274 C hap te rei g h t

Transcription

1 274 C hap te rei g h t Sampling Distributions n most Six Sigma projects involving enumerative statistics, we deal with samples, not populations. We now consider the estimation of certain characteristics or parameters of the distribution from the data. The empirical distribution assigns the probability lin to each Xi in the sample, thus the mean of this distribution is (8.54) The symbol X is called "Xbar." Since the empirical distribution is determined by a sample, X is simply called the sample mean. The sample variance is given by (8.55) This equation for S2 is commonly referred to as the unbiased sample variance. The sample standard deviation is given by (8.56) Another sampling statistic of special interest in Six Sigma is the standard deviation of the sample average, also referred to as the standard error of the mean or simply the standard error. This statistic is given by (8.57) As can be seen, the standard error of the mean is inversely proportional to the square root of the sample size. That is, the larger the sample size, the smaller the standard deviation of the sample average. This relationship is shown in Fig t can be seen that averages of n = 4 have a distribution half as variable as the population from which the samples are drawn. Binomial Distribution Assume that a process is producing some proportion of nonconforming units, which we will call p. f we are basing p on a sample we find p by dividing the number of nonconforming units in the sample by the number of items sampled. The equation that will tell us the probability of getting x defectives in a sample of n units is shown by Eq. (8.58). (8.58) This equation is known as the binomial probability distribution. n addition to being useful as the exact distribution of nonconforming units for processes in continuous production, it is also an excellent approximation to the cumbersome hypergeometric probability distribution when the sample size is less than 10% of the lot size.

2 Pro c e s s B e hay i 0 r C h art s 275 Distribution of Population distribution ---- FGURE 8.34 Effect of sample size on the standard error. Example of Applying the Binomial Probability Distribution A process is producing glass bottles on a continuous basis. Past history shows that 1 % of the bottles have one or more flaws. f we draw a sample of 10 units from the process, what is the probability that there will be 0 nonconforming bottles? Using the above information, n = 10, P =.01, and x = O. Substituting these values into Eq. (8.58) gives us p(o) = qoo.01 (1-0.01)10-0 = 1 x 1 X = = 90.4% Another way of interpreting the above example is that a sampling plan "inspect 10 units, accept the process if no nonconformances are found" has a 90.4% probability of accepting a process that is averaging 1 % nonconforming units. Example of Binomial Probability Calculations Using Microsoft Excel" Microsoft Excel has a built-in capability to analyze binomial probabilities. To solve the above problem using Excel, enter the sample size, p value, and x value as shown in Fig Note the formula result near the bottom of the screen. Poisson Distribution Another situation encountered often in quality control is that we are not just concerned with units that don't conform to requirements, instead we are concerned with the number of nonconformances themselves. For example, let's say we are trying to control the quality of a computer. A complete audit of the finished computer would almost certainly reveal some nonconformances, even though these nonconformances might be of minor importance (for example, a decal on the back panel might not be perfectly straight). f we tried to use the hypergeometric or binomial probability distributions to evaluate sampling plans for this situation, we would find they didn't work because our

3 276 C hap te rei g h t 8NOMDST "'J x.j ;; =BNOMDST(B4.B1,B2,FALSE) A B c 1 D 1 E 1 F G 1 n 10 t--- r-l P x 0.- r--i- 5 P(x) i--- ~ -BNOMDST FALSE] ~ Number..:.15 " j] Tnials 1B1 j] = Pnj bability..:.s 1B2 j] = 0.Q1 -- r-11- Cumula,tive FALSE ~ = FALSE,J '" 0.91J Returns the individjal term binomial distribution probability, t ~ Nlfllbe:r --.:s is the rumber off successes in trials. Formula result "'0, FGURE 8.35 Example of finding binomial probability using Microsoft Excel. OK Cal'l(le'l lot or process would be composed of 100% nonconforming units. Obviously, we are interested not in the units per se, but in the non-conformances themselves. n other cases, it isn't even possible to count sample units per se. For example, the number of accidents must be counted as occurrences. The correct probability distribution for evaluating counts of non-conformances is the Poisson distribution. The pdf is given in Eq. (8.59). p(x) = 11 x:;~ (8.59) n Eq. (8.59), 11 is the average number of nonconformances per unit, x is the number of nonconformances in the sample, and e is the constant approximately equal to P(x) gives the probability of exactly x occurrences in the sample. Example of Applying the Poisson Distribution A production line is producing guided missiles. When each missile is completed, an audit is conducted by an Air Force representative and every nonconformance to requirements is noted. Even though any major nonconformance is cause for rejection, the prime contractor wants to control minor nonconformances as well. Such minor problems as blurred stencils, small burrs, etc., are recorded during the audit. Past history shows that on the average each missile has 3 minor nonconformances. What is the probability that the next missile will have 0 nonconformances? We have 11 = 3, x = O. Substituting these values into Eq. (8.59) gives us p(0)=3 0 e- 3 =lxo =5% O! 1 n other words, 100% - 5% = 95% of the missiles will have at least one nonconformance.

4 Process Behavior Charts 277 The Poisson distribution, in addition to being the exact distribution for the number of non-conformances, is also a good approximation to the binomial distribution in certain cases. To use the Poisson approximation, you simply let /.l = np in Eq. (8.59). Juran (1988) recommends considering the Poisson approximation if the sample size is at least 16, the population size is at least 10 times the sample size, and the probability of occurrence p on each trial is less than 0.1. The major advantage of this approach is that it allows you to use the tables of the Poisson distribution, such as in Appendix 7. Also, the approach is useful for designing sampling plans. Example of Poisson Probability Calculations Using Microsoft Excel Microsoft Excel has a built-in capability to analyze Poisson probabilities. To solve the above problem using Excel, enter the average and x values as shown in Fig Note the formula result near the bottom of the screen. Hypergeometric Distribution Assume we have received a lot of 12 parts from a distributor. We need the parts badly and are willing to accept the lot if it has fewer than 3 nonconforming parts. We decide to inspect only 4 parts since we can't spare the time to check every part. Checking the sample, we find 1 part that doesn't conform to the requirements. Should we reject the remainder of the lot? This situation involves sampling without replacement. We draw a unit from the lot, inspect it, and draw another unit from the lot. Furthermore, the lot is quite small, the sample is 25% of the entire lot. The formula needed to compute probabilities for this POSSON A ~ m ea n 21x 3 + ip(x) 6l POSSON... J X,,~ =POSSON( B2.B1,O) B led E 3 o t2.b1.0) 1 7 ~= o 8 Mea nfb-l ~jg~ - : = cumulative 10 jj = FALSE 11 = Reams the POisson distr ibution, 13 l - 14 Cumulative s a logica l value: for the curnulative Poisson probability, use TRUE; for the Poisson probabllit:,t mass function, use F.il.LSE E [1J1 Formul1a result =0, OK Cancel.J6 1! FGURE 8.36 Example of finding Poisson probability using Microsoft Excel. F G

5 278 Chapter Eight procedure is known as the hypergeometric probability distribution, and it is shown in Eq. (8.60). (8.60) n Eq. (8.60), N is the lot size, m is the number of defectives in the lot, n is the sample size, x is the number of defectives in the sample, and P(x) is the probability of getting exactly x defectives in the sample. Note that the numerator term c ::-~m gives the number of combinations of non-defectives while C;Z is the number of combinations of defectives. Thus the numerator gives the total number of arrangements of samples from lots of size N with m defectives where the sample n contains exactly x defectives. The term C~ the denominator is the total number of combinations of samples of size n from lots of size N, regardless of the number of defectives. Thus, the probability is a ratio of the likelihood of getting the result under the assumed conditions. For our example, we must solve the above equation for x = 0 as well as x =, since we would also accept the lot if we had no defectives. The solution is shown as follows. C 12-3 C x 1 P(O)= =--=0255 q C12-3C 3 84 x P(l)= =--=-=0509 q P(l or less) = P(O) + P(l) Adding the two probabilities tells uoooos the probability that our sampling plan will accept lots of 12 with 3 nonconforming units. The plan of inspecting 4 parts and accepting the lot if we have 0 or 1 nonconforming has a probability of = 0.764, or 76.4%, of accepting this "bad" quality lot. This is the "consumer's risk" for this sampling plan. Such a high sampling risk would be unacceptable to most people. Example of Hypergeometric Probability Calculations Using Microsoft Excel Microsoft Excel has a built-in capability to analyze hypergeometric probabilities. To solve the above problem using Excel, enter the population and sample values as shown in Fig Note the formula result near the bottom of the screen (0.509) gives the probability for x = 1. To find the cumulative probability you need to sum the probabilities for x = 0 and x = 1 etc. Normal Distribution The most common continuous distribution encountered in Six Sigma work is, by far, the normal distribution. Sometimes the process itself produces an approximately normal distribution, other times a normal distribution can be obtained by performing a mathematical transformation on the data or by using averages. The probability density function for the normal distribution is given by Eq. (8.61). (8.61)