Question

At time t 0, 21 identical components are tested. The lifetime distribution of each is exponential with parameter A. The experimenter then leaves the test facility unmonitored. On his return 24 hours later, the experimenter immediately terminates the test after noticing that y 12 of the 21 components are still in operation (so 9 have failed). Derive the mle of A. [Hint: Let Y the number that survive 24 hours. Then Y Bin(n, p). What is the mle of p? Now notice that p P(X, 24), where X, is exponentially distributed. This relates A to p, so the former can be estimated once the latter has been.] (Round your answer to four decimal places.) =

Answer 1

Soluton P PIxi 2u -P/i 24 F1-F(24,2) -24л 2-1- -24A Now, Calculate A by applyug iawance prineiples . -24 Pe tate en both nde -Ln ( f) > 24 herefore The nven, ythe numbes tue Sunvive 24 hons. Then Yn Bin(n,p mle bor 24 2n (m) 24 O.0233 2 o.0233An