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DOI 10.1007/s00170-003-1870-0 O R I G I N A L A R T I C L E

W.L. Pearn · Chien-Wei Wu · H.C. Chuang

Procedures for testing manufacturing precision C

p

based on ( ¯

X

, R) or ( ¯X, S)

control chart samples

Received: 31 March 2003 / Accepted: 8 July 2003 / Published online: 23 September 2004  Springer-Verlag London Limited 2004

Abstract Process precision index Cp has been widely used in the manufacturing industry for measuring process potential and precision. Estimating and testing process precision based on one single sample have been investigated extensively. In this paper, we consider the problem of estimating and testing pro-cess precision based on multiple samples taken from( ¯X, R)or

( ¯X, S )control chart. We first investigate the statistical

proper-ties of the natural estimator of Cpand implement the hypothesis testing procedure. We then develop efficient MAPLE programs to calculate the lower confidence bounds, critical values, and

p-values based on m samples of size n. Based on the test, we

develop a step-by-step procedure for practitioners to use in deter-mining whether their manufacturing processes are capable of re-producing products satisfying the preset precision requirement. Keywords Critical value· Lower confidence bound · Process precision index· p-value · Testing hypothesis

1 Introduction

Process precision index Cp has been proposed in the manufac-turing industry for measuring process potential and precision ([1–4], among others). The precision index Cp is designed to provide numerical measures on process potential (product qual-ity consistency) in meeting the preset process precision (process W.L. Pearn (u)

Department of Industrial Engineering & Management, National Chiao Tung University,

Hsinchu, 300 Taiwan, R.O.C. E-mail: roller@cc.nctu.edu.tw C.-W. Wu

Department of Business Administration, Feng Chia University,

407 Taiwan, R.O.C. H.C. Chuang

Department of Industrial Engineering & Management, Chung Hua University,

Hsinchu, 300 Taiwan, R.O.C.

variation relative to the manufacturing tolerance) requirements. The precision index Cpis defined as

Cp=

USL− LSL

6σ (1)

where USL and LSL are the upper and the lower specification limits, and σ is the process standard deviation of the quality characteristic. The precision index Cp is primarily designed to monitor process data that are taken independently, from a normal process under statistical control. Using the index under other pro-cess conditions, without proper modifications, would certainly give severely inaccurate measurement on process precision.

The use of the precision index Cpand other capability indices was first explored within the automotive industry. Ford Motor Company initially used Cpto keep track of the process perform-ance. Recently, the manufacturing industries have been making an extensive effort to implement statistical process control (SPC) in their plants and supply bases. Capability indices derived from SPC have received increasing usage not only in capability assess-ments, but also in the evaluation of purchasing decisions. Capabil-ity indices are becoming the standard tools for qualCapabil-ity reporting, particularly, at the management level around the world. Proper understanding and accurate estimating of capability indices are essential for the company to maintain a capable supplier.

2 Estimating Cp

based on control chart samples

For applications where the data are collected as one single sam-ple, Pearn et al. [3] considered an unbiased estimator of Cp. They showed that the unbiased estimator is the UMVUE (uniformly minimum variance unbiased estimator) of Cp. They also pro-posed an efficient test for Cp based on one single sample, and showed that the test is the UMP (uniformly most powerful) test. Kirmani et al. [5] considered the estimation ofσ and the preci-sion index Cp for cases where the data are collected as multiple samples. Pearn and Yang [6] proposed an unbiased estimator of

Cp for multiple samples, and showed that the unbiased estima-tor is the UMVUE of Cp, which is asymptotically efficient. Pearn

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and Yang [6] also developed an efficient test for Cpfor cases with

multiple samples, and showed that the proposed test is indeed the UMP test.

For applications where routine-based data collection plans are implemented, a common practice on process control is to es-timate the process precision by analyzing past “in control” data. Consider m preliminary multiple samples (subgroups) each of size n taken from the control chart samples. To estimateσ we typically use either the sample standard deviation or the sample range. The control chart can be used as a monitoring device or logbook to show the effect of changes in the process perform-ance. We note that a process may be in control but not necessarily operating at an acceptance level. Thus, management interven-tion is required either to improve the process capability, or to change the manufacturing requirements to ensure that the prod-ucts meet the minimum acceptable level. We remark that the process must be stable in order to produce a reliable estimate of process capability. If the process is out of control in the early stages of process capability analysis, it will be unreliable to esti-mate process capability. The top priority is to find and eliminate the assignable causes in order to bring the process into an in control state.

2.1 Estimating Cpbased on( ¯X, R) samples

If m samples each of size n from the ( ¯X, R) control chart are available, let R1,n be the range of a sample of size n and ¯Rm,n

be the average range in m samples of each size n. Then the mean and variance of the relative range ¯Rm,n/σ are given by

E¯Rm,n/σ= ER1,n/σ  = d2 (2) Var¯Rm,n/σ=Var(R1,n) 2 = d32 m (3)

where d2and d3are functions of n, which are available in

qual-ity control books and literature (see Pearson’s Table A [7]). Thus, the estimated process capability precision by the range method can be expressed as ˆCp(R)=USL− LSL 6ˆσR , ˆσR= ¯Rm,n d2 . (4)

Table 1. Coefficients of distribution for multiple samples with m= 5(5)25, n = 2(1)10, and α = 0.01, 0.025, 0.05

m= 5 m= 10 m= 15 m= 20 m= 25 n d2 d3 c v c v c v c v c v 2 1.128 0.853 1.191 4.582 1.160 8.973 1.149 13.351 1.144 17.727 1.141 22.101 3 1.693 0.888 1.739 9.317 1.716 18.414 1.708 27.505 1.705 36.594 1.702 45.682 4 2.059 0.880 2.096 13.923 2.078 27.616 2.071 41.304 2.068 54.992 2.067 68.679 5 2.326 0.864 2.358 18.359 2.342 36.483 2.337 54.603 2.334 72.723 2.332 90.842 6 2.534 0.848 2.562 22.565 2.548 44.893 2.543 67.218 2.541 89.542 2.540 111.866 7 2.704 0.833 2.730 26.586 2.717 52.932 2.713 79.276 2.710 105.620 2.709 131.963 8 2.847 0.820 2.871 30.380 2.859 60.519 2.855 90.656 2.853 120.793 2.852 150.929 9 2.970 0.808 2.992 34.022 2.981 67.803 2.977 101.581 2.975 135.359 2.974 169.137 10 3.078 0.797 3.099 37.532 3.088 74.822 3.085 112.110 3.083 149.398 3.082 186.685

If m= 1, the cumulative distribution function of the range from a normal distribution is F(x) = P  R1,n σ ≤ x  = n ∞  −∞ [Φ(x + t) − Φ(t)]n−1φ(t)dt , for t > 0 (5)

whereΦ(·) and φ(·) are the cumulative distribution function and probability density function of the standard normal distribution N(0, 1). Furthermore, using the first two moments of the average range, Patnaik [8] has shown that ¯Rm,n/σ is distributed

approxi-mately as cχv/v, where χ2

vis the chi-square distribution withv

degree of freedom and c andv are constants which are functions of the first two moments of the range as follows:

E  ¯Rm,n/σ  =√c v √ 2Γ v + 1 2  /Γv 2  (6) Var  ¯Rm,n=c2 v  v − 2 Γ v + 1 2  /Γv 2 2 . (7) The values of the mean and variance in Eq. 2 and Eq. 3 are known from the coefficients of the mean and variance of the aver-age range, d2and d3. By letting Eq. 6 equal Eq. 2 and Eq. 7 equal

Eq. 3, we obtain the values of c and v, which are solutions to the above system of equations. Table 1 displays the values of d2

,and d3and the corresponding c andv for multiple samples with m= 5(5)25 and n = 2(1)10.

In the early days of control chart usage, the range method of estimatingσ was employed to simplify the arithmetic associated with control chart operation. With modern computer software and hand-held calculators for control chart operation, this is not a consideration, and other methods could be used. If the sample size is relatively small, the range method yields almost as good an estimator of varianceσ2as does the usual sample variance S2. The relative efficiency (RE) of the range method to S2 is shown in Table 2 for various sample sizes [4]. For values of n≥ 10, the range method loses efficiency rapidly, since it ignores all the in-formation in the sample between the maximum and minimum values. However, for the small sample sizes often employed on variables control charts (n= 4, 5, or 6), it is entirely satisfactory.

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Table 2. The relative efficiency of the range method to S2

n 2 3 4 5 6 10

RE 1.000 0.992 0.975 0.955 0.930 0.850

2.2 Estimating Cpbased on( ¯X, S ) samples

If m samples each of size n from the( ¯X, S ) control chart are available, Kirmani et al. [5] considered measuring the process precision index Cp, and the natural estimator ˆCp defined as the

following: ˆCp(S )=USL− LSL 6ˆσS , ˆσS= 1 εn−1¯S , where (8) ¯S = 1 m m i=1 Si, Si=   1 n− 1 n j=1  Xij− ¯Xi 2   1/2 and εn−1=  2 n− 1 Γ [n/2] Γ [(n − 1)/2].

The ¯Xi and Si represent the sample mean and sample

stan-dard deviation of the ith sample, andεn−1is denoted by c4in the

general quality control literature. Kirmani et al. [5] showed that under the normality assumption, the statistic ¯S is approximately

distributed as the normal distribution. That is, ¯S−n− 1εn−1  (n−1)1−ε2 n−1  m ∼ N (0, 1) . (9)

This is particularly true in situations where reasonable tight control of the process variability is needed so that moderately large subgroups (n> 10) are required. In this case, the S-chart is preferred to the R-chart. We note that the expressions for the distribution of ˆCp(S )obtained in Kirmani et al. [5], Kocher-lakota [9], and Kotz and Lovelace [10] need to be modified. In fact, they addressed the distribution ofˆσSas

ˆσS∼  σ,σm21− ε 2 n−1 ε2 n−1  . (10) Hence, we have ˆCp(S )  1+ N  0,1− ε 2 n−1 2n−1 −1 Cp : (11)

The estimator ˆCp(S )is biased, and its probability density func-tion (PDF) can be obtained, and expressed as the following, for

x> 0, which is a function of Cp. g(x) =Cp 2πkx −2exp  −(Cp/x − 1)2 2k2  (12) k=    1 − ε2n−1 2n−1 , and εn−1=  2 n− 1 Γ(n/2) Γ [(n − 1)/2].

Figures 1–4 display the PDF plots of ˆCp(S )for various sam-ple sizes of m= 10, 15, 20, 25, n = 5, with Cp= 1.00, 1.33, 1.50,

1.67, and 2.00 (from left to right in plot). Figures 5–8 display the PDF plots of ˆCp(S ) for various sample sizes of m= 15, n = 3, 6, 9, 12, with Cp= 1.00, 1.33, 1.50, 1.67, and 2.00 (from left to

right in plot). We note that (i) for fixed sample size m and n, the variance of ˆCp(S )increases as Cpincreases, (ii) for fixed n and

Fig. 1. PDF plot of ˆCp(S)with m= 10 and n = 5, for various values of Cp

Fig. 2. PDF plot of ˆCp(S)with m= 15 and n = 5, for various values of Cp

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Fig. 4. PDF plot of ˆCp(S)with m= 25 and n = 5, for various values of Cp

Fig. 5. PDF plot of ˆCp(S)with m= 15 and n = 3, for various values of Cp

Fig. 6. PDF plot of ˆCp(S)with m= 15 and n = 6, for various values of Cp

Cp, the variance of ˆCp(S )decreases as m increases, and (iii) for

fixed m and Cp, the variance of ˆCp(S )decreases as n increases. Lower confidence bound on Cp. Since ˆCp (denote either ˆCp(R)

or ˆCp(S )) is subject to sampling error, it is desirable to construct a confidence interval to provide a range, which contains the true

Cpwith high probability. For cases where multiple samples taken

Fig. 7. PDF plot of ˆCp(S)with m= 15 and n = 9, for various values of Cp

Fig. 8. PDF plot of ˆCp(S)with m= 15 and n = 12, for various values of Cp

from( ¯X, R) control chart at various points in time are available, by Patnaik’s approximate distribution of the average range, the 100(1 − α)% lower confidence bound CL(R)can be constructed, which satisfies PCp≥ CL(R)= 1 − α = P  ˆσR σCL(R) ˆCp(R)  = P  ¯Rm,n σd2CL(R) ˆCp(R)   P  χv≥ √ vd2CL(R) c ˆCp(R)  . (13)

Thus, we can obtain that √vd 2CL(R) c ˆCp(R) = χv,α, or the ratio CL(R) ˆCp(R) = cvd 2  χ2 v,α, (14) whereχ2

v,α is the lowerα-th percentile of the chi-square

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of CL(R)/ ˆCp(R)depends onv, c, d2, andα. The values of v, c, d2 are determined from the number of samples m and sample

size n. We refer to this ratio CL(R)/ ˆCp(R) as lower confidence factors. Table 3 displays the lower confidence factors for multi-ple sammulti-ples with m= 5(5)25, n = 2(1)10, and α = 0.01, 0.025, 0.05. From Table 3 we observe that for fixed sample size n and

α, the lower confidence factors CL(R)/ ˆCp(R) increases as the number of samples m increases. As an example, for fixed n= 5, we have m= 5 with CL(R)/ ˆCp(R)= (0.636, 0.689, 0.735),

m= 15 with CL(R)/ ˆCp(R)= (0.784, 0.817, 0.845), and m = 25

with CL(R)/ ˆCp(R)= (0.831, 0.857, 0.879). On the other hand,

for a fixed number of samples m andα, the lower confidence fac-tors increase as the sample size n increases. This phenomenon can be explained easily. Since the estimation is usually more accurate as the total collected sample increases, we need only a smaller penalty of ˆCp(R)to account for the smaller uncertainty in the estimation.

For cases where m multiple samples of size n are available due to sampling from( ¯X, S ) control chart at various point in time, Kirmani et al. [5] constructed the 100(1 − α)% lower con-fidence bound as the following:

CL(S )= ˆCp(S )1+ Zα    1 − ε2n−1 2n−1 . (15)

Table 3. Lower confidence factors CL(R)/ ˆCp(R)for multiple samples with m= 5(5)25, n = 2(1)10, and α = 0.01, 0.025, 0.05

m 5 10 15 20 25 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 2 0.327 0.406 0.482 0.495 0.563 0.624 0.578 0.637 0.690 0.630 0.683 0.730 0.667 0.715 0.758 3 0.503 0.570 0.631 0.637 0.689 0.735 0.700 0.744 0.783 0.738 0.777 0.811 0.765 0.800 0.831 4 0.586 0.645 0.697 0.700 0.744 0.783 0.753 0.790 0.822 0.785 0.817 0.845 0.807 0.836 0.862 5 0.636 0.689 0.735 0.738 0.777 0.811 0.784 0.817 0.845 0.812 0.841 0.865 0.831 0.857 0.879 6 0.670 0.718 0.760 0.763 0.798 0.829 0.805 0.834 0.860 0.830 0.856 0.879 0.848 0.871 0.891 7 0.695 0.740 0.779 0.781 0.814 0.843 0.820 0.847 0.871 0.843 0.867 0.888 0.860 0.881 0.900 8 0.714 0.756 0.793 0.795 0.826 0.853 0.831 0.857 0.879 0.853 0.876 0.895 0.869 0.889 0.906 9 0.729 0.769 0.804 0.706 0.835 0.861 0.840 0.865 0.886 0.861 0.883 0.901 0.876 0.895 0.911 10 0.741 0.780 0.813 0.815 0.843 0.867 0.848 0.871 0.891 0.868 0.888 0.906 0.882 0.900 0.916

Table 4. Lower confidence factors CL(S)/ ˆCp(S)for multiple samples with m= 10(5)25, n = 2(1)15, and α = 0.01, 0.025, 0.05

m 10 15 20 25 α n 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 2 0.443 0.532 0.607 0.545 0.618 0.679 0.606 0.669 0.722 0.648 0.704 0.751 3 0.615 0.676 0.728 0.686 0.735 0.778 0.728 0.771 0.808 0.756 0.795 0.828 4 0.689 0.738 0.780 0.746 0.786 0.821 0.780 0.815 0.845 0.803 0.835 0.861 5 0.733 0.775 0.811 0.782 0.816 0.846 0.811 0.841 0.866 0.831 0.858 0.881 6 0.762 0.800 0.832 0.806 0.836 0.863 0.832 0.858 0.881 0.849 0.873 0.894 7 0.783 0.818 0.847 0.823 0.851 0.875 0.847 0.871 0.892 0.863 0.885 0.903 8 0.800 0.832 0.859 0.837 0.863 0.885 0.858 0.881 0.900 0.873 0.894 0.911 9 0.813 0.843 0.868 0.847 0.872 0.892 0.868 0.889 0.907 0.882 0.901 0.917 10 0.824 0.852 0.876 0.856 0.879 0.899 0.876 0.895 0.912 0.889 0.906 0.921 11 0.833 0.860 0.882 0.864 0.886 0.904 0.882 0.901 0.917 0.895 0.911 0.926 12 0.841 0.866 0.888 0.870 0.891 0.908 0.888 0.906 0.921 0.900 0.916 0.929 13 0.848 0.872 0.893 0.876 0.896 0.912 0.893 0.910 0.924 0.904 0.919 0.932 14 0.854 0.877 0.897 0.881 0.900 0.916 0.897 0.913 0.927 0.908 0.922 0.935 15 0.860 0.882 0.901 0.885 0.904 0.919 0.900 0.916 0.930 0.911 0.925 0.937

Table 4 displays the lower confidence factors of ˆCp(S ) for multiple samples with m= 10(5)25, n = 2(1)15, and α = 0.01, 0.025, 0.05. For example, with input parameters m= 10, n = 5, risk α = 0.05 and ˆCp(S )= 1.520, the program gives the lower confidence bound of Cpas 1.233. Or, by simply checking

Table 4, we obtain the lower confidence factor 0.811. Multiply-ing the lower confidence factor 0.811 by the ˆCp(S )= 1.520 we

obtain 1.233. Thus, it is ensured that with 95 percent confidence, the process precision is no less than 1.233, or Cp≥ 1.233. Other

values of ˆCp(S )< CL(S ) will support the null hypothesis that the process is incapable. Hence, the process is capable for any value of the required Cpthat is greater than the lower confidence

bound. It is noted that for fixed sample size n andα, the lower confidence factors CL(S )/ ˆCp(S )increase as the number of sam-ples m increases. And for a fixed number of samsam-ples m andα, the lower confidence factors increase as the sample size n increases. The results are consistent with the lower confidence bound ob-tained by the range method, and the explanation is the same as above.

An efficient MAPLE computer program for solving the corresponding Eq. 15 and calculating the lower confidence bound CL(S ) is listed below. The input parameters are set to: the upper specification limit USL= 12, the lower spe-cification limit LSL= 4, number of samples m = 10, the

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sample size n= 4, the collected data d[data], and the risk

α = 0.01. The program calculates the process precision ˆCp(S )

as 0.742, and finds the lower confidence bound as CL(S )= 0.511.

Maple program for lower confidence bounds CL(S ).

#input values of parameter U(USL), L(LSL), d[data], m, n, z; #Shat:= Sigmahat; Cphat:= ˆCp(S ); CL:= lower confidence bound >U:=12; L:=4; d[1]:=[10,5,7,9]; d[2]:=[5,8,7,7]; d[3]:=[7,6,6,8]; d[4]:=[5,7,9,5]; d[5]:=[6,5,8,9]; d[6]:=[10,9,11,4]; d[7]:=[4,5,10,6]; d[8]:=[6,6,7,9]; d[9]:=[8,10,6,7]; d[10]:=[9,6,11,8]; n:=4; m:=10; alpha:=0.01; z:=icdf[normal[0,1](alpha)]; v:=matrix([seq(d[i],i=1..10)]); sample:=[seq([seq(v[i,j],j=1..4)],i=1..10)]; s:=evalf(sum(describe[standarddeviation](t[i]),i=1..10)/10): epsilon:=evalf(( ((2/(n−1))ˆ(1/2))∗(GAMMA(n/2))/(GAMMA((n−1)/2)) )): Shat:=evalf(s/epsilon); Cphat:=evalf((U−L)/(6∗Shat)); CL:=evalf(Cp∗(1+z∗((1−epsilonˆ2)/(m∗epsilonˆ2))ˆ(1/2))); The output is:

U:=12 L:=4 n:=4 m:=10 alpha:=0.01 Shat:=1.79692 Cphat:=0.74201 CL:=0.51129

3 Testing C

p

based on multiple control chart samples

Cases where the data are collected as one single sample of size

n have been discussed by Kane [1]. In this case, Chou et al. [11]

give tables for lower confidence limit on Cp when σ is esti-mated by the sample standard deviation S. Whenσ is estimated by range divided by d2, Li et al. [12] also give tables for the

lower confidence limit on Cp. Now, we are interested in the test of process precision Cp based on( ¯X, R) or ( ¯X, S ) control chart samples. To test whether the process meets the precision requirement, we consider the following testing hypothesis with H0: Cp≤ C (the process is incapable), versus the alternative H1:

Cp> C (the process is capable). Thus, we may consider the test

φ(x) = 1 if ˆC

p> C0, andφ(x) = 0, otherwise. The test φ

re-jects the null hypothesis if ˆCp> C0, with type I errorα(C0) = α,

the chance of incorrectly judging an incapable process as a capa-ble one.

Table 5. Critical values C0(R)for Cp= 1.00, with m = 5(5)25, n = 2(1)10, and α = 0.01, 0.025, 0.05

m 5 10 15 20 25 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 2 3.058 2.463 2.075 2.020 1.776 1.603 1.730 1.570 1.449 1.587 1.464 1.370 1.499 1.399 1.319 3 1.988 1.754 1.585 1.570 1.451 1.361 1.429 1.344 1.277 1.355 1.287 1.233 1.307 1.250 1.203 4 1.706 1.550 1.435 1.429 1.344 1.277 1.328 1.266 1.217 1.274 1.224 1.183 1.239 1.196 1.160 5 1.572 1.451 1.361 1.355 1.287 1.233 1.276 1.224 1.183 1.232 1.189 1.156 1.203 1.167 1.138 6 1.493 1.393 1.316 1.311 1.253 1.206 1.242 1.199 1.163 1.205 1.168 1.138 1.179 1.148 1.122 7 1.439 1.351 1.284 1.280 1.229 1.186 1.220 1.181 1.148 1.186 1.153 1.126 1.163 1.135 1.111 8 1.401 1.323 1.261 1.258 1.211 1.172 1.203 1.167 1.138 1.172 1.142 1.117 1.151 1.125 1.104 9 1.372 1.300 1.244 1.416 1.198 1.161 1.190 1.156 1.129 1.161 1.133 1.110 1.142 1.117 1.098 10 1.350 1.282 1.230 1.227 1.186 1.153 1.179 1.148 1.122 1.152 1.126 1.104 1.134 1.111 1.092

3.1 Testing Cpbased on( ¯X, R) samples

When the estimated process capability precision by the range method from the( ¯X, R) control chart samples, the critical value

C0(R)can be obtained by finding the appropriate value satisfying

the following equation:

PˆCp(R)≥ C0(R)|Cp= C  = α = P  d 3ˆσR≥ C0(R)  = P  ¯Rm,n σd2 C0(R) d 3σ   P  χv≤ √ vd2 cC0(R) C  . (16)

In fact, the critical value C0(R)can be found and expressed as the following: C0(R)= √ vd2 cχ2 v,α C. (17)

Table 5 displays the critical values C0(R) for precision

re-quirement Cp= 1.00 with m subgroups of size n, and various risksα = 0.01, 0.025, and 0.05. We see that the critical value

C0(R)is proportional to the precision requirement C. Hence, we

need only to calculate C0(R)for the Cp= 1.00 case. For general

Cp= C (common requirements as 1.33, 1.67, 2.00), we obtain the corresponding critical values by multiplying C to the critical values C0(R)with Cp= 1.00. For instance, if the required preci-sion requirement C, is set to 1.33, with m= 10 subgroups of size

n= 5, and various risks α = 0.01, 0.025, and 0.05, then the

cor-responding critical values are C0(R)= (1.355, 1.287, 1.233) × 1.33 = (1.802, 1.712, 1.640). It is noted that for fixed α and sam-ple size n, the critical value C0(R) increases as the number of samples m decreases, and for fixedα and m the critical value

C0(R) increases as the n decreases. It can be understood

in-tuitively, since the estimation error is potentially larger as the total sample size m× n is smaller. It is reasonable that we need a larger C0(R) to claim that the process is capable. Under the same conditions, the p-value corresponding to ˆcp(R), a specific

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value calculated from the sample data, can be calculated as p-value= P  ˆCp(R)≥ ˆcp(R)|Cp= C  = P  ¯Rm,n σd2 ˆcp(R) d 3σ   P  χv≤ √vd 2 cˆcp(R)C  = G  √ vd2 cˆcp(R)C 2 , (18)

where G(·) is the cumulative distribution of the chi-square distri-bution withv degree of freedom.

3.2 Testing Cpbased on( ¯X, S ) samples

If the ( ¯X, S ) control chart is available, then the critical value

C0(S )can be obtained by finding the appropriate value satisfying

the following equation:

PˆCp(S )≥ C0(S )|Cp= C  = α = 1 − c0  0 Cp √ 2πkx −2exp  −(Cp/x − 1)2 2k2  dx. (19)

In fact, the critical value C0(S )can be found and expressed as

the following, with Zαrepresenting the lower 100α% percentage point of the standard normal distribution, N(0, 1),

C0(S )= C 1+ Zα  1−ε2 n−1 2 n−1 , where εn−1=  2 n− 1 Γ [n/2] Γ [(n − 1)/2]. (20)

An efficient MAPLE computer program is developed to cal-culate Eq. 20 , thereby obtaining the critical value C0(S )for given m, n,α. The program is listed below, with input parameters set

to: C= 1.00, m = 10, n = 10, and α = 0.01. The program gives the critical value C0(S )= 1.213.

Table 6. Critical values C0(S)for Cp= 1.00, m = 10(5)25, n = 2(1)15, and α = 0.01, 0.025, 0.05

m 10 15 20 25 α n 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 0.01 0.025 0.05 2 2.256 1.881 1.647 1.833 1.619 1.473 1.649 1.495 1.385 1.543 1.421 1.331 3 1.626 1.472 1.373 1.459 1.360 1.285 1.374 1.297 1.238 1.322 1.258 1.208 4 1.451 1.354 1.281 1.340 1.272 1.218 1.282 1.227 1.184 1.245 1.198 1.161 5 1.365 1.290 1.233 1.279 1.225 1.182 1.233 1.189 1.154 1.204 1.166 1.136 6 1.313 1.251 1.202 1.241 1.196 1.159 1.202 1.165 1.135 1.177 1.145 1.119 7 1.277 1.223 1.181 1.215 1.175 1.143 1.181 1.148 1.121 1.159 1.130 1.107 8 1.250 1.202 1.165 1.195 1.159 1.130 1.165 1.135 1.111 1.145 1.119 1.098 9 1.230 1.187 1.152 1.180 1.147 1.121 1.152 1.125 1.103 1.134 1.110 1.091 10 1.213 1.174 1.142 1.168 1.137 1.113 1.142 1.117 1.096 1.125 1.103 1.085 11 1.200 1.163 1.133 1.158 1.129 1.106 1.134 1.110 1.091 1.118 1.097 1.080 12 1.189 1.154 1.126 1.149 1.122 1.101 1.126 1.104 1.086 1.112 1.092 1.076 13 1.179 1.146 1.120 1.142 1.116 1.096 1.120 1.099 1.082 1.106 1.088 1.073 14 1.171 1.140 1.115 1.135 1.111 1.092 1.115 1.095 1.079 1.102 1.084 1.070 15 1.163 1.134 1.110 1.130 1.107 1.088 1.110 1.091 1.075 1.097 1.081 1.067

Maple program for critical value C0(S ).

> #input parameter values C, n, m, z, alpha.

#C0:= critical value; with (statevalf): C:=1.00; n:=10; m:=10; alpha:=0.01; z:=icdf[normal[0,1](alpha)]; epsilon:=((2/(n−1))ˆ(1/2))∗(GAMMA(n/2))/(GAMMA((n−1)/2)); C0:=evalf(C/(1+z∗((1−epsilonˆ2)/(m∗epsilonˆ2))ˆ(1/2))); The output is:

C:=1.00 n:=10 m:=10 alpha:=0.01 z:=−2.326 epsilon:=0.972659 C0:=1.213075

It can be seen that the critical value C0(S ) is proportional to the precision requirement C. Hence, we tabulate the critical values C0(S )with m= 10(5)25, subgroups of size n = 2(1)15,

and various risks α = 0.05, 0.025, and 0.01 for the preci-sion requirement Cp= 1.00 case, displayed in Table 6. For general Cp= C (common requirements as 1.33, 1.67, 2.00), we obtain the corresponding critical values by multiplying

C to the critical values C0(S ) with Cp= 1.00. For instance, if the required precision requirement C, is set to 1.33, with

m= 15 subgroups of size n = 10, and various risks α = 0.01,

0.025, and 0.05, then the corresponding critical values are

C0(S )= (1.168, 1.137, 1.113) × 1.33 = (1.553, 1.512, 1.480). It

is noted that for fixed α and sample size n, the critical value

C0(S ) increases as the number of samples m decreases, and

for fixed α and m the critical value C0(S ) increases as n

de-creases. Again, it can be explained that since the estimation error is potentially larger based on the smaller total sample size, we need a larger C0(S ) to claim that the process is capable.

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a specific value calculated from the sample data, can be ex-pressed as p-value= PˆCp(S )> ˆcp(S )|Cp= C  (21) = 1 − ˆcp(S ) 0 Cp √ 2πkx −2exp  −(Cp/x − 1)2 2k2  dx, where k=    1 − ε2n−1 2 n−1 , εn−1=  2 n− 1 Γ [n/2] Γ [(n − 1)/2].

An efficient MAPLE computer program is developed based on Eq. 21 , to calculate the p-value corresponding to the hypoth-esis test, H0: CpC versus H1: Cp> C. For given sample data

of m subgroups of size n, and a specific value of ˆCp calculated from the sample data, the program reads the input (an example) with m= 15, n = 8, C = 1.00, and ˆCp(S )= 1.204; the program gives the p-value= 0.00785.

Maple program for the p-value.

> #input parameter values Cphat, C, n, m; > # Cphat= ˆCp(S ); Cphat:=1.204; C:=1.00; m:=15; n:=8; epsilon:=((2/(n−1))ˆ(1/2))∗(GAMMA(n/2))/(GAMMA((n−1)/2)); b:=((1−epsilonˆ2)/(m∗epsilonˆ2))ˆ0.5; f(x):=C∗(exp(−((C/x−1)ˆ2)/(2∗bˆ2)))/((2∗Pi)ˆ0.5∗b∗(xˆ2)); p_value:=1−int(f(x),x=0..Cp);

The output is: p_value:=0.00785

A procedure for testing process precision. To judge if a given

process meets the preset precision requirement we first deter-mine the value of C, the preset precision requirement, and the

α-risk (the chance of wrongly concluding an incapable

pro-cess as capable). Checking the appropriate table (or running the program), we may obtain the critical value C0 based on given

values ofα-risk, C, and m samples of size n. If the estimated value ˆCp is greater than the critical value C0 ( ˆCp> C0), then

we may conclude that the process meets the precision require-ment Cp> C. Otherwise, we do not have sufficient informa-tion to conclude that the process meets the present precision requirement. In this case, we would believe that Cp≤ C. In the following, we develop a practical step-by-step procedure for testing process precision. The practitioners (engineers) can use the procedure in their in-plant applications to obtain reliable decisions.

Step 1: Decide the definition of “capable” (common requirement values of C include 1.00, 1.33, 1.50, 1.67, and 2.00), and theα-risk (normally set to 0.01, 0.025, or 0.05), the chance of wrongly concluding an incapable process as capable.

Step 2: Estimate the process precision Cpfrom the past “in con-trol” data by using either the sample range defined in Eq. 4 or the sample standard deviation method defined in Eq. 8.

Step 3: Check the appropriate table (or run the attached com-puter program) to find the critical value C0 based on the

specifiedα-risk, C, and m samples of size n.

Step 4: Conclude that the process is capable (Cp> C) if ˆCp value is greater than the critical value C0(i.e., ˆCp> C0).

Otherwise, we do not have enough information to con-clude that the process is capable.

4 An application to chip resistors

Consider a resistor manufacturing process making certain types of chip resistors. The chip resistor is developed applying the sur-face mount technology, which impels the electronic component to be made like a chip. Designed for surface mount applications, this style is generally mounted with the resistor element face up. Attachment may be made by use of conductive epoxy or sol-der. For a solder attachment, pre-tinned chips with nickel barriers are recommended. For an epoxy attachment, terminations are generally Pd/Ag or Pt/Ag alloys and Au terminations are avail-able upon request. The additional surface area provided by the wraparound style offers improved mechanical performance, as well as better thermal efficiency.

The chip resistor is made with a metal glaze layer screened on a high ceramic body. Its miniature size can be made compact on printed circuit board, and it has excellent mechanical strength and electrical stability. We investigate a specific chip resistor process taken from a factory located in Taiwan, with manufac-turing specifications of USL= 12.0 Ω, and LSL = 11.5 Ω. Sup-pose the minimal precision requirement for this process is set to

Cp= 1.33. The collected data of 15 subgroups of size 10 are dis-played in Table 7. Figure 9 shows the individual observation plot of each sample with respect to the two-sided specifications.

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Table 7. The collected 15 samples each of 10 observations Sample 1 11.709 11.809 11.693 11.806 11.740 11.806 11.736 11.789 11.661 11.793 Sample 2 11.715 11.650 11.719 11.774 11.797 11.766 11.867 11.757 11.733 11.774 Sample 3 11.873 11.714 11.745 11.765 11.743 11.788 11.753 11.772 11.708 11.749 Sample 4 11.713 11.743 11.746 11.773 11.819 11.741 11.719 11.724 11.777 11.739 Sample 5 11.730 11.695 11.752 11.767 11.809 11.748 11.760 11.743 11.698 11.756 Sample 6 11.672 11.720 11.724 11.700 11.765 11.739 11.792 11.759 11.727 11.637 Sample 7 11.780 11.684 11.825 11.750 11.756 11.685 11.695 11.760 11.777 11.729 Sample 8 11.741 11.784 11.682 11.721 11.800 11.759 11.691 11.678 11.724 11.736 Sample 9 11.800 11.757 11.794 11.812 11.736 11.683 11.776 11.774 11.817 11.749 Sample 10 11.673 11.819 11.680 11.774 11.740 11.685 11.749 11.748 11.685 11.702 Sample 11 11.710 11.684 11.744 11.655 11.725 11.667 11.734 11.838 11.686 11.743 Sample 12 11.741 11.709 11.807 11.829 11.678 11.847 11.737 11.751 11.868 11.755 Sample 13 11.765 11.732 11.694 11.743 11.805 11.655 11.774 11.687 11.811 11.692 Sample 14 11.726 11.719 11.842 11.681 11.751 11.697 11.675 11.723 11.794 11.654 Sample 15 11.752 11.767 11.727 11.782 11.768 11.829 11.611 11.833 11.800 11.716

As mentioned earlier, in order to make the estimation of these capability indices meaningful, it is necessary to check whether the manufacturing process is under statistical control. For these collected 15 samples with each of sample size n= 10, the rela-tive efficiency of the range method is about 85%. Therefore,

Fig. 10. a ¯X control chart of the process, b S control chart of the process

we suggested using the ( ¯X, S ) chart for retrospectively test-ing whether the process is in control, which are displayed in Fig. 10a,b. The( ¯X, S ) control charts show that all the sample points are within the control limits without any special pattern, and the process is justified to be well in control. Therefore, we consider the process stable and so we proceed with the capability measurement.

The overall sample mean =X= 11.7448 and the sample

standard deviation ¯S= 0.0490 are first calculated. Then, the

value of the estimated ˆσS= ¯S/c4= 0.0504 and ˆCp(S )= (USL −

LSL)/6× ˆσS



= 1.6534 are calculated from the 15 samples of each size 10. With risk α = 0.01, the minimal precision requirement for this process is set to Cp= 1.33. We check

Table 6 and find the corresponding critical value as 1.168 × 1.33 = 1.553. Therefore, in this case the estimated value ˆCp(S ), based on the S-chart sample data, is 1.6534, which is greater than the obtained critical value, 1.553. The corresponding p-value is also found to be 0.00075. We therefore conclude with 99% confidence that the chip resistors manufacturing process satisfies the requirement of Cp> 1.33, which is

con-sidered satisfactory and reliable in terms of product quality (originally set by the product designers or the manufacturing engineers).

5 Conclusions

Process precision index Cp has been widely used in the

manu-facturing industry for measuring process potential and product precision. Estimating and testing process precision based on one single sample has been investigated extensively. In this paper, we considered the problem of estimating and testing process preci-sion based on multiple samples taken from the( ¯X, R) or ( ¯X, S ) control charts. We investigated the statistical properties of the natural estimator of Cp(use either the sample standard deviation

or the sample range method), and implemented the statistical hy-pothesis testing. We also developed efficient MAPLE programs to calculate the lower confidence bounds, the critical values, and the p-values based on m samples of size n. Based on the test, we developed a practical procedure for the practitioners to use for

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their applications. Engineers can use the proposed testing pro-cedure to determine whether their manufacturing processes are capable of reproducing products satisfying the preset precision requirement.

References

1. Kane, VE (1986) Process capability indices. J Qual Technol 18:41– 52

2. Rado LE (1989) Enhance product development by using capability in-dexes. Qual Prog 22(4):38–41

3. Pearn WL, Lin GH, Chen KS (1998) Distributional and inferential properties of the process accuracy and process precision indices. Com-mun Stat: Theory & Methods 27(4): 985–1000

4. Montgomery DC (1985) Introduction to statistical quality Control. Wi-ley, New York

5. Kirmani SNUA, Kocherlakota K, Kocherlakota S (1991) Estimation of

σ and the process capability index based on subsamples. Commun Stat:

Theory & Methods 20:275–291

6. Pearn WL, Yang YS (2003) Distributional and inferential properties of the estimated precision index Cpbased on multiple samples. Qual

Quant 37:443–453

7. Pearson ES (1932) The percentage limits for the distribution of range in samples from a normal population. Biometrika 24: 404–417

8. Patnaik PB (1950) The use of mean range as an estimator of variance in statistical tests. Biometrika 37:78–87

9. Kocherlakota S (1992) Process capability index: recent developments. Sankhya Indian J Stat 54:352–369

10. Kotz S, Lovelace, CR (eds)(1998) Process capability indices in theory and practice. Arnold, London

11. Chou YM, Owen DB, Borrego SA (1990) Lower confidence limits on process capability indices. J Qual Technol 22:223–229

12. Li H, Owen DB, Borrego SA (1990) Lower confidence limits on pro-cess capability indices based on the range. Commun Stat: Simul and Comput 19(1):1–24

數據

Table 1. Coefficients of distribution for multiple samples with m = 5(5)25, n = 2(1)10, and α = 0.01, 0.025, 0.05
Table 2. The relative efficiency of the range method to S 2
Fig. 4. PDF plot of ˆ C p (S) with m = 25 and n = 5, for various values of C p
Table 4. Lower confidence factors C L (S) / ˆC p (S) for multiple samples with m = 10(5)25, n = 2(1)15, and α = 0.01, 0.025, 0.05
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