3
$\begingroup$

I have a system with two components, component A and component B. Let $X$ be the random variable representing time between failures of component A and $X$ is exponentially distributed. Let $E[X]=m$. Now we want to test the following hypothesis:

null hypothesis: $m\le m_0$ where $m_0$ is some given constant alternate hypothesis: $m>m_0$

To test the hypothesis we need samples of $X$ which are not available. But, we have the information that A has a failure rate lower than B (time between failures for B is also exponentially distributed). MTTF of component B is $n_0$. Can we use samples generated from B and use as samples of A to test the above hypothesis?

$\endgroup$
  • 1
    $\begingroup$ You know that the failure rate $h_A$ of A is smaller than the failure rate $h_B$ of B. Since the failures of B are exponentially distributed with mean $n_0$, we have that $h_B=1/n_0$. Since X is also exponentially distributed,$$h_A=\frac{1}{E[X]}=\frac{1}{m}<h_B=\frac{1}{n_0},$$ that is, $m>n_0$. So if $n_0$ is a known quantity, then the null hypothesis $m\leq m_0$ is trivially false if $m_0\leq n_0$ and untestable if $n_0<m_0$ since samples of B have no information about A. If $n_0$ is not known, test null hypothesis $n_0\leq m_0$ and if you reject it, then reject $m\leq m_0$ also. $\endgroup$ – Dilip Sarwate Nov 10 '11 at 20:17
1
$\begingroup$

Let A have MTTF of $X\sim\text{Exp}(\mu)$

$H_0: \mu\leq \mu_0$
$H_1: \mu> \mu_0$

Let B have MTTF of $Y\sim\text{Exp}(\nu)$

We know $\nu<\mu$ (B has a higher failure rate, so shorter lifetime)

Hence, we can test

$H_0: \nu\leq\mu_0$
$H_1: \nu> \mu_0$

and if we reject that, we can reject it for A. (If B's lifetime is at least $\mu_0$, so is A's.)

So how do we test it for B?

The likelihood ratio test is equivalent to using the statistic $T=\bar{y}$, which under the equality part of the null is $\sim \text{Gamma}(n,\mu_0/n)$ (for the shape-scale parameterization), and we will reject for large values of $T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.