Statistical Analysis Of Constant Stress Accelerated Life Testing Under Multiply Type-Ⅱ Censoring When It Follows Weibull

First, with the development of science and technology, the high dependability products are getting more and more, the test can not meet the need of this kind , so we adopt and accelerate life test. It is generally more difficult to deal with the lacked data because various kinds of reasons will often meet the phenomenon that the test data will be lacked in the course of testing. When the life of product follows Weibull distribution, this paper has disscussed the statistics analytical methods of multiply Type-II censored data which comes from constant strss accelerated life testing. And has imitated comparing to various kinds of ones that estimated finely.Basic research question1 Add and test going on permanently as follows, choose 1 stress levels at firstS₁ < S₂ < ... < S_lThese l stress levels are higher than the normal stress level S₀ , Then collect n_i products at random which are placed on stress S_i, Test stops when the losing efficiency counts and reachs straight integer r_i. Set invalid data of order of the products as in test0 = t_ir0 < t_ir1 < t_ir2 < ... < t_iri i= 1,2,..., l Consider situation that test data lacks, is it set up i pieces of stress test data of competenc only left tok_i there is no harm in, Its fetching value is successively0 = t_iri0 < t_iri1 < t_iri2 < ... < t_iriki-1 irik_i , i= 1,2,..., l A stress under the assuming of the above S_i likelihood function of the test data and logarithm likelihood function are respectively underIn Li = ï¿¡ [in m - m In m + (m - 1) In *iri. - (^f-)+ jE (r,0+1) -ry - l)ln{exP [-(^)?] - exp [-(^Because tests are separate under every stress level, so likelihood function of test data and logarithm likelihood function are respectivelyAsk and simply lead the parameter separately, the great likelihood that can get the parameter is estimated .2 The BLUE of three parametersSince i pieces of stress test data of competence only have ki left one,0_ï¿¡. <ï¿¡?. is equal.But because of experimental randomness, the di might not be self-same, so it is asked to appear I pieces of di to demand , tries to get a common a.Sincei i iE(a) = J2*li=land cfi, ? â–  â– , di are separate , thenVar(a) = Var{ Utilize and ride the sub law Lagrangianly to getAmong them var(&i) = l^mC2 ? so1m = —. athe estimation of a, b uses Gauss-Markov model to get that the BLUE of a, b isS GH-IM t a - EG-P 'among themS GH-IM t EM-IH a - EG-P ' ° BGM2then3 TheABLUEofthreeparametersWhenrii > 25,let 1 (py)) = -qij \gqij, j = 1, ? ? ?, hi,i = 1, ? ? ?, I,G(-),g(-) are the function and density function of EV(0,1) respectively. Utilize Taylor to launch gettingThe same festival method getting...