92 Variables and process variation and when n = 4, SE = ␴/2, i.e. half the spread of the parent distribution of individual items. SE has the same characteristics as any standard deviation, and normal tables may be used to evaluate probabilities related to the distribution of sample averages. We call it by a different name to avoid confusion with the population standard deviation. The smaller spread of the distribution of sample averages provides the basis for a useful means of detecting changes in processes. Any change in the process mean, unless it is extremely large, will be difficult to detect from individual results alone. The reason can be seen in Figure 5.5a, which shows the parent distributions for two periods in a paint filling process between which the average has risen from 1000 ml to 1012 ml. The shaded portion is common to both process distributions and, if a volume estimate occurs in the shaded portion, say at 1010 ml, it could suggest either a volume above the average from the distribution centred at 1000 ml, or one slightly below the average from the distribution centred at 1012 ml. A large Figure 5.5 Effect of a shift in average fill level on individuals and sample means. Spread of sample means is much less than spread of individuals Variables and process variation 93 number of individual readings would, therefore, be necessary before such a change was confirmed. The distribution of sample means reveals the change much more quickly, the overlap of the distributions for such a change being much smaller (Figure 5.5b). A sample mean of 1010 ml would almost certainly not come from the distribution centred at 1000 ml. Therefore, on a chart for sample means, plotted against time, the change in level would be revealed almost immediately. For this reason sample means rather than individual values are used, where possible and appropriate, to control the centring of processes. The Central Limit Theorem What happens when the measurements of the individual items are not distributed normally? A very important piece of theory in statistical process control is the central limit theorem. This states that if we draw samples of size n, from a population with a mean µ and a standard deviation ␴, then as n increases in size, the distribution of sample means approaches a normal distribution with a mean µ and a standard error of the means of ␴/ ͱ ස n. This tells us that, even if the individual values are not normally distributed, the distribution of the means will tend to have a normal distribution, and the larger the sample size the greater will be this tendency. It also tells us that the Grand or Process Mean X will be a very good estimate of the true mean of the population µ. Even if n is as small as 4 and the population is not normally distributed, the distribution of sample means will be very close to normal. This may be illustrated by sketching the distributions of averages of 1000 samples of size four taken from each of two boxes of strips of paper, one box containing a rectangular distribution of lengths, and the other a triangular distribution (Figure 5.6). The mathematical proof of the Central Limit Theorem is beyond the scope of this book. The reader may perform the appropriate experimental work if (s)he requires further evidence. The main point is that, when samples of size n = 4 or more are taken from a process which is stable, we can assume that the distribution of the sample means X will be very nearly normal, even if the parent population is not normally distributed. This provides a sound basis for the Mean Control Chart which, as mentioned in Chapter 4, has decision ‘zones’ based on predetermined control limits. The setting of these will be explained in the next chapter. The Range Chart is very similar to the mean chart, the range of each sample being plotted over time and compared to predetermined limits. The development of a more serious fault than incorrect or changed centring can lead to the situation illustrated in Figure 5.7, where the process collapses from form A to form B, perhaps due to a change in the variation of material. The ranges of the samples from B will have higher values than 94 Variables and process variation Figure 5.6 The distribution of sample means from rectangular and triangular universes Figure 5.7 Increase in spread of a process Variables and process variation 95 ranges in samples taken from A. A range chart should be plotted in conjunction with the mean chart. Rational subgrouping of data We have seen that a subgroup or sample is a small set of observations on a process parameter or its output, taken together in time. The two major problems with regard to choosing a subgroup relate to its size and the frequency of sampling. The smaller the subgroup, the less opportunity there is for variation within it, but the larger the sample size the narrower the distribution of the means, and the more sensitive they become to detecting change. A rational subgroup is a sample of items or measurements selected in a way that minimizes variation among the items or results in the sample, and maximizes the opportunity for detecting variation between the samples. With a rational subgroup, assignable or special causes of variation are not likely to be present, but all of the effects of the random or common causes are likely to be shown. Generally, subgroups should be selected to keep the chance for differences within the group to a minimum, and yet maximize the chance for the subgroups to differ from one another. The most common basis for subgrouping is the order of output or production. When control charts are to be used, great care must be taken in the selection of the subgroups, their frequency and size. It would not make sense, for example, to take as a subgroup the chronologically ordered output from an arbitrarily selected period of time, especially if this overlapped two or more shifts, or a change over from one grade of product to another, or four different machines. A difference in shifts, grades or machines may be an assignable cause that may not be detected by the variation between samples, if irrational subgrouping has been used. An important consideration in the selection of subgroups is the type of process – one-off, short run, batch or continuous flow, and the type of data available. This will be considered further in Chapter 7, but at this stage it is clear that, in any type of process control charting system, nothing is more important than the careful selection of subgroups. Chapter highlights ᭹ There are three main measures of the central value of a distribution (accuracy). These are the mean µ (the average value), the median (the middle value), the mode (the most common value). For symmetrical distributions the values for mean, median and mode are identical. For asymmetric or skewed distributions, the approximate relationship is mean – mode = 3 (mean–median). 96 Variables and process variation ᭹ There are two main measures of the spread of a distribution of values (precision). These are the range (the highest minus the lowest) and the standard deviation ␴. The range is limited in use but it is easy to understand. The standard deviation gives a more accurate measure of spread, but is less well understood. ᭹ Continuous variables usually form a normal or symmetrical distribution. The normal distribution is explained by using the scale of the standard deviation around the mean. Using the normal distribution, the proportion falling in the ‘tail’ may be used to assess process capability or the amount out-of-specification, or to set targets. ᭹ A failure to understand and manage variation often leads to unjustified changes to the centring of processes, which results in an unnecessary increase in the amount of variation. ᭹ Variation of the mean values of samples will show less scatter than individual results. The Central Limit Theorem gives the relationship between standard deviation (␴), sample size (n), and Standard Error of Means (SE) as SE = ␴/ͱn. ᭹ The grouping of data results in an increased sensitivity to the detection of change, which is the basis of the mean chart. ᭹ The range chart may be used to check and control variation. ᭹ The choice of sample size is vital to the control chart system and depends on the process under consideration. References Besterfield, D. (2000) Quality Control, 6th Edn, Prentice Hall, Englewood Cliffs NJ, USA. Pyzdek, T. (1990) Pyzdek’s Guide to SPC, Vol. One: Fundamentals, ASQC Quality Press, Milwaukee WI, USA. Shewhart, W.A. (1931 – 50th Anniversary Commemorative Reissue 1980) Economic Control of Quality of Manufactured Product, D. Van Nostrand, New York, USA. Wheeler, D.J. and Chambers, D.S. (1992) Understanding Statistical Process Control, 2nd Edn, SPC Press, Knoxville TN, USA. Discussion questions 1 Calculate the mean and standard deviation of the melt flow rate data below (g/10 min): 3.2 3.3 3.2 3.3 3.2 3.5 3.0 3.4 3.3 3.7 3.0 3.4 3.5 3.4 3.3 3.2 3.1 3.0 3.4 3.1 3.3 3.5 3.4 3.3 3.2 Variables and process variation 97 3.2 3.1 3.5 3.2 3.3 3.2 3.6 3.4 2.7 3.5 3.0 3.3 3.3 2.4 3.1 3.6 3.6 3.5 3.4 3.1 3.2 3.3 3.1 3.4 2.9 3.6 3.6 3.5 If the specification is 3.0 to 3.8g/10 min, comment on the capability of the process. 2 Describe the characteristics of the normal distribution and construct an example to show how these may be used in answering questions which arise from discussions of specification limits for a product. 3 A bottle filling machine is being used to fill 150 ml bottles of a shampoo. The actual bottles will hold 156 ml. The machine has been set to discharge an average of 152 ml. It is known that the actual amounts discharged follow a normal distribution with a standard deviation of 2 ml. (a) What proportion of the bottles overflow? (b) The overflow of bottles causes considerable problems and it has therefore been suggested that the average discharge should be reduced to 151 ml. In order to meet the weights and measures regulations, however, not more than 1 in 40 bottles, on average, must contain less than 146 ml. Will the weights and measures regulations be contravened by the proposed changes? You will need to consult Appendix A to answer these questions. 4 State the Central Limit Theorem and explain how it is used in statistical process control. 5 To: International Chemicals Supplier From: Senior Buyer, Perpexed Plastics Ltd Subject: MFR Values of Polyglyptalene As promised, I have now completed the examination of our delivery records and have verified that the values we discussed were not in fact in chronological order. They were simply recorded from a bundle of Certificates of Analysis held in our Quality Records File. I have checked, however, that the bundle did represent all the daily deliveries made by ICS since you started to supply in October last year. Using your own lot identification system I have put them into sequence as manufactured: 98 Variables and process variation 1) 4.1 2) 4.0 3) 4.2 4) 4.2 5) 4.4 6) 4.2 7) 4.3 8) 4.2 9) 4.2 10) 4.1 11) 4.3 12) 4.1 13) 3.2 14) 3.5 15) 3.0 16) 3.2 17) 3.3 18) 3.2 19) 3.3 20) 2.7 21) 3.3 22) 3.6 23) 3.2 24) 2.9 25) 3.3 26) 3.0 27) 3.4 28) 3.1 29) 3.5 30) 3.1 31) 3.2 32) 3.5 33) 2.4 34) 3.5 35) 3.3 36) 3.6 37) 3.2 38) 3.4 39) 3.5 40) 3.0 41) 3.4 42) 3.5 43) 3.6 44) 3.0 45) 3.1 46) 3.4 47) 3.1 48) 3.6 49) 3.3 50) 3.3 51) 3.4 52) 3.4 53) 3.3 54) 3.2 55) 3.4 56) 3.3 57) 3.6 58) 3.1 59) 3.4 60) 3.5 61) 3.2 62) 3.7 63) 3.3 64) 3.1 I hope you can make use of this information. Analyse the above data and report on the meaning of this information. Worked examples using the normal distribution 1 Estimating proportion defective produced In manufacturing it is frequently necessary to estimate the proportion of product produced outside the tolerance limits, when a process is not capable of meeting the requirements. The method to be used is illustrated in the following example: 100 units were taken from a margarine packaging unit which was ‘in statistical control’ or stable. The packets of margarine were weighed and the mean weight, X = 255 g, the standard deviation, ␴ = 4.73 g. If the product specification demanded a weight of 250 ± 10 g, how much of the output of the packaging process would lie outside the tolerance zone? Figure 5.8 Determination of proportion defective produced Variables and process variation 99 The situation is represented in Figure 5.8. Since the characteristics of the normal distribution are measured in units of standard deviations, we must first convert the distance between the process mean and the Upper Specification Limit (USL) into ␴ units. This is done as follows: Z = (USL – X)/␴, where USL = Upper Specification Limit X = Estimated Process Mean ␴ = Estimated Process Standard Deviation Z = Number of standard deviations between USL and X (termed the standardized normal variate). Hence, Z = (260 – 255)/4.73 = 1.057. Using the Table of Proportion Under the Normal Curve in Appendix A, it is possible to determine that the proportion of packages lying outside the USL was 0.145 or 14.5 per cent. There are two contributory causes for this high level of rejects: (i) the setting of the process, which should be centred at 250 g and not 255 g, and (ii) the spread of the process. If the process were centred at 250 g, and with the same spread, one may calculate using the above method the proportion of product which would then lie outside the tolerance band. With a properly centred process, the distance between both the specification limits and the process mean would be 10 g. So: Z = (USL – X )/␴ =(X – LSL)/␴ = 10/4.73 = 2.11. Using this value of Z and the table in Appendix A the proportion lying outside each specification limit would be 0.0175. Therefore, a total of 3.5 per cent of product would be outside the tolerance band, even if the process mean was adjusted to the correct target weight. 2 Setting targets (a) It is still common in some industries to specify an Acceptable Quality Level (AQL) – this is the proportion or percentage of product that the producer/customer is prepared to accept outside the tolerance band. The characteristics of the normal distribution may be used to determine the target maximum standard deviation, when the target mean and AQL are 100 Variables and process variation specified. For example, if the tolerance band for a filling process is 5 ml and an AQL of 2.5 per cent is specified, then for a centred process: Z = (USL – X )/␴ =(X – LSL)/␴ and (USL – X )=(X – LSL) = 5/2 = 2.5 ml. We now need to know at what value of Z we will find (2.5%/2) under the tail – this is a proportion of 0.0125, and from Appendix A this is the proportion when Z = 2.24. So rewriting the above equation we have: ␴ max = (USL – X )/Z = 2.5/2.24 = 1.12 ml. In order to meet the specified tolerance band of 5 ml and an AQL of 2.5 per cent, we need a standard deviation, measured on the products, of at most 1.12 ml. (b) Consider a paint manufacturer who is filling nominal one-litre cans with paint. The quantity of paint in the cans varies according to the normal distribution with a standard deviation of 2 ml. If the stated minimum quality in any can is 1000 ml, what quantity must be put into the cans on average in order to ensure that the risk of underfill is 1 in 40? 1 in 40 in this case is the same as an AQL of 2.5 per cent or a probability of non-conforming output of 0.025 – the specification is one- sided. The 1 in 40 line must be set at 1000 ml. From Appendix A this probability occurs at a value for Z of 1.96␴. So 1000 ml must be 1.96␴ below the average quantity. The process mean must be set at: (1000 + 1.96␴) ml = 1000 + (1.96 ϫ 2) ml = 1004 ml This is illustrated in Figure 5.9. A special type of graph paper, normal probability paper, which is also described in Appendix A, can be of great assistance to the specialist in handling normally distributed data. 3 Setting targets A bagging line fills plastic bags with polyethylene pellets which are automatically heat-sealed and packed in layers on a pallet. SPC charting of Variables and process variation 101 the bag weights by packaging personnel has shown a standard deviation of 20 g. Assume the weights vary according to a normal distribution. If the stated minimum quantity in one bag is 25 kg what must the average quantity of resin put in a bag be if the risk for underfilling is to be about one chance in 250? The 1 in 250 (4 out of 1000 = 0.0040) line must be set at 25 000 g. From Appendix A, Average – 2.65␴ = 25 000 g. Thus, the average target should be 25 000 + (2.65 ϫ 20) g = 25 053 g = 25.053 kg (see Figure 5.10). Figure 5.9 Setting target fill quantity in paint process Figure 5.10 Target setting for the pellet bagging process . 3.2 24) 2.9 25) 3.3 26) 3.0 27) 3 .4 28) 3.1 29) 3 .5 30) 3.1 31) 3.2 32) 3 .5 33) 2 .4 34) 3 .5 35) 3.3 36) 3.6 37) 3.2 38) 3 .4 39) 3 .5 40 ) 3.0 41 ) 3 .4 42) 3 .5 43 ) 3.6 44 ) 3.0 45 ) 3.1 46 ) 3 .4 47). 3.6 44 ) 3.0 45 ) 3.1 46 ) 3 .4 47) 3.1 48 ) 3.6 49 ) 3.3 50 ) 3.3 51 ) 3 .4 52 ) 3 .4 53 ) 3.3 54 ) 3.2 55 ) 3 .4 56 ) 3.3 57 ) 3.6 58 ) 3.1 59 ) 3 .4 60) 3 .5 61) 3.2 62) 3.7 63) 3.3 64) 3.1 I hope you can make use. as manufactured: 98 Variables and process variation 1) 4. 1 2) 4. 0 3) 4. 2 4) 4. 2 5) 4. 4 6) 4. 2 7) 4. 3 8) 4. 2 9) 4. 2 10) 4. 1 11) 4. 3 12) 4. 1 13) 3.2 14) 3 .5 15) 3.0 16) 3.2 17) 3.3 18) 3.2 19)