Hypothesis test for a non-central parameter

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I would like to get a better understanding of how to conduct hypothesis tests generally. So rather than pick a test for a parameter where the mean seems like the good or obvious test, say that I want to test for a parameter that controls something else, like the spread. So for instance, suppose I have this symmetric distribution controlled by $\lambda$:

$$f(x) = \frac{\lambda}{2}e^{-\lambda|x|}$$

and I take $n$ independent observations from it. I'd like to conduct a test for $\lambda$ over the parameter space $\Omega=(0,\infty)$.

As I understand it this is an under-determined question because you have a free choice about what to select as the null and alternative hypotheses, right? So let's say that I want to test the null hypothesis $H_0: (\lambda_0,\infty)$ against $H_a:(0,\lambda_0]$ for some fixed choice of $\lambda_0$.

Since $\lambda$ basically controls the spread of the distribution, and the distribution has mean 0 for any choice of $\lambda$, then the mean is a poor statistic to base a hypothesis test on. We could compute the variance, and intuitively that seems like a reasonable choice, but then I have two questions:

First, are you always just guided by a kind of intuitive guess for the statistic that you pick?

Second, if we choose the variance then we have to compute the sample variance and discover its distribution, right? That is to say, if the sample is $X_1, ..., X_n$ then we need to find the distribution of

$$S = \frac{\sum (X_i-\mu)^2}{n-1}$$

Certainly $\mu=0$ and if we can find the distribution of $\sum X_i^2$ then we can find the distribution of $S$. But in this case that seems hard, even though the distribution seems pretty simple. Although finding the distribution of any $X_i^2$ takes a little work using the distribution technique, I can't see a way through finding the distribution of a sum of them.

So although I have a number of intermediate questions, my main question is: How do you test this hypothesis? It seems like the question should have an answer ... but I'm not even certain it does.