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A Test For Weibull IFR/DFR Alternatives Based On Type-2 With Replacement Censored Samples

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A Test For Weibull IFR/DFR Alternatives Based On Type-2 With Replacement Censored Samples
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  JIRSS (2006)Vol. 5, Nos. 1-2, pp  || - || A Test For Weibull IFR/DFR AlternativesBased On Type-2 With Replacement CensoredSamples K. Muralidharan Department of Statistics, Faculty of Science, The M. S. University of Bar-oda, Vadodara 390 002 India. (lmv muraliyahoo.com) Abstract.  This article presents a test based on quadratic form usingType-2 with replacement-censored sample for testing exponentialityagainst weibull IFR/DFR alternative. The percentile points and pow-ers are simulated. The proposed test is compared with that of Bainand Engelhardt (1986) test. An example based on Type-2 censoringis also discussed. 1 Introduction The weibull distribution is defined by the pdf  f  ( x ; θ,β  ) =  β θx β  − 1 e − x β /θ , x >  0 , θ,β >  0 (1.1)This distribution is quite popular as a life testing model and formany other applications where a skewed distribution is required. Thismodel includes the exponential distribution with constant failure rate Received: 2006 Key words and phrases:  Conditional distribution, nuisance parameter, per-centile points, power of the test, shape parameter, weibull distribution.  2  Muralidharan (CFR) for β   = 1 and provide an increasing failure rate (IFR) for β >  1and decreasing failure rate (DFR) for  β <  1. Hence test for  β   is of interest.Thoman et.al. (1969) have considered the problem of testing of hypothesis regarding the shape parameter based on complete sam-ples. Bain and Engelhardt (1986) have proposed a modified versionof Thoman et.al. (1969) test statistic whose asymptotic distributionis approximated to a chi-squared distribution. Lawless (1982), Bainand Engelhardt (1991) have discussed the problem of estimation andtesting of shape parameter under Type-2 without replacement cen-soring and Type-1 censoring schemes. Very recentlty Muralidharanand Shanubhogue (2004) have obtained conditional test for WeibullDFR alternatives based on without replacement censored scheme.But there is less work done in the case of Type-2 with replacementcensoring scheme because of complexity of finding distributions of the statistics obtained. In this article we propose a computationallysimple test for testing  H  0  :  β   = 1 against  H  1  :  β >  1 (or  H  1  :  β <  1)using Type-2 with replacement censored samples. The proposed testis also compared with that of Bain and Engelhardt (1986) test. 2 Derivation of the test Many times observations of failures are naturally occurring in order.In this case, it is convenient to terminate the experiment after ob-serving the first  r  failures from  n  units by replacing each failed itemwith a new item. In this section we derive a test statistic based on aType-2 with replacement sample and study its properties.Let X  (1)  ≤ X  (2)  ≤···≤ X  ( r ) , r ≤ n , be a Type-2 with replacement-censored samples of a complete sample of size  n  from (1.1). Then the joint density is given by f  X  ( x ; θ,β  ) ∝ f  ( x (1) ) f  ( x (2) − x (1) ) ...f  ( x ( r ) − x ( r − 1) )[ F  ( x ( r ) )] n − 1 . Let  Y  i  =  X  β i  , i  = 1 , 2 ,...,n  then  Y  i  follows an exponential distribu-tion with density function f  Y  i ( y ; θ ) = (1 /θ ) e − y/θ , y >  0 ,θ >  0 .  (2.1)Let  Y  (1)  ≤  Y  (2)  ≤ ··· ≤  Y  ( r )  be the corresponding type-2 withreplacement-censored samples. Then f  Y   ( y ; θ ) = ( n/θ ) r e − ny ( r ) /θ , 0  < Y  (1)  ≤ Y  (2)  ≤···≤ Y  ( r )  < ∞ .  (2.2)  A Test For Weibull IFR/DFR Alternatives ...  3For known  β  ,  T   =  Y  ( r )  is the complete sufficient statistic for  θ . Bymaking the transformation  Z  i  =  Y  ( i ) − Y  ( i − 1) ,  Y  (0)  = 0,  i  = 1 , 2 ,...,r ,We get   ri =1 Z  i  =  Y  ( r ) . Hence (2.2) reduces to f  Z  ( z ; θ ) = ( n/θ ) r e − n  ri =1  z i /θ = r  i =1 nθe − nz i /θ . Therefore  Z  i ’s are i.i.d exponential with parameter ( θ/n ) and hence Y  ( r )  is gamma with parameter ( θ/n ) and  r . Then the conditional pdf of   Y  (1) ,Y  (2) ,...,Y  ( r − 1)  given  T   =  t  is obtained as f  ( Y  (1) ,Y  (2) ,...,Y  ( r − 1) | T   =  t ) = Γ( r ) t r − 1 ,  (2.3)0  < Y  (1)  ≤ Y  (2)  ≤···≤ Y  ( r − 1)  < t It is seen that this conditional density does not depend on the nui-sance parameter  θ . Hence we derive the test statistic for testing  H  0 versus  H  1  by treating the observations have come from (2.3). For this,we consider the quadratic form  Q  = ( Y  − µ 0 ) ′ Σ − 10  ( Y  − µ 0 ), where  Y  ′ =( Y  (1) ,Y  (2) ,...,Y  ( r − 1) ),  µ 0 ′ = ( µ 1 ,µ 2 ,...,µ r − 1 );  µ i  =  E  h 0 [ Y  ( i ) | T   =  t ]and Σ 0  = (( σ ij )) is the conditional variance-covariance matrix of   Y  given  T   =  t  computed under  H  0 . They can be obtained as follows:We know that  Z  i ’s are i.i.d exponential random variables withmean ( θ/n ) and   ri =1 Z  i  =  t . Then the pdf of   Z  i  given  T   =  t  is f  Z  i | T  ( z i | t ) = ( r − 1) t  1 −  z i t  r − 2 ,  0  < z i  < t  (2.4)and f  Z  i ,Z  j | T  ( z i ,z  j | t ) = ( r − 1)( r − 2) t  1 −  z i t  −  z  j t  r − 3 ,  0  < z i  + z  j  < t. According to the theorem 1.6.7 of Reiss (1989), the distribution of thevector ( Z  1 | T,Z  2 | t,...,Z  r | T  ) ′ is the same as that of ( V  1 ,V  2 ,...,V  r ) ′ where  V  ’s are the spacings of a random sample of size  r  from theuniform distribution on (0 , 1). Further, the author has given theasymptotic distribution of this vector of spacings and other relatedresults (see corollary 1.6.10 of Reiss (1989)).From (2.4) we obtain the moments of   Z  i | t . Since  Y  ( i )  =   i j =1 z  j ,under  H  0 µ i  =  E  H  0 ( Y  ( i ) | t ) = i   j =1 E  ( Z   j | T   =  t ) =  itr  (2.5)  4  Muralidharan σ ii  =  V  H  0 ( Y  ( i ) | t )=  V  H  0 ( i   j =1 Z   j | T   =  t )= i   j =1 V  H  0 ( Z   j | T   =  t ) − i   j  = k V  H  0 ( Z   j | T   =  t ) / ( r − 1)=  i ( r − i ) v ( r − 1)  ,  (2.6)and σ il  =  COV  H  0 ( Y  ( i ) ,Y  ( l ) | t )= i   j =1 l  k =1 COV  ( Z   j ,Z  k | T   =  t )= i   j =1 V  H  0 ( Z   j | T   =  t ) −  1( r − 1) i   j =1 l   j  = k =1 V  H  0 ( Z   j | T   =  t )=  i ( r − l ) v ( r − 1)  ,  (2.7) i  = 1 , 2 ,...,r − 1;  l  = 1 , 2 ,...,r − 1,  i < l  and v  =  V  H  0 ( Z   j | T   =  t ) = ( r − i ) t 2 r 2 ( r  + 1) . Using (2.6) and (2.7), the variance-covariance matrix Σ 0  of ( Y  | T   =  t )under  H  0  isΣ 0  =  vr − 1  r − 1  r − 2  r − 3  ···  1 r − 2 2( r − 2) 2( r − 3)  ···  2 r − 3 2( r − 3) 3( r − 3)  ···  3... ... ... ... ...1 2 3  ···  r − 1  . Using Graybill (1969, pp.181), we get the inverse of Σ 0  asΣ − 10  =  r 2 ( r  + 1) t 2  2 /r  − 1 /r  0  ···  0 − 1 /r  2 /r  − 1 /r  ···  00  − 1 /r  2 /r  ···  0... ... ... ... ...0 0 0  ···  2 /r   A Test For Weibull IFR/DFR Alternatives ...  5Substituting the values of   µ i  and Σ − 10  in the expression for  Q  and onsimplification, we obtain the test statistic as Q  = 2 r ( r  + 1) r − 1  i =1  Y  ( i ) t  Y  ( i ) t  −  Y  ( i +1) t  + ( r 2 − 1) . Replacing  t  by its corresponding random variable  T  , we suggest thetest statistics as Q ∗ = 2 r ( r  + 1) r − 1  i =1 W  i ( W  i − W  i +1 ) + ( r 2 − 1) ,  (2.8)where  W  i  = ( Y  ( i ) /Y  ( r ) ). The mean and variance of the test statisticsunder  H  0  is given by E  ( Q ∗ ) =  r − 1 and  V  ( Q ∗ ) = 4 r 2 ( r − 1)( r  + 2)( r  + 3) . Since we could not find the expression for  E  ( Q ∗ ) under  H  1 , weobserved the direction of the test statistics by simulating its values fordifferent  n  and  β   and under different censoring proportion  p (=  r/n ),0  < p <  1. They are presented in Table 1.Table 1:  E  ( Q ∗ ) for different values of   β  . β   0.4 0.6 0.8 1.0 1.2 1.4 1.6 2.0 n 10 15.36 10.92 8.85 8.01 7.72 8.28 9.78 10.7113.30 9.66 7.59 7.02 6.89 7.19 8.39 9.0420 34.18 22.77 18.35 17.03 17.09 18.47 23.71 27.3529.63 20.86 16.46 15.02 15.18 15.92 20.72 23.2430 53.23 35.41 28.49 26.02 26.27 28.99 39.58 45.0747.58 31.49 25.33 23.06 23.45 25.34 33.40 39.37(The first value corresponds to 10% censoring and second valuecorresponds to 20% censoring).The table shows that  E  H  1 ( Q ∗ )  > E  H  0 ( Q ∗ ) for 0  < β <  1 and forall  n . For  β >  1,  E  H  1 ( Q ∗ ) decreases below  E  H  0 ( Q ∗ ) for some rangeof   β   and then increases for  β >  1 . 4. Thus the test procedure is toreject  H  0  for large values of   Q ∗ in the above ranges. 3 Simulation study We obtain the upper tail percentile points of the distribution of   Q  byMonte Carlo method by generating 5000 random samples of different
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