Using the Central Limit Theorem to form CI from Gamma distributed random variables Firstly, I am studying the basic concepts of statistics and so any explanations, advice and suggestions are more than appreciated.
I am doing a question in my note, not sure am I doing correct or not. The question:
Let
$\overline X$
denote the mean of a sample of size 25 from a gamma distribution
with $\alpha=4$ and $\beta>0$. Use the Central Limit Theorem to find a 95% confidence interval for μ, the mean of the gamma distribution.
I don't know what is $\beta$, so I have done the following:
$E(X)=\frac{4}{\beta}$, $var(X)=\frac{4}{\beta^2}$
$$\frac{4}{\beta}\pm1.645\frac{\sqrt \frac{4}{\beta^2}}{\sqrt n}$$
May I ask is it correct? I am also thinking it could be:
$$\frac{4}{\beta}\pm1.645\frac{\sqrt \frac{4}{n\beta^2}}{\sqrt n}$$
because the variance is $\sigma^2/n$ instead of $\sigma^2$ when applying the CLT?
Which one is correct? Also, the question says $\beta >0$, is it useless information? Is there anyway to find $\beta$? 
Thanks!!
 A: Ignore for the moment that you're working with Gamma random variables. Let $Z$ be a standard normal random variable; i.e., $Z \sim N(0,1)$. Then, using a table, or some numerical tools, you can find a value $a_\ast \in \mathbb{R}$ such that
$$
\mathbb{P}(-a_\ast \leq Z \leq a_\ast) = 0.95
$$
I believe this threshold value is $a_\ast \approx 1.96$, rather than $1.645$ as you have written; the $1.645$ corresponds to a $90\%$ confidence interval (i.e., with $0.90$ replacing $0.95$ as I have written above). 
Now, given a sequence of iid random variables $X_1, ..., X_n$ with true mean $\mu$ and true variance $\sigma^2$, the Central Limit Theorem, put very roughly, tells us that for reasonably large $n$, we can use the approximation
$$
\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \approx Z
$$
where as usual, $\bar{X} = \frac{1}{n}\sum\limits_{i=1}^n X_i$. Before moving forward, let's understand the ratio on the lefthand side. Rewriting, we have
$$
\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{\bar{X} - \mathbb{E}[\bar{X}]}{\sqrt{\operatorname{Var(\bar{X})}}}
$$
Let's call this ratio $Y$. Subtracting the mean in the numerator of $Y$ ensures $\mathbb{E}[Y] = 0$ (this is called "centering"); the standard deviation in the denominator ensures $\operatorname{Var}(Y) = 1$ (this is called "normalizing"). So, looking again at the Central Limit Theorem, we are trying to compare $\bar{X}$ to the standard normal random variable $Z$, and what we do to make this comparison is adjust $\bar{X}$ (forming the ratio $Y$) so that it has the same mean and variance as $Z$.
With that covered, let's go back to the confidence interval. We previously said that 
$$
\mathbb{P}(-1.96 \leq Z \leq 1.96) = 0.95
$$
and, by the Central Limit Theorem, $\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \approx Z$. Make the substitution,
$$
0.95 = \mathbb{P}(-1.96 \leq Z \leq 1.96) \approx \mathbb{P}\left( -1.96 \leq \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \leq 1.96\right)
$$
From here, we solve for the mean $\mu$ to find
$$
\mathbb{P}\left(\bar{X} -1.96 \cdot \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X} + 1.96 \cdot \frac{\sigma}{\sqrt{n}} \right) \approx 0.95
$$
In other words, there is approximately a $95\%$ chance that the interval $\bar{X} \pm 1.96 \cdot \dfrac{\sigma}{\sqrt{n}}$ contains the true mean $\mu$. So, if you were to run this experiment many many times and kept creating confidence intervals in this way, you'd find that approximately $95%$ of those confidence intervals contained the correct mean $\mu$.
Now, knowing what the random variables $X_i$ are (in your case Gamma random variables) and knowing what $n$ is (in your case $25$), you can fill in the blanks for $\sigma = \sqrt{\operatorname{Var}(X_i)}$ and $n$. Note that the point is usually to try to create an interval to "catch" the true mean $\mu$ with a measured average $\bar{X}$ since the true mean is usually unknown and what you are hoping to approximate. 
