Suppose $Y_1, Y_2, ... , Y_n$ denotes a random sample from a uniform distribution on the interval $(-\theta, 4\theta)$, where $\theta \gt 0$ us an unknown parameter.
Assume that n is sufficiently large, find a pivotal quantity in terms of $\overline{Y}$and $\theta$. Use this pivotal quantity to derive a formula of:
(i) an approximate 95% confidence interval for $\theta$.
(ii) an approximate 95% confidence interval for $\theta^2$.
What I have tried so far:
(i) We know that $Y$ ~ $Unif(-\theta, 4\theta)$, so $\mu = 1.5\theta$ and $\sigma^2 = \frac{25\theta^2}{12}$. However, we also know that n is sufficiently large to apply the central limit theorem, so $Y$ ~ $N(1.5\theta, \frac{25\theta^2}{12})$
From that, we can create the pivotal quantity: $$\frac{\overline{Y} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{\overline{Y} - 1.5\theta}{\frac{5\theta}{\sqrt{12n}}} $$
So now we can set the problem up as $P(Z_{0.025} \le \frac{(\overline{Y} - 1.5\theta)\sqrt{12n}}{5\theta} \le Z_{0.975})$
After going through the math, I ended up with $P((\frac{-\sqrt{12n}}{9.8} + \frac{2}{3})\overline{Y} \ge \theta \ge (\frac{\sqrt{12n}}{9.8} + \frac{2}{3})\overline{Y})$ as my interval. I'm not really sure if that is correct. It seems a bit messy.
(ii) For this part, I was having difficulty getting started. I'm not really sure how to go about finding a pivotal quantity for $\theta^2$, or if I'm just supposed to use the above pivotal quantity, which wouldn't make sense to me.
I know that to be a pivotal quantity, it has to be a function of my random variable Y and my parameter $\theta$. And I also know that the distribution of my pivotal quantity has to be parameter free. But I'm otherwise stuck at this point.
Any help or direction would be greatly appreciated.
First note that $Z_{0.025}=-Z_{0.975}=-1.96$ and solve the inequaalities $$-1.96 \le \frac{(\overline{Y} - 1.5\theta)\sqrt{12n}}{5\theta} \le 1.96$$ Multiplying by $5\theta$ and dividing by $\sqrt{12n}$, obtain $$-\frac{1.96\cdot 5\theta}{\sqrt{12n}} \le \overline{Y} - 1.5\theta \le \frac{1.96\cdot 5\theta}{\sqrt{12n}}$$ Add $1.5\theta$ to all parts: $$-\frac{1.96\cdot 5\theta}{\sqrt{12n}}+1.5\theta \le \overline{Y} \le \frac{1.96\cdot 5\theta}{\sqrt{12n}}+1.5\theta$$ Take $\theta$ out of brackets: $$\theta\left(\frac32-\frac{9.8}{\sqrt{12n}}\right) \le \overline{Y} \le \theta\left(\frac32+\frac{9.8}{\sqrt{12n}}\right)$$ Solving both inequalities with respect to $\theta$ for sufficiently large $n$ which guarantee both brackets are positive, we have: $$\frac{\overline{Y}}{\frac32+\frac{9.8}{\sqrt{12n}}}\le\theta\le \frac{\overline{Y}}{\frac32-\frac{9.8}{\sqrt{12n}}} $$ Note that $$\frac{1}{\frac32\pm\frac{9.8}{\sqrt{12n}}}\neq \frac23\pm\frac{\sqrt{12n}}{9.8}$$
To get confidence interval for $\theta^2$, simply square both ends of this interval. The only problem is that $\overline{Y}$ can be negative but for $n$ sufficiently large probabililty of this event is negligible since $\overline{Y}\to 1.5\theta >0$ in probabililty as $n\to\infty$.