Find a pivotal quantity and use it to approximate a 95% confidence interval 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. 
 A: 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 $n\to\infty$. 
