Qual. Reliab. Engng. Int. 15: 321–326 (1999)
MAKING DECISIONS IN ASSESSING PROCESS CAPABILITY
INDEX C
pk
W. L. PEARN1∗AND K. S. CHEN2
1Department of Industrial Engineering and Management, National Chiao Tung University, Hsinchu, Taiwan, Republic of China 2Department of Industrial Engineering and Management, National Chin-Yi Institute of Technology, Taichung, Taiwan, Republic
of China
SUMMARY
Process capability indices Cp, Cpkand Cpmhave been used in manufacturing industries to provide a quantitative measure of process potential and performance. The formulae for these indices are easy to understand and straightforward to apply. However, since sample data must be collected in order to calculate these indices, a great degree of uncertainty may be introduced into capability assessments owing to sampling errors. Currently, most practitioners simply look at the value of the index calculated from the sample data and then make a conclusion on whether the given process meets the capability (quality) requirement. This approach is not reliable, since sampling errors are ignored. Cheng (Qual. Engng., 7, 239–259 (1994)) has developed a procedure involving estimators of Cp and Cpmfor practitioners to use to determine whether a process meets the capability requirement or not. However, no procedure for Cpkwas given, because difficulties were encountered in calculating the sampling distribution of the estimator of Cpk. In this paper we use a newly proposed estimator of Cpk to develop a procedure for practitioners to use so that decisions made in assessing process capability are more reliable. Copyright1999 John Wiley & Sons, Ltd.
KEY WORDS: process capability indices; non-central t distribution; critical values; power of the test;α risk; capability requirement
INTRODUCTION
Understanding processes and quantifying process performance are essential for any successful quality improvement initiative. The relationship between the actual process performance and the specification limits or tolerance may be quantified using appropriate process capability indices. Three capability indices commonly used in manufacturing industries are Cp, Cpk and Cpm. These indices, providing numerical
measures of whether a production process meets predetermined specification limits, have been defined as Cp= USL− LSL 3σ Cpk= min USL− µ 3σ , µ − LSL 3σ Cpm= USL− LSL 3pσ2+ (µ − T )2
∗Correspondence to: W. L. Pearn, Department of Industrial Engineering and Management, National Chiao Tung University, 1001 Ta Hsueh Road, Hsinchu, Taiwan 30050, Republic of China.
where USL is the upper specification limit, LSL is the lower specification limit, µ is the process mean, σ is the process standard deviation (overall process variability) and T is the target value. The formulae for these indices are easy to understand and straightforward to apply. However, in order to calculate these indices, sample data must be collected. Therefore a great degree of uncertainty may be introduced into capability assessments owing to sampling errors. Currently, most practitioners simply look at the value of the estimators calculated from the sample data and then make a conclusion on whether the given process meets the capability (quality) requirement or not. This approach is highly unreliable, since sampling errors have been ignored. Chen [1] has developed a procedure (using estimators of Cp and Cpm) for practitioners to use to determine if a process
satisfies the targeted quality condition. However, no procedure for Cpkwas given, because difficulties were
encountered in calculating the sampling distribution of the estimator of Cpk. In this paper we use an estimator
of Cpk proposed by Pearn and Chen [2] to develop
a simple procedure for practitioners to use so that
CCC 0748–8017/99/040321–06$17.50 Received 1 March 1998
decisions made in assessing process capability are more reliable.
ESTIMATION OF Cpk
Three estimators have been proposed to estimate the
Cpk value, namely (a) Bissell’s estimator ˆCpk0 [3], (b)
the natural estimator ˆCpk [4] and (c) the
Bayesian-like estimator ˆCpk00 [2]. Bissell’s estimator assumes the knowledge of P(µ ≥ m) = 0 or 1, where
m = (USL + LSL)/2. If µ ≥ m, then ˆCpk0 =
(USL − ¯X)/3S; otherwise, ˆC0
pk = ( ¯X − LSL)/3S.
Kotz et al. [4] investigated a different estimator of Cpk
which is defined as ˆCpk= min{(USL − ¯X)/3S, ( ¯X −
LSL)/3S}, where ¯X = (Pni=1Xi)/n and S = {(n −
1)−1Pni=1(Xi− ¯X)2}1/2are conventional estimators
of µ and σ which may be obtained from a stable
process. Both estimators ˆCpk0 and ˆCpk are biased, but
Kotz et al. [4] showed that the variance of ˆCpk is
smaller than that of Bissell’s estimator.
Pearn and Chen [2] considered a Bayesian-like estimator ˆCpk00 to relax Bissell’s assumption on the process mean. The evaluation of the estimator ˆCpk00
only requires the knowledge of P(µ ≥ m) = p
or P(µ < m) = 1 − p, where 0 ≤ p ≤ 1,
which may be obtained from historical information on a stable process. Clearly, if P(µ ≥ m) = 0 or 1, then the estimator ˆCpk00 reduces to Bissell’s estimator. The estimator is defined as ˆCpk00 = {d −( ¯X −m)IA(µ)}/3S,
where IA(µ) = 1 if µ ∈ A, IA(µ) = −1 if µ 6∈ A,
and A= {µ|µ ≥ m}.
Pearn and Chen [2] showed that under the assumption of normality the distribution of the estimator 3n1/2ˆCpk00 is tn−1(δ), a non-central t with n−1 degrees of freedom and non-centrality parameter
δ = 3n1/2C
pk. The probability density function can be
expressed as f(x) = 3n1/2 2n/20 n− 1 2 (π(n − 1)]1/2 Z ∞ 0 y(n−2)/2 ×exp −y+ 9n[xy1/2(n − 1)−1/2− Cpk]2 2 ! dy
Pearn and Chen [2] also showed that by adding the well-known correction factor bf to the estimator
ˆC00
pk, where bf= [2/(n − 1)]
1/20[(n − 1)/2]{0[(n −
2)/2]}−1, an unbiased estimator ˜Cpk = bfˆC00pkcan be
obtained. They also showed that the variance of ˜Cpk
Table 1. Quality conditions and Cpkvalues
Quality condition Cpkvalue Inadequate Cpk< 1.00 Capable 1.00 ≤ Cpk< 1.33 Satisfactory 1.33 ≤ Cpk< 1.50 Excellent 1.50 ≤ Cpk< 2.00
Super 2.00 ≤ Cpk
is smaller than those of ˆCpk0 and ˆCpk. Therefore in
this paper we will use the unbiased estimator ˜Cpk to
develop a simple procedure, similar to those described in References [1] and [5], for the index Cpk.
TEST HYPOTHESIS
A process is called ‘inadequate’ if Cpk < 1.00:
this indicates that the process is not adequate with respect to the production tolerances; either the process variation (σ2) needs to be reduced or the process mean (µ) needs to be shifted closer to the target value. A process is called ‘capable’ if 1.00 ≤
Cpk < 1.33: this indicates that caution needs to be
taken regarding the process distribution; some process control is required. A process is called ‘satisfactory’ if 1.33 ≤ Cpk < 1.50: this indicates that the process
quality is satisfactory; material substitution may be allowed and no stringent quality control is required. A process is called ‘excellent’ if 1.50 ≤ Cpk < 2.00.
Finally, a process is called ‘super’ if Cpk ≥ 2.00.
Table1summarizes the five quality conditions and the corresponding Cpkvalues.
To determine whether a given process meets the capability requirement and runs under the desired quality condition, we can consider the following statistical test hypothesis. The process meets the capability (quality) requirement if Cpk > C, and fails
to meet the capability requirement if Cpk≤ C: H0: Cpk≤ C
H1: Cpk> C
The critical value C0is determined by
p{ ˜Cpk> C0|Cpk= C} = α p{bfˆCpk00 > C0|Cpk= C} = α p ˆC00 pk> C0 bf Cpk= C = α p ( 3n1/2ˆCpk00 > 3n 1/2C 0 bf Cpk= C ) = α
p ( tn−1(δc) > 3n1/2C0 bf ) = α
whereδc= 3n1/2C. Hence we have
3n1/2C0 bf
= tn−1,α(δc)
where tn−1,α(δc) is the upper α is the upper α quantile
of the tn−1(δc) distribution, or C0=
bf
3n1/2tn−1,α(δc)
The power of the test can be computed as
π(Cpk) = p{ ˜Cpk> C0|Cpk} = p{bfˆCpk00 > C0|Cpk} = p ˆC00 pk> C0 bf Cpk = p ( 3n1/2ˆCpk00 > 3n 1/2C 0 bf Cpk ) = p ( tn−1(δ) > 3n1/2C0 bf ) whereδ = 3n1/2Cpk. MAKING DECISIONS
Tables2(a)–2(d) display critical values C0 for C =
1.00, 1.33, 1.50 and 2.00 respectively with sample
sizes n = 10(5)250 and α risk = 0.01, 0.025 and 0.05. The computer program (using SAS) generating the tables is available from the authors. To determine if the process meets the capability (quality) requirement, we first determine C and theα risk. Then we calculate the value of the estimator ˜Cpkfrom the sample. From
the appropriate table we find the critical value C0
based onα risk, C and sample size n. If the estimated value ˜Cpk is greater than the critical value C0, then
we conclude that the process meets the capability (quality) requirement. Otherwise, we do not have sufficient information to conclude that the process meets the present capability requirement.
The procedure
1. Determine the value of C (normally chosen from Table 1), the desired quality condition, and the α risk (normally set to 0.01, 0.025 or 0.05), the chance of incorrectly accepting an incapable process (which does not meet the quality requirement) as a capable process (which meets the quality requirement).
Table 2(a). Critical values C0 for C = 1.00, n = 10(5)250 and
α = 0.01, 0.025, 0.05 n α = 0.01 α = 0.025 α = 0.05 10 1.957 1.175 1.541 15 1.686 1.529 1.411 20 1.556 1.436 1.343 25 1.477 1.377 1.299 30 1.422 1.337 1.269 35 1.383 1.306 1.246 40 1.352 1.283 1.228 45 1.327 1.264 1.213 50 1.307 1.248 1.201 55 1.290 1.235 1.190 60 1.275 1.223 1.181 65 1.262 1.213 1.174 70 1.251 1.205 1.167 75 1.241 1.197 1.161 80 1.233 1.190 1.155 85 1.225 1.184 1.150 90 1.217 1.178 1.145 95 1.211 1.173 1.141 100 1.205 1.168 1.137 105 1.199 1.163 1.134 110 1.194 1.159 1.131 115 1.189 1.155 1.127 120 1.185 1.152 1.125 125 1.181 1.148 1.122 130 1.177 1.145 1.119 135 1.173 1.142 1.117 140 1.170 1.140 1.115 145 1.166 1.137 1.113 150 1.163 1.135 1.111 155 1.160 1.132 1.109 160 1.157 1.130 1.107 165 1.155 1.128 1.105 170 1.152 1.126 1.104 175 1.150 1.124 1.102 180 1.148 1.122 1.100 185 1.145 1.120 1.099 190 1.143 1.118 1.098 195 1.141 1.117 1.096 200 1.139 1.115 1.095 205 1.138 1.114 1.094 210 1.136 1.112 1.093 215 1.134 1.111 1.092 220 1.132 1.110 1.090 225 1.131 1.108 1.098 230 1.129 1.107 1.088 235 1.128 1.106 1.087 240 1.126 1.105 1.086 245 1.125 1.103 1.086 250 1.124 1.102 1.085
Table 2(b). Critical values C0for C = 1.33, n = 10(5)250 and α = 0.01, 0.025, 0.05 n α = 0.01 α = 0.025 α = 0.05 10 2.569 2.255 2.208 15 2.216 2.012 1.859 20 2.046 1.891 1.771 25 1.943 1.815 1.714 30 1.873 1.762 1.675 35 1.822 1.723 1.645 40 1.782 1.693 1.622 45 1.750 1.668 1.603 50 1.724 1.648 1.587 55 1.702 1.631 1.574 60 1.683 1.616 1.562 65 1.666 1.604 1.552 70 1.652 1.592 1.543 75 1.639 1.582 1.535 80 1.628 1.573 1.528 85 1.618 1.565 1.522 90 1.608 1.558 1.516 95 1.600 1.551 1.511 100 1.592 1.545 1.506 105 1.585 1.539 1.501 110 1.578 1.534 1.497 115 1.572 1.529 1.493 120 1.567 1.524 1.489 125 1.561 1.520 1.486 130 1.556 1.516 1.483 135 1.551 1.512 1.480 140 1.547 1.509 1.477 145 1.543 1.505 1.474 150 1.539 1.502 1.471 155 1.535 1.499 1.469 160 1.532 1.496 1.467 165 1.528 1.493 1.465 170 1.525 1.491 1.462 175 1.522 1.488 1.460 180 1.519 1.486 1.458 185 1.516 1.484 1.457 190 1.513 1.481 1.455 195 1.511 1.479 1.453 200 1.508 1.477 1.452 205 1.506 1.475 1.450 210 1.504 1.474 1.448 215 1.501 1.472 1.447 220 1.499 1.470 1.446 225 1.497 1.468 1.444 230 1.495 1.467 1.443 235 1.493 1.465 1.442 240 1.492 1.464 1.440 245 1.490 1.462 1.439 250 1.488 1.461 1.438
Table 2(c). Critical values C0 for C = 1.50, n = 10(5)250 and
α = 0.01, 0.025, 0.05 n α = 0.01 α = 0.025 α = 0.05 10 2.887 2.535 2.281 15 2.490 2.263 2.091 20 2.300 2.126 1.992 25 2.185 2.041 1.929 30 2.106 1.983 1.885 35 2.049 1.939 1.852 40 2.004 1.905 1.826 45 1.969 1.878 1.804 50 1.939 1.855 1.787 55 1.915 1.836 1.772 60 1.894 1.819 1.759 65 1.875 1.805 1.748 70 1.859 1.793 1.738 75 1.845 1.781 1.729 80 1.832 1.771 1.721 85 1.821 1.762 1.714 90 1.811 1.754 1.707 95 1.801 1.746 1.701 100 1.792 1.740 1.696 105 1.784 1.733 1.691 110 1.777 1.727 1.686 115 1.770 1.722 1.682 120 1.764 1.717 1.678 125 1.758 1.712 1.674 130 1.752 1.707 1.670 135 1.747 1.703 1.667 140 1.742 1.699 1.664 145 1.737 1.695 1.661 150 1.733 1.692 1.658 155 1.729 1.688 1.655 160 1.725 1.685 1.652 165 1.721 1.682 1.650 170 1.717 1.679 1.648 175 1.714 1.677 1.645 180 1.711 1.674 1.643 185 1.708 1.671 1.641 190 1.705 1.669 1.639 195 1.702 1.667 1.637 200 1.699 1.664 1.635 205 1.696 1.662 1.634 210 1.694 1.660 1.632 215 1.691 1.658 1.630 220 1.689 1.656 1.629 225 1.687 1.654 1.627 230 1.684 1.653 1.626 235 1.682 1.651 1.625 240 1.680 1.649 1.623 245 1.678 1.647 1.622 250 1.676 1.646 1.621
Table 2(d). Critical values C0for C = 2.00, n = 10(5)250 and α = 0.01, 0.025, 0.05 n α = 0.01 α = 0.025 α = 0.05 10 3.826 3.361 3.026 15 3.302 3.002 2.776 20 3.050 2.821 2.645 25 2.899 2.710 2.562 30 2.795 2.633 2.504 35 2.270 2.575 2.461 40 2.661 2.531 2.426 45 2.614 2.495 2.399 50 2.576 2.465 2.376 55 2.543 2.440 2.356 60 2.516 2.418 2.339 65 2.492 2.399 2.324 70 2.471 2.383 2.311 75 2.452 2.368 2.300 80 2.435 2.355 2.289 85 2.420 2.343 2.289 90 2.407 2.332 2.271 95 2.394 2.323 2.264 100 2.383 2.313 2.256 105 2.372 2.305 2.250 110 2.363 2.297 2.243 115 2.354 2.290 2.238 120 2.345 2.283 2.232 125 2.337 2.277 2.227 130 2.330 2.271 2.223 135 2.323 2.266 2.218 140 2.317 2.261 2.214 145 2.311 2.256 2.210 150 2.305 2.251 2.206 155 2.299 2.247 2.203 160 2.294 2.242 2.199 165 2.289 2.238 2.196 170 2.284 2.235 2.193 175 2.280 2.231 2.190 180 2.276 2.227 2.187 185 2.272 2.224 2.185 190 2.268 2.221 2.182 195 2.264 2.218 2.180 200 2.260 2.215 2.177 205 2.257 2.212 2.175 210 2.253 2.209 2.173 215 2.250 2.207 2.171 220 2.247 2.204 2.169 225 2.244 2.202 2.167 230 2.241 2.199 2.165 235 2.238 2.197 2.163 240 2.236 2.195 2.161 245 2.233 2.193 2.159 250 2.230 2.191 2.158
Figure 1. OC curves for C= 1.00, α = 0.01 and n = 10(40)250 (top to bottom in plot)
2. Calculate the value of the estimator ˜Cpkfrom the
sample.
3. Check Tables2(a)–2(d)to find the corresponding
C0based onα, C and sample size n
4. Conclude that the process meets the capability requirement if ˜Cpkis greater than C0. Otherwise,
we do not have enough information to conclude that the process meets the capability requirement.
To accelerate the calculations of the estimator ˜Cpk,
we have provided values of the correction factor bf
for various sample sizes n = 10(5)250 (see Table3). Figure1plots the OC curves(β = 1 − π(Cpk) versus Cpk value) for the quality conditions with C set to
1.00,α risk = 0.01 and sample sizes n = 10(40)250.
AN EXAMPLE
Consider the following example taken from bopro, a manufacturer and supplier in Taiwan exporting high-end audio speaker components including rubber edge, Pulux edge, Kevlar cone, honeycomb and many others. The production specifications for a particular model of Pulux edge are the following: USL= 5.95, LSL = 5.65, T = 5.80. The quality requirement was defined as ‘Satisfactory’ (Cpk > 1.33). A total
of 90 observations were collected which are displayed in Table4.
To determine whether the process is ‘Satisfactory’, we first calculate d = (USL − LSL)/2 = 0.15, m =
(USL + LSL)/2 = 5.80, sample mean ¯X = 5.830
and sample standard deviation S= 0.023. To calculate the value of the estimator ˜Cpk, we need to determine
the value of IA(µ), which requires the knowledge of P(µ ≥ m) or P(µ < m). The historical information
Table 3. Values of bffor various sample sizes n n bf n bf n bf n bf n bf n bf n bf 10 0.914 45 0.983 80 0.990 115 0.993 150 0.995 185 0.996 220 0.997 15 0.945 50 0.985 85 0.991 120 0.994 155 0.995 190 0.996 225 0.997 20 0.960 55 0.986 90 0.992 125 0.994 160 0.995 195 0.996 230 0.997 25 0.968 60 0.987 95 0.992 130 0.994 165 0.995 200 0.996 235 0.997 30 0.974 65 0.988 100 0.992 135 0.994 170 0.996 205 0.996 240 0.997 35 0.978 70 0.989 105 0.993 140 0.995 175 0.996 210 0.996 245 0.997 40 0.981 75 0.990 110 0.993 145 0.995 180 0.996 215 0.996 250 0.997
Table 4. Collected sample data (90 observations)
5.88 5.83 5.84 5.80 5.89 5.81 5.84 5.83 5.82 5.83 5.81 5.82 5.85 5.81 5.81 5.81 5.84 5.82 5.80 5.84 5.86 5.87 5.82 5.87 5.80 5.81 5.85 5.84 5.83 5.86 5.81 5.81 5.82 5.83 5.85 5.80 5.86 5.82 5.86 5.83 5.80 5.77 5.82 5.85 5.84 5.82 5.85 5.81 5.86 5.79 5.84 5.83 5.80 5.83 5.81 5.83 5.81 5.85 5.83 5.88 5.82 5.87 5.80 5.82 5.83 5.81 5.84 5.79 5.85 5.85 5.84 5.84 5.80 5.82 5.84 5.85 5.86 5.81 5.81 5.85 5.86 5.81 5.81 5.83 5.85 5.85 5.82 5.83 5.86 5.81
we can determine the value of IA(µ) = 1 or −1 using
available random number tables.
Suppose the generated two-digit random number is 65, then we have IA(µ) = 1. Checking the value of bffrom Table3, we obtain bf = 0.992. Thus ˜Cpk = bfˆCpk00 = bf(d − ¯X + m)/3S = 1.890. Assume the α
risk is 0.05. We find the critical value C0= 1.516 from
Table2(b)based on C = 1.33, α = 0.05 and sample size n= 90. Since ˜Cpkis greater than the critical value C0, we conclude that the process is ‘Satisfactory’.
REFERENCES
1. S. W. Cheng, ‘Practical implementation of the process capability indices’, Qual. Engng., 7, 239–259 (1994) 2. W. L. Pearn and K. S. Chen, ‘A Bayesian-like estimator of
Cpk’, Commun. Statist.—Simul. Comput., 25, 321–329 (1996) 3. A. F. Bissell, ‘How reliable is your capability index?’, Appl.
Statist., 39, 331–340 (1990)
4. S. Kotz, W. L. Pearn and N. L. Johnson, ‘Some process capability indices are more relaible than one might think’, Appl. Statist., 42, 55–62 (1993)
5. S. W. Cheng, ‘Is the process capable? Tables and graphs in assessing Cpm’, Qual. Engng., 4, 563–576 (1992)
Authors’ biographies:
W. L. Pearn is a Professor of Operations Research
and Quality Management in the Department of Industrial Engineering and Management, National Chiao Tung University, Hsinchu, Taiwan, ROC. He received his MS degree in statistics and PhD degree in operations research from the University of Maryland, College Park, MD, USA. He started his career with California State University, Fresno, CA, USA as an Assistant Professor. He also worked for AT&T Bell Laboratories at Switch Network Control and Process Quality Centers. Currently, Professor Pearn is also a consultant to several companies in Taiwan manufacturing audio speaker components.
K. S. Chen received his MS degree in statistics from
National Cheng Kung University and PhD degree in quality management from National Chiao Tung University, Taiwan, ROC. Currently, he is an Associate Professor in the Department of Industrial Engineering and Management, National Chin-Yi Institute of Technology, Taichung, Taiwan, ROC.
