Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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−$\alpha$)% 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 |$\mu_0$) = 0.05 (for a test of $\mu$= $\mu_0$ versus $\mu$< $\mu_0$) for $\mu$. Determine the range of values of $\mu$ 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 $\mu$.)

I don't seem to understand what they mean by 'solve the equations'. Do I have to find a specific value for T or compute Pr(T ≥ 622 |$\mu_0$), Pr(T ≤ 622 |$\mu_0$) and compare with 0.05? I believe I will get the second part after I understand this bit.

share|cite|improve this question
up vote 1 down vote accepted

I also found the wording somewhat difficult.

For the first problem, you need to find the value of $\mu_0$ such that an exponentially distributed random variable $T$ with mean $\mu_0$ has probability $0.05$ of being $\ge 622$. You probably know what the preceding sentence means. But to make sure, we sketch the calculation.

An exponentially distributed rv with mean $\mu_0$ has density function $\frac{1}{\mu}e^{-t/\mu_0}$ (for $t\ge 0$), and therefore cdf $1-e^{-t/\mu_0}$. So the probability that $T\ge 622$ is $e^{-622/\mu_0}$.

Set this equal to $0.05$ and solve for $\mu_0$.

For the second problem, we want to find $\mu_0$ such that an exponential $T$ with mean $\mu_0$ is $\le 622$ with probability $0.05$.

The range for $\mu_0$ that you will get from the two calculations turns out to be very large. Basically that says you can't learn much about the mean lifetime of lightbulbs by observing a single one.

share|cite|improve this answer
It now seems rather simple! Thank you! – keksain Feb 23 '13 at 20:08
You are welcome. Confidence intervals, and tests at significance level $\alpha$, are quite subtle notions. It will take some time before you are fully comfortable with them. – André Nicolas Feb 23 '13 at 20:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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