I am working on problem 10.101 in Wackerly Mathematical Statistic 7th ed. I will post the problem here for people who do not have the text:
Suppose that $Y_1,....,Y_n$ denote a random sample from a population having an exponential distribution with mean $\theta$.
a) Derive the most powerful test for $H_0:\theta = \theta_0$ against $H_a: \theta = \theta_a$ where $\theta_a < \theta_0$.
I will jump straight to the point in my problem that I am doing wrong. I set $$\frac {L(\theta_0)} {L(\theta_a)} < k$$ This is by definition of Neyman Pearson Lemma. I simplified $\frac {L(\theta_0)} {L(\theta_a)} < k$ to $$(\frac {\theta_a}{\theta_0})^n*e^{-\sum y_i * (\frac 1 {\theta_0} - \frac 1 {\theta_a})} < k$$ Now when I take the log and solve for $\sum y_i$ I get the following: $$-\sum y_i * (\frac 1 {\theta_0} - \frac 1 {\theta_a}) < \ln(k(\frac {\theta_a}{\theta_0})^n)$$ Now I divide by the negative and the rest to get $\sum y_i$. This then becomes $$\sum y_i > -\frac {\ln(k(\frac {\theta_a}{\theta_0})^n)} {\frac 1 {\theta_0} - \frac 1 {\theta_a}}$$
The answer has the opposite inequality sign as above. I called the right hand side $c$. Please let me know how they got the opposite inequality sign.
Observe that $\dfrac{1}{\theta_0}-\dfrac{1}{\theta_a}<0$, or $\color{blue}{-\left(\dfrac 1 {\theta_0} - \dfrac 1 {\theta_a}\right)} \color{red}{>}0$. Consequently, $$\color{blue}{-\left(\frac 1 {\theta_0} - \frac 1 {\theta_a}\right)}\sum_{i=1}^{n} y_i < \ln\left(k\left(\frac {\theta_a}{\theta_0}\right)^n\right) \implies \sum_{i=1}^{n} y_i < -\frac{\ln\left(k\left(\frac {\theta_a}{\theta_0}\right)^n\right)}{\frac 1 {\theta_0} - \frac 1 {\theta_a}}.$$