$X$ is a random variable and its PDF is:
$$ f(x;θ) = \begin{cases} θe^{-θx}, & x > 0 \\\ 0, & x \le 0 \end{cases} $$
Now, I would like to find the confidence interval regarding parameter $θ$
$$ S(x) = \{θ|0 < θ \le \dfrac{c}{x}\} $$
$c$ is a constant that satisfies $P(θ \in S(X)) = 1 - α$ (α is significance level).
I have found the expectation and variance based on the continuous definition case.
$$ E[X] = \dfrac{1}{θ}, \space V[X] = \dfrac{1}{θ^2} $$
Also, I have tried to find the confidence interval based on the common case that I saw from a textbook.
$$ \begin{align*} & -Φ(α) \le \dfrac{\bar{X} - 1/θ}{\sqrt{1/θ^2n}} \le Φ(α) \\ & -\bar{X} - \dfrac{Φ(α)}{θ\sqrt{n}} \le -\dfrac{1}{θ} \le \dfrac{Φ(α)}{θ\sqrt{n}} - \bar{X} \\ & \bar{X} + \dfrac{Φ(α)}{θ\sqrt{n}} \ge \dfrac{1}{θ} \ge \bar{X} - \dfrac{Φ(α)}{θ\sqrt{n}} \\ & \dfrac{θ\sqrt{n}}{θ\sqrt{n}\bar{X} + Φ(α)} \le θ \le \dfrac{θ\sqrt{n}}{θ\sqrt{n}\bar{X} Φ(α)}\\ \end{align*} $$
Yet, this quite different from what I expect to see, how can I derive the confidence interval without sampling?
Based on the advices from Henry,
$$ \begin{align*} P(θ \in S(X)) & = P(θ \le \dfrac{c}{X}) \\ & \equiv P(X \le \dfrac{c}{θ}) = F(\dfrac{c}{θ}) = \int_0^{c/θ}θe^{-θx}dx = 1 - e^{-c} = 1 - α \\ \therefore c = -\ln{α} \end{align*} $$