Introduction to Probability and Statistics for Engineers and Scientists
584 Chapter 14*:Life Testing whereas the corresponding probability for a smoker is, by the same reasoning, P{A-year-old smoker r ...
14.3The Exponential Distribution in Life Testing 585 and so, by independence, the joint probability density ofXij,j=1,...,ris fX ...
586 Chapter 14*:Life Testing and so ∂ ∂θ logL(x 1 ,...,xr,i 1 ,...,ir)=− r θ + ∑r i= 1 xi θ^2 + (n−r)xr θ^2 Equating to 0 and so ...
14.3The Exponential Distribution in Life Testing 587 In general, we have Y 1 =nX(1) Y 2 =(n−1)(X(2)−X(1)) .. . Yj=(n−j+1)(X(j)−X ...
588 Chapter 14*:Life Testing That is, 2τ/θhas a chi-square distribution with 2rdegrees of freedom. Hence, P { χ 12 −α/2,2r< 2 ...
14.3The Exponential Distribution in Life Testing 589 SOLUTION Since 2τ/θ 0 =3,600/150=24, thep-value is p-value=P{χ 402 ≤ 24 } = ...
590 Chapter 14*:Life Testing Thus, for instance, if in Example 14.3b the true mean life was 120 hours, then the expectation and ...
14.3The Exponential Distribution in Life Testing 591 From Equation 14.3.8, we obtain that the likelihood of the datar,x 1 ,...,x ...
592 Chapter 14*:Life Testing increases inθ(why?). Hence, ifθ<θL, thenPθ{N(T)≤r}<PθL{N(T)≤r}= α 2 ifθ>θU, thenPθ{N(T)≥r} ...
14.3The Exponential Distribution in Life Testing 593 EXAMPLE 14.3c If a one-at-a-time sequential test yields 10 failures in the ...
594 Chapter 14*:Life Testing SOLUTION This is a one-sided test of H 0 :θ≥25 versus H 1 :θ< 25 The relevant probability for de ...
14.3The Exponential Distribution in Life Testing 595 Hence, ∂ ∂θ logf(i 1 ,...,ir,x 1 ,...,xr)=− r θ + ∑r 1 xi+(n−r)T θ^2 Equati ...
596 Chapter 14*:Life Testing which time they had been on test for respective timesy 1 ,...,ys. The likelihood of this outcome wi ...
14.3The Exponential Distribution in Life Testing 597 foregoing we see that f(data|λ)=Kλre−λt If we suppose prior to testing, tha ...
598 Chapter 14*:Life Testing 3.9, 4.6, 5.8. The 10 items that did not fail had, at the time the test was terminated, been on tes ...
14.4A Two-Sample Problem 599 Hence, by the equivalence of the gamma and chi-square distribution it follows that 2 θ 1 ∑n i= 1 Xi ...
600 Chapter 14*:Life Testing SOLUTION The value of the test statisticX/Yis 42/34=1.2353. To compute the prob- ability that anF-r ...
14.5The Weibull Distribution in Life Testing 601 Weibull (1, 0.5) x 012345 1.2 1.0 0.8 0.6 0.4 0.2 0.0 Weibull (1, 2) x 012345 1 ...
602 Chapter 14*:Life Testing or, equivalently, αˆ= n ∑n i= 1 x βˆ i n+βˆlog (n ∏ i= 1 xi ) = nβˆ ∑n i= 1 x βˆ i logxi ∑n i= 1 x ...
14.5The Weibull Distribution in Life Testing 603 could conclude that yi≈βlogx(i)+logα, i=1,...,n (14.5.4) We could then chooseαa ...
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