Consider a population of independent light bulbs with an exponential lifetime distribution with mean $\mu$. It is claimed that their expected lifetime is 1000 hours. A definition of a 100(1−)% confidence interval obtained from an observation t0 is the set of all $\mu_0$ which are not rejected in a test of a null hypothesis $\mu$= $\mu_0$ against an alternative hypothesis $\mu$$\neq$$\mu_0$.

One particular light bulb fails after 622 hours. Solve the equations of the two significance probabilities Pr(T ≥ 622 |$\mu_0$) = 0.05 (for a test of $\mu$= $\mu_0$ versus $\mu$> $\mu_0$) and Pr(T ≤ 622 | 0) = 0.05 (for a test of $\mu$= $\mu_0$ versus $\mu$< $\mu_0$) for $\mu$. Determine the range of values of such that both of the probabilities Pr(T ≥ 622 | $\mu$) and Pr(T ≤ 622 |$\mu$) are at least 0.05. (This range gives an equi-tailed 90% confidence interval for $\mu$.)

What inference may be drawn from this interval about the claim that the expected lifetime is 1000 hours?


1 Answer 1


From wikipedia, http://en.wikipedia.org/wiki/Exponential_distribution, the probability density function (pdf) of an exponential distribution is

$$f(x;\lambda) = \begin{cases} \lambda e^{-\lambda x}, & x \ge 0, \\ 0, & x < 0. \end{cases}$$

where $λ > 0$ is the parameter of the distribution, often called the rate parameter. The distribution is supported on the interval $[0, ∞)$. If a random variable $X$ has this distribution, we write $X \tilde{} Exp(λ)$.

The expected value of an exponentially distributed random variable $X$ with rate parameter $λ$ is given by

$$\mathrm{E}[X] = \frac{1}{\lambda}. \!$$

So knowing the mean, we know the distribution's parameter.

Then, you want to compute the critical values of this test. The cumulative distribution function is given by

$$F(x;\lambda) = \begin{cases} 1-e^{-\lambda x}, & x \ge 0, \\ 0, & x < 0. \end{cases}$$

which should be easy enough to invert: for a given cumulative probability, find the $x$ (value on the x-axis).


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