Hypothesis testing with an exponential distribution

exponential distributionhypothesis testingprobabilityprobability distributionsstatistics

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

Best Answer

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:

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

Related Solutions

Finding the $p$-value of a test with exponential data

Yes. Distribution $\mathsf{Exp}(\lambda=3)$ with mean $1/3$ tends to give smaller values than $\mathsf{Exp}(\lambda=1)$ with mean $1.$ So you are correct to reject $H_0: \lambda = 3$ against $H_0: \lambda=1$ for large observed values $X.$

In particular, the critical value for a test at level 5% is $c = 0.999.$ That is, you would reject $H_0$ for observed $X \ge c.$

c = qexp(.95, 3);  c
[1] 0.9985774

As you say, for observed value $X = 0.30,$ the P-value is $0.466 > 0.05,$ so you do not reject.

1 - pexp(.3, 3)
[1] 0.4065697

If $H_a: \lambda = 1$ is true, then the probability of rejecting based on a single observation is $P(X \ge c\,|\,\lambda=1) = 0.3584.$ With only one observation, it is not surprising to have poor power.

1 - pexp(qexp(.95, 3), 1)
[1] 0.3684031

In the figure below, the critical value $c = 0.999$ is at the vertical red dotted line and the observed value $X = 0.5$ at the solid vertical line.

A test with only one observation may seem counter-intuitive on account of its poor power. By contrast, with $n = 10$ exponential observations, the sample mean has $\bar X_{10}\sim\mathsf{Gamma}(\mathrm{shape}=10, \mathrm{rate}=30)$ under $H_0:\lambda=3$ and $\bar X_{10}\sim\mathsf{Gamma}(10,10)$ under $H_a:\lambda=1.$ These results can be derived using moment generating functions.

Then the critical value for a test at the 5% level is $c=0.523$ and the power of this test using $\bar X_{10}$ as test statistic is about $95.9\%.$

c = qgamma(.95,10,30); c
[1] 0.5235072
1 - pgamma(c, 10, 10)
[1] 0.9588243

The corresponding graph showing null and alternative distributions of $\bar X_{10}$ and the critical value is as follows:

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