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.


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.

  • $\begingroup$ It now seems rather simple! Thank you! $\endgroup$ – keksain Feb 23 '13 at 20:08
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Feb 23 '13 at 20:18

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