I have the following problem:

Given the data $X_1, X_2, \ldots, X_{15}$ which we consider as a sample from a distribution with a probability density of $\exp(-(x-\theta))$ for $x\ge\theta$.

We test the $H_0: \theta=0$ against the $H_1: \theta>0$. As test statistic $T$ we take $T = \min\{x_1, x_2, \ldots, x_{15}\}$ . Big values for $T$ indicate the $H_1$. Assume the observed value of $T$ equals $t=0.1$.

What is the p-value of this test?

Hint: If $X_1, X_2,\ldots,X_n$ is a sample from an $\operatorname{Exp}(\lambda)$ distribution, than $\min\{X_1, X_2,\ldots,X_n\}$ has an $\operatorname{Exp}(n\lambda)$ distribution.

The solution says 0.22.

I know that the first question you have to ask youself regarding the p-value is:

"What is the probability that the H0 would generate a sample θ>0?"

So I assume H0 is true and take θ = 0. The probability-density function becomes:

f(x) = Exp(-x). I take up the hint, so I make it f(x) = Exp(-nx)

This is where I get stuck. I don't know how to proceed with the information given:

Assume the observed value of T equals t=0.1.

Can I have feedback on this problem?

Thanks, Ter

  • $\begingroup$ Please try to format your questions properly using MathJax. $\endgroup$ Jun 11, 2020 at 14:00
  • $\begingroup$ Related Q&A. $\endgroup$
    – BruceET
    Jun 4, 2021 at 5:54

2 Answers 2


If yuo are familiar with models having a Monotone Likelihood Ratio, in this case the p-value can be easily calculated (under $H_0$) in the following way:

$e^{-\frac{15}{10}}\approx 0.22$

  • $\begingroup$ Hi, where does the 10 come from in the denominator? $\endgroup$
    – Tim
    Jun 11, 2020 at 10:34
  • $\begingroup$ $e^{-15\cdot0.1}=e^{-\frac{15}{10}}$ $\endgroup$
    – tommik
    Jun 11, 2020 at 10:35

You know that the test statistic under the null hypothesis has distribution $$T\sim \operatorname{Exp}(n)=\operatorname{Exp}(15)$$ The weight of the tail of this distribution is $$\operatorname{P}(T>t)=\exp(-15t)$$

We reject null at significance level $$\alpha \leq \operatorname{P}(T>t)=\exp(-15t)$$

p-value is maximum significance level at which we reject null, so $$\text{p-value}=\exp(-15t)=\exp(-15\times 0.1)\approx 0.22$$


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