Uniformly Most Powerful test for normal distribution Let ${Y_1,...,Y_n}$ be independent random variables and $Y_i$~$N(\beta x_i, 1)$ where $x_1,...,x_n$ are fixed known constants, and $\beta$ is an unknown parameter.  
I'm trying to find a uniformly most powerful level $\alpha$ test for 
$$H_0: \beta=0 \quad \text{vs} \quad H_1:\beta>0$$
First I found the Likelihood Ratio statistic
$$\begin{align}\frac{L(0)}{L(\beta_1)}&=\frac{e^{-\frac{1}{2}\sum^n_{i=1}(y_i-0x_i)^2}}{e^{-\frac{1}{2}\sum^n_{i=1}(y_i-\beta_1x_i)^2}}
\\[15pt] & = \frac{e^{-\frac{1}{2}\sum^n y_i^2}}{e^{-\frac{1}{2}(y_i^2-2y_i\beta_1x_i+\beta_1^2x_i^2)}}  
\\[15pt] & =e^{\beta_1\sum^n y_ix_i-\frac{1}{2}\beta_1^2\sum^nx_i^2}\end{align}$$
Then the rejection region of the LRT is
$$e^{\beta_1\sum^n y_ix_i-\frac{1}{2}\beta_1^2\sum^nx_i^2} \le k \\$$
Now moving the constants to the right side
$$\\[15pt]\Rightarrow\beta_1\sum^n y_ix_i-\frac{1}{2}\beta_1^2\sum^nx_i^2 \;\le\; \text{ln}(k)$$
$$\\[15pt]\Rightarrow\sum^ny_ix_i \;\le\; \frac{\text{ln}(k)+1/2\sum^n\beta_1^2x_i^2}{\beta_1}$$
Here is where I get confused because I'm not sure what to leave on the left side. From what I have read I'm supposed to move constants to the right. I'm thinking that only $y_i$'s should remain on the left. Since $x_i$'s are constants shouldn't I move it?
The steps that follow to find the rejection region by using the distribution of the left of the inequality, then equating to a quantile, puzzle me because of the $x_i$
I'm not even sure I'm going about this correctly or if there are more straight forward methods.
 A: First of all, you lose a sign:
\begin{align}\frac{L(0)}{L(\beta_1)}&=\frac{e^{-\frac{1}{2}\sum^n y_i^2}}{e^{-\frac{1}{2}(y_i^2-2y_i\beta_1x_i+\beta_1^2x_i^2)}}  
\\[15pt] & =e^{-\beta_1\sum^n y_ix_i+\frac{1}{2}\beta_1^2\sum^nx_i^2}\end{align}
The rejection region looks as 
$$
-\beta_1\sum_{i=1}^n y_ix_i+\frac{1}{2}\beta_1^2\sum_{i=1}^nx_i^2 \leq k.
$$
Since only $y_i$ are random variables here, and $k$ should be found so that test has level $\alpha$, simplify rejection region:
$$
\sum_{i=1}^n y_ix_i \geq k_0.
$$
Let us find $k_0$ from type 1 error $\alpha_1=\alpha$. We need to know the distribution of l.h.s. under $H_0$:
$$
\sum_{i=1}^n y_ix_i \sim N\bigl(0, \sum_{i=1}^n x_i^2\bigr).
$$
Then
$$
\alpha_1=\mathbb P_{H_0}\left(\sum_{i=1}^n y_ix_i \geq k_0\right)=P_{H_0}\left(\frac{\sum_{i=1}^n y_ix_i}{\sqrt{\sum_{i=1}^n x_i^2}} \geq \frac{k_0}{\sqrt{\sum_{i=1}^n x_i^2}}\right)=1-\Phi\left(\frac{k_0}{\sqrt{\sum_{i=1}^n x_i^2}}\right)=\alpha
$$
if 
$$
\frac{k_0}{\sqrt{\sum_{i=1}^n x_i^2}} = z_{1-\alpha},\ \quad k_0 = z_{1-\alpha}\cdot\sqrt{\sum_{i=1}^n x_i^2},
$$
where $z_{1-\alpha}$ is the $(1-\alpha)^{\text{th}}$ quantile of standard normal distribution.
Finally, we've got rejection region of level $\alpha$ test for testing $H_0$ under the simple alternative $H_1: \beta=\beta_1>0$: 
$$
\sum_{i=1}^n y_ix_i \geq z_{1-\alpha}\cdot\sqrt{\sum_{i=1}^n x_i^2}.
$$
Since this test 
1) is the most powerful test for any $\beta_1>0$,
2) does not depend on $\beta_1>0$ at all
is is the uniformly most powerful test for testing $H_0$ under the composite alternative $H_1: \beta>0$.
