The life length $T$ of a LED light can be assumed to be exponentially distributed, with probability density function $$f_{\theta}(t)=\frac{1}{\theta}e^{-\frac{t}{\theta}}1\{t\ge0\},$$
so that $\theta$ is the expectation of $T$. A person buys one LED light, and records the life length for that light to be $t=1000$ hours. Construct a lower $95$% confidence interval for the expected life length of $\theta$.
Please see my (wrong) solution below followed by some questions/queries.
As $T\backsim\mathrm{exp}(\theta)$, then
$$P(T\le t)=F_{\theta}(t)=1-e^{-\frac{t}{\theta}}$$
Let $Z=\frac{T}{\theta}$, then
$$F_{Z}(t)=P(Z\le t)=P(\frac{T}{\theta}\le t)=P(T\le t\theta)=F_{\theta}(t\theta)=1-e^{-\frac{t\theta}{\theta}}=1-e^{-t}.$$
Thus $Z\backsim\mathrm{exp}(1)$. We want a lower confidence interval for the expected life length of $\theta$.
\begin{align*} 1-\alpha &= P(Z\in [\lambda_{\alpha},\infty))\\ &= P(Z\ge \lambda_{\alpha})\\ &= P(\frac{T}{\theta}\ge \lambda_{\alpha})\\ &= P(\theta\le \frac{T}{\lambda_{\alpha}}).\\ \end{align*}
This answer confuses me? Ordinarily I would just plug in $\alpha=0.05$ to get the interval but I assume I have made a mistake for the following reasons
1) My answer would give the confidence interval for $\theta$ in the interval from 0 to infinity. Should not the lower confidence interval be for $\theta$ in the interval from some constant to infinity?
2) I have not made use of the given information "A person buys one LED light, and records the life length for that light to be $t=1000$ hours". Not sure why or how to use this info.
I also have a question regarding my choice of $Z=\frac{T}{\theta}$. I did this because it was done in an example in the textbook and it seems to work in this case. I don't know however why it works? What is the thought process one should go through to "choose" a $Z$?
You are moving in the right direction, but you do need to take the observation that one LED from the population of interest lasted $X = 1000$ hours.
Here is how to make a 90% two-sided confidence interval (CI) for the exponential mean $\theta$ based on the observation $X = 1000$ from the distribution $\mathsf{Exp}(\theta),$ with the density function you specify. First, observe that $X/\theta \sim \mathsf{EXP}(1).$
For suitable $L$ and $U,$ we have $$P(L < X/\theta < U) = P(1/U < \theta/X < 1/L) = P(X/U < \theta < X/L) = .90.$$ In particular, we choose $L$ and $U$ to be the 5th and 95th percentiles, respectively, of $\mathsf{EXP}(1).$
Thus, a two-sided 90% CI for $\theta$ is of the form $(X/U, X/L).$ Now, you can make the appropriate modification to get the required one-sided 95% CI.