# Hypothesis testing for mean

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?

• 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. – Dilip Sarwate Nov 10 '11 at 20:17

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 $T$.