) - \frac{t}{\mu}}{\sigma \sqrt{t}\mu^{-\frac{3}{2}}} \overset{d}{\longrightarrow} \mathcal{N}(0,1) \text{ as } t \rightarrow \infty \end{equation*} where $0 < \mu < \infty$ is the expected value of the interarrival times $\{X_n\}_{n \in \mathbb{N}}$ and $0 < \sigma^2 < \infty$ is their variance.
From this I would like to derive an asymptotic prediction interval for $N(t)$. From the above result it is clear that \begin{equation*} P(\frac{t}{\mu} - z_{1-\frac{\alpha}{2}} \sigma \mu^{-\frac{3}{2}} \sqrt{t} \leq N(t) \leq \frac{t}{\mu} + z_{1-\frac{\alpha}{2}} \sigma \mu^{-\frac{3}{2}} \sqrt{t}) \overset{t \rightarrow \infty}{\longrightarrow} 1-\alpha, \end{equation*} which gives me such an interval. However in my case I do not know $\mu$, nor $\sigma$. I can use consistent estimators $\hat{\mu}$ and $\hat{\sigma}$ and apply Slutsky's Theorem to get \begin{equation*} \frac{N(t) - \frac{t}{\mu}}{\hat{\sigma} \sqrt{t} \hat{\mu}^{-\frac{3}{2}}} \overset{d}{\longrightarrow} \mathcal{N}(0,1) \text{ as } t \rightarrow \infty \end{equation*} but I do not know how to get rid of the $\mu$ in the numerator. Has anybody got any suggestions?