Suppose $Y_1, Y_2,...,Y_n$ denotes a random sample from a uniform distribution on the interval $(-\theta, 4\theta)$, where $\theta \gt 0$ is an unknown parameter.
For a general n, find a pivotal quantity in terms of $Y_{(n)}$ and $\theta$. Use this pivotal quantity to derive a formula of a 95% confidence interval of $\theta$.
What I have tried so far:
We know that $Y$ ~ $Unif(-\theta, 4\theta)$. So I let $U = Y_{(n)}$ and $X = \frac{U}{\theta}$
$X_1, X_2,...,X_n$ are i.i.d and ~$Unif(-1, 4)$, and $X_{(n)} = \frac{Y_{(n)}}{\theta}$
Using order statistics, I find:
$$ F_{x_{(n)}} = P(X_{(n)} \le x) = (F_{x}(x))^n$$
Given that $X_i$ ~ $Unif(-1, 4)$, it follows that:
$$(F_{x}(x))^n = (\frac{x+1}{5})^n$$
Now I want the pdf:
$$f_x(x) = \frac{n}{5}(\frac{x+1}{5})^{n-1}$$
This is where I get stuck. I am fairly certain I will want to integrate from here to find my end points to finish the problem. But I'm not sure what my bounds are or how to find them. So I am left with:
$$\frac{n}{5^n}\int (x+1)^{n-1} dx = 0.05$$
Any help from this point would be appreciated.