Let $X$~$Exp(\lambda)$ then $f$($x$;$\lambda$) $=$ $\lambda$e$^{-\lambda x}$ $I$$_{(0,\infty)}$($x$). Obtain confidence interval for the mean of the population.
To do this, we need to remember a few facts about the gamma distribution. More specifically, If $Y=X_1+X_2+\cdots+X_n$, where the $X_i$'s are independent $Exp(\lambda)$ random variables then $Y \sim Gamma(n, \theta)$. Thus, the random variable $Q$ is defined as $Q=\theta (X_1+X_2+\cdots+X_n)$ has a $ Q \sim Gamma(n, 1)$ distribution. Let us define $\gamma_{p,n}$ as follows. For any $p \in [0,1]$ and $n \in \mathbb{N}$ we define $\gamma_{p,n}$ as the real value for which $P(Q > \gamma_{p,n})=p,$ where $ Q \sim Gamma(n, 1)$ the Confidence interval is $\frac{\gamma_{1-\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n}, \frac{\gamma_{\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n}$. Is this right?
Observe that
$$Y=2\lambda\Sigma_i X_i\sim \chi_{(2n)}^2$$
Now $Y$ is your pivotal qty so first calculate the 2 quantiles of $Y$ with paper tables or calculator and then you can easy find the confidence bounds for $\frac{1}{\lambda}$ solving the double inequality in $\frac{1}{\lambda}$.
Thus, setting $a,b$ the 2 quantiles of the $\chi_{(2n)}^2$ you get
$$a<2\lambda \Sigma_i X_i<b$$
That is
$$\frac{2\Sigma_i X_i}{b}<\frac{1}{\lambda}<\frac{2\Sigma_i X_i}{a}$$