Suppose that the distribution of lifetimes of TV tubes can be modelled by an exponential distribution with mean $\lambda$ so $f(x|\lambda)$ = $\frac{1}{\lambda}$ $e^\frac{-x}{\lambda}$, $x>0$. Under usual production conditions the mean lifetime is $50$ hours but if a fault occurs in the process the mean lifetime drops to $40$ hours A random sample of $20$ tube lifetimes is to taken in order to test the hypothesis $H_0:\lambda = 50$ vs $H_1:\lambda = 30$ and the following statistics is observed $\sum_{i=0}^n X_i = 680$. Use Neyman–Pearson lemma to find the most powerful test with signficance level $\alpha = 0.05$.
My attempt was to start with the joint probability ratio under $H_0$ and $H_1$ $$\frac{\frac{1}{50^{20}}e^{\frac{-\sum_{i=1}^{20} x_i}{50}}}{ \frac{1}{30^{20}}e^{\frac{-\sum_{i=1}^{20} x_i}{30}}} .$$
We reject $H_0$ when $(\frac{1}{\lambda_1} -\frac{1}{\lambda_0}) \sum_{i=1}^n X_i <k$.
Since in the parenthesis we have a positive quantity should we reject $H_0$?
I do not understand the importance of the data $40$ since the alternative hypothesis is $30$ and not $40$.
Is this the right way to solve this exercise? I am trying to understand Neyman–Pearson lemma and any help would be appreciate.