# Using Neyman-Pearson Lemma to show that the most powerful test has critical region of the form:

A certain type of electronic component is susceptible to instantaneous failure at any time. The lifetime, T, of such as component can be modelled by an exponential distribution with probability density function $$\begin{cases} \frac{1}{\theta}e^{-t/\theta} , & t>0 \\ 0, & \text{otherwise} \end{cases}$$ $\theta$ is the expected lifetime of the components. $t_1,t_2,...,t_{81}$ are observed lifetimes for a random sample of size 81 of these components.

Consider testing the hypotheses $$H_0 : \theta = 10 \quad against \quad H_1 : \theta = 15$$

Use the Neyman-Pearson Lemma to show that the most powerful test has critical region of the form $$\begin{cases} t: \frac{\sum_{i=1}^{81} t_i}{81} = \bar t > k \end{cases}$$

So far, I've got this: According to my notes, the most powerful size $\alpha$ test has critical region: $$C = t : \frac{\mathrm{L}(t;H_0)}{\mathrm{L}(t;H_1)} \le k$$

$\mathrm{L}(\theta;t_1,...,t_{81}) = \prod_{i=1}^{81} f(t_i,\theta) = \prod_{i=1}^{81} \frac{1}{\theta}e^{-t/\theta} = \frac{1}{\theta^{81}}e^{\frac{-\sum_{i=1}^{81} t_i}{\theta}}$

Now If I were to let $\theta = H_0 \ and \ H_1$, I would just get this: $$\frac{\frac{1}{10^{81}}e^{\frac{-\sum_{i=1}^{81} t_i}{10}}}{ \frac{1}{15^{81}}e^{\frac{-\sum_{i=1}^{81} t_i}{15}}}$$

Which doesn't really get me anywhere? Where am I going wrong?

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– NCh
Apr 20 '17 at 21:08

Your setup is: reject if $\frac{\frac{1}{10^{81}}e^{\frac{-\sum_{i=1}^{81} t_i}{10}}}{ \frac{1}{15^{81}}e^{\frac{-\sum_{i=1}^{81} t_i}{15}}} \le k$ for some $k$ sufficiently small. Ultimately, $k$ will be determined by your null hypothesis, so I'll just reduce it to your "form" and then you explicitly find $k$ so that $P ( C \le k ; H_0) = \alpha$.
Multiplying by constants (and writing generic constant $k$ at each step) gives $\frac{e^{\frac{-\sum_{i=1}^{81} t_i}{10}}}{ e^{\frac{-\sum_{i=1}^{81} t_i}{15}}} \le k$, which means $e^{ x (- \frac{1}{10} + \frac{1}{15})} \le k \iff e^{-x/30} \le k$, where $x = \sum_{i=1}^{81} t_i$.
Now, you ask for what values of $x$ is this satisfied -- i.e. when is $e^{-x/30}$ sufficiently small? The answer is for $x$ large, so our rejection region is $\sum_{i=1}^{81} t_i \ge k$ where $k$ is determined as above.
Also, dividing by sample size won't change inequalities, so this is the same as having rejection region $\frac{1}{81} \sum_{i=1}^{81} t_i \ge k$. You'll get the same answer either way.