# calculating a confidence intervall using the central limit theorem

I have a collection of random variables $Y_1, ... , Y_n$ which can be written as $Y_i = \mu + X_i$. The $X_i$ are independent and identically distributed with density $f(x) = \frac{3}{4}(1-x^2)1_{[-1,1]}(x)$.

I want to calculate the approximate two-sided $99$% confidence interval for $\mu$ using the central limit theorem.

I am unsure how to tackle the problem. The central limit theorem is about the standardized sum of random variables. But how do I define the confidence interval and how is it connected to the theorem?

I used the given density to calculate the expectation values and variances for the $X_i$. I got $$E[X_i] = 0\\ Var[X_i] = \frac{1}{5}$$ and therefor $$E[Y_i] = \mu\\ Var[Y_i] = \frac{1}{5}$$

I also know how I get the corresponding values for a sum of variables.

## 1 Answer

Hint: The approximate confidence interval is

$$\large{\left[\overline y-z_{(1-\frac{\alpha}{2})\cdot \frac{\sigma}{\sqrt n}} ; \ \overline y+z_{(1-\frac{\alpha}{2})\cdot \frac{\sigma}{\sqrt n}} \right]}$$

Therefore you have to insert the values for $\sigma$ and $\alpha=1-0.99=0.01$ (significance level). The value for $z_{(1-\frac{\alpha}{2})}$ can be looked up at a table of the standard normal distribution.