Five identical components are run to failure

PHYSICS · Multi-part · Fermi University Physics · Scientific Practice & Experimental Methods · Mathematical Methods for Physics · Data analysis and model fitting in the laboratory

Error, Chance and the Machine, Ch Tests, intervals and Bayesian inference

Problem

Five identical components are run to failure. Their lifetimes x1,…,x5x_1,\ldots,x_5, in hours, are independent, each with the exponential density f(x;τ)=τ−1e−x/τf(x;\tau)=\tau^{-1}e^{-x/\tau} for x≥0x\ge0, where τ\tau is the mean lifetime. The two simple hypotheses are H0:τ=1.0 hH_0:\tau=1.0\ \mathrm{h} and H1:τ=2.0 hH_1:\tau=2.0\ \mathrm{h}. ((a)) Form the likelihood ratio Λ(x)=f1(x)/f0(x)\Lambda(\mathbf{x})=f_1(\mathbf{x})/f_0(\mathbf{x}) for the five lifetimes. Show that it depends on the data only through the total lifetime S=x1+⋯+x5S=x_1+\dots+x_5 and increases with SS, and state the form of the most powerful test. ((b)) The characteristic function of one lifetime with mean τ\tau is (1−iτt)−1(1-i\tau t)^{-1}. Show that when the true mean is τ\tau, the variable 2S/τ2S/\tau follows χ102\chi^2_{10}. ((c)) The upper 55 per cent point of χ102\chi^2_{10} is 18.30718.307. Find the cutoff cc, in hours, such that rejecting H0H_0 when S>cS>c has size 0.050.05. ((d)) Using Q(χ2;ν)=e−χ2/2∑i=0ν/2−1(χ2/2)i/i!Q(\chi^2;\nu)=e^{-\chi^2/2}\sum_{i=0}^{\nu/2-1}(\chi^2/2)^i/i!, find the power of this test against H1H_1.

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