The following is a homework problem and I am not really sure where to begin or how find what the question is asking.

Suppose that one observation from the exponential pdf $f_{y}(y)=\lambda e^{-\lambda y}$, $y>0$ is to be used to test $H_0: \lambda=1$ versus $H_1: \lambda <1$. The decision rule calls for the null hypothesis to be rejected if $y \geq \ln 10$. Find $\beta$ as a function of $\lambda$.


Recall that we make a Type II error if we fail to reject the null hypothesis when it is in fact false. The answer depends on how false it is. So it depends on (is a function of) the true value of $\lambda$.

We will reject the null hypothesis if the result is $\ge \ln 10$. So we will fail to reject if the result is $\lt \ln 10$.

Let $Y$ be an exponentially distributed random variable with parameter $\lambda$. then $$\Pr(Y\lt \ln 10)=\int_0^{\ln 10}\lambda e^{-\lambda t}\,dt.$$ Integrate. Actually, we don't quite need to, if we know the cdf of the exponential. We get $1-e^{(\ln 10)\lambda}$. This can be expressed alternately as $1-10^{-\lambda}$.

Remark: Note that the probability of Type II error is very high, unless $\lambda$ is quite close to $0$. But one cannot expect much from a statistical test based on a single observation!

The probability of Type I error is, by way of contrast, fairly low. For suppose that $\lambda=1$. Then $\Pr(Y\ge \ln 10)=\frac{1}{10}$.


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