Reference : Chapter 15 of Ebeling (The simple case of r = n) Statistical Inference (SI) When the Underlying Distribution is Weibull

Transcription

1 Referece : Chapter 5 of Ebelig (The simple case of r ) Statistical Iferece (SI) Whe the Uderlyig Distributio is Weibull Maghsoodloo Recall that by SI we mea estimatio ad test of hypothesis. We will cover oly the case of miimum life 0, ad therefore, the uderlyig TTF pdf is give by: f(t) t (t/ ) (t/ ) ( t) (t / ) e ( )( t) e. We cosider two experimetal life testig situatios: () No Cesorig, () Failure Data with Cesorig. The experimeter should bear i mid that if data shows the miimum life is ot zero (for example if the value of the st -order statistic were equal to 3000 cycles, the obviously the miimum life is ot zero), the such data ca easily be coverted to 0 by subtractig, say 0.90t() 0700 cycles, from all times to failure ad treatig the resultig data with miimum life equal to zero i order to use the above -parameter Weibull pdf. The experimeter should bear i mid that if data shows that the miimum life is ot zero (for example if the value of the st order statistic were equal to 3000 cycles to failure, the obviously the miimum life is ot zero), the such data ca easily be coverted to t 0 0 by subtractig, say 0.90t () 0700 cycles (you may eve try 0.85t () ), from all times to failure ad treatig the resultig data with miimum life equal to zero i order to use the above parameter Weibull pdf. For more accurate estimate of miimum life t 0 see pp. 4 ad 43 of Ebelig, where for the expoetial (β ) t 0 Max [t () t (), 0], this is basically the Eq. 5.5 of Ebelig o p. 43. For β e > (IFR), use Eq. (5.3) of Ebelig o his page 4, t (t t t )/(t t t ), where t t (), 0 j j t t () is the th order statistic, ad t j for the Weibull is jth order statistic, where j p ad p is always rouded up to the ext higher positive iteger. The value of p is give i Ebelig s Eq. (5.4) o his p. 43 as p I almost all cases, the results of these more accurate formulas come very close to 0.85t() t 0 0.9t(). Further, Ebelig (p. 4) states that the MLE of t 0 Miimum life is equal to t (). Whe TTF, T, has the 3 parameter Weibull W(,,), my recommedatio is to make a simple trasformatio of the data to t T i order to reduce the 3 parameter to the parameter Weibull. 30

2 Failure Data Without Cesorig (Complete Samples, i.e. r ) I this case we have uits o test ad our objective is to test all of them to failure i order to obtai the ordered failure times t, t,..., t ad the use these observed failure istaces to obtai the MLEs of ad. Note that this testig situatio is applicable oly to compoets that have moderately large icreasig hazard rate (fuctio); otherwise, the testig duratio will become cost prohibitive. I will provide the MLF ad its log ad the resultig two partial derivatives wrt ad, but I am doig it oly for the record. You will be resposible oly for kowig how to apply the results, which I will outlie i a step-by-step procedure. ML Estimatio Procedure Recall that the Pr elemet of the i th failure time, ti, is give by hece the LF (likelihood fuctio) is give by (t L(, ) i / ) ti e ( ) i Takig the atural log of (09a) leads to [ t i i ] e (t i / ) i t (t i / ) i e dti ad (09a) L(, ) [l() l()] + ( ) l(t i ) ( ti / ) (09b) i The partial derivative of L(, ) wrt is give by i L(, )/ / [ i (t ) ] / + The solutio to equatio (0) is the MLE of give below. i i (t ) i Set to 0 (0) / ti i (a) Equatio (a) is the result that we eed to obtai the poit MLE of the characteristic life tc, but the difficulty lies i the fact that uless we st obtai the poit ML estimate of the slope, the we will ot be able to compute from Eq. (a). Next we will obtai the MLE of by partially differetiatig L(, ) wrt. 3

3 L(, )/ l() + l() + l(t i ) i (t / ) ] l(t i ) [ i i i i (t i/ ) l(t i/ ) Set to 0 (b) Equatios (a&b) will have to be solved simultaeously i order to obtai the ML estimates of ad. Ufortuately, o closed-form solutio will ever exist for ad. Therefore, the solutios have to be obtaied thru trial/ad error that will make both partial derivatives L(, )/ i (0) ad L(, )/ i (b) almost equal to zero. I will ow go thru the Example 5.7 o pages of E. A. Elsayed ( d Ed.) i a step-by step procedure, ad compare aswers with his solutio ad that of Miitab s. Step : For the Example 5.7 o pages of E. A. Elsayed ( d Ed.), 0 diodes were tested to failure uder accelerated coditios. The istaces to failure i his example are ti 3000, 36000, 40000, 44000, 50000, 5000, 5500, 54000, 57000, ad miutes (i,,, 0); however, o iformatio about the load factor Af is provided. Matlab computatios yield x S 0 0 ti miutes, S i 0 (ti x) i 9 CSS 9 ( ) miutes. (Notice that Elsayed lists hours for the uits of x ad s.) Thus the sample coefficiet of variatio of TTF is give by cv S/ x 0.704% < <, which clearly implies that the uderlyig TTFS distributio is ot expoetial. Step. Sice the hazard fuctio of a Weibull is decreasig oly durig the early-life cycle of products (i.e., oly durig the RE growth cycle or bur-i period the slope lies i the iterval 0 < < ), the for our example >. Equatio (5.3) o page 304 of E. A. Elsayed (d Ed.) gives a rough approximatio of the Weibull slope i terms of the correspodig sample cv..05 cv (Equatio 5.3 of Elsayed) The above equatio gives us a startig poit for ML estimatio, i.e., we estimate our iitial slope as.05/ I have used the values i my Table i my Chapters,3&4, p.0, to 3

4 obtai a better regressio model that relates the Weibull slope to the cv. The Miitab output is give below. Regressio Aalysis: Beta versus x /CV The regressio equatio is Beta x, where x /CV, 5. Predictor Coef SE Coef T P Costat x S R-Sq 99.9% R-Sq(adj) 99.9% Aalysis of Variace Source DF SS MS F P Regressio Residual Error Total Uusual Observatios Obs x Beta Fit SE Fit Residual St Resid R R deotes a observatio with a large stadardized residual Results for: WBeta.MTW Regressio Aalysis: Beta versus x /CV The regressio equatio is Beta x, 5 0 Predictor Coef SE Coef T P Costat x S R-Sq 00.0% R-Sq(adj) 00.0% Aalysis of Variace Source DF SS MS F P Regressio E Residual Error Total Uusual Observatios Obs x Beta Fit SE Fit Residual St Resid R R R deotes a observatio with a large stadardized residual 33

5 The estimate of from the above regressio models are, respectively, / , or / These are fairly close to the value of 4.83 obtaied from Cohe (965), Techometrics 7, pp , give i equatio (5.3) of Elsayed. So, I will settle for a rough estimate of 5.00 ad will use this to obtai a rough value of from my equatio (a), which is miutes. These two estimates ( 5.00) make the RHS of equatio (b) equal to Sice L(, )/ i equatio (b) is a decreasig fuctio of, the we must icrease the value of i order to reduce the RHS of (b) close to zero. Through the Excel-Solver, I foud that at ad miutes, the RHS of (b) is close to 0 5. As a matter of fact, the estimate is very close to that of Elsayed s i the middle of his page 306. Therefore, the MLE of ad to 6 decimals are ad , which lead to R(t) (t/ ) e. As a result, a poit ML estimate of the RE fuctio at miutes is give by R(40000 mi) Elsayed s aswer o his page , which is cosistet (to 3 decimals) with We have completed poit ML estimatio for a W(0,, ) for a complete sample, ad ow it is time to perform further SI by obtaiig a 95% CI for both parameters ad. Agai sice the Weibull tc (characteristic life) is defiitely a LTB type parameter (ad thus o cocer o the high side), my persoal egieerig preferece would be to obtai the oe-sided CI for, ( L < ), ad a -sided CI for the parameter : L U. The reader must be cogizat that throughout these otes, we are settig the cofidece coefficiet, for the sake of coveiece, at 0.95; further, the statistical etities L, L, ad U are ideed (correlated) radom variables before experimetatio for obtaiig failure data. However, after experimetal data are gathered ad the eeded sample statistics are computed, the o loger L, L, ad U are radom variables but they are simply real-valued umbers ad hece the determiistic itervals o loger have 95% pr of cotaiig the true values of ad. Oce these two CIs are obtaied, the they may be used to coduct tests of hypotheses about the parameters ad. For example, if the data of the example 34

6 5.7 of Elsayed provides the 95% CI : , the we caot reject the ull hypothesis H0 : 6.5 at the LOS 0.05 because the hypothesized value of lies iside the 95% CI : Cofidece Iterval for the Weibull Slope (or Shape) It is well kow that a exact CI for a parameter ca be obtaied oly if the frequecy fuctio of the correspodig poit estimator is exactly kow. Equatio (b) clearly shows that o closed-form solutio exists for the poit estimator. I am almost certai that the exact SMD (or pdf) of for small to moderate values of r (r < 50) is ot kow, but from statistical theory the frequecy fuctio of all ML estimators (MLEs) asymptotically approach ormality with asymptotic mea equal to the correspodig parameter ad the asymptotic variace equal to /I(), where the sample iformatio I() is described below. Therefore, the asymptotic 95% CI for parameter is give by Z 0.05 se( ), where Z I am also fairly certai that o exact closed-form formula exists for the V( ) ad hece the exact se( ) caot be computed, ad as a result all of the followig developmets are simply approximatios, ad oly for moderately large values of r failures the approximate CIs are expected to be fairly accurate. Thus, our ext objective is to obtai a rough estimate of the se( ). This will require uderstadig the cocept of statistical efficiecy. Efficiecy We first show that E[L(t; )/] is always zero for ay parameter, where t [t t... t] represet a complete sample of failure times. E[L(t; )/] [ L(t; )/ ] f(t; )dt f(t; ) [ ] f(t; )dt f(t; ) d dθ f(t; θ)dt d dθ () 0 V[L(t; )/] E{[L(t; )/] }, where we are usig the fact that the joit pdf of t [t t... t] is give by f(t; ) f(t i; ). i Let be a MLE of (or ay estimator of ); the, the COV[, L(t; )/] E[ L(t; )/] 35

7 [ L(t; )/ ] f(t; )]dt Rt d [d f(t; )]dt d dθ E( ) d dθ [ + B( )] + B( ), where B( ) is the amout of bias i the estimator ad B( ) d dθ B( ). For coveiece of otatio, let L(t; )/ for which we just have show that E( ) 0 ad COV(, ) E( ) + B( ). It is well kow that 0, where is the correlatio coefficiet betwee ay two radom variables. Thus, 0 [COV(, )] V( ) ρ (θ, ξ ) 0 COV (θ,ξ ) V(θ)V(ξ ) V( ) < 36 The quatity V( ) V[L(t; )/] E( ) I() i the deomiator of (a) is called the (a) iformatio i the sample (about ); sice, COV(, ) E( ) + B( ), the (a) reduces to [ B( )] I( ) V( ) < (b) The iformatio iequality (b) is called the Cramer-Rao iequality i the field of statistics. It provides the greatest lower boud (glb) for the variace of ay estimator i the uiverse. Simply put, there exists o estimator i the uiverse whose variace is less tha the glb [+ B (θ)] I(θ) Further, oly those estimators whose variace is equal to the Cramer-Rao s glb are called efficiet. Before discussig a Example, we eed to show that V( ) V[L(t; )/] E( ) I() is also equal to E( /) as illustrated below. [ l(f(t; )] [ f (t; ) f(t; ) ], where f(t; ) [f(t; )]. Thus, ff (f ) f (3) f f Applyig the expected-value operator to both sides of (3), ad usig the fact that E( f f ) f ( )fdt f d d f(t; )dt d d () 0, we obtai V() E(/)..

8 This last result also shows that I() V() E[ l(f(t; )] E[ L(t; )] I (4) I geeral, if i ( i,,..., m) are the MLEs of m parameters with the log likelihood fuctio L(x;,,..., m), the the (i, j) th elemet of the iformatio matrix, I, is give by Iij E[ L(x;,,..., m)/ij]. It ca the be show that the asymptotic covariace matrix of the vector [... m ] is give by the iverse of the iformatio matrix I, i.e., COV( ) COV[... m ] I. I will start the procedure by st obtaiig the exact Cramer-Rao s glb for the V( ); recall from iequality (b) that [+ B (β)] I(β) V( ) <. From statistical theory it is well kow that MLEs are asymptotically ubiased, ad hece for large r, I(β) V( ) <, where I() E[ L(, )/ ]. I partially differetiated equatio (b) to obtai L(, )/, which is provided below. L(, )/ / L(, )/ / + (t i/ ) [l(t i/ )] i (t i/ ) [l(t i/ )] i I() E[ L(, )/ ] / + E (t i/ ) [l(t i/ )] I (5) i Eq. (5) clearly shows that the iformatio o i the sample is directly proportioal to the umber of failures because as icreases, the amout of iformatio about icreases. Further, i ivolves a very complicated mathematical expectatio o the RHS of equatio (5), give by E (t i) [l(t i / )]. Applyig the Expected-Value operator to the term iside the i summatio i order to obtai a exact result is impossible, at least to the capability of this author. 37

9 Eve simulatio may ot help because both parameters ad are ukow ad therefore the simulatio procedure has to start with specified values of these parameters of the Weibull ad hece the simulatio result for E (t i) [l(t i / )] would deped o the iputted values of ad i. Havig stated the simulatio problems, there is hope because oce failure data are obtaied, the the MLEs of ad ca be obtaied, ad hece the simulatio ca get started with these MLEs. This last computer simulatio procedure i order to estimate the se(of ay estimator) is called Bootstrappig i the field of Statistics. Bootstrap Estimatio Procedure Step : Recall that the MLEs of ad (to 3 decimals) for the Example 5.7 are ad miutes. First assume that the uderlyig distributio of the failure data is W(0, , 5.997), i.e., assume that the cdf is actually F(t) (t/ ) e. Recall from statistical theory that all cotiuous cdfs i the uiverse are uiformly distributed over the iterval [0, ], i.e., F(t) ~ U(0, ). This is because df for all cotiuous F(t). 0 Step : Use the cdf from step with the aid of a computer to geerate a radom bootstrap sample of r ( i this case) t, t,.... t. Use these sample results to obtai the poit MLE estimate of ad, deoted by * ad *, as outlied above. Step 3: Repeat step roughly B 500 times (i.e., a simulatio ru size of at least 500, where B 500 is my recommedatio). By ow we will have roughly 500 bootstrap estimates (, * *,..., * 500 * ) ad (, *,..., * 500 ). Next compute the bootstrap averages * B * i i ad B *. Step 4: The Bootstrap estimate of the se( ) is give by S B * * i B i [ ] ad similarly 38

10 for the se( ), i.e., S B * * i B i [ ]. You may wish to replace B i the divisors by B, if that is your preferece. Note that the above Bootstrap procedure ca be applied to all estimates whether the uderlyig distributio family is kow or ot. Whe the uderlyig distributio family is ukow, it will become more difficult. Now that I have outlied how to obtai reliable poit estimates of the stadard errors by a computer simulatio, we have to get back to the problem at had where we have to obtai rough estimate of se( ) W/O the use of a computer. Eq. (b & 5) show that the glb for the V( ) is give by [B ( )] E (t i/ ) [l(t i/ )] i V( ) <. (6) If we assume that r is sufficietly large i order to igore the amout of bias i the MLE, the the umerator of (6) reduces to. To get the procedure for estimatio of v( ) started from iequality (6), we first assume that v( ) β /, where the two dots o the equality imply very rough approximatio, ad r whe the failure data are ot cesored. Note that v( ) β /r is cosistet with equatio (5.37) atop page 308 of Elsayed with the elemet i the d row ad d colum of I, except for the multiplier c, which for the case of r is equal to from Table 5. o page 308 of Elsayed, reproduced below. Therefore, for the Example 5.7 data o pp of Elsayed, we have a better approximatio for v( ) c β / 0.608( ) / se( ).4786 cv( ) Therefore, lower 95% cofidece limit for the parameter is give by e , which is fairly close to Miitab s aswer of Aother way to estimate the value of the se( ) is simply computig the glb of the iequality (6) usig the give data, which I will ow proceed to do. From Eq. (6), v( ) E (t i/ ) [l(t i/ )] i 39 / (t i/ ) [l(t i/ )] i

11 Table 5. of E. A. Elsayed (o his page 308, p proportio of the sample that is cesored; p failed proportio) p c c c ; this Cramer-Rao glb (CRGLB) is very close to the actual v( ) We ca ow use the 95% CI: to perform further SI o the parameter. For example, suppose we wish to test the ull -sided hypothesis H0 : 4.00 at the preassiged LOS Sice the hypothesized value of, 0 4, is iside this 95% CI, the we caot reject H0: 4.00 at the % LOS. However, if we were to test H0 : 0.00 at 0.05, the we have sufficiet evidece to reject H0: 0.00 because lies outside the 95% CI I other words, our 95% CI: has provided all possible 5%-level tests of hypotheses regardig the parameter. Our ext objective is to use the MLE poit estimator of, which is give by /β β ti, ad its se to obtai a lower oe-sided 95% CI for the characteristic life tc. i To this ed, we st eed to compute the value of the sample iformatio o give by I I() E[ L(, )/ ] ( + ) β E( ti i ) β θ β θ ( ) E( t i ). Therefore, i from Cramer-Rao s iequality the glb[v( )] ( ) E( t i ) i. Equatio (0) shows that L(, )/ / + β Set to ti 0, or + ( ) β t i 0. i i 40

12 Isertig ( ) β t i ito glb[v( )] i ( ) E( t i ) i, we obtai v( ) ( ) ( / ) (7a) It is iterestig to otice that E t i i E(t i ) i (t/ ) t (t/ ) e dt, which is 0 cosistet with ML estimatio. The v( ) i Equatio (7a) is idetical to the term i the st row ad colum of the I matrix give i equatio (5.37) of Elsayed, except for the multiplier c, which is equal to listed i his Table 5.. Thus, for the case of ucesored data (i.e., complete sample), ( / ) v( ).09 For the failure data of the Example 5.7 of Elsayed, v( ).087 ( ) / se( ) hours. Hece, the value of the asymptotic 95% lower (7b) cofidece limit is L se( ) <. This 95% CI provides ucoutably ifiite umber of right-tailed tests o the parameter of the type H0: 0 versus the alterative H : > 0. For example, if we were to test H0: hours vs H: > 50000, the our CI does ot provide sufficiet evidece at the 5% level to reject H0 ad hece we will be uable to coclude that > This is due to the fact that the hypothesized value hours lies iside the 95% CI: <. O the other had, our 95% CI will provide sufficiet evidece to coclude, at the 5% level, that the value of tc exceeds hours. Computig the Stadard Error usig Fisher s Iformatio-Matrix The Fisher s iverse of iformatio matrix, I, for the -parameter Weibull is give by V( ) Cov(, ) I Cov(, ) V( ) 4

13 E. A. Elsayed recommeds the followig variace-covariace matrix i order to compute the variaces of the two estimators ad Î c /( ) c /. (5.37 of E. A. Elsayed) c / c / For example, if a sample is 70% completed to failure, or 30% cesored (i.e., p r/ 0.70), the Table 5. of Elsayed shows that v( ).4473 (θ / β), v( ) c β /.5β /, ad cov(, ) /. The above I clearly shows that the two Weibull estimators ad are ot idepedet, i.e., they are correlated. However, Miitab first obtais the Local-Fisher s Ifo-matrix as follows: F I I I I I( ) I(, ) I(, ) I( ) E( L/ ) E( L/ ) E( L/ ) E( L/ ) where E( L/ ) ( / )l( ) ( / ) ti l(t i ) Î. Miitab uses the iverse of the above approximated Fisher s matrix to obtai the se s of ad, which takes the correlatio betwee the Weibull estimators ito accout. I will ow do so for the Example 5.7 of E. A. Elsayed. Î [ L(, )/ ] i ( β / θ) , Î Î ad Î + (t i / ) [l(t i / )] i + t i [l(t i / )] i ; F Upo ivertig this last Local-Fisher s matrix, Î(, ) we obtai F Thus, the se( ) 4

14 ad the se( ) , which are idetical to those of Miitab s to 5 decimals. The Example 5.8 o page 309 of Elsayed. This example provides a complete sample with times to failure of 0 idetical uits. Sice the uit of TTF is ot specified, I will assume the times TF were measured i hours. The failure data are 0,, 4, 5, 6, 7, 30, 35, 4, 5 hours, which give mttf , S 0.07, ad cvt 33.04% < <. The Miitab output o page of my otes yields / as the iitial value of. Next, usig the MS-solver ad solvig (a) ad (b) simultaeously yields These MLEs are cosistet with those of Elsayed s listed i the middle of his page 309. Usig v( ) (β) / yields v( ) (Elsayed s aswer is 0.65) ad the se( ) Thus, the asymptotic 95% CIs are provided i the Excel file that has already bee ed to you. Further, Elsayed uses the more exact SMD of (cr)(β / β) (+p ) which is χ, where p c(r-) proportio of the sample that has failed ad ( p) represets the cesored proportio of the sample), ad c [(+p ) pc]. Table 5. o page 308 of Elsayed gives the values of cij for p 0.0(0.0). The 95% lower cofidece limit for the characteristic life is give by.645se( ), where se( ) c (θ / β) ( / ) L < at the 95% cofidece level. Obtaiig More Exact 95% Lower Oe-Sided CI for the Weibull W(0,, ) Characteristic Life ad Reliability Fuctio Sice the characteristic life, tc, is a LTB type populatio parameter, the there is absolutely o cocer o the high side ad hece, as before, we will obtai oly a lower 95% CI for of the form L <. As always, before experimetatio L is a rv (radom variable), but after data have bee gathered ad a umerical value computed for L, the the determiistic iterval L < is o loger radom ad its pr of cotaiig the true value of 43

15 reduces to 0 or. I will discuss the CI estimatio for the ucesored sample case first. This estimatio is easier tha CI estimatio for a cesored data. We will first obtai the lower oe-sided CI for, followed by a lower -sided CI for R(t). Sice there does ot exist a closed-form solutio for the ML estimator, the it will always be impossible to use mathematical statistics to obtai the precise SMD of the statistic θ / θ. However, Bai, L. J. ad Egelhardt, M. (99), d Editio, New York, Marcel Dekker, used simulatio to obtai the approximate percetiles of the SMD of the statistic U β l(θ / θ), which they tabulate o their page 30 for differet sample sizes ad cofidece levels 0.0, 0.05, 0.0, 0.5, 0.50, 0.75, 0.90, 0.95, ad I do ot kow why the tabulatio was ot doe for 0.005, 0.99, ad Sice, we are iterested oly i developig a 95% lower oe-sided CI for, the our iterest ceters oly at the value of I took Bai ad Egelhardt s 95 percetiles of U, for 5 to 0, give i their Table 4A o their page 30 ad used Miitab to obtai the followig models for the percetiles u0.95 ad u û 0.95() /.3533 / / 3 (8a) û 0.975() / 30.98/ / 3 (8b) Miitab reported a R Model 99.6% for the above models ad I tried to improve them to obtai early a 00% value of R Model by addig / 4, but the models would ot improve ay further. I tested the values of u 0.95() from equatio (8a) agaist those give i equatio (5.48) of Elsayed o page 3 as U / e, both Elsayed s ad mie agree (to 3 decimals) with those listed i Table 4A of Bai ad Egelhardt (o their page 30) for 5 to 0. The Example 5.8 o page 309 of Elsayed provides a complete sample for 0 with times to failure as 0,, 4, 5, 6, 7, 30, 35, 4, 5 hours (ote that uits are ot provided i this example ad I am guessig that it is hours to failure). Isertig 0 i equatio (09a) gives u 0.95(0) (B&E report U i their Table 4A o page 30 for 0). Therefore, the P [β l(θ / θ) ] Rearragig the iequality iside this last brackets yields 44

16 u /( ) e L (0.95) 0.95 (9) Isertig the values of , , ad u 095. ( 0 ) ito equatio (9) yields L( 095. ) Thus, we are 95% cofidet that the true value of lies i the iterval hours <. This CI is ot as coservative as that of Elsayed s 6.0 hours <, give o his page 33, because of the fact that Elsayed used the ubiased estimates of ad, which he computed to be.8 ad θ Sice his 95% lower boud for is more coservative, the you may wish to also use the ubiased estimates of ad, as he did. Usig equatio (9), θ L(0.975) u 0.975/( ) e A poit MLE of R(t) is give by R(t) (t/ ) e. For example, R(5hours) Due to the fact that the RE fuctio of a Weibull is ot a mootoically icreasig fuctio of the slope (recall that the Weibull RE icreases with icreasig up to the characteristic life (t/ ad the decreases with icreasig ). Hece, RL(t) will ot equal to L ) e L ad as a result i order to obtai a lower 95% CI for R(5hours), we eed to have some idea about the SMD of R(5hours). As stated before, the SMD of all MLEs approach ormality (as ) with mea equal to the correspodig parameter, amely R(5), ad a variace whose asymptotic value is equal to /I[R(t)]. I surmise that I[R(t)] E[ L(, )/R ] caot be computed directly. Thus, Bai ad Egelhardt (p. 7) provide a empirical formula to compute the asymptotic variace of R(t), but their formula does ot provide a coservative glb for R(t) as show i their Table 7A (pp ). Thus, I took the liberty, through trial/error, to provide a revised versio of their formula, give below, so that the ormal approximatio will provide a more coservative value of RL(t) for almost all. v[ R(t)] R (t)[l(/ R(t)] { l(l(/ R(t)) )+ 0.70[l(l(/ R(t)))] }/ (0) Isertig the value of R(5hours) ito equatio (0) gives v[ R(5)] 0.003, ad the se[ R(5)] ad.645se( R(5)) , ad hece RL(5 hours) R(5) 45

17 Z0.05se[ R(5)] R(5) <. Thus, we are 95% cofidet that the reliability fuctio at t 5 hours exceeds Estimatio of Weibull Parameters usig Weibull Graph Paper (WGP) ad Least-Squares. t ( ) Recall that because R(t) e t, t, the l[r(t)] ( ) l[/r(t)] t ( ) t l{l[/r(t)]} l( ) l{l[/r(t)]} [l(t ) l( )] l(t ) l( ) 46 l(t ) C, where C l( ) is a costat. Lettig y l[l(r )], x l(t ) yields y βx+cwhich represets a lie with slope ad Y-itercept C. This is why the shape parameter is also called the slope of the Weibull distributio. For the case of zero miimum life, x l(t) ad the y-itercept is C l( ) βl( / θ). Because all Weibull distributios have a RE value of e at their characteristic life tc, the Weibull graph paper (WGP) has two ordiates, where the right scale gives l[l(r )] ad the opposig left scale gives F(t). Thus, y l[l(r )] is the right coordiate scale ad its value at t is equal to y() l[l(e ) ] l[l(e)] l() 0 while the correspodig left ordiate is obtaied from 0 l[l(/r)] l[l( )] F l( ) F e 0 e F e F 0.63 The 63.% failure poit o the F left scale correspods to the characteristic life o the abscissa because l[l(/r())] 0 [see Figure. of Kapur ad Laberso (977) o their p. 96] reproduced o the ext page. Yet as aother example, l[l(/r)] o the right scale of figure. implies that l( ) e F F 0.69 F the WGP left scale will correspod F to roughly 30.78% cumulative failure, etc. Fially, the RE fuctio of the Weibull for the same values of ad is a icreasig fuctio of the slope up to the characteristic life ad the becomes a decreasig fuctio of for t values beyod. As a example of usig the WGP to

18 estimate the parameters of a W(0,, ), see the Example 5.5 o pages of Ebelig ad my Excel file o my website that obtais the LSE of followed by. The 95% glb o RE at 40 hours for the example 5.5 of Ebelig is roughly give by , while we are 90% cofidet that 30% of the uits will fail withi the iterval (0, 87 hours). The 5 failure-times of Ebelig s Example 5.5 are 3 hours, 5, 74, 90, ad 0. Chapter 5 Summary. I order to obtai MLE of Weibull ad, solve the two Eqs. / ti i ad L(, )/ l() + l(t i ) [ (t i / ) ] i i l() + l(t i ) i simultaeously i order to obtai i ad (t i/ ) l(t i/ ) Set to 0. The, the poit estimate of RE at time t is give by [(t / ) ] R(t) e. I geeral, the elemets of Fisher s Ifo-matrix are Î β β βθ β +)θ [( ti ], Î i ( β ) β ( β ) [ ] t β i i i i /θ + θ βl(θ) βθ (t l(t i ), ad Î + (t i / ) [l(t i / )] i + t i [l(t i / )]. These will i reduce further for MLE, but have to be used as is for LSQ estimatio.. Better Approximatio for CI limits o ad whe < 0. Bai ad Egelhardt state that a better approximatio for cesored ad complete samples is that the SMD of cr where the costat c ( / ) p( p ) c p approximately follows a with c(r ) df,. Table 5. o page 308 of Elsayed (page 3 of my otes) provides the values of c for differet ad cesorig proportio 47

19 Figure. of Kapur ad Lamberso (977) from their page 96 48

20 49

21 50

22 p. For the Example 5.5 of Ebelig, 5, ti : 3, 5, 74, 90, 0 hr, p, c ( ) ( / ) ~ df (5 ) ; ad ( / ) ] 0.95 L ; P[ ad u û 0.95() /.3533 / / 3 û 0.95(5).47658, L (0.95) e u 0.95 /( ) e /( ) Note that the above L 0.70 is troublesome because the baselie distributio is IFR Weibull ad < implies DFR. Miitab gives the 95% CI for as From their help meu, I got the formulas L z cv(β) e 0.05 z cv(β) ad u e 0.05 I iverted Miitab s formulas to ascertai that the SMD of l( / ) must be approximately ormal with mea zero ad stdev of roughly equal to se( ), as depicted below. The Figure shows that the P[.96se( ) l( / ).96se( ) ] 0.95 P[.96cv( ) l( / ).96cv( ) ] cv( ).96cv( ) P[ e e ] 0.95 For the data of Example 5.5 of Ebelig,.6455, se( ) 0.93 cv( ) L e.3356 ad u cv( ) e.96

23 u e Miitab s aswers: Roughly se(betahat) se( ) l( / ) To obtai the 95% glb o, use the SMD of U l( / ) with the approximate percetiles û 0.95() /.3533 / / 3 ad For 5, 0.95 P [ û 0.975() / 30.98/ / 3 û (5) P [ l( / ) ] 0.95 l( / ).4765/ ] 0.95 ; P [( / ) e ] 0.95 P [ e < ] 0.95 L e hours It turs out the -sided 95% CI o ca also be obtaied from a similar formula as i the case of the shape parameter. That is, L z 0.05 cv(θ) ad u e 5 z cv(θ) 0.05 e

24 L e ad u To obtai the glb o R(40 hours), we use the fact that for > 0 v[ R(t) ] R(40) R (t)[l(/ R(t) ] { l(l(/ R(t) ) )+ 0.70[l(l(/ R(t) ))] }/.6455 [(40/ ) ] e v[ R(40)] se [R(40)] R L(40) Table 5. of E. A. Elsayed (o his page 308, p proportio of the sample that is cesored; p failed proportio) p c c c Fially, it will be best to compute the se s of the Weibull estimators by ivertig the estimated Fisher s Iformatio-Matrix Î I I, i.e., Î v( ) cov(, ) ; this procedure takes the I I cov(, ) v( ) correlatio betwee ad ito accout. The eeded formulas for Îij are give o p. 43 of my otes. The formulas for the ormally approximated CIs are L e Z cv(β) 0.05, u Z 0.05 cv(β) e, ad similarly for the parameter. 53