A random sample of size $n = 64$ from a Poisson random variable with unknown
mean yields a sample mean of $\bar{x} = 8.3$. Use the central limit theorem to find a$95\%$ condence interval for $\mu$.
I'm confuse on this problem.
Gives an $\bar{x}=8.3$ which is a sample mean.
after manipulating the CLT i came up with
$\bar{x}\pm t_{\frac{\alpha}{2}*\frac{S}{\sqrt{n}}}$
where $\bar{x}=8.3$, $n=64$, degrees of freedom $=63$, $t_{\frac{\alpha}{2}}=1.998$
How do I get $S$? is it equal to $\bar{x}$?
I thought i could come from the $poisson(64)\sim N(64, 8)$ with $S=8$ or
$\bar{x} \sim N(1, 1/64)$ with $S= \frac{1}{64}$
Any ideas of how to get $S$?
Am I even approaching the problem correctly?
Is it an Normal distribution and not a T distribution?
Best Answer
Assume $X_1, \ldots, X_n \text{ i.i.d. } \sim \text{Poisson}(\mu)$, then by classic CLT, it follows that $$\frac{\sum_{i = 1}^nX_i - n \mu}{\sqrt{n \mu}} \Rightarrow N(0, 1),$$ or equivalently, $$\frac{\sqrt{n}(\bar{X} - \mu)}{\sqrt{\mu}} \Rightarrow N(0, 1).$$ Based on this, when $n = 64$, an approximate $95\%$ confidence interval for $\mu$ can be obtained by solving $$\left|\frac{8(8.3 - \mu)}{\sqrt{\mu}}\right| \leq z_{0.975} = 1.96$$ for $\mu$ (the $t$ quantiles appeared in your attempt was unjustified). The result is $[7.62351,\, 9.03652]$.