Calculate significance level and power for exponential distribution

Question

My textbook is unclear about how to do this with exponential distributions (it spends the whole time talking about normal ones), so I'm asking this here to see if I got it right.

The question mentions a statistical test such that:

$H_{0} : \lambda = 2$

$H_{1} : \lambda > 2$

$H_{0}$ is rejected if $X > 3$

(There's only one data point).

a) Calculate the significance level.

Assuming $H_{o}$ is true, we have that $\frac{1}{\lambda} = 2$ and $\lambda = \frac{1}{2}$.

$$P(X > 3\mid H_0) = 1 \ - P(X \leq 3)$$

$$1 - e^{-\lambda (3)} = 1 - e^{-1.5} = 0.77686983985$$

b) Calculate the power level when $\lambda = 2.5$

\begin{align} & P(X > 0.7769\mid\lambda = 2.5) \\[10pt] = {} & 1 - e^{-\lambda(0.7769)} = 1 - e^{-2.5(0.7769)} = 0.8566 \end{align}

c) If the actual value of $\lambda$ is $2.5$, is this procedure going to make a good decision?

Yes, because the power is high?

Why I think I'm missing something: I haven't used the info $H_1 : \lambda > 2$ given in the question

EDIT: I don't know why I said $\frac{1}{\lambda} = 2$. I was thinking about something else. Here's the alternative answer:

Significance level

\begin{align} & P(X > 3\mid H_0) = 1 - P(X \leq 3) \\[10pt] = {} & 1 - e^{-3(2)} = 0.9975 \end{align}

Power

$$P(X > 0.9975\mid\lambda = 2.5)$$

$$1 - e^{-2.5(0.9975)} = 0.9174$$

Answer 1

As stated the problem makes no sense, and not surprisingly, you have made some mistakes in dealing with exponential probabilities. I will do my best to write something helpful.

Suppose $H_0: \mu = 2$ and $H_a: \mu > 2.$ Then it makes sense to reject if $X > 3.$

Because large values of $X$ would be associated with large parameter $\mu$ and small values of rate $\lambda = 1/\mu,$ it would make no sense to reject for large $X$ if the null and and alternative hypotheses were $H_0: \lambda = 2$ and $H_a: \lambda > 2.$

Then the significance level is $$P(\text{Rej}\,|\,\mu = 2) = P(X > 3\,|\,\mu = 2) = e^{-\lambda(3)} = e^{-1.5} = 0.2231.$$

In R statistical software, the exponential functions use the rate $\lambda = 1/\mu$ as parameter and pexp is an exponential CDF. So the computation in R is:

 1 - pexp(3, rate=1/2)
 [1] 0.2231302
 exp(-1.5)
 [1] 0.2231302

The power against the alternative that $\mu = 2.5$ is $$P(\text{Rej}\,|\,\mu = 2.5) = P(X > 3\,|\, \mu= 2.5) = e^{-\lambda(3)} = e^{-(1/2.5)(3)} = e^{-1.2} = 0.3012.$$

 1 - pexp(3, rate=0.4)
[1] 0.3011942
exp(-1.2)
[1] 0.3011942

The figure below shows the null distribution with $\mu = 2,\,$ $(\lambda = 0.5)$ (in blue) and the particular alternative distribution with $\mu = 2.5,\,$ $(\lambda = 0.4)$ (in red). The rejection region is the interval $(3, \infty).$ Rejection probabilities are areas under the curves to the right of the vertical dotted line. The probability of rejection is a little higher under the alternative. This is a difficult testing situation because there is so little difference between the null and alternative distributions.

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Note: Tests based on only one observation are seldom of practical use. Instead, suppose $n = 100,\,$ $H_0: \mu = 2,\,$ $H_a: \mu > 2.$ Then it makes sense to reject $H_0$ when $\bar X > 2.34,$ which gives significance level about $\alpha = 0.05.$

Using moment generating functions, one can show that $\bar X \sim \mathsf{Gamma}(shape = 100, scale = n\lambda),$ where $\lambda = 1/\mu.$ Then $E(\bar X) = \mu$ and $SD(\bar X) = \mu/\sqrt{n} = \mu/10.$

The significance level is 5% and the power against the alternative $\mu = 2.5,\, \lambda = 0.4$ is about 73%.

1 - pgamma(2.34, 100, 100*.5)
[1] 0.04997338
1 - pgamma(2.34, 100, 100*.4)
[1] 0.7325898

A plot of density curves of the distribution of $\bar X$ for $\mu = .5$ (null hypothesis) and $\mu = .4$ (specific alternative) are much different, so that a practical test is possible.