A new car is assumed to have a lifetime of 120 months. A set of 15 cars is teste

Question

A new car is assumed to have a lifetime of 120 months. A set of 15 cars is tested and the life of each car is normally distrubuted with unknown mean and assumed standard deviation of 10 months.Assume that the sample mean lifetime comes out to be 135 months.

a) Find the posterior mean and variance of the true mean lifetime.

b) An analysis of potential profits and losses indicates that the car should be marketed if the probability that the mean life exceeds 125 months is at least 0.99. Should the car be marketed?

Explanation / Answer

What's not clear from the description is the prior information about mu. Given that no prior for mu is provided I shall assume that no prior information is available and assume a very non-informative prior. That is, I shall assume that, a priori, mu has a uniform distribution on the real line. In this case, the posterior distribution for mu is a normal distribution with posterior mean equal to the sample mean xbar and posterior variance (sigma^2)/n. a) Hence, the posterior mean is 135 months. The posterior variance is 100/15 = 6.6667 (4 d.p.) months^2. b) We need to compute the posterior probability that the mean lifetime exceeds 125 months. This is P(X > 125 | data), where X| data is N(135, 6.6667). Let Z= (X-35)/sqrt(6.6667). Then Z is N(0,1). P(X > 125 | data) = P(Z > [(125-135)/sqrt(6.6667)]) = 1 - Phi( - 3.8730) = 1- 0.000054 = 0.999946. As this is larger than 0.99 the car should be marketed.

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