kasibug1v

2022-09-09

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.

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.

Johnathon Mcmillan

Beginner2022-09-10Added 7 answers

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.

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.

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