# "Equations of significance probabilities Consider a population of independent light bulbs with an exponential lifetime distribution with mean μ. 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=!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 versusmu< mu_0) for mu. Determine the range of values of μ such that both of the probabilities Pr(T >= 622 |mu) and Pr(T =< 622 |mu) are at least 0.05. (This

Equations of significance probabilities
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 to is the set of all ${\mu }_{0}$ which are not rejected in a test of a null hypothesis ${\mu }_{0}$ against an alternative hypothesis $\ne$.
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 ${\mu }_{0}$ versus ${\mu }_{0}$) and Pr(T "" 622 |${\mu }_{0}$ = 0.05 (for a test of ${\mu }_{0}$ versus ${\mu }_{0}$) 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 $\ge$ 622 |${\mu }_{0}$), Pr(T $\le$ 622 |${\mu }_{0}$) and compare with 0.05? I believe I will get the second part after I understand this bit.
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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 $\le 622$ 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.