Find a sufficient statistic for $\theta$ and show that a UMP test of $H_0: \theta = 6$ against $H_1:\theta <6$ is based on this statistic. [duplicate]

1. (20%) Let $X_1, \dotsc, X_n$ be an iid sample from a distribution with pdf $f(x;\theta) = \theta x^{\theta-1}, 0< x< 1$, zero elsewhere, where $\theta >0$. Find a sufficient statistic for $\theta$ and show that a UMP test of $H_0: \theta = 6$ against $H_1:\theta <6$ is based on this statistic.

I used the exponential family form to get that the summation of $\ln(x)$ is a sufficient statistic for $\theta$, but I do not know how to find a UMP Test based on this statistic. I believe it has something to do with likelihood ratios. Thanks in advance for your help.

My question is different from that question because I do not already have a MP test, these are different numbers, the pdf is different, the significance level is not given, and the alternative hypothesis is an inequality.

marked as duplicate by Seyhmus Güngören, NCh, dantopa, Claude Leibovici, user223391 May 19 '17 at 17:16

Fisher's factorization theorem says $T(X_1,\ldots,X_n)$ is sufficient for a family of distributions if the joint density can be written as $$g_\theta(T(x_1,\ldots,x_n)) h(x_1,\ldots,x_n)$$ and only the first factor depends on $\theta$. In this case the joint density is $\theta^n (x_1\cdots x_n)^{\theta-1},$ so you can take $T(x_1,\ldots,x_n)$ to be the product $x_1\cdots x_n$ and $h(x_1,\ldots,x_n)$ to be $1$. (For most of the families of distributions considered in introductory courses you get $h$ constant; an exception is the Poisson distribution.)
For any particular positive number $\theta < 6$ you have a likelihood ratio $$\frac{L_0}{L_1} = \frac{6^n (x_1\cdots x_n)^{6-1}}{\theta^n(x_1\cdots x_n)^{\theta-1}} = \left( \frac 6 \theta \right)^n (x_1\cdots x_n)^{6-\theta}.$$ The likelihood ratio test rejects the null hypothesis if this ratio is too small. Since $6-\theta>0,$ this ratio gets smaller as the product $x_1\cdots x_n$ gets smaller, or equivalently, if the sum of the logarithms gets smaller. The Neyman–Pearson lemma says that test is more powerful than any other test regardless of the value of $\theta\in(0,6).$