Given a distribution $X~Pois(λ)$ with unknown parameter $λ=E(X)$ with $14$ random observations given from this distribution.
How do I calculate a confidence Interval at $95\%$ for $λ?$ Do I have to use T-Student or normal distribution? (I personally think I have to use T-Student, since $λ$ is unknown and n is small).
I tried it in the following way:
$λ =$ Mean([14 Values]) = 8
$pt(0.975, 13) = 0.8263$
Confidence interval: $(λ - 0.8263\sqrt{(λ/n)}, λ + 0.8263\sqrt{(λ/n)})$
Is my approach correct?
Thank you
No, Student t distribution is used when you are estimating the mean of a Normal model with unknown variance. Here the model is not normal. It's a poisson one.
$n=14$ is great enough to use a Gaussian approximation for the interval.
Using CLT we know that, with $n$ great enough, $\overline{X}_n\dot{\sim} N(\lambda;\frac{\lambda}{n})$ thus the pivotal quantity is the following
$$\frac{\overline{X}_n-\lambda}{\sqrt{\frac{\lambda}{n}}}$$
First approximation
Estimate standard deviation with $\sqrt{\frac{\overline{X}_n}{n}}$
Then solving in $\lambda$
$$-1.96\leq \frac{8-\lambda}{\sqrt{\frac{8}{14}}}\leq 1.96$$
That is
$$\lambda \in[6.52;9.48]$$
Second approximation
Solve in $\lambda$ the following double inequality
$$-1.96\leq \frac{8-\lambda}{\sqrt{\frac{\lambda}{14}}}\leq 1.96$$
getting, after some algebraic calculations
$$8+\frac{1.96^2}{28}\pm\sqrt{\Bigg(\frac{1.96^2}{28}\Bigg)^2+8\frac{1.96^2}{14}}$$
That is
$$\lambda\in[6.65;9.63]$$