"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

kasibug1v 2022-09-09 Answered
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− α)% confidence interval obtained from an observation to is the set of all μ 0 which are not rejected in a test of a null hypothesis μ 0 against an alternative hypothesis .
One particular light bulb fails after 622 hours. Solve the equations of the two significance probabilities Pr(T " " 622 | μ 0 ) = 0.05 (for a test of μ 0 versus μ 0 ) and Pr(T " " 622 | μ 0 = 0.05 (for a test of μ 0 versus μ 0 ) for μ. Determine the range of values of μ such that both of the probabilities Pr(T " " 622 | μ) and Pr(T " " 622 | μ) are at least 0.05. (This range gives an equi-tailed 90% confidence interval for μ.)
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 | μ 0 ), Pr(T 622 | μ 0 ) and compare with 0.05? I believe I will get the second part after I understand this bit.
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Answers (1)

Johnathon Mcmillan
Answered 2022-09-10 Author has 7 answers
I also found the wording somewhat difficult.
For the first problem, you need to find the value of μ 0 such that an exponentially distributed random variable T with mean μ 0 has probability 0.05 of being 622. You probably know what the preceding sentence means. But to make sure, we sketch the calculation.
An exponentially distributed rv with mean μ 0 has density function 1 μ e t / μ 0 (for t 0), and therefore cdf 1 e t / μ 0 . So the probability that T 622 is e 622 / μ 0 .
Set this equal to 0.05 and solve for μ 0 .
For the second problem, we want to find μ 0 such that an exponential T with mean μ 0 is 622 622 with probability 0.05.
The range for μ 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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