# Hypothesis testing with an exponential distribution

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

• Please try to format your questions properly using MathJax. Jun 11, 2020 at 14:00
• Jun 4, 2021 at 5:54

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:

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

• Hi, where does the 10 come from in the denominator?
– Tim
Jun 11, 2020 at 10:34
• $e^{-15\cdot0.1}=e^{-\frac{15}{10}}$ 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 $$\text{p-value}=\exp(-15t)=\exp(-15\times 0.1)\approx 0.22$$