Paper Info
Title
A Bayes test of homogeneity of several means for one parameter exponential populations
Abstract
The problem of testing the homogeneity of k (k ⩾ 2) mean time between failures (MTBF) vs. multiple t-slippage alternatives (1 ⩽t < k) has been considered for the one parameter exponential distribution. The exact null distribution of a Bayes test has been derived for selected values of k and t . Several approximations for computing the percentage points of the test statistic are also investigated. Simulated examples are included to illustrate the test. Keywords ANOVA Bonferroni inequality Beta distribution Mean time between failures 1 Introduction A one way analysis of variance (ANOVA) procedure is concerned with testing the homogeneity of several means against the alternative that the means are different. A multiple slippage test is for testing the null hypothesis of homogeneity of k means, against alternatives that some subset of size t (1⩽ t ⩽ k ) of the populations has a larger common mean than the remaining populations. The problem of testing a single slippage ( t =1) was first considered by Karlin and Truax [5] , who gave a decision theoretic formulation of the problem, and derived a class of symmetric Bayes rules. Hall and Kudo [2] , Hall et al. [3] also considered the single slippage testing problem. Singh [8] generalized the results of [5] for t ⩾2 and derived a class of invariant Bayes rules for testing the null hypothesis of homogeneity against the slippage alternatives. Hsiao et al. [4] computed the power of a non-parametric slippage test for location and mean. Singh et al. [10] gave approximations to the power of the likelihood ratio tests for slippage alternatives. In this paper we derive the null distribution of the test statistic for selected values of k and t , and also consider several approximations for computing the percentage points of the test statistic. We also provide some numerical examples. 2 Problem formulation Let π 1 ,…, π k be k exponentially distributed populations with mean time between failures (MTBF), θ i , where i =1,2,…, k (⩾2). The problem of testing homogeneity of k exponential populations H 0 =θ 1 =θ 2 =…=θ k =θ ( say ) vs. H 1 = H 0 is false has been investigated by Mathai [6] , Singh [9] and Nagarsenker [7] among others. In this paper, we have considered the problem of testing homogeneity: H 0 :θ 1 =θ 2 =⋯=θ k =θ (say) vs. the slippage alternative (1) H j :θ j = θ+Δ, j∈J, 1−θ, j∉J, where J ∈ S , and S={I⊂{1,2,…,k} such that the number of elements in I=t},1⩽t⩽k, t is the number of populations which have slipped (i.e., have larger MTBF). The problem is to test the hypothesis (1), X ={(x i 1 <x i 2 <x i 3 <⋯<x i r ): i=1,2,…,k} a type 2 censored sample from, π i ∼exp( θ i ), i =1,2,… k . The sufficient statistic in this case is known [1] to be (2) U i = ∑ j=1 r x i j +(n−r)x i r , i=1,2,…,k which has the distribution f u i (u i ;θ)= 1 θ r i Γ r u r−1 i e iu i /θ i , u i >0, θ i >0. Singh [8] under certain assumptions derived symmetric Bayes rule φ 0 for the above distribution as (3) φ 0 ( u )=1, if u ∈R 0 = u : max J∈S u j u <c , (4) Φ J ( u )=1, if u ∈R C 0 and u i ∈I= u [k−t+1] ,…,u [k] , i=1,2,…,k, (5) u j = ∑ i∈I u i t ∑ k i=1 u i k . In order to select Bayes rules from the above class of rules, we need to find C such that (6) Prob (Reject H 0 /H 0 is true )=α. For determining C , we need the null-distribution of (7) V= max i∈I u i ∑ k i=1 u i = u [k−t+1] +⋯+u [k] u [1] +u [2] +⋯+u [k] . In this paper, we derive the distribution of Eq. (7) for particular k and t . We also give some approximations, and numerical simulations. Comparisons with exact values show these approximations to be quite good. 3 Distribution theory In order to carry out the test given by Eqs. (3)-(5) , the null distribution (8) V= k t u k−t+1 +⋯+u k u 1 +u 2 +⋯+u k is needed. For a type 2 censored sample u 1 <u 2 <⋯<u k , the j.p.d.f. can be written as (9) f u 1 ⋯u k = k! ∏ i=1 k f u i ,θ i , u 1 <⋯<u k , 0, otherwise . Let (10) v 1 =u 1 ,v 2 =u 1 +u 2 ,⋯,v k =u 1 +u 2 +⋯+u k . Using Eqs. (10) and (9) gives the j.p.d.f. of Eq. (8) as we can compute the j.p.d.f. at v 1 ,…,v k as (11) f v v = k! Γ r k v 1 v 2 −v 1 v 3 −v 2 ⋯ v k −v k−1 r−1− v k /r θ . Using MACSYMA package, we verified that (12) ∫ 0 ∞ ∫ 2v 1 ∞ ∫ 2v 2 −v 1 ∞ ⋯ ∫ 2v k−1 −v k−1 ∞ f(·) d v k … d v 1 =1. In order to find the distribution of V given by Eq. (8) , the j.p.d.f. of U u−k+1 U k is needed. We could only derive the exact p.d.f of V for k =3, r =1, t =1. Since exact null p.d.f. could not be derived in the general case, we used following approximations: 1. Non-parametric approach to estimate cut-off points of U (via simulation). 2. Fitting a beta p.d.f (using method of moments). 3. Using Bonferroni's inequality. To check the accuracy of these approximations, we computed the exact cut-off points for the special case k =3, t =1, r =1 and compared these exact values with the values obtained from numerical approximations. 3.1 Exact distribution for k =3, t =1, r =1 As stated in earlier sections, the test considered in our problem rejects the null hypothesis H 0 if (13) v= max 1⩽i⩽3 u i ∑ 3 j=1 u j ⩾C. Let Y 1 = u [1] u [1] +u [2] +u [3] , Y 2 = u [2] u [1] +u [2] +u [3] . Then (14) f y 1 ,y 2 (Y 1 ,Y 2 )= Γ (k)=(k−1)! 0<y 1 <1, 0<y 2 <1, y 1 +y 2 <1, 0 otherwise . The c.d.f. of V works out to be: (15) F v (V)=P(Y 1 ⩽v,Y 2 ⩽v,Y 1 +Y 2 ⩾1−v)= 0 v< 1 3 , (3v−1) 2 1 3 <v< 1 2 , (6v−3v 2 −2) 1 2 <v<1, 1 v>1, and the cut-off points for the above exact distribution ( Appendix A ) can be calculated solving following quadratic equations: (3v−1) 2 =α, (6v−3v 2 −2)=α, which gives v= 1+ α 3 1 3 <v< 1 2 , v=1− 1− 2+α 3 , 1 2 <v<1. For α=0.01,0.05,0.5,0.90,0.95, and 0.99 the exact cut-off points are computed and shown in the next section. 4 Approximations 4.1 Non-parametric method Following are the steps of the simulation experiment. (1) Generate x 1 , x 2 ,…, x n ∼ gamma distribution with parameter r and 1 using IMSL subroutine rngam, for each population Π i , i =1,2,…, k . (2) Sort x 1 , x 2 ,…, x n . (3) Compute sufficient statistic U i =∑x i + n−r x r i=1,2,…,k (4) Sort U i (5) Compute statistic V i =(U [k−t+1] +⋯+U [k] ) ∑ k i=1 U [i] (6) Sort values V i (7) Calculate the c.d.f. F ̂ (V)= #(V i ⩽v) n (8) Invert F ̂ for α=0.1,0.05,0.1,0.5,0.9,0.95,0.99. Steps (1)–(8) were repeated 10,000 times for each run. 4.2 Fitting a beta p.d.f. The null p.d.f. of V in Eq. (13) can be approximated by the Beta ( α , β ) p.d.f. The parameters α and β are estimated by the method of moments: (16) α ̂ = v (m 2 − v ) v 2 −m 2 , (17) β= (1− v )(m 2 − v ) ( v 2 −m 2 ) , where m 2 =∑ v 2 / n . 4.3 Bonferroni inequality Our problem is to find C such that prob H 0 max j∈J ∑ j∈J u j ∑ k i=1 u i >C =α or (18) prob H 0 max j∈J ∑ j∈J u j ∑ k i=1 x j ⩽C =1−α. Since the exact null distribution of the test statistic is not available, we will use the Bonferroni inequality [11] . We write Eq. (18) as (19) P H 0 ⋂ j ∑ j∈J U j ∑ n i=1 U i ⩽C ⩾1−α which simplifies to P H 0 ⋂ J ∑ j∈J U j U 1 +U 2 +…+U n ⩽C ⩾1− ∑ J (1−p J )⩾1−α⇔1−p= α k t ⇒p=1− α k t , where p J =P ∑ j∈J U i ∑ n i=1 U i ⩽C =p+P(V⩽c) under H 0 , with V ∼beta( t , k − t ). Thus the cut-off point C can be found from P(V⩽C)=1− α k t . 4.4 Comparison of the above methods We used Monte Carlo simulate to compare the above three approximations. The steps in simulation are as follows: 1. Generate u i from Beta ( rt , r ( k − t )) 2. Compute max u i /∑u i 3. For different α find the value of 1− α k t such that P max u i ∑u i ⩽C =1− α k t . Table 1 shows the non-parameter approximation and the Bonferroni's inequality approximation work reasonably well, and yield better results, than the ones given by the Beta approximation. 5 Examples Example 1 ( H 0 true) . In this case we used k=5, n=r=10, t=3, Δ=1. Generated and sorted U i ∼ Gam(10,1) are: (a) 7.11, 7.87, 8.00, 12.71, 12.89 from these (20) ∑ i=3 5 U i =33.59 and ∑ i=1 5 U i =48.58 (21) ⇒V= ∑ 5 i=3 U i ∑ 5 i=1 U i =0.69. H 0 will be rejected if the above value of the statistic calculated from the sample exceeds the cut-off value C . The values of C calculated from the Boneferroni's inequality and the non-parameter approximations are: (a) Using Bonferroni's inequality (22) 90% cut−off =0.752, 95% cut−off =0.766, 99% cut−off =0.791. (b) Using non-parametric (23) 90% cut−off =0.749, 95% cut−off =0.765, 99% cut−off =0.795. Thus H 0 is not rejected at these α -levels. Example 2(a) ( H 1 true) . (a) K =5, r =3, t =3, Δ =2. Generated and sorted U i ∼ Gam(10,1) are: 8.25, 8.65, 14.27, 20.06, 20.89 (24) gives ∑ k=1 3 U i =60.23. ∑ k=1 5 U i =77.15 gives V=0.781. Example 2(b). For another set the generated and sorted U i are: 8.79, 11.97, 11.97, 19.35, 30.57, 44.63 (25) which gives ∑ i=1 3 U i =94.56. ∑ k=1 5 X i =115.33 and V=0.819. Both Eqs. (24) and (25) when compared to Eqs. (22) and (23) suggest to reject H 0 at α =0.05 and accept H j ={ π 1 , π 2 , π 3 }. Appendix A 1. Null distribution of the test statistic k =3, r =1, t =1 X 1 ,X 2 ,X 3 ∼ exp (1)⇒f(x 1 x 2 x 3 )= e −Σx i , 0<x i <∞ set Y 1 = X 1 X 1 +X 2 +X 3 , Y 2 = X 2 X 1 +X 2 +X 3 , s=X 1 +X 2 +X 3 . Inverse transform is X 1 =SY 1 , X 2 =SY 2 , X 3 =Sk(1−Y 1 −Y 2 ). Jacobian of transformation: J= ∂ X 1 ∂ Y 1 ∂ X 1 ∂ Y 2 ∂ X 1 ∂ s ∂ X 2 ∂ Y 1 ∂ X 2 ∂ Y 2 ∂ X 2 ∂ s ∂ X 3 ∂ Y 1 ∂ X 3 ∂ Y 2 ∂ X 3 ∂ s = s 0 Y 1 0 s Y 2 −s −s 1−Y 1 −Y 2 , |J|= S S(1−Y 1 −Y 2 )+SY 2 +Y 1 S 2 = S 2 −S 2 Y 1 +S 2 Y 1 = S 2 =S 2 , f(Y 1 ,Y 2 ,S)= e −s ·|J|= e −s s 2 , 0<Y 1 <1, αY 2 <1, Y 1 +Y 2 <1, 0<s<∞,⇒f(Y 1 ,Y 2 )= ∫ 0 ∞ e −s s 2 d s=Γ 3 =21. It can be shown that for any k ⩾2: f(Y 1 ,Y 2 ,…,Y k )=Γ(k)=(k−1)!. 2. Derivation of exact distribution for k =3 (proof of Eq. (15) ) (A.1) f v (V)=P(Y 1 ≤v,Y 2 ≤v,Y 1 +Y 2 ≥1−v)= 0, v< 1 3 ( a ), (3v−1) 2 , 1 3 <v< 1 2 ( b ), (6v−3v 2 −2), 1 2 <v<1 ( c ), 1, v>1 ( d ). Proof. (a) and (d): follow from the fact that 1/3< max U i /∑ 3 1 U i ⩽1. (b) and (c): (see Fig. 1 Fig. 2 given below). References 1 L.J. Bain, Statistical Analysis of Reliability and Life Testing Models, Marcel Dekker, Inc., 1978. 2 I.J. Hall A. Kudo On slippage tests-I: a generalization of the Neyman–Pearsons lemma Annals of Mathematical Statistics 39 1968 1693 1699 3 I.J. Hall A. Kudo N.C. Yeh On slippage tests-II: similar slippage tests Annals of Mathematical Statistics 39 1968 2029 2037 4 R.-R. Hsiao, C.E. McCulloch, R.-B. Wu, The power of some nonparametric test statistics for the location and mean slippage problems, Journal of Statistical Planning and Inference (1990) 391–402 5 S. Karlin D. Truax Slippage problems Annals of Mathematical Statistics 31 1960 296 324 6 A.M. Mathai On the non-null distribution of test statistics connected with exponential populations, Communications in Statistics Theory and Methods 8 1979 47 55 7 P.B. Nagarsenker On a test of equality of several exponential survival distributions Biometrika 67 1980 475 478 8 A.K. Singh Testing multiple slippages Canadian Journal of Statistics 6 1978 201 218 9 Singh, Anita, On nun-null distribution of likelihood ratio criterion for testing homogeneity of several exponential populations, Journal of Organizational Behaviour and Statistics 2 (1985) 177–185 10 B. Singh M. Schell F.T. Wright Approximations to the powers of the likelihood ratio tests; the loop ordering and the slippage alternatives Communications in Statistics Simulation and Computation 24 1995 91 109 11 Tone, Y.L., Probability inequalities in multivariate distributions, Academic Press, New York, 1980
Year
DOI
Venue
2000
10.1016/S0096-3003(98)10075-9
Applied Mathematics and Computation
Keywords
Field
DocType
bayes test,parameter exponential population,beta distribution,mean time between failures,anova,exponential distribution,mean time between failure
Mean time between failures,Homogeneity (statistics),Exponential function,Statistics,Bonferroni inequality,Mathematics,Bayes' theorem,Beta distribution
Journal
Volume
Issue
ISSN
108
1
Applied Mathematics and Computation
Citations 
PageRank 
References 
1
0.45
0
Authors
2
Name
Order
Citations
PageRank
Alok Pandey113.15
Ashok K. Singh212.14