Is $\hat{\theta} = 2\bar{X}$ a sufficient estimator for $\theta$? How can I prove it?

Question

Given a random sample $\underline{X}_{(n)}$ of a random variable $X\sim U(0,\theta)$, is $\hat{\theta} = 2\bar{X}$ a sufficient estimator for $\theta$ ?
Given $T(\underline{X}_{(n)}) = 2\bar{X}$, I tried through $f_{\underline{X}_{(n)}\mid T=t}(\underline{x}_{(n)})$ and try to see that $\frac{f_{\underline{X}_{(n)}}(\underline{x}_{(n)})}{f_{T}(t)}$ given $T(\underline{x}_{(n)}) = t$ does not depend on $\theta$ , but I cannot find the distribution of the parameter. I tried going from $X \sim U(0,\theta)$ to $Y \sim U(0,1)$ by a linear transformation, fruitlessly.

Answer 1

No, it's not. Suppose $\theta=1000$ and $\overline X=1,$ with sample size $100.$ Then the probability that at least one observation in the sample is more than $10$ is positive. But if all is as above except that $\theta=8,$ then that probability is $0.$ In other words, the conditional distribution of the data given $\overline X$ depends on $\theta.$ Thus $\overline X$ is not a sufficient statistic for this family of distributions.