Let $X_1, X_2,…, X_n$ be i.i.d. $U(−\theta,\theta)$. Show that $Z_n \to N(0,\sqrt{5/9})$ in distribution, where $Z_n =\frac{\sqrt{\frac{1}{n}}\sum_{k=1}^{n}X_k}{\sqrt[4]{\frac{\sum_{k=1}^{n}X_{k}^{3}+\sum_{k=1}^{n}X_{k}^{4}}{n}}}$
My solution so far: This should just be a direct application of CLT, WLLN, and Slutsky's theorem. The top term converges by the CLT to $N(0,\sigma^2)$, but I'm not sure what $\sigma^2$ is. Is it just $Var[U(-\theta,\theta)]$? In the denominator, $\frac{\sum_{k=1}^{n}X_{k}^{3}+\sum_{k=1}^{n}X_{k}^{4}}{n}$ converges to $E[X_1^3]+E[X_1^4]$ by the WLLN. The last step is to apply Slutsky's Theorem. Since $X_i$ are $U(-\theta,\theta)$, $f(x)=\frac{1}{b-a}=\frac{1}{2\theta}$. Then $$E[X]=\int_{-\theta}^{\theta}\frac{1}{2\theta}xdx=0\\E[X^2]=\int_{-\theta}^{\theta}\frac{1}{2\theta}x^2dx=\frac{1}{3}{\theta}^3\\ E[X^3]=\int_{-\theta}^{\theta}\frac{1}{2\theta}x^3dx=0\\E[X^4]=\int_{-\theta}^{\theta}\frac{1}{2\theta}x^4dx=\frac{1}{5}{\theta}^5 $$
So $\operatorname{Var} [U(-\theta,\theta)]=\frac{1}{3}{\theta}^3$. At this point I can't get the algebra to work out. I don't know if incorrectly computed the moments.