Applying Karlin-Rubin I have trouble understanding how to apply Karlin-Rubin
My situation: $X_1 ... X_5\sim Poisson(\lambda)$ independent. My hypothesis test is
$$H_0: \lambda \leq 3$$
$$H_1: \lambda > 3$$
I checked that $\sum_i x_i$ is a sufficient MLR statistic for Poisson. Find a UMP level $\alpha = 0.05$ test.
Complementary data:
\begin{matrix}
Observation: & 1 & 2 & 3 & 4 & 5 \\
Count:& 2 & 5 & 4 & 7 & 5
\end{matrix}
Theorem: Consider testing $H_0:\theta \leq \theta_0$ vs. $H_1:\theta > \theta_0$ and consider $T$ to be a sufficient statistic for $\theta$ with MLR property on $\theta$. Then for any $t_0$, the test that rejects $H_0$ iff $T>t_0$ is a UMP level $\alpha$ test, where $\alpha = P_{\theta_0}(T>t_0)$
My attempt:
I know $\sum x_i \sim Poisson(5\lambda)$ so I can write
$$\alpha = 0.05  = \Pr_{\lambda = 3}\left(\sum x_i > t_0\right)$$
$$ = 1 - CDF_{poisson(15)}(t_0)$$
I don't really understand what $t_0$ is in the definition and how to arrive to the test. I know the CDF of poisson so that's not a problem. I just don't really understand what the last sentence of the theorem is saying and how to apply it.
EDIT:
From looking through some examples it seems like the test is just $I(\sum_i > t_0)$ where $1$ means to rejects $H_0$. And you just solve for $t_0$ above. Is that it?
 A: Here is a rationale for the test that I hope you will find intuitive:
A good estimate of $\lambda$ is sample mean $\bar x = \frac{\sum_{i=1}^n x_i}{n} = T/n = 23/5 = 4.6.$ This estimate is larger than the hypothetical $\lambda_0 = 3.$ So you may wonder
whether the random quantity $\bar x = \hat \lambda$ is sufficiently larger
than $\lambda_0 = 3$ to reject $H_0$ in favor of the alternative. The
question is how much larger is large enough for 'statistical significance'.
Testing at level $\alpha = 0.05,$ you are willing to make a mistake (specifically, a Type I Error) 5% of the time you do such a test.
So you seek a 'critical value' $t_0$ such that $$P(T \ge t_0\,|\,n\lambda = n\lambda_0) = P(n\hat \lambda \ge t_0\,|\,n\lambda = 15) \le .05.$$
This amounts to supposing that $T \sim \mathsf{Pois}(15)$ and finding
$t_0$ with $P(T \ge t_0) \le .05$ or $P(T < t_0) \ge .95.$ 
Because of the discrete nature of the Poisson distribution, it is not
generally possible to find a critical value that corresponds precisely
to $\alpha = .05.$
In R statistical software:
 qpois(.95, 15)
 ## 22
 ppois(22, 15)
 ## 0.9672558
 ppois(21, 15)
 ## 0.9468936

So it seems that $t_0 = 22$ is the smallest critical value that gives
$\alpha = .05$ or smaller. In other words, reject if $\hat \lambda = \bar X \ge 22/5 = 4.4.$
Notes: (a) In practice, people sometimes use a normal approximation. Here we might try
$$P(T \ge t_0) = P\left(Z= \frac{T - n\lambda_0}{\sqrt{n\lambda_0}}
\ge \frac{t_0 - 15}{\sqrt{15}} \right) \approx .05,$$
where $Z$ is standard normal. Thus $\frac{t_0 - 15}{\sqrt{15}} \approx 1.645$
and $t_0 \approx 21.4.$ (This method sweeps the discreteness of the Poisson
distribution 'under the rug' and the approximation may be inaccurate for
Poisson rates of small or moderate size.)
(b) It may be worthwhile to ask with what probability we reject $H_0$ if
the true rate is $\lambda = 5$, which is the 'power' of the test against
alternative $H_a: \lambda_a = 5.$ Then we find 
$$P(T \ge 22\,|\,\lambda =n5 = 25) = 1 -P(T \le 21\,|\,\lambda = 25) = 0.7527.$$
 1-ppois(21, 25)
 ## 0.7527012

In the plots below, the rejection region is to the right of the vertical dashed line.

