Chiao Tung University, Taiwan b Department of Industrial Management, National Taiwan

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1 This article was downloaded by: [National Chiao Tung University 國立交通大學 ] On: 6 April 14, At: :59 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: Registered office: Mortimer House, Mortimer Street, London W1T 3JH, UK Journal of Applied Statistics Publication details, including instructions for authors and subscription information: Multivariate Capability Indices: Distributional and Inferential Properties W. L. Pearn a, F. K. Wang b & C. H. Yen a a Department of Industrial Engineering and Management, National Chiao Tung University, Taiwan b Department of Industrial Management, National Taiwan University of Science and Technology, Taiwan Published online: 5 Oct 7. To cite this article: W. L. Pearn, F. K. Wang & C. H. Yen (7) Multivariate Capability Indices: Distributional and Inferential Properties, Journal of Applied Statistics, 34:8, , DOI: 1.18/ To link to this article: PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francis makes every effort to ensure the accuracy of all the information (the Content ) contained in the publications on our platform. However, Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Any opinions and views expressed in this publication are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. Terms & Conditions of access and use can be found at

2 Journal of Applied Statistics Vol. 34, No. 8, , October 7 Multivariate Capability Indices: Distributional and Inferential Properties Downloaded by [National Chiao Tung University ] at :59 6 April 14 W. L. PEARN,F.K.WANG &C.H.YEN Department of Industrial Engineering and Management, National Chiao Tung University, Taiwan Department of Industrial Management, National Taiwan University of Science and Technology, Taiwan ABSTRACT Process capability indices have been widely used in the manufacturing industry for measuring process reproduction capability according to manufacturing specifications. Properties of the univariate processes have been investigated extensively, but are comparatively neglected for multivariate processes where multiple dependent characteristics are involved in quality measurement. In this paper, we consider two commonly used multivariate capability indices MC p and MC pm,to evaluate multivariate process capability. We investigate the statistical properties of the estimated MC p and obtain the lower confidence bound for MC p. We also consider testing MC p, and provide critical values for testing if a multivariate process meets the preset capability requirement. In addition, an approximate confidence interval for MC pm is derived. A simulation study is conducted to ascertain the accuracy of the approximation. Three examples are presented to illustrate the applicability of the obtained results. KEY WORDS: value Multivariate capability index, lower confidence bound, hypothesis testing, critical Introduction Process capability indices have been widely used in the manufacturing industry for measuring process reproduction capability according to its manufacturing specifications. In current practice, suppliers are often required to provide their process capability of the product to the customers in the supply chain partnership. Process capability indices can also be used as the benchmarking for quality improvement activities. Capability indices, C p, C pk and C pm, have been proposed to evaluate process performance but restricted to cases with single engineering specification. A large number of papers have dealt with the statistical properties and the estimation of these univariate indices. Kotz & Lovelace (1998) provided a review of these indices in their textbook. Kotz & Johnson () provided a compact survey and commented on some 17 publications on process capability indices during the years 199 to. Process capability is defined to be the range over which the measurements of a process vary when the process variation is due to random causes only. Process capability indices provide an effective measure of process performance. Engineers can realize whether the product Correspondence Address: F. K. Wang, Department of Industrial Management, National Taiwan University of Science and Technology, Taiwan. fukwun@mail.ntust.edu.tw Print/ Online/7/ Taylor & Francis DOI: 1.18/

3 94 W. L. Pearn et al. Downloaded by [National Chiao Tung University ] at :59 6 April 14 meets its specifications using process capability indices. In real applications, manufactured products often have multiple quality characteristics. That is, the process capability analysis involves more than one engineering specification. For this reason, multivariate methods for assessing process capability are proposed. Wang et al. () reviewed three multivariate methods (Taam et al., 1993; Chen 1994; and Shahriari et al., 1995) and compared their capability for four problems. Although some multivariate capability indices have been proposed, research on the statistical properties of those multivariate capability indices has received little attention. It is needed to investigate the statistical properties for the multivariate process capability indices for practical purposes. In this paper, the statistical properties for the multivariate process indices MC p and MC pm are investigated. The next section describes the processes with multiple characteristics. The estimator of MC p and its properties are presented in the third section. Confidence interval, lower confidence bound, and hypothesis testing for MC p are presented in the fourth section. An approximate confidence interval and lower confidence bound for MC pm and a simulation study to ascertain the accuracy of the approximate confidence interval for MC pm are presented in the fifth section. Three examples are chosen to illustrate the proposed methodology in the sixth section. Conclusions are made in the final section. Multivariate Capability Indices Traditionally, process capability is confined to single characteristic of the product. In many cases, a manufactured product is described by more than one characteristic. That is, manufactured items are confined to several different characteristics for adequate description of their quality. Each of those characteristics must satisfy certain specifications. The assessed quality of a product depends on the combined effects of these characteristics, rather than on their individual values. For example, automobile paint needs to have a range of light reflective abilities and a range of adhesion abilities. A paint that satisfies one criterion but not the other may be undesirable. Those characteristics are related through the compositions of the paint. It is therefore only natural to consider a bivariate characterization of this paint. As for the tolerance region about multiple characteristics, we often take an ellipsoidal region or a rectangular region. For more complex engineering specifications, the tolerance region will be very complicated. For instance, a drawing of a connecting rod in a combustion engine consists of crank-bore inner diameter, pin-bore inner diameter, rod length, bore truelocation and so on. In multivariate processes, we usually assume that the observations X have a multivariate normal distribution N v (μ, ), where v is the dimension of variables, μ is the mean vector and represents the variance covariance matrix of X. In addition, T is the target vector, X is the sample mean vector and S is the sample covariance matrix. We will focus on the multivariate process indices MC p and MC pm (Taam et al., 1993). The multivariate capability index MC p is defined as vol.(modified tolerance region) vol.[(x μ) 1 (X μ) k(q)] = vol.(modified tolerance region) MC p = (πχv,.9973 )v/ 1/ [Ɣ(v/ + 1)] 1 (1) where k(q) is the 99.73th percentile of the χ distribution with v degrees of freedom, is the determinant of, and Ɣ( ) is the gamma function. Also the multivariate capability index MC pm is defined in the following, where D =[1 + (μ T) 1 (μ T)] 1/. MC pm = MC p D ()

4 Multivariate Capability Indices 943 Note that, MC pm is less than 1, the process is not close to the specified tolerance region. On the other hand, MC p is larger than 1, which only indicates that the process variation is smaller than the specified range of variation. Estimation of MC p An estimator of MC p can be expressed as Downloaded by [National Chiao Tung University ] at :59 6 April 14 ˆMC p = vol.(modified tolerance region) vol.(estimated 99.73% process region) = vol.(modified tolerance region) (πχv,.9973 )v/ S 1/ [Ɣ(v/ + 1)] 1 (3) where S is the sample variance-covariance matrix and S is the determinant of S. According to the equation (1), ˆMC p can be expressed as MC p ( S / ) 1/. Let X = (X 1,X,...,X n ) represents an n-dimensional vector of measurements from a multivariate normal distribution with mean vector μ = (μ 1,μ,...,μ v ), T is the vector of target values, is the process variance-covariance matrix. Using the following theorem, we can derive the distribution of ˆMC p. Theorem 1 The distribution of the generalized variance S of a sample X 1,X,...,X n from N v (μ, ) is the same as the distribution of /(n 1) v times the product of v independent factors, the distribution of the ith factor being the χ distribution with n i degrees of freedom. For the proof of Theorem 1, see Anderson (3, p. 68). From the above theorem, we can have that S / is the distribution of χn 1 χ n χ n v /(n 1)v. Let y = χn 1 χ n χ n v. Then, we have MC p ˆMC p y (n 1) v Using the transformation method, the probability density function of ˆMC p can be expressed as [ ] f(x)= f Y [g 1 (x)] d MC dx (x) = f p (n 1) v Y (n 1)v MC p x x 3 for x> (4) When v = 1, the probability density function of ˆMC p is equivalent to the pdf of Ĉ p. So, we have Cp y Ĉp (n 1) where y = χ n 1. From equation (4) and the pdf of y is the pdf of Ĉ p is given by f(x)= f Y (y) = (1/)(n 1)/ y (n 1)/ 1 e y/ Ɣ(n 1/) (n 1) (n 1)/ C p Ɣ(n 1/) (n 3)/ ( Cp x ) e [(n 1)/] (C p/x) for x> (5)

5 944 W. L. Pearn et al. When v =, we have MC p ˆMC p y (n 1) where y = χ n 1 χ n. It can be shown that χ n 1 χ n (χ n 4 ) /4 (see Corollary 1 in Appendix). Thus, we have MC p z ˆMCp 4(n 1) where z = (χ n 4 ). Using the transformation method, the pdf of z is Downloaded by [National Chiao Tung University ] at :59 6 April 14 f(z)= (1/)(n 1) z n/ e z/ for z> Ɣ(n ) Similarly, from equation (4) and the pdf of z, the pdf of ˆMC p is given by f(x)= When v = 3, we have (n 1)(n ) MC p Ɣ(n ) ( ) n 1 MCp e (n 1) (MC p/x) x MC p ˆM C p y where y = χ (n 1) 3 n 1 χ n χ n 3. Let z 1 = χn 1 χ n (χ n 4 ) /4 and z = χn 3. Then, we have MC p ˆMC p z 1z (n 1) 3 for x> (6) Let u = z 1 z and v = z 1. Using the transformation method, the joint pdf of u v is f (u, v) = (1/)(n 1)/ v 1/ u (n 5)/ e ( v+u/v) Ɣ(n )Ɣ (n 3/) Then, the pdf of u is given by for u, v > f U (u) = (1/) (n 1)/ v 1/ u (n 5)/ e ( v+u/v) dv for u> Ɣ(n )Ɣ (n 3/) Similarly, from equation (4) and the pdf of u, the pdf of ˆMC p is given by f(x)= (n 1)3 (1/) (n/) (3/) MC p Ɣ(n )Ɣ (n 3/) ( ) n MCp x v 1/ e [ v+(n 1) 3 (MC p /x) /(v)] dv for x> (7) Since the hth moment of a χ distribution with v degrees of freedom is h Ɣ(v/ + h)/ɣ(v/) and the moment of a product of independent variables is the product of the

6 Multivariate Capability Indices 945 moments of the variables, the hth moment of S / can be obtained as vh v Ɣ [1/(n i) + h] E( S / ) h i=1 = (n 1) vh v Ɣ [1/(n i)] i=1 (8) Now we can derive the rth moment of ˆMC p according to the equation (8) and ˆMC p = MC p ( S / ) 1/. Thus, we have E( ˆMC r p ) = E(MCr p ( S / ) r/ ) = MC r p E( S / ) r/ (9) Downloaded by [National Chiao Tung University ] at :59 6 April 14 Now, we can substitute r = 1 and h = 1/ into the equations (8) and (9), respectively. Then, we have v/ v Ɣ [1/(n i) 1/] i=1 E( ˆMC p ) = MC p (n 1) v/ v Ɣ [1/(n i)] i=1 ( ) n 1 v/ Ɣ [1/(n v) 1/] = MC p = 1 MC p (1) Ɣ [1/(n 1)] b v where b v = ( n 1) v/ Ɣ [1/(n 1)] Ɣ [1/(n v) 1/] is a correction factor, so that b v ˆMC p is an unbiased estimator of MC p. Again, we can substitute r = and h = 1 into the equations (8) and (9), respectively. Then, we have E( ˆMC p ) = MC p v v Ɣ [1/(n i) 1] i=1 (n 1) v v Ɣ [1/(n i)] From the equations (1) and (11), we have the variance of ˆMC p as Var( ˆMC p ) = MC p v i=1 v Ɣ [1/(n i) 1] i=1 (n 1) v v Ɣ [1/(n i)] i=1 (11) [ ] 1 MC p (1) b v When v = 1, according to the equations (1) and (1), we can find that the expectation and variance of ˆMC p are equal to those of Ĉ p (see Kotz & Lovelace, 1998). That is, the expectation and variance of Ĉ p can be obtained as and n 1 Ɣ [1/(n )] E(Ĉ p ) = MC p Ɣ [1/(n 1)] = 1 C p b 1 [ n 1 Var(Ĉ p ) = n 3 1 ] b1 Cp where b 1 = /(n 1) [Ɣ [(1/)(n 1)]/Ɣ [(1/(n ))]] is a correction factor, so that b 1 Ĉ p is an unbiased estimator of C p. When v =, according to the equations (1) and

7 946 W. L. Pearn et al. (1), the expectation and variance of ˆMC p can be obtained as E( ˆMC p ) = MC p n 1 n 3 = 1 b MC p and Var( ˆMC p ) = MC p (n 1) (n 3) (n 4) = 1 n 4 1 b MC p where b = (n 3)/(n 1) is a correction factor, so that b ˆMC p is an unbiased estimator of MC p. When v = 3, according to the equations (1) and (1), the expectation and variance of ˆMC p can be obtained as Downloaded by [National Chiao Tung University ] at :59 6 April 14 and ( ) n 1 3/ E( ˆMC p ) = MC p [ Var( ˆMC p ) = MC p (n 1) 3 (n 3)(n 4)(n 5) [ (n 1) 3 = (n 3)(n 4)(n 5) 1 b3 Ɣ [1/(n 3) 1/] Ɣ [1/(n 1)] = 1 b 3 MC p ] (n 1)3 (n 3) Ɣ (1/n ) Ɣ (1/n 3/) ] MC p where b 3 = (n 3)/(n 1) /(n 1) (Ɣ(1/n 3 ))/(Ɣ( 1 n )) is a correction factor, so that b 3 ˆMC p is an unbiased estimator of MC p. Confidence Intervals and Hypothesis Testing for MC p Confidence Interval and Lower Confidence Bound for MC p Since ˆMC p = MC p ( S / ) 1/, like other statistics, is subject to the sampling variation, it is critical to compute an interval to provide a range that includes the true MC p with high probability. Based on the definition, a 1(1 α)% confidence interval for MC p can be established as: ( ) S 1/ P {L MC p U} =1 α P {L ˆMC p U} =1 α P { P P L ˆMC p L ˆMC p { L (n 1) v ˆMC p ( ) } S 1/ U = 1 α ˆMC p χn 1 χ n χ n v U (n 1) v ˆMC p = 1 α } χn 1 χ n χ n v U (n 1) v ˆMC p = 1 α Let y = χ n 1 χ n χ n v, then we have U (n 1) v / ˆMC p L (n 1) v / ˆMC p f Y (y)dy = 1 α

8 Multivariate Capability Indices 947 So, we have F 1 Y (α/) = (L (n 1) v )/ ˆMCp and F 1 Y (1 α/) = (U (n 1) v )/ ˆMCp, where F Y (z) = z f(y)dy. Thus, a 1(1 α)% confidence interval for MC p can be obtained as: F ˆMC 1 Y (α/) p (n 1), F ˆMC 1 Y (1 α/) v p (13) (n 1) v Downloaded by [National Chiao Tung University ] at :59 6 April 14 Furthermore, a 1(1 α)% lower confidence bound for MC p can be obtained as F ˆMC 1 Y (α) p (n 1) v (14) When v = 1, a 1(1 α)% confidence interval and lower confidence bound for C p are given by χ n 1,α/ Ĉ p (n 1), Ĉ χn 1, 1 α/ χ p n 1,α and Ĉ p (15) (n 1) (n 1) When v =, we can find that the distribution of χn 1 χ n is equal to (χ n 4 ) /4. Thus, a 1(1 α)% confidence interval and lower confidence bound for MC p are given by (χ n 4,α/ ˆMC ) p 4 (n 1), (χn 4,1 α/ ˆMC ) (χ p and n 4,α ˆMC ) 4 (n 1) p (16) 4 (n 1) When v = 3, we can find that the distribution of χn 1 χ n χ n 3 can be expressed as y = (χn 4 ) /4 χn 3. Now, let y 1 (χn 4 ) /4 and y χn 3. So, we have y = y 1y. Let w = y 1. Using the transformation method, the probability density function of y is given by f Y (y) = = f y1,y (y 1,y )dy 1 (1/) (n 1)/ y 1/ 1 y (n 5)/ e y1/ Ɣ(n ) Ɣ[(n 3)/] 1 (y /y 1 ) dy 1, y< Thus, a 1(1 α)% confidence interval and lower confidence bound for MC p are given by F ˆMC 1 Y (α/) p (n 1), F ˆMC 1 Y (1 α/) 3 p and F ˆMC 1 Y (α) (n 1) 3 p (17) (n 1) 3 where F Y (y) = y (1/) (n 1)/ x 1/ y (n 5)/ e x1/ (y/x) dx dy Ɣ(n ) Ɣ[(n 3)/] Efficient Mathematica programs are developed to obtain equation (17). These programs are available by sending an request to Wang.

9 948 W. L. Pearn et al. Hypothesis Testing for MC p In hypothesis testing, we determine whether or not a hypothesized value of a parameter is true or not, based on the sample taken and the parameter estimate derived from it. That is, we are trying to find out where the estimated capability is relative to either true capability, hypothesized capability, or how different the estimated and true capabilities are. To do this, we estimate an index value, compare it to a lower bound c, and compute the so-called p-value. The quantity p refers to the actual risk of incorrectly concluding that the process is capable for a particular test. In general, we want p-value to be no greater than.5. To test whether a given process is capable, we may consider the following statistical hypothesis testing: Downloaded by [National Chiao Tung University ] at :59 6 April 14 H : MC p c (process is not capable) H 1 : MC p >c (process is capable) where c is the standard minimal criteria for MC p. The critical value, c, can be determined as: { } MC p P ˆMC p >c MC p = c =α P (χn 1 χ n χ n v)/(n 1) v >c MC p = c = α. Let y = χn 1 χ n χ n v, then we have { } { c P >c c } = α P y/(n 1) v y/(n 1) v >c = α P { y< (n 1)v c } c = α F 1 Y (α) = (n 1)v c. c Thus, the critical value can be expressed as = α (n 1) v c /c f Y (y)dy c = c (n 1) v F 1 Y (α) (18) When v = 1 and y = χn 1, the critical value is equal to n 1 c χn 1,α. When v = and y = χ n 1 χ n (χ n 4 ) /4, the critical value is equal to c 4(n 1) (χ n 4,α ).

10 Multivariate Capability Indices 949 When v = 3 and y = χn 1 χ n χ n 3, the critical value is equal to c (n 1) 3 F 1 Y (α), where F Y (y) = y (1/) (n 1)/ x ( 1/) y (n 5)/ e x1/ y/x dxdy Ɣ(n ) Ɣ((n 3)/) From the definition of c, it is obvious that the value of ˆMC p must be higher than the original target value for the true MC p. The amount of difference required depends on the sample size, n. The power of the test, β,isgivenby Downloaded by [National Chiao Tung University ] at :59 6 April 14 β(mc p ) = P { ˆMC p >c MC p } MC p = P >c MC p (χ n 1 χ n χ n v)/(n 1) v { = P y< (n } 1)v MC p MC c p where y is the probability density function of χn 1 χ n χ n v. From the above equation, we can obtain the operating characteristic (OC) curve for MC p, which plots the true value of 1 β(mc p ) against MC p for two situations: (1) n = 3, c = 1.33; () n = 7, c = When v = 1, and 3, several operating characteristic (OC) curves for MC p are shown in Figures 1 5. From these operating characteristic curves, we found that some interesting results are as follows. (i) From these graphs, it is obvious that when the c is larger, the chance of incorrectly concluding the process is not capable is smaller. (ii) When n and c of the two are the same, then their OC curves are similar regardless of v = 1, v = orv = 3. (iii) When c is smaller, the chance of incorrectly concluding the process is not capable will be smaller. (19) Figure 1. Operating characteristic curve for MC p when v = 1

11 95 W. L. Pearn et al. Downloaded by [National Chiao Tung University ] at :59 6 April 14 Figure. Operating characteristic curve for MC p when v = Figure 3. Operating characteristic curve for MC p when v = 3 Figure 4. Operating characteristic curve for MC p when n = 3 Approximate Confidence Intervals for MC pm An Approximate Confidence Interval and Lower Confidence Bound for MC pm An estimator of MC pm can be expressed as ˆMC pm = ˆMC p ˆD = MC p ( S / ) 1/ = MC p D ( S / ) 1/ ˆD D ˆD = MC pm Ḓ ( ) S 1/ D ()

12 Multivariate Capability Indices 951 Downloaded by [National Chiao Tung University ] at :59 6 April 14 where Since we have where Figure 5. Operating characteristic curve for MC p when n = 7 ˆD = [ ( ) ] n 1/ 1 + (X T) S 1 (X T) n 1 ˆMC pm = MC pm Ḓ ( ) S 1/ D and (MC pm / ˆMC pm ) D = ˆD = ( ) S 1/ ( MC pm / ˆMC pm ) D = ˆD D =[1 + (μ T) 1 (μ T)], ˆD = S χ n 1 χ n χ n v (n 1) v ( ) S = ˆD χ n 1 χ n χ n v (n 1) v [ ( ) ] n 1 + (X T) S 1 (X T) n 1 Based on the definition, a 1(1 α)% confidence interval formc pm is given as follows: P { L MC pm U } ( ) ( ) L = 1 α P D MC pm ˆMC pm ˆMC pm ( ) D U D ˆMC pm = 1 α ( ) ( L χ P D ˆD n 1 χn χ ) n v ˆMC pm (n 1) v ( ) U D ˆMC pm = 1 α

13 95 W. L. Pearn et al. ( ) L P D (n 1) v ˆD χn 1 ˆMC χ n χ n v pm ( ) U D (n 1) v ˆMC pm = 1 α Let z = ˆD χn 1 χ n...χ n v. The above equation can be rewritten as U (n 1) v D / ˆMC pm L (n 1) v D / ˆMC pm f Z (z)dz = 1 α. Downloaded by [National Chiao Tung University ] at :59 6 April 14 So, we have F 1 Z (α/) = L (n 1) v D and F 1 ˆMCpm Z (1 α/) = U (n 1) v D ˆMCpm where F Z (z) = z f Z(z)dz. Thus, a 1(1 α)% confidence interval and lower confidence bound for MC pm are given by F ˆMC 1 Z (α/) pm (n 1) v D, ˆMC pm (1 α/) and F ˆMC 1 Z (α) (n 1) v D pm (1) (n 1) v D F 1 Z In fact, D and τ are unknown values, we can use ˆD and ˆτ to estimate the values of D and τ, where [ ( ) n 1/ ˆD = 1 + (X T) S 1 (X T)] and ˆτ = n(x T) S 1 (X T) n 1 Thus, an approximate 1(1 α)% confidence interval and lower confidence bound for MC pm are given by ˆMC pm ˆF z 1 (α/) (n 1) v ˆD, ˆMC pm ˆF z 1 (1 α/) (n 1) v ˆD and ˆF z ˆMC 1 (α) pm (n 1) v ˆD () From Corollary in theappendix, the pdf of z can be found. When v =, an approximate 1(1 α)% confidence interval and lower confidence bound for MC pm are given by ˆF z ˆMC 1 (α/) ˆF pm (n 1) ˆD, z ˆMC 1 (1 α/) pm (n 1) ˆD and ˆF z ˆMC 1 (α) pm (n 1) ˆD (3) where ˆF Z (z) = z 1 i= 1/e ( 1/( ˆτ )) xɣ(n )Ɣ[(n )/] (z/x)(n 4)/ e z/x ( ˆτ /) i (x 1) i Ɣ(n/ + i) i!ɣ(i + 1)x 1/n+i dxdz for z

14 Multivariate Capability Indices 953 When v = 3, an approximate 1(1 α)% confidence interval and lower confidence bound for MC pm are given by ˆF z ˆMC 1 (α/) ˆF pm (n 1) 3 ˆD, z ˆMC 1 (1 α/) pm (n 1) 3 ˆD and ˆMC pm ˆF z 1 (α) (n 1) 3 ˆD (4) where Downloaded by [National Chiao Tung University ] at :59 6 April 14 ˆF Z (z) = z 1 i= (1/) (n 1)/ x 1/ e x z/(wx) e ( (1/) ˆτ) (z/w) (n 5)/ Ɣ(n )Ɣ[(n 3)/] wɣ[(n 3)/] ( ˆτ /) i (w 1) i+1/ Ɣ(n/ + i) dxdwdz i!ɣ(i + 3/)w n/+i Simulation Study In order to ascertain the performance of the confidence interval and the lower confidence bound in equation (3), a simulation study was conducted. In this study, random samples of size 5, 45 and 65 were generated from the multivariate normal distribution with a plethora of combinations of μ, and MC pm. The specification limits were assumed to be, without loss of generality, LSL 1 = 1, T 1 = 13, USL 1 = 16, LSL = 1, T = 13, USL = 14. For each combination, 1 random samples were generated and, for each of these samples, the corresponding confidence intervals and lower confidence bound were assessed. The proportion of times that each of these limits contains the actual value of the index was recorded. The frequency of coverage for the limit is a binomial random variable with p =.95 and N = 1. Thus, a 99% confidence interval for the coverage proportion is.95 ± /1. Hence, the limits from.93 to.968 are the critical values for a statistical test at the 99% confidence level of the hypothesis that the p = value is.95. The obtained results are summarized in Table 1. More specifically, Table 1 presents that the observed coverage of 95% confidence interval as well as the observed coverage of the lower confidence bound are within the nominal interval at 99% confidence level. Thus, we can ascertain the performance of the confidence interval and the lower confidence bound in equation (3). Illustration Examples Three examples were chosen to illustrate the proposed methodology. The first example was used to illustrate a process with two variables. The other two examples were used to illustrate a process with three variables. Example 1 Chen (1994) discussed a bivariate normal example and employed Sultan (1986) bivariate process data (n = 5). Of particular interest were the Brinell hardness (H) and the tensile strength (S) of a process. The specification limits for H and S were set at (11.7, 41.3) and (3.7, 73.3), respectively. The center of the specifications was μ T =[177, 53]. The sample

15 954 W. L. Pearn et al. Table 1. Observed coverage of 95% confidence limits n = 5 n = 45 n = 65 μ MC pm CI LCB CI LCB CI LCB [ ] χ, χ, Downloaded by [National Chiao Tung University ] at :59 6 April 14 [ ] [ ] [ ] χ, χ, χ, χ, χ, χ, χ, χ, χ, χ, χ, χ, χ, χ,.9973 Note: CI = confidence interval; LCB = lower confidence bound. mean vector and sample covariance matrix were X T =[177., 5.3] and S = [ ] Using the discussion in the second and third sections, we can derive the estimated value, expectation, variance, confidence interval, lower confidence bound and critical value for MC p. From the data, we have χ,.9973 = and S = Then, we have ˆMC p = π [( )/] [( )/] π S 1/ χ,.9973 = 1.78 Since v = and n = 5, the expectation and variance of ˆMC p can be calculated as (5/) MC p,(48/847) MC p, respectively. According to the equation (16), a 95% confidence interval for MC p is calculated as 1.78 (χ46,.5 ) (χ46,.975, =[1.499,.3985] In addition, a 95% lower confidence bound for MC p is (χ46, ) =

16 Multivariate Capability Indices 955 To judge whether this process meets the present capability requirement, we consider a statistical hypothesis testing for MC p :H : MC p 1 versus H 1 : MC p > 1. According to equation (18), the critical value is obtained as 4(5 1) c = 1 = ) (χ 46,.5 Downloaded by [National Chiao Tung University ] at :59 6 April 14 Since ˆMC p = 1.78 > 1.568, we can conclude that the MC p is larger than 1 at 95% confidence level. It implies that this process variation is smaller than the specified range of variation. From the data, we have ˆD = 1.8 and ˆτ = Thus, the ˆMC pm can be calculated as 1.78/1.8 = According to equation (3), an approximate 95% confidence interval for MC pm is calculated as [ (4) 1.8, (4) 1.8 ] =[1.583,.58] Also, according to equation (3), an approximate 95% lower confidence bound for MC pm is (4) 1.8 = Therefore, we can conclude that the MC pm is larger than 1 at 95% confidence level. It implies that this process is close to the specified target. Example A previous study (Wang & Chen, 1998) presented a trivariate quality control involving the joint control of the depth (D), the length (L), and the width (W) of a plastic product from a multivariate normality. Fifty observations were collected from a plastic production line. The specified limits for D, L, and W were set at [.1,.3], [34.5, 35.1] and [34.5, 35.1], respectively. The 3D tolerance region is illustrated in Figure 6. The specification of the target value is T T =[., 34.8, 34, 8]. The sample mean vector and sample covariance matrix for 5 observations were X T =[.16, 34.7, 34.77] and S = From the data, we have χ 3,.9973 = and S = Then we have ˆMC p = 4/3π (./) (.6/) (.6/) S 1/ (π χ 3,.9973 )3/ [Ɣ(.5)] 1 =.98 Figure 6. Tolerance region for example

17 956 W. L. Pearn et al. Since v = 3 and n = 5, the expectation of ˆMC p can be calculated as MC p and.3885 MC p, respectively. According to equation (17), a 95% confidence interval for MC p is calculated as [ ] ,.98 =[1.9136, ] In addition, a 95% lower confidence bound is Downloaded by [National Chiao Tung University ] at :59 6 April = To judge whether this process meets the present capability requirement, we consider a statistical hypothesis testing for MC p :H : MC p 1 versus H 1 : MC p > 1. According to the equation (18), the critical values 49 3 c = = Since ˆMC p =.98 > , we can conclude that the MC p is larger than 1 at 95% confidence level. It implies that this process variation is smaller than the specified range of variation. From the data, we have ˆD =.348 and ˆτ = Thus, the ˆMC pm can be calculated as.98/.348 = According to equation (4), an approximate 95% confidence interval for MC pm is calculated as [ ] (49) 3.348, =[.84, 1.8] (49) Also, according to equation (4), an approximate 95% lower confidence bound for MC pm is (49) =.858 Therefore, we can conclude that the lower bound for MC pm is not larger than 1 at 95% confidence level. It implies that this process is not close to the specified target. Example 3 Taam et al. (1993) and Karl et al. (1994) discussed a geometric dimensioning and tolerancing (GD&T) drawing that specifies a target value for a pin diameter corresponding to the midpoint of allowable pin sizes and allowable perpendicularity of the pin depending on its size. The specifications require a pin diameter between 9 and 11 tenths of an inch (all units in tenths of an inch) and the center line of the pin to be within a cylinder of diameter.5 at maximum material condition (MMC, i.e., maximum pin diameter), increasing to a cylinder.5 diameter at Least Material Condition (LMC, i.e. minimum pin diameter). Therefore, the tolerance of perpendicularity depends on the pin diameter: for a pin with a diameter of 9, the allowable perpendicular tolerance zone is a.5 diameter cylinder, whereas for a pin diameter of 11, the allowable perpendicular tolerance is a.5 diameter cylinder. A pin meeting this specification will fit a gage with an 11.5 diameter hole. These GD&T specifications result in a three-dimensional tolerance region in the shape of a frustum, as illustrated

18 Multivariate Capability Indices 957 Downloaded by [National Chiao Tung University ] at :59 6 April 14 Figure 7. Tolerance region and modified tolerance region for example 3 in Figure 7. The center of the specification is T T =[,, 1]. The sample mean vector and sample covariance matrix for 7 observations were X T =[.14,.6, 1.586] and S = From the data, we have χ3,.9973 = and S =.19. Then we have ˆMC p = 4/3π [(11 9)/] (.5/) (.5/) S 1/ (π χ 3,.9973 )3/ [Ɣ(.5)] 1 = Now, v = 3 and n = 7, the expectation and variance of ˆMC p can be calculated as 1.57 MC p and.5685 MC p, respectively. According to equation (17), a 95% confidence interval for MC p is calculated as [ ] , =[1.579,.613] In addition, a 95% lower confidence bound is = To judge whether this process meets the present capability requirement, we consider a statistical hypothesis testing for MC p :H : MC p 1 versus H 1 : MC p > 1. According to the equation (18), the critical values c = = Since ˆMC p = > 1.343, we can conclude that the MC p is larger than 1 at 95% confidence level. It implies that this process variation is smaller than the specified range of variation. From the data, we have ˆD = and ˆτ = Thus, the ˆMC pm can be

19 958 W. L. Pearn et al. calculated as 1.775/1.437 = According to equation (4), an approximate 95% confidence interval for MC pm is calculated as [ ] (69) , =[1.19,.383] (69) Also, according to equation (4), an approximate 95% lower confidence bound for MC pm is Downloaded by [National Chiao Tung University ] at :59 6 April (69) = Therefore, we can conclude that the MC pm is larger than 1 at 95% confidence level. It implies that this process is close to the specified target. Conclusions Processes with multiple quality characteristics often occur in manufacturing industries. Multivariate capability indices such as MC p and MC pm have been proposed to measure process reproduction capability according to the corresponding multiple specifications. In this paper, we obtained the probability density function of the estimated ˆMC p. We constructed lower confidence bounds for ˆMC p, and developed the corresponding hypothesis testing for MC p. In addition, we derived an approximate lower confidence bound for MC pm for v =, 3. A simulation study was conducted to ascertain the accuracy of the approximation. Three examples are given to illustrate the obtained results. Although we only provided the distribution of MC p for v = 1,, 3, using the variable transformation technique we could obtain the sampling distribution of MC p for v 4. Practitioners can use the proposed procedure to determine whether their process meets the preset capability requirement, and so make reliable decisions. References Anderson, T. W. (3) An Introduction to Multivariate Statistical Analysis, 3rd edn (New York: Wiley). Chen, H. (1994) A multivariate process capability index over a rectangular solid tolerance zone, Statistica Sinica, 4, pp Karl, D. P., Morisette, J. & Taam, W. (1994) Some applications of a multivariate capability index in geometric dimensioning and tolerancing, Quality Engineering, 6, pp Kotz, S. & Johnson, N. L. () Process capability indices a review, 199, Journal of Quality Technology, 34, pp. 19. Kotz, S. & Lovelace, C. R. (1998) Process Capability Indices in Theory and Practices (London: Arnold). Shahriari, H., Hubele, N. F. & Lawrence, F. P. (1995) A multivariate process capability vector, Proceedings of the Fourth Industrial Engineering Research Conference, Institute of Industrial Engineers, pp Sultan, T. L. (1986) An acceptance chart for raw material of two correlated properties, Quality Assurance, 1, pp Taam, W., Subbaiah, P. and Liddy, J. W. (1993) A note on multivariate capability indices, Journal of Applied Statistics,, pp Wang, F. K. & Chen, J. (1998) Capability index using principal component analysis, Quality Engineering, 11, pp Wang, F. K., Hubele, N. F., Lawrence, F. P., Miskulin, J. D. & Shahriari, H. () Comparison of three multivariate process capability indices, Journal of Quality Technology, 3, pp

20 Multivariate Capability Indices 959 Appendix Corollary 1 If χn 1 and χ n are independently distributed, then χ n 1 χ n (χn 4 ) /4. is distributed as Proof Let x 1 χn 1 and x χn. The joint pdf of x 1 and x is given by Downloaded by [National Chiao Tung University ] at :59 6 April 14 f x1,x (x 1,x ) = (1/)(n 1)/ x n/ 3/ 1 e x 1/ Ɣ[(n 1)/] (1/)(n )/ x n/ e x / Ɣ[(n )/] Let z 1 = x 1 and z = x 1 x. Using the transformation method, the solution is x 1 = z 1 and x = z /4z 1, and the Jacobian of the transformation is J = So, we find that the joint pdf of z 1 z is 1 z 4 z z 1 z = z. z 1 1 f z1,z (z 1,z ) = (1/)(n 1)/ z n/ 3/ 1 e z 1/ (1/)(n )/ (z /4z 1) n/ e (z /4z 1)/ Ɣ[(n 1)/] Ɣ[(n )/] z, z 1,z z 1 Then, the marginal density function of z is obtained as follows: f z (z ) = where (1/) (n 1)/ z n/ 3/ 1 e z 1/ (1/)(n )/ (z /4z 1) n/ e (z /4z 1)/ z dz 1 Ɣ[(n 1)/] Ɣ[(n )/] z 1 = C 1 z n 3 z 1/ 1 e z 1/ (z /4z 1)/ dz 1, z C 1 = (1/) n 9/ Ɣ[(n 1)/] Ɣ[(n )/] Let h(z ) = z 1/ 1 e z 1/ (z /4z1)/ dz 1. Hence, h (z ) = ( z /4z 1 ) z 1/ 1 e z 1/ (z /4z1)/ dz 1.Now,letz /4z 1 = w. Using the transformation method, we find that h (z ) = ( 1 ) w 1/ e w/ (z /4w)/ dw = ( 1 ) h(z ) The above equation gives h(z ) = e ( z /+C ), where C is a constant. Thus, the pdf of z is given as the following, where C 3 = C 1 e C. Therefore, we have z χn 4,f z (z ) = C 1 e C z n 3 e z/ = C 3 z (n 4)/ 1 e z/, z.

21 96 W. L. Pearn et al. Theorem Let T = n( X μ ) S 1 ( X μ ), where X = (x 1,x,...,x n ) be a sample from N(μ, ) with mean vector μ = (μ 1,μ,...,μ v ) and covariance matrix v v, and μ is the vector of target values. The distribution of T n 1 n v v is a non-central F with v and n v degrees of freedom and non-centrality parameter τ = n( X μ ) 1 ( X μ ). Downloaded by [National Chiao Tung University ] at :59 6 April 14 Proof See Anderson (3, pp ). It can be shown that the pdf of T is given by e (1/)τ (n 1)Ɣ[1/(n v)] (See Anderson, 3, p. 186.) (τ /) [ i t /(n 1) ] (1/)v+i 1 Ɣ(1/n + i) i= i!ɣ(1/v + i) [ 1 + t /(n 1) ] (1/)n+i (A1) Corollary (1) For v =, if z = χn 1 χ n ˆD, where ˆD =[1 + n/n 1( X μ ) S 1 ( X μ )], then the pdf of z is f Z (z) = 1 (1/)e (1/)τ z/w wɣ(n )Ɣ[(n )/] (z/w)(n 4)/ e (τ /) i (w 1) i Ɣ(n/ + i) dw for z i!ɣ(i + 1)w (1/)n+i i= () For v = 3, if z = χ n 1 χ n χ n 3 ˆD, where ˆD =[1 + n/n 1( X μ ) S 1 ( X μ )], then the pdf of z is f Z (z) = 1 i= (1/) (n 1)/ x 1/ 1 e x 1 z/(wx 1 ) e (1/)τ (z/w) (n 5)/ Ɣ(n )Ɣ[(n 3)/] Ɣ[(n 3)/] (τ /) i (w 1) i Ɣ(n/ + i) i!ɣ(i + 3/)w 1/n+i dx 1 dw for z Proof (1) From ˆD =[1 + n/(n 1)( X μ ) S 1 ( X μ )] and Theorem, we find that ˆD = 1 + (1/(n 1))T.

22 Multivariate Capability Indices 961 Let y = 1 + (1/(n 1))T. Using the transformation method and the equation (A1), the pdf of y is obtained as follows: Downloaded by [National Chiao Tung University ] at :59 6 April 14 f Y (y) = f T ((n 1)(y 1)) (n 1) e ( 1/)τ (τ /) i (y 1) i Ɣ(n/ + i) = (n 1) (n 1)Ɣ[(n )/] i!ɣ(i + 1)y (1/)n+i = e (1/)τ Ɣ[(n )/] i= i= (τ /) i (y 1) i Ɣ(n/ + i) i!ɣ(i + 1)y (1/)n+i for y 1 Let x = χn 1 χ n. From Corollary 1, we find that x = (χ n 4 ) /4. That is, the pdf of x is f X (x) = (1/)x(n 4)/ e x for x> Ɣ(n ) Now, let z = xy and w = y. Using the transformation method, the solution is x = z/w and y = w, and the Jacobian of the transformation is J = 1/w z/w 1 = 1/w. So, we find that the joint pdf of zw is f ZW (z, w) = f XY (z/w, w) (1/w) = (1/)(z/w)(n 4)/ e z/w wɣ(n ) e (1/)τ (τ /) i (w 1) i Ɣ(n/ + i) for z,w >1 Ɣ[(n )/] i!ɣ(i + 1)w 1/n+i i= Then, the marginal density function of z is obtained as follows: (1/)(z/w) (n 4)/ e z/w f Z (z) = e (1/)τ 1 wɣ(n ) Ɣ[(n )/] (τ /) i (w 1) i Ɣ(n/ + i) dw i!ɣ(i + 1)w 1/n+i i= 1/e (1/)τ z/w = 1 wɣ(n )Ɣ[(n )/] (z/w)(n 4)/ e (τ /) i (w 1) i Ɣ(n/ + i) dw for z i!ɣ(i + 1)w (1/)n+i i= () From ˆD =[1 + n/(n 1)( X μ ) S 1 ( X μ )] and Theorem, we find that ˆD = 1 + (1/(n 1))T. Let y = 1 + (1/(n 1)T ). Using the transformation method and the equation (A1), the pdf of y is obtained as follows: f Y (y) = f T ((n 1)(y 1)) (n 1) e (1/)τ (τ /) i (y 1) i+1/ Ɣ(n/ + i) = (n 1) (n 1)Ɣ[(n 3)/] i!ɣ(i + 3/)y (1/)n+i = e (1/)τ Ɣ[(n 3)/] i= i= (τ /) i (y 1) i+1/ Ɣ(n/ + i) i!ɣ(i + 3/)y 1/n+i for y 1

23 96 W. L. Pearn et al. Let x 1 = χ n 1 χ n and x = χ n 3. From Corollary 1, we find that x 1 = (χ n 4 ) /4. That is, the pdf of x 1 is f x1 (x 1 ) = (1/)x(n 4)/ 1 e x 1 Ɣ(n ) for x 1 > Now, let x = x 1 x and u = x 1. Using the transformation method, the solution is x 1 = u and x = x/u, and the Jacobian of the transformation is J = 1 1/u x/u = 1/u. So, we find that the joint pdf of xu is f XU (x, u) = f X1 X (u, x/u) 1 u Downloaded by [National Chiao Tung University ] at :59 6 April 14 = (1/)(n 1)/ u 1/ x (n 5)/ e u x/u Ɣ(n ) Ɣ[(n 3)/] Then, the marginal density function of x is x,u < (1/) (n 1)/ u 1/ x (n 5)/ e u x/u f X (x) = du, x Ɣ(n ) Ɣ[(n 3)/] Again, let z = xy and w = y. Using the transformation method, the solution is x = z/w and y = w, and the Jacobian of the transformation is the J = 1/w z/w 1 = 1/w. So, the joint pdf of zw is obtained as follows: f ZW (z, w) = f XY (z/w, w) (1/w) (1/) (n 1)/ u 1/ (z/w) (n 5)/ e u z/wu = du Ɣ(n ) Ɣ[(n 3)/] e (1/)τ (τ /) i (w 1) i+1/ Ɣ(n/ + i) 1 Ɣ[(n 3)/] i!ɣ(i + 3/)w (1/)n+i w = i= i= (1/) (n 1)/ u 1/ e u z/(wu) Ɣ(n )Ɣ[(n 3)/] du e 1/τ (z/w) (n 5)/ wɣ[(n 3)/] (τ /) i (w 1) i+1/ Ɣ(n/ + i) i!ɣ(i + 3/)w 1/n+i for z,w >1 Then, the marginal density function of z is f Z (z) = 1 i= (1/) (n 1)/ u 1/ e u z/(wu) Ɣ(n )Ɣ[(n 3)/] e 1/τ (z/w) (n 5)/ wɣ[(n 3)/] (τ /) i (w 1) i+(1/) Ɣ(n/ + i) i!ɣ(i + 3/)w (1/)n+i dudw for z.