CONTROL CHARTS FOR VARIABLES

UNIT CONTOL CHATS FO VAIABLES Structure.1 Introuction Objectives. Control Chart Technique.3 Control Charts for Variables.4 Control Chart for Mean(-Chart).5 ange Chart (-Chart).6 Stanar Deviation Chart (S-Chart).7 Process Capability Analysis.8 Summary.9 Solutions/Answers.1 INTODUCTION In Unit 1, you have learnt that the control chart is the most important tool for process control. With the help of control chart, we quicly etect the occurrence of assignable causes of variation an can tae corrective action to eliminate them. Several control charts may be esigne for ifferent situations. Each chart has its own fiel of applications an its own avantages an isavantages. However, all control charts have some characteristics in common an are interprete in almost the same manner. Generally, quality characteristics are of two types (variables an attributes). So control charts are broaly classifie into two categories: 1. Control charts for variables (Control charts for measurable characteristics). Control charts for attributes (Control charts for non-measurable characteristics) In this unit, you stuy the construction an interpretation of control chart for variables. We introuce the general technique of constructing a control chart in Sec... We also iscuss control charts for variables, i.e., control chart for mean an control chart for variability in Secs..3 to.6. Finally, we introuce the process capability analysis, which is use to chec whether the process is able to meet the specifications or not in Sec..7. The control charts for attributes are taen up in Units 3 an 4. Objectives After stuying this unit, you shoul be able to: explain ifferent types of control charts for variables; ecie which control chart for variables to use in a given situation; escribe the proceure for constructing a control chart; construct an interpret the control chart for process mean (-chart); 33

Process Control construct an interpret the control chart for process variability (-chart an S-chart); efine the concepts of specification limits, natural tolerance limits an process capability.. CONTOL CHAT TECHNIQUE Joseph Moses Juran (December 4, 1904-February 8, 008) was a omanianborn American engineer an management consultant. He is principally remembere for quality an quality management. There are two types of control charts: one for measurable characteristics nown as control charts for variables an secon for non-measurable characteristics nown as control charts for attributes. The technique for rawing a control chart is the same for all types. The main steps are as follows: 1. Select the Quality Characteristic In any manufacturing plant (small or big) of a prouct, there exist a large number of quality characteristics in the prouct. A single prouct/item/unit usually has several quality characteristics such as weight, length, with, strength, thicness, etc. It is, therefore, impossible to construct a control chart for each characteristic because it is rawn for controlling a single quality characteristic. So it is necessary to mae a juicious selection of quality characteristic. For example, a cricet ball has many quality characteristics such as weight, iameter, surface, elasticity, etc. A manufacturer of cricet ball woul lie to fin out which of these quality characteristics, shoul be given priority for quality control. While selecting a quality characteristic, generally, we give higher priority to the one that causes more non-conforming (efective) items an increases cost. For this, we use Pareto iagram. It is a bar iagram of the quality characteristics of a prouct an was use for the first time by Joseph Moses Juran for quality control. For rawing a Pareto iagram, the frequency of non-conformities (efects) of each quality characteristic such as weight, iameter, surface, etc. is note. In Pareto iagram, quality characteristics are taen along the -axis an the percentage of efects occurring in the quality is taen along the Y-axis. For example, consier the ata for quality characteristics for 00 cricet balls given in Table 1. Table 1: Quality Characteristics Data for a Sample of 00 Cricet Balls S. No. Quality Characteristic Frequency of Defects Percentage of Defects 1 Weight 90 45 Diameter 70 35 3 ough Surface 0 10 4 Elasticity 10 5 5 Other 10 5 Total 00 100 34

The Pareto iagram rawn from the above ata is as follows: Percentage of Defects 50 45 40 35 30 5 0 15 10 5 0 Y Weight Diameter Surface Elasticity Other Quality Characteristic Control Charts for Variables For rawing the bar chart, please recall Unit 14 of MST-001 entitle Founation in Mathematics an Statistics. Fig..1: Pareto iagram for ata of Table 1. From the Pareto iagram shown in Fig..1, we observe that the weight an iameter are serious problems accounting for most of the non-conforming (efective) cricet balls prouce. ough surface occurs less frequently, elasticity an other problems are relatively rare. So for this problem, we shoul select quality characteristics, weight an iameter of the balls for quality control.. Select the Type of Control Chart The choice of control chart epens upon the measurement, quality characteristic an the cost involve. For selection of control chart, we first select the category of the chart to be use as variables or attributes. If the characteristic to be controlle is measureable such as weight, length, iameter, etc., we use the control chart for variables. In the example of cricet balls, we have selecte weight an iameter for control an both are measurable. So we will use control charts for variables for both quality characteristics. If a characteristic to be controlle is non-measureable, e.g., colour, surface roughness, etc., we use the control chart for attributes. If many characteristics nee control, it is too costly to use control charts for variables because these charts are use for controlling only one characteristic at a time. Therefore, each characteristic has its variables chart. In such situations, a single chart for attributes can be use in place of all the charts for variables. After selecting the category of the chart to be use (variables or attributes), we select the actual chart to be use. For this, we nee to now which chart can be use for a given situation an also the purpose of the control chart. We shall explain this aspect while escribes each control chart. 3. Selection of ational Subgroups We now that control charts for quality characteristics are rawn with the help of sample ata. Now, the question may arise how are the ata collecte for analysis an controlling processes? For this Walter A. Shewhart evelope an introuce the concept of rational subgroups for control charts. He suggeste that the ifferences between rational

Process Control ecall the proceure of rawing a sample from a population (process), which has been explaine in Unit 1 of MST-005 (Statistical Techniques). For statistic an stanar error please recall Unit 1 of MST-004. The proceure of rawing a sample is calle sampling. subgroups are an inication of process changes (assignable causes) while ifferences within rational subgroup are an inication of inherent variability, i.e., chance causes. So the subgroups are selecte in such a way that we are able to ifferentiate between assignable causes an chance causes. If assignable causes are present in the process, they will show up as ifferences between the subgroups rather than ifferences within a subgroup. A rational subgroup is a small set of items/units that are prouce uner similar conitions within a relatively short time, i.e., the variation within the subgroup is only ue to chance causes. Generally, in inustries, the wor sample is use for a subgroup. So we use wors, sample an subgroup, in this course. 4. Size of Subgroup (Sample Size) To provie maximum homogeneity within a subgroup/sample, the size of the sample shoul be as small as possible. However, a sample size of four or five units/items is quite common in the inustry. It is also vali on statistical grouns. As we now from sampling istribution theory, the istribution of the sample mean is nearly normal for samples of four or more, even though the samples are taen from a non-normal population. This fact is helpful in interpretation of control chart limits. When a sample of size 5 is use there is ease in the computation of the average. When we have to mae the control chart more sensitive, samples of size 10 or 0 are use. This is because the stanar error of a statistic is inversely proportion to sample size, i.e., as the sample size increases, the stanar error ecreases. Therefore, 3limits (upper an lower control limits) will lie closer to the centre line. However, if the items prouce are estroye uner inspection or are expensive, a small sample of size or 3 is use. 5. Frequency of Subgroups (Number of Samples) There is no har an fast rule for the number of subgroups/samples, but it shoul be such that it is enough to etect tren, pattern of level shifts an process changes. To ecie the number of samples, we nee to eep in min that the cost of taing samples must be balance with the ata obtaine. There are two ways of taing samples: i) Taing larger samples at less frequent intervals, or ii) Taing smaller samples at more frequent intervals. It is best to tae samples more frequently in the beginning of the process. As the processes are brought into control, frequency of sampling can be reuce. For example, in a firm, the analyst may tae a sample every hour at the beginning of the process. If the process remains uner statistical control for two or three ays, the frequency of sampling may be reuce to every or 3 hour. If the process continuously remains uner statistical control for a few wees, the frequency may be further reuce. However, if ifficulties are encountere in eeping the process uner statistical control, samples can be taen more frequently for inspection. 6. Design the Forms for Data Collection 36

After taing the ecision about the quality characteristic, subgroup/sample size an number of subgroups, we esign the ata recoring form. The form for ata recoring shoul be esigne in accorance with the control chart to be use. A typical ata recoring form for the -chart is shown below: Control Charts for Variables Name of the Chart: Plot No.: Pin: Part Name: Operation: SPEC: Operator: Machine: Gauge: Subgroup Number 1 Date Time Measurements 1 3 4 Unit of Measurement: Mean SD Comment 10:00 5 5 50 51 51.5 0.83 1:00 50 53 5 53 5.00 1. 17/5 3 0:00 64 60 63 61 6.00 1.58 4 04:00 51 5 50 53 51.50 1.1 5 10:00 50 5 51 50 50.75 0.83 6 1:00 50 50 5 51 50.75 0.83 18/5 7 0:00 5 51 53 50 51.50 1.1 8 04:00 5 53 5 55 53.00 1. 9 10 1:00 5 5 54 5 5.50 0.87 19/5 11 0:00 5 53 54 49 5.00 1.87 New temporary operator 10:00 4 41 40 43 41.50 1.1 Ba material 1 04:00 5 51 54 51 5.00 1. 13 10:00 51 51 5 5 51.50 0.50 14 1:00 50 50 51 5 50.75 0.83 0/5 15 0:00 50 51 53 51 51.5 1.09 16 04:00 5 50 49 53 51.00 1.58 17 10:00 5 51 54 51 5.00 1. 18 1:00 51 51 50 5 51.00 0.71 1/5 19 0:00 5 50 53 45 50.00 3.08 0 04:00 50 49 50 48 49.5 0.83 Total 107.50 3.68 Generally, a ata recoring form is ivie into two parts. The top part of the form contains information about part name, operation machine, gauge use, unit of measurement an specifications. The remaining segment of the form contains information about subgroup/sample number, the ate an time when the sample was selecte an raw values of the observations. A column for comment is use to incorporate remar about the process. 7. Determination of Trial Centre Line an Control Limits You have learnt in Unit 1 that the quality characteristic can be escribe by a probability istribution. In most cases, it follows a normal istribution or can be approximate by a normal istribution. You have learnt in Sec. 14. 37

Process Control 38 of Unit 14 of MST-003 (Probability Theory) that the probability of a normally istribute ranom variable () that lies between 3 an 3 is 0.9973 where an are the mean an stanar eviation of ranom variable. That is, P 3 3 0.9973 The probability that the ranom variable lies outsie the limits 3 is 1 0.9973 = 0.007, that is, it is very small. It means that if we consier 100 samples, then most probably 0.7 items fall outsie 3 limits. Therefore, if an observation falls outsie 3σ limits, it is logical to suspect that something might have gone wrong. For this reason, the control limits are set up using 3 limits. Suppose M is a sample statistic (e.g., mean, range, proportion of efectives, etc.) that measures some quality characteristic of interest. Further suppose that µ M an σ M are the mean an stanar error (stanar eviation) of the sample statistic M, respectively. Then the centre line an control limits for controlling the quality characteristic are given by: M Centre line (CL) E M Upper control limit (UCL) 3 Lower control limit (LCL) 3 8. Constructing a Control Chart M M M The centre line an control limits are ifferent for ifferent charts. We shall explain the metho of etermining these limits while escribing each type of chart. After obtaining centre line an control limits, we construct the control chart. In a control chart, the statistic (e.g., mean, range, number of efects, etc.) is taen along the Y-axis an the sample number or time is taen along the -axis. We represent the centre line by a soli line an the control limits by otte lines. We plot the value of statistic for each sample against the sample number an consecutive sample points are joine by line segments. 9. Drawing Preliminary Conclusions from the Control Chart After constructing the control chart, we raw preliminary conclusions from it. We chec whether the plotte sample points lie on or in between the upper an lower control limits or some of them lie outsie these control limits. If one or more points lie outsie the control limits, we inicate each such point by rawing a otte circle aroun it. The control limits on this chart are calle trial control limits. ecall Sec.1.5 of Unit 1. If all sample points lie on or in between the upper an lower control limits an there is no unnatural patterns of variation, the chart inicates that the process is uner statistical control. Even if a process is uner statistical control, small variations may exist ue to machine performances, operator performance, material characteristics, etc. Such small variations are consiere to be a part of a stable process. It means that only chance causes are present in the process an no assignable causes are present. M

When a control chart inicates that the process is uner statistical control, perioic samples are taen from the process to etermine whether it remains uner statistical control continuously. When we assure that the process has been uner statistical control after the analysis of the preliminary ata, the trial control limits are use to control future prouction. Then,,, etc. can be consiere as representative of the process an their values are taen as stanar values. However, when more ata accumulates, the limits may be revise from time to time or whenever necessary. To ensure that the process remains uner statistical control, perioic samples are taen once a wee, once a month or once every 5, 50 or 100 items. Most processes are not uner statistical control at the time of the initial analysis, i.e., some points lie outsie the control limits or there are long runs or unusual patterns of variation. Then the control chart inicates that the process is not uner statistical control. Some assignable causes are present in the process. We investigate the reasons of assignable causes an tae corrective action to eliminate them from the process. Suppose these are ue to raw materials supplie by venor, we eliminate the cause by choosing a venor or maintaining the quality of raw material at venor s point. ectification is mae by eleting the out-of-control points an calculating the revise centre line an control limits for the chart. These revise limits are nown as the revise control limits. This proceure is continue till the process is being uner statistical control. Control Charts for Variables So far you have stuie the general technique for constructing a control chart. You have learnt how a control chart is constructe an how to etermine whether a process is uner statistical control or out of control on the basis of the control chart. We now iscuss the control charts for variables..3 CONTOL CHATS FO VAIABLES In any prouction process, it is not possible to ensure that all items prouce are alie in respect of certain characteristics. Some variations are always present in the items. These variations may be ue to raw material quality, unsille wor force, machines, faulty equipment, etc. The existence of variation in proucts affects the quality of the prouct. These variations may be ue to chance causes or assignable causes. So the aim of Statistical Process Control is to trace the sources of such variation an as far as possible try to eliminate them. If the quality characteristic in which we have to control the variation is measurable, we use control chart for variables. In quality control, the term variable means the quality characteristic which can be measure, e.g., iameter of ball bearings, length of refills, weight of cricet balls, etc. The control charts base on measurements of quality characteristics are calle control charts for variables. When we eal with a measurable quality characteristic, it is necessary to control the central tenency (average) as well as the ispersion (variability) of the quality characteristic or the process. For central tenency, we usually apply mean an for variability we calculate range or stanar eviation. Therefore, there are ifferent control charts for variables for controlling the mean an variability of the process. The most frequently use control charts for variables are: 39

Process Control 1. Control chart for mean. Control chart for variability We generally use control chart for mean ( -chart, rea as -bar chart) to control the mean quality level or process mean. We iscuss this chart with examples in Secs..4. To monitor an control variability process, we use -chart or S-chart. But -chart is most wiely use in quality control as compare to S-chart because in quality control, we generally use samples of sizes less than 10. We now that for n 10, the range an stanar eviation reflect almost the same information about the variability. However, the range is easier to calculate an interpret in comparison to the stanar eviation. Experience reveals that process variability also affects prouct quality. When variation in the prouction process is high, that is, the prouce items have a wier range of values then the quality of the prouct is assume to be poor. So to improve the quality of the prouct we must reuce the variability. The following charts are use to control process variability: 1. ange chart (-chart), an. Stanar eviation chart (S-chart). We iscuss each of these charts with examples in Secs..5 an.6. You may now lie to chec your unerstaning of ifferent types of control charts by answering the following exercise. E1) Choose the correct option from the following: i) The control chart for variables is/are the ii) a) -chart b) -chart c) S-chart ) all of these The control chart for process mean is the a) -chart b) -chart c) S-chart ) p-chart iii) The control chart for process variability is the a) -chart b) -chart c) c-chart ) p-chart We now iscuss the control chart for the mean..4 CONTOL CHAT FO MEAN (-CHAT) If we want to control the process mean, in any prouction process, we use the control chart for mean, i.e., the -chart.with the help of the -chart, we monitor the variation in the mean of the samples that have been rawn from time to time from the process. We plot the sample means instea of iniviual measurements on the control chart because sample means are calculate from n iniviuals an give aitional information. The -chart is more sensitive than the iniviual chart for etecting the changes in the process mean. We now explain various steps for constructing the -chart. Steps Involve in the Construction of the -chart The main steps for the construction of the -chart are as follows: Step 1: We select the measurable quality characteristic for which the -chart has to be constructe. We have iscusse how to select the quality characteristic in Sec... 40

Step : Then we ecie the subgroup/sample size as explaine in Sec... Generally, four or five items are selecte in a sample an twenty to twenty five samples are collecte for the -chart. Step 3: After eciing the size of the sample an the number of samples, we select sample units/items ranomly from the process so that each unit/item has an equal chance of being selecte. Step 4: We measure the quality characteristic ecie in Step 1 for each selecte item/unit in each sample. Errors in measurement may be ue to many reasons. These may be ue to: i) use of faulty machine, ii) iii) ifferent methos of taing measurements, inept use of instruments by unsille or inexperience wor force, etc. Therefore, care shoul be taen that measurements are mae by experts, using stanar instruments. Step 5: Calculate the sample mean for each sample. Suppose we measure the quality characteristic, say, for each unit/item of the sample. Also suppose 1,,..., n are the measurements for the units of a sample of size n. Then sample mean is given by n 1 1 n (1) 1 n i n i 1 where i represents the measurement of the quality characteristic of the i th unit. Suppose there are samples each of size n. Then we calculate the sample mean for each sample by using equation (1). Suppose the means for 1 st, n,, th samples are represente by 1,,...,, respectively. Step 6: Set the control limits To fin out whether the process is uner statistical control or out-ofcontrol, we set up the control limits. The 3σ control limits for the -chartare given by Centre line = E Lower control limit (UCL) = E 3SE Upper control limit (UCL) = E 3SE where Ean SE are the mean an the stanar error of the sampling istribution of, respectively. For getting the control limits of the -chartwe nee the sampling istribution of the sample mean. From Sec.. of Unit of MST-004, we now that if 1,,..., n is a sample of size n rawn from a normal population (istribution) with mean µ an variance σ, the Control Charts for Variables ecall the proceure of rawing a sample from a population (process) which has been explaine in Unit 1 of MST-005 (Statistical Techniques). 41

Process Control sample mean is normally istribute with mean µ an variance σ /n. It means that, if i ~ N(, ) then ~ N(, / n), an Var E We also now that SE() () n V ar() SE Var (3) n n Therefore, the centre line an control limits for the -chart are given as follows: I n Centre line (CL) = E (4a) Lower control limit (LCL) = E 3SE 3 / n p r or LCL = A... (4b) a cupper control limit (UCL) = E 3SE 3 / n t i or UCL = A (4c) c ewhere A 3 n is a constant an epens on the size of the sample., It has been tabulate for various sample sizes in Table I given at the en of this bloc. t he values of µ an σ are not nown. Therefore, these are estimate from the samples, which are taen when the process is thought to be uner statistical control. We use their best possible estimates. In Unit 5 of MST-004, we have seen that the best estimators of the population mean (µ) an population stanar eviation (σ) of normal population are the sample mean an sample stanar eviation, respectively. So in this case, the best estimator of the process mean (µ) is the gran mean (), that is, the mean of all sample means given by 4 1 1, (5) 1 i i 1 To calculate the gran mean, we first calculate the mean of each sample from equation (1). Then we calculate the gran mean from equation (5). Now, we nee the best estimate of the unnown stanar eviation σ. It can either be estimate by sample range () or sample stanar eviation (S). So we have two cases: Case I: When stanar eviation is estimate by the sample range. In this case, the stanar eviation σ is estimate as follows: ˆ (6)

where represents the mean of all sample ranges given by 1 1, (7) 1 i i 1 Here i is the sample range an is given by i max min (8) where max an min represent the maximum (or largest) an minimum (or smallest) measurements of the quality characteristic of a sample, respectively. Also is a constant an epens on the size of the sample. It has been tabulate for various sample sizes in Table I given at the en of this bloc. The control limits for the -chartwhen µ is estimate by an σ is estimate by are given by: Control Charts for Variables Centre line (CL) = ˆ (9a) Lower control limit (LCL) = 3 3 ˆ ˆ n n or LCL = A (9b) 3 3 Upper control limit (UCL) = ˆ ˆ A n n or UCL = A (9c) 3 where A is a constant an epens on the size of the n sample. It has been also tabulate for various sample sizes in Table I given at the en of this bloc. Case II: When stanar eviation is estimate by the sample stanar eviation. In this case, the stanar eviation σ is estimate as follows: ˆ S / c 4 (10) where S represents the mean of all sample stanar eviations given by 1 1 S S S,S S (11) 1 i i 1 Here S i is the sample stanar eviation given by n 1 S (1) i i i n 1 i 1 43

Process Control Therefore, the control limits for the -chartwhen µ is estimate by an σ is estimate by S are given by Centre line (CL) = ˆ Lower control limit (LCL) = 3 3 S ˆ ˆ n n c 4 (13a) or LCL = A3S (13b) Upper control limit (UCL) = 3 3 S ˆ ˆ n n c or UCL = A3S (13c) 3 where A3 is a constant an epens on the size of the c n 4 sample. It has been also tabulate for various sample sizes in Table I given at the en of this bloc. 4 Note: Generally, for the -chart, we select samples of four or five units/items. So for this chart, we use the sample range to estimate the stanar eviation of the process instea of the sample stanar eviation because the range is easier to calculate as compare to the stanar eviation. Moreover, it reflects almost the same information about variability as the stanar eviation. Step 7: Construct the -chart After setting the centre line an control limits, we construct the -chart as explaine in Step 8 of Sec... It means that we tae the sample number on the -axis (horizontal scale) an the sample mean () on the Y-axis (vertical scale). We plot the value of sample mean for each sample against the sample number an then join the consecutive sample points by line segments. Step 8: Interpret the result If all sample points on the plot lie on or in between upper an lower control limits, the process is uner statistical control. In this situation, only chance causes are present in the process. If one or more points lie outsie the control limits the control chart alarms (inicates) that the process is not uner statistical control. Some assignable causes are present in the process. To bring the process uner statistical control, we investigate the assignable causes an tae corrective action to eliminate them. Once the assignable causes are eliminate from the process, we elete the out-of-control points (samples) an calculate the revise centre line an control limits for the -chart by using the remaining samples. These limits are nown as revise control limits. To calculate the revise limits for the -chart,we first calculate an as follows: 44

A f t an e r f i n i n g i i1 j1 i i1 j1 where the number of iscare samples, the sum of means of iscare samples, an j1 j1 j i j the sum of ranges of iscare samples. j (14) (15) Control Charts for Variables After fining the an, we reconstruct the centre line an control limits, replacing by (9a to 9c) as follows: Centre line (CL) = an by in equations (16a) Lower control limit (UCL) = A (16b) Upper control limit (UCL) = A (16c) If we use the sample stanar eviation to estimate the process stanar eviation, then the revise control line an control limits are given as follows: Centre line (CL) (17a) Lower control limit (UCL) A3S (17b) Upper control limit (UCL) A3S (17c) where S S i i1 j1 S j an Sj the sum of stanar eviations of iscare samples. j1 (18) Note: Generally, we control the process mean an the process variability at the same time. So we use both an -charts or an S-charts together. So far, we have iscusse various steps involve in the construction of the -chart. Let us now tae up some examples to illustrate this metho. 45

Process Control Fig.. The CL, UCL an LCL for the -chart when μ an σ are unnown are CL= UCL= A LCL = A Example 1: A statistical quality controller uses the -chart for monitoring a quality characteristic of a prouct. If the process mean (µ) an process stanar eviation (σ) are 100 an 5, respectively, fin the centre line an control limits for the -chart. It is given that A = 1.5 for n = 4. Solution: Here we are given that 100, 5 an A 1.5 Therefore, we can calculate the centre line an control limits for the -chart using equations (4a to 4c) as follows: Centre line (CL) 100 Upper control limit (UCL) A 100 1.5 5 100 7.5 9.5 Lower control limit (LCL) A 100 1.5 5 100 7.5 107.5 Example : An automatic machine (shown in Fig..) is use to fill an seal 0 ml tube of meicine. The - control chart has been use to monitor this process. The process is sample in samples of four an the values of an are compute for each sample. After 5 samples, 55 an 90. Estimate the process mean an stanar eviation. Compute the control limits for the -chart. It is given that =.088 an A = 0.79 for n = 4. Solution: It is given that 5, n 4, 55, 90, =.088 an A = 0.79. The process mean is estimate by the gran sample mean, which is calculate as follows: 1 1 ˆ 55 1 5 The process stanar eviation is estimate by sample range, which is calculate as follows: 1 1 1 1 ˆ 90 1.74.088 5 Since the process average (µ) an process variability (σ) are unnown in this case, we use equations (9a to 9c) to calculate the centre line an the control limits for the -chart.these are given by CL = 1 UCL = A 1 0.79 3.6 3.644 LCL = A 1 0.79 3.6 18.3756 Example 3: A process of proucing ball bearings is starte. For monitoring the outsie iameter of the ball bearings, the quality controller taes the sample of five ball bearings at 10.00 AM, 1.00 PM,.00 PM, 4.00 PM an 6.00 PM an measures the outsie iameter (in mm) of each selecte ball bearing (Fig..3). The results of the test over a 4-ay prouction perio are as follows: Fig..3

Day Sample Number Time Observations 1 3 4 5 Control Charts for Variables Monay Tuesay Wenesay Thursay 1 10.00AM 5 5 50 51 51 51. 1.00PM 50 53 5 53 51 51.8 3 3.00 PM 54 51 50 5 53 5.0 4 4 4.00PM 56 55 53 55 53 54.4 3 5 6.00 PM 51 5 50 53 53 51.8 3 6 10.00AM 50 5 51 50 51 50.8 7 1.00PM 50 50 5 51 53 51. 3 8.00 PM 5 51 53 50 50 51. 3 9 4.00PM 5 53 5 55 53 53.0 3 10 6.00 PM 51 51 50 51 5 51.0 11 10.00AM 5 5 54 5 5 5.4 1 1.00PM 49 48 50 50 51 49.6 3 13.00 PM 5 53 54 49 5 5.0 5 14 4.00PM 5 51 54 51 54 5.4 3 15 6.00 PM 51 51 5 5 51 51.4 1 16 10.00AM 50 50 51 5 51 50.8 17 1.00PM 50 51 53 51 53 51.6 3 18.00 PM 5 50 49 53 50 50.8 4 19 4.00PM 5 51 54 51 51 51.8 3 0 6.00 PM 51 51 50 5 5 51. Total 103.4 56 Calculate an raw the centre line an control limits of the -chart. Draw the conclusion about the process, assuming assignable causes for any out-of-control point. If the process is out-of-control, calculate the revise centre line an control limits. Solution: Since the process average (µ) an process variability (σ) are unnown, we use equations (9a to 9c) for etermining the centre line an control limits for the -chart. From Table I given at the en of this bloc, we have A = 0.577 for n = 5. To calculate the centre line an control limits, we first calculate the values of an using equations (5) an (7) as follows: 1 1 103.4 51.6 0 1 1 56.80 0 Putting the values of, an A in equations (9a to 9c), we have CL = 51.6 47

Process Control The CL, UCL an LCL for the -chart when μ an σ are unnown are CL = UCL= A LCL = A UCL = A 51.6 0.577.80 53.36 LCL = A 51.6 0.577.80 50.004 We now construct the -chart by taing the sample number on the -axis an the average iameter of the ball bearing () on the Y-axis as shown in Fig..4. Average Diameter of the Ball Bearing 55 54 53 5 51 50 49 Y 1 3 4 5 6 7 8 9 10 11 1 13 14 15 16 17 18 19 0 Sample Number UCL = 53.36 CL = 51.6 LCL= 50.004 Fig..4: The -chart for the average iameter of the ball bearings. Formulas for calculating the an are i i1 j1 i i1 j1 j where the number of iscare samples, j1 j j an the sum of means of iscare sample, an j1 i the sum of ranges of iscare samples. Interpretation of the result From Fig..4, we observe that the points corresponing to samples 4 an 1 lie outsie the control limits. Therefore, the process is out-of-control an some assignable causes are present in the process. To bring the process uner statistical control, it is necessary to investigate the assignable causes an tae corrective action to eliminate them. After eliminating the assignable causes from the process, we elete the out-of-control points (4 an 1 samples) an calculate the revise centre line an control limits for the -chart using the remaining samples. For revise limits of the -chart, we first calculate the an by using equations (14) an (15). In our case, 0,, 54.4 49.6 104 an 3 3 6. j j1 j1 i j i1 j1 103.4 104 51.578 an 0 i j i1 j1 56 6.778 0 We calculate the revise centre line an control limits of the -chart using equations (16a to 16c) as follows: j CL = 51.578

LCL = A 51.578 0.577.778 49.975 Control Charts for Variables UCL = A 51.578 0.577.778 53.181 You shoul now chec your unerstaning about the -chart by solving the following exercises. E) Choose the correct option from the following: i) The control limits for the -chart are a) A b) 3 c) B ) A ii) If one point of the -chart lies outsie the control limits, the process is a) in control b) out-of-control c) both (a) an (b) iii) The -chart is use for a) variables b) attributes c) efects ) efectives E3) Samples of size 5 are taen from a manufacturing process at regular intervals. A normally istribute quality characteristic is measure an an are calculate for each sample. After 0 samples, we have 6.40 an 0.0877. Compute the centre line an control limits for the -chart. E4) A factory was proucing electrical bulbs. To test whether the process was uner statistical control or not with respect to the average life of the bulbs, 16 samples of 3 bulbs were rawn at regular intervals an the life of each sample bulb was measure. The measurements are given below: Sample Number Life of Bulb (in hour) () Total 1 115 650 1090 893 964.00 688 540 1074 304 767.33 3 876 134 1187 3408 1135.00 4 75 1070 847 646 880.67 5 900 846 154 3005 1000.00 6 1080 790 871 747 913.67 7 95 1087 116 317 1055.00 8 741 786 578 113 701.67 9 1005 144 1140 3398 119.67 10 130 1005 654 899 963.00 11 100 870 859 940 976.33 1 745 964 1050 771 919.67 13 96 117 100 3311 1099.33 14 781 876 981 65 879.33 15 110 1350 750 335 1073.33 16 880 975 768 639 874.33 49

Process Control Test whether the process is uner statistical control or not by rawing the -chart when the average life an the stanar eviation of the life of the bulbs are i) nown to be 1000 hour an 15 hour, respectively. ii) unnown. It is given that A = 1.73 an A = 1.03 for n = 3. In Sec..4, you have stuie the -chartan learnt when it is use an how it is constructe. We now iscuss the charts which are use for controlling the process variability. We first tae up the -chart..5 ANGE CHAT (-CHAT) You have learnt in Sec..3 that we use -chart or S-chart to control the process variability an the -chart is wiely use in quality control. The unerlying logic an basic form of an -chart are similar to the -chart. The primary ifference between the -chart an the -chart is that instea of plotting the sample means an monitoring their variation, we plot sample range an monitor the variation in sample ranges. To get the control limits for the -chart, we nee to now the sampling istribution of the range. However, the erivation of the sampling istribution of the range is beyon the scope of this course. So we state the result without proof. If 1,,...,n is a ranom sample of size n taen from a normal population with mean µ an variance σ, the mean an variance of sampling istribution of range are given as follows: an E() (19) Var() 3 (0) where an 3 are constants an epen on the size of the sample. These constants have been tabulate for various sample sizes in Table I given at the en of this bloc. We also now that SE() Var() Therefore, SE() Var() (1) 3 3 Although the sampling istribution of range is not a normal istribution, it is common in practice to use the 3 limits. These limits assure that the probability of an observation outsie these limits is very small. Therefore, we efine CL, LCL an UCL for the -chart as follows: 50

Centre line (CL) = E() (a) Control Charts for Variables Lower control limit (LCL) = E() 3 SE() 33 or LCL = ( 3 3) D1 (b) Upper control limit (UCL) = E() + 3 SE() 33 or UCL ( 3 3) D (c) where D1 33 an D 33 are constants an epen on the size of the samples. These constants have been tabulate for various sample sizes in Table I given at the en of this bloc. In practice, the value of σ is not nown. Therefore, it is estimate from the samples, which are taen when the process is thought to be uner statistical control. Since the -chart is use for small sample size (<10), σ is estimate by sample range (). If 1,,...,n is a sample of size n, we estimate σ from equation (6) as: ˆ So the centre line an control limits for the -chart when σ is estimate by are given as: Centre line (CL) = E() ˆ E 3SE ˆ ˆ 3 Lower control limit (LCL) 3 3 (3a) 33 3 3 or LCL 1 D3 (3b) E 3SE ˆ ˆ 3 Upper control limit (UCL) 3 3 33 3 3 or UCL 1 D4 (3c) 3 3 3 3 where D3 1 an D4 1 are constants an epen on the size of the sample. These constants have been tabulate for various sample sizes in Table I given at the en of this bloc. After setting the centre line an the control limits, we construct the -chart by taing the sample number on the -axis (horizontal scale) an the sample range () on the Y-axis (vertical scale). We represent the centre line by soli line an UCL an LCL by otte lines. We plot the value of sample range for each sample against the sample number. The consecutive sample points are joine by line segments. Interpretation of the result 51

Process Control If all sample points lie on or between the upper an lower control limits, the -chart inicates that the process is uner statistical control. Otherwise, it is out-of-control. To bring the process uner statistical control, it is necessary to investigate the assignable causes an tae corrective action to eliminate them. For this, we elete the out-of-control points an calculate the revise centre line an control limits as we i for the control chart of process mean. We Centre Line (CL) = (4a) Lower control limit (LCL) = D3 (4b) Upper control limit (UCL) = D4 (4c) efine the revise limits for the -chart as follows: The following points shoul be ept in min while using the an -charts: In orer to etermine whether a process is uner statistical control, we control both process mean an process variability. For that, we use the an -charts together. We first control the process variability. So we analyse the -chart before the -chart. If the -chart inicates that the process variability is out-of-control, we first control the process variability by investigating the assignable causes an taing corrective action to eliminate them. Only then we analyse the -chartbecause when the -chart is brought uner statistical control, many assignable causes for the -chart are eliminate automatically. If the -chart inicates that the process variability is uner statistical control, but the -chart inicates that the process mean is out-of-control, we continue to use the -chart, but revise the -chart.the centre line an control limits of the -chart are revise by eliminating these points. As a result, the position of the control limits in the revise -chart shift. But the istance of control limits remains the same as in out-of-control case because we o not calculate again. If both an -charts inicate that the process variability an process mean are uner statistical control, an can be consiere as representative of the process an their values are taen as stanar values. The following examples will help you unerstan how to construct an interpret the an -charts together. Example 4: A statistical quality controller uses the an -charts together for monitoring the quality characteristic of a prouct. Samples of size 5 are taen from the manufacturing process at regular intervals. A normally istribute quality characteristic is measure an the an are calculate for each sample. After 0 samples have been analyse, we have 6.40 an 0.0877. Compute the centre line an control limits for the an -charts. Solution: Here we are given that n 5, 0, 6.40, 0.0877 5

Note that the process mean (µ) an process variability (σ) are unnown in this case. Therefore, we use equations (9a to 9c) an (3a to 3c) for calculating the centre an control limits for the -chart an the -chart, respectively. Control Charts for Variables From Table I, we have A 0.577, D3 0,D4.114 for n = 5 Substituting the values of, an A in equations (9a to 9c), we get the centre line an control limits of the -chartas follows: CL 6.40 UCL A 6.40 0.5770.0877 6.451 LCL A 6.40 0.5770.0877 6.349 Similarly, substituting the values of, D3 an D 4 in equations (3a to 3c), we get the centre line an control limits of the -chart as follows: CL = 0.0877 UCL = D4.1140.0877 0.185 LCL = D3 0 0.0877 0 Example 5: A mil company uses automatic machines to fill 500 ml mil pacets. A quality control inspector inspecte four pacets for each sample at given time intervals an measure the weight of each fille pacet. The ata for 0 samples are shown in the following table: The CL, UCL an LCL for the -chart when μ an σ are unnown are CL = UCL= A The CL, UCL an LCL for the -chart when σ is unnown are CL = UCL= D4 LCL = D3 LCL = A Sample Number Weight of Fille Mil Pacet (in ml) 1 500 50 500 500 506.67 0 500 490 50 530 503.33 40 3 490 550 570 540 536.67 80 4 510 50 500 50 510.00 0 5 510 480 490 490 493.33 30 6 50 500 50 500 513.33 0 7 50 510 530 510 50.00 0 8 530 490 50 500 513.33 40 9 510 490 500 510 500.00 0 10 50 50 490 50 510.00 30 11 50 500 510 500 510.00 0 1 480 500 50 510 500.00 40 13 530 510 50 510 50.00 0 14 500 510 510 500 506.67 10 15 490 50 500 510 503.33 30 16 50 500 530 500 516.67 30 17 50 560 490 510 53.33 70 18 500 490 500 510 496.67 0 19 50 500 530 500 516.67 30 0 500 490 500 490 496.67 10 Total 10196.67 600 53

Process Control The CL, UCL an LCL for the -chart when σ is unnown are CL = UCL= D4 LCL = D3 Using the an -charts together, raw the conclusion about the process by assuming assignable causes for any out-of-control points. If the process is out-of-control, calculate the revise centre line an control limits to bring the process uner statistical control. Solution: Since the an -charts are to be use together, we raw an analyse the -chart first. Here, the variability of the process (σ) is unnown. Therefore, we use equations (3a to 3c) for calculating the centre an control limits for the -chart From Table I, we have A 0.79, D3 0 an D4.8 for n 4 We first calculate the value of for the centre line an the control limits from the given ata using equation (7) as follows: 1 1 600 30 0 Substituting the values of, D3 an D4 in equations (3a to 3c) we get CL = 30 UCL = D4.8 30 68.46 LCL = D3 030 0 We now construct the -chart by taing the sample number on the -axis an the sample range () of the mil pacets on the Y-axis as shown in Fig..5. 90 Y Sample ange of the Mil Pacets 80 70 60 50 40 30 0 10 UCL = 68.46 CL = 30 LCL = 0 0 1 3 4 5 6 7 8 9 10 11 1 13 14 15 16 17 18 19 0 Sample Number Interpretation of the result Fig..5: The -chart for mil pacets. From Fig..5, we observe that the points corresponing to samples 3 an 17 lie outsie the control limits. Therefore, the process variability is out-of-control an some assignable causes are present in the process.

For calculating the revise control limits, we first elete the out-of-control points an calculate the for the remaining samples. In this example, 0,, 80 70 150 j1 j i1 j1 600 150 450 5 0 18 j We calculate the revise centre line an control limits of the -chart using equations (4a to 4c) as follows: CL = 5 UCL =D4.8 5 57.05 LCL = D3 0 5 0 Fig..6 shows the revise -chart. 60 Y UCL = 57.05 Control Charts for Variables Formula for calculating the is i i1 j1 where the number of iscare samples, an j1 i sum of ranges of iscare samples. j Sample ange of the Mil Pacets 50 40 30 0 10 CL = 5.0 LCL = 0 0 1 3 4 5 6 7 8 9 10 11 1 13 14 15 16 17 18 Interpretation of the result Fig..6: The revise -chart. The revise -chart (shown in Fig..6) inicates that all points lie within the control limits. Hence, we say that the process variability is uner statistical control. Note: If one or more points lie outsie the revise control limits, we calculate the revise control limits for the -chart again. This is continue until the process is brought uner statistical control. After controlling the process variability, we stuy the process mean. For this we calculate the control limits of the -chart for the remaining samples. We first calculate the from equation (14). In this example, Sample Number 0,, 536.67 53.33 1060 j1 j

Process Control j i1 j1 10196.67 1060 9136.67 507.593 The CL, UCL an 0 18 LCU for the -chart when out-of-control We now calculate the revise centre line an control limits of the -chart using samples are elete equations (16a to 16c) as follows: given by CL = UCL= LCL = A A CL = 507.593 UCL = A 507.593 0.79 5 55.818 LCL = A 507.593 0.79 5 489.368 We can construct the -chart by taing sample the number on the -axis an the average weight of the mil pacets on the Y-axis as shown in Fig..7. Sample Average Weight of the Mil Pacets 530 55 50 515 510 505 500 495 490 485 480 Y 1 3 4 5 6 7 8 9 10 11 1 13 14 15 16 17 18 Sample Number UCL = 55.818 CL = 507.593 LCL = 489.368 489.37 Interpretation of the result Fig..7: The -chart. Since no point lies outsie the control limits of the -chart (shown in Fig..7), it inicates that the process mean is uner statistical control. You may now lie to construct the -chart an the -chart for practice. E5) Choose the correct option from the following: i) The upper control limit for the -chart is ii) a) D4 b) D3 c) ) D4 Which control chart is use for controlling the process variability? a) -chart b) -chart c) p-chart ) c-chart

E6) A high-voltage power station supplies a nominal output voltage of 50 V. A ranom sample of five outputs is taen each ay an the mean an range calculate for all samples are tabulate below: Control Charts for Variables Day 1 3 4 5 6 7 8 60 40 50 60 90 40 60 50 0 0 60 30 0 60 50 30 Day 9 10 11 1 13 14 15 16 40 300 40 60 50 70 50 40 50 40 50 0 0 60 40 Test whether the process is uner statistical control or not by rawing the an -charts. In this section, you have learnt about the -chart an how it is constructe. We now iscuss the S-chart which is also use for controlling process variability..6 STANDAD DEVIATION CHAT (S-CHAT) use chart -chart. In Secs..4 an.5, you have stuie the control charts for process mean (-chart) an process variability (-chart). Although the -chart is easy to calculate an explain, for large sample size (usually greater than 10) it oes not give a correct picture of the process variability. In Unit of MST-00 entitle Descriptive Statistics, you have stuie that the range epens only on maximum an minimum values of the ata whereas the stanar eviation epens on all observations. As the number of observations increases, the range of the ata fluctuates more than the stanar eviation. So, when the number of observations increases ( 10), we use stanar eviation chart (Schart) instea of the range chart (-chart). To get the control limits for the S-chart, we nee to now the sampling istribution of the stanar eviation (S). Since the erivation of the sampling istribution of S is beyon the scope of this course, we state the result without proof. If 1,,...,n is a ranom sample of size n rawn from a normal istribution with mean µ an variance, an E(S) c 4 (5) Var(S) 1 c 4 (6) where c 4 is a constant an epens on the size of the sample. It has been tabulate for various sample sizes in Table I given at the en of this bloc. We also now that SE() = Var() SE(S) Var(S) 1 c 1 c (7) 4 4 57

Process Control Although the sampling istribution of the stanar eviation is not normal, it is common to use the 3 limits. These limits assure that the probability of an observation outsie these limits is very small. Therefore, we can calculate the CL, LCL an UCL for the S-chart as follows: Centre line (CL) E S 4 Lower control limit (LCL) E S 3SE S or LCL c4 3 1 c4 Upper control limit (UCL) E(S) 3SE(S) or UCL c4 3 1 c4 c (8a) c 3 1 c 4 4 B5 (8b) c 3 1 c 4 4 B6 (8c) where B c 3 1 c an B c 3 1 c are constants an epen 5 4 4 6 4 4 on the size of the sample. These constants have been tabulate for various sample sizes in Table I given at the en of this bloc. In practice, the value of σ is not nown. Therefore, it is estimate from the samples which are taen when the process is thought to be uner statistical control. Since the S-chart is use for large sample size ( 10), σ is estimate by the sample stanar eviation (S) as we have escribe in Case II of Sec..4. Hence, the control limits for the S-chart when σ is estimate by the average sample stanar eviation are given by S E S c c S (9a) c CL ˆ 4 4 LCL E S 3SE S or LCL UCL E S 3SE S or where UCL 4 c ˆ 3ˆ 1 c 4 4 S S c4 3 1 c4 c 4 c 3 1 1 c 4 S B3S (9b) 4 c4 c ˆ 3ˆ 1 c 4 4 S S c4 3 1 c4 c 4 c 3 1 1 c 4 S B4S (9c) 4 c4 3 3 B 1 1 c an B 1 1 c are constants an epen on 3 4 4 4 c4 c4 the size of the sample. These constants have been tabulate for various sample sizes in Table I given at the en of this bloc. The construction of the S-chart is similar to the -chart. We tae the sample number on the -axis (horizontal scale) an the sample stanar eviation on the Y-axis (vertical scale). We represent the centre line by soli line an UCL an LCL by otte lines. We also plot the value of sample stanar eviation 58

for each sample against the sample number. The consecutive sample points are joine by line segments. Interpretation of the result If all sample points lie on or in between the upper an lower control limits, the S- chart inicates that the process is uner statistical control. Otherwise the process is out-of-control. To bring the process uner statistical control it is necessary to investigate the assignable causes an tae corrective action to eliminate them. For this, we elete the out-of-control points (samples) an calculate the revise centre line an control limits as was one in the -chart. These limits are nown as the revise control limits. For revise limits for the S-chart, we first calculate S by using the following formula: Control Charts for Variables S S i1 j1 where the number of iscare samples, an S j (30) Sj the sum of stanar eviations of iscare samples. j1 We reconstruct the centre line an control limits of the S-chart by replacing S by S in equations (9a to 9c) as follows: Centre Line (CL) = S Lower control limit (LCL) = B3S Upper control limit (UCL) = B4S (30a) (30b) (30c) Let us illustrate the construction an interpretation of the S-chart with the help of an example. Example 6: The mean an stanar eviation of 15 samples of size 1 for a prouction process are given in the following table: Sample No Mean Stanar Deviation Sample No Mean Stanar Deviation 1 9.5 1. 9 8.75.1 7.5 1.5 10 9.75 3.5 3 10.0.6 11 7.5.6 4 8.5 3. 1 10.5 1.1 5 7.5 1.9 13 9.5. 6 9.0.6 14 10.0 1.6 7 10.5 5.4 15 10.5 3. 8 9.5.4 Determine the control limits for the an S-charts. Comment on whether the process variability an process mean are uner statistical control or not. Use the quality control factors A 3 = 0.886, B 3 = 0.354, B 4 = 1.646 for n = 1. 59

Process Control The CL, UCL an LCL for the -chart when σ is estimate by S are CL = UCL= A S 3 LCL = A S The CL, UCL an LCL for the S-chart when σ is estimate by S are CL = S UCL= B4S LCL = B3S 3 Solution: Here, we want to control the process variability an process mean. Since the sample size is greater than 10, we use S an -charts. Since sample size n > 10, we use equations (13a to13c) an (9a to 9c) for calculating the centre an control limits for the -chart an S-chart, respectively. It is given that n 1, 15,A 3 0.886, B3 0.354 an B4 1.646 We first calculate the values of an S for centre line an control limits using equations (5) an (11) as follows: 1 1 9.5 7.5... 10.5 1 137 9.133 15 15 S 1 S 1 1. 1.5... 3. 1 37.1.473 15 15 To chec whether the process is uner statistical control or not, we first raw an analyse the S-chart. Substituting the values of S, B3 an B4 in equations (9a to 9c), we get the centre line an control limits of the S-chart as follows: CL = S.473 UCL B4S 1.646.473 4.071 LCL B3S 0.354.473 0.875 We construct the S-chart by taing the sample number on the -axis an the sample stanar eviation (S) on the Y-axis as shown in Fig..8. 6 Y Sample Stanar Deviation 5 4 3 1 0 UCL = 4.071 CL =.473 LCL= 0.875 1 3 4 5 6 7 8 9 10 11 1 13 14 15 Sample Number Interpretation of the result Fig..8: The S-chart for Example 6. The S-chart shown in Fig..8 inicates that the point corresponing to sample 7 lies outsie the upper control limits. Therefore, the process variability is

out-of-control an some assignable causes are present in the process. To bring the process uner statistical control it is necessary to investigate the assignable causes an tae corrective action to eliminate them. Control Charts for Variables We calculate the centre line an control limits of the -chart as follows: CL = 9.133 UCL A3S 9.133 0.886.473 11.34 LCL A3S 9.133 0.886.473 6.94 We construct the -chart by taing the sample number on the -axis an the sample average () on the Y-axis as shown in Fig..9. 1 11 Y UCL = 11.34 Sample Average 10 9 8 7 6 CL = 9.133 LCL = 6.94 1 3 4 5 6 7 8 9 10 11 1 13 14 15 Sample Number Fig..9: The -chart for Example 6. Interpretation of the result The -chart shown in Fig..9 inicates that all the sample points lie within the control limits. So the process mean is uner statistical control. Hence, with the help of S an -chart we conclue that the process variability is not uner statistical control, whereas the process average is uner statistical control. You may lie to chec your unerstaning about the S-chart by answering the following exercise. E7) In a factory, quality is controlle using mean ( ) an stanar eviation (S) charts. Twenty samples are chosen, each having 10 units for which = 645 an S = 6. Determine the 3σ limits for the an S-charts. It is given that A 3 = 0.97, B 3 = 0.84 an B 4 = 1.716 for n = 10. So far we have iscusse one aspect of statistical process control, that is, to ensure that the process is uner statistical control with the help of control chart. If a process is uner statistical control, it merely inicates that there are no assignable causes of variations an that it is stable. Once a process is uner statistical control, the next step is to chec whether the process is able to meet

Process Control the specifications or not. For this, we use process capability analysis, which we iscuss in the next section..7 POCESS CAPABILITY ANALYSIS Variation in manufacture proucts is inevitable an it is a fact of nature an inustrial life. Even if the prouction process is well esigne or carefully maintaine, no two objects are ientical. The ifference in two items/units may range from very large to very small or it may even be unetectable epening on the sources of variation. Sometimes, it is possible that the process is uner statistical control, but the customer is not satisfie with the prouct. Such a situation may arise when the prouct oes not conform to its specifications. Then we nee to carry out the process capability analysis to etermine whether a prouct conforms to its specifications. To conuct the process capability stuy, it is important to istinguish between the specification limits of a prouct, the control limits of the process an the natural tolerance limits of the prouct. We now explain these limits. Specification Limits The limits specifie by the consumer for the value of a variable characteristic of the prouct or set by the management or the manufacturing engineers at the esign an evelopment stages of the prouct are calle specification limits. Thus, the specification limits are etermine externally. The upper value of the specification limits is calle upper specification limit (USL) an the lower value of the specification limits is calle lower specification limit (LSL). If a prouct fails to meet the specifications, it is calle a efective prouct. This means that any prouct manufacture outsie these limits woul have to be scrappe an rewore. Therefore, the process shoul be set in a way that these limits are met (if possible) by all proucts. For example, the weight of a cricet ball may be set by the customer or by the manufacturing engineers as 160.0 ± 3.0 gram. Then the weight of a cricet ball prouce by the manufacturer can vary from 157.0 (= 160.0 3.0) gram to 163.0 (= 160.0 + 3.0) gram. Thus, the USL is 163.0 gram an LSL is 157.0 gram. If any ball prouce by the manufacturer has weight greater than 163.0 gram or less than 157.0 gram, it may have to be scrappe. Control Limits In Sec.., you have learnt about the control limits for a process. You now that the control limits are usually etermine by samples collecte from a process. If a sample point lies outsie the control limits, the process is sai to be out-of-control. However, even in this case the iniviual prouct may not necessarily be efective because in the control charts, we plot a statistic such as the sample mean, range instea of iniviual measurement of the quality characteristic. It means that a single prouct that falls outsie the control limits will neither cause the process to be out-of-control nor efective. Natural Tolerance Limits The natural tolerance limits (NTLs) are the limits usually etermine by iniviual measurements of the characteristic of the prouct manufacture. These limits represent the natural variability of proucts in the process an are set at the 3σ limits from the mean, i.e., 3 where µ an σ are the process 6

mean an stanar eviation, respectively, for iniviual measurements of the characteristic of the proucts prouce. Thus, 3 is nown as the upper natural tolerance limit (UNTL) an 3 is nown as the lower natural tolerance limit (LNTL). If µ an σ are unnown, their values are estimate by the gran sample mean an range or stanar eviation, respectively, as we have escribe in Sec..4 for the -chart. It is not that there is no relationship between the specification limits, control limits an natural tolerance limits just because the specification limits are specifie by the customer an the control limits an natural tolerance limits are riven by the variability of the process. We now istinguish between the specification limits of a prouct, the control limits of the manufacturing process an the natural tolerance limits of the prouct with the help of an example. Example 7: A cricet ball manufacturing company wants to control the weight of the ball. Twenty-five samples, each of size 4 were collecte. The sum of averages an sum of sample ranges were foun to be 4010 gram an 7 gram, respectively. If the customer specifie the weight of the ball as 160 ± 3.0 gram, fin the estimates of process mean an stanar eviation. Also fin the specification limits, control limits an natural tolerance limits. It is given that, =.088 an A = 0.79 for n = 4. Solution: It is given that 5, n 4, 4010, 7, =.059 an A = 0.79 The process mean is estimate by the gran sample mean which is calculate as follows: 1 1 ˆ 4010 160.4 5 The process stanar eviation is estimate by the sample range which is calculate as follows: 1 1 1 1 ˆ 7 1.399.059 5 It is given that the customer specifie the weight of the ball as: 160.0 ± 3.0 gram. Therefore, The targete or aime value (µ 0 ) = 160.0 Upper specification limit (USL) = 160.0 3.0 163.0 Lower specification limit (USL) = 160.0 3.0 157.0 Since the manufacturer has to control the mean weight of the balls, we use -chart. Here μ an σ are unnown, therefore, we can calculate the centre line an control limits for the -chart using equations (9a to 9c) as follows: Control Charts for Variables Centre line (CL) = 160.4 Upper control limit (UCL) = A 160.4 0.79.88 16.50 Lower control limit (LCL) = A 160.4 0.79.88 158.30 63

Process Control The natural tolerance limits can be obtaine as follows: Process mean ( ) ˆ 160.4 Upper natural tolerance limit (UNTL) ˆ 3ˆ 160.4 31.399 164.597 Lower natural tolerance limit (LNTL) ˆ 3ˆ 160.4 31.399 156.03 So far we have explaine the specification limits, control limits an natural tolerance limits an basic ifference among them. We now iscuss process capability. Process Capability Process capability is a measure of the ability of a process to meet specifications. It tells us how goo an iniviual prouct is. Even though the process may seem to be uner statistical control, it is necessary to see whether the process is capable of manufacturing proucts within the specific limits or not. This can be one by process capability analysis. Process capability may be efine as the sprea of a quality characteristic measurement (i.e., natural variability) when the process is uner statistical control (stable). If the stanar eviation of the stable process is σ, the process capability is efine as follows: Process capability = 6 This inicates that proucts can be manufacture within ± 3σ limits. Generally, stanar eviation of a process is not nown an we estimate by range or stanar eviation as follows: S ˆ or ˆ c 4 To test whether the process is capable of manufacturing the prouct within the specific limits, we compare process capability with tolerance. Tolerance is efine as the ifference between specification limits, i.e., Tolerance = USL LSL When we compare process capability with tolerance, the following situations may arise: Case I: 6 USL LSL In this case, the process is capable of proucing proucts of esire specifications, even though the process is uner statistical control or out-of-control. Case II: 6 UCL LSL In this case, the process is capable of meeting specifications only if no shift in the process mean or a change in process variability taes place. Case III: 6 USL LSL In this situation, the proucts o not meet the esire specifications, even though the process is uner statistical control. The relation of process capability with the specification limits can be unerstoo from the following example of your aily life. 64

Consier a car an a garage (Fig..10). The garage efines the specification limits an the car efines the output of the process. If the car is only a little bit smaller than the garage, we can par it right in the mile of the garage (centre of the specification). If the car is wier than the garage, it oes not matter if we have it centre; it will not fit. If the car is a lot smaller than the garage, it oes not matter if we par it exactly in the mile; it will fit an we have plenty of room on either sie. If we have a process in control with little variation, we shoul be able to (par the car easily within the garage) manufacture a prouct that meets its specifications as well as customer requirements. Now, we tae up an example base on process capability. Example 8: A manufacturing process prouces a certain type of bolt of mean iameter inche with a stanar eviation of 0.05 inche. The lower an upper specification limits of the process are 1.90 an.05 inche. Calculate the process capability. Does it appear that the manufacturing process is capable of meeting the specification requirements? Solution: We now the process capability (PC) of any process is efine as PC 6 It is given that where is the stanar eviation of the process., 0.05, USL.05 an LSL 1.90 PC 6 60.05 0.30 Now we have to test whether the process is capable of meeting the require specifications. We first calculate tolerance an then compare PC with the tolerance: Tolerance USL LSL.05 1.90 0.15 Since process capability > tolerance, the proucts o not meet the esire specifications. Now, you can chec your unerstaning of process capability by answering the following exercise. Control Charts for Variables Fig..10 E8) An automatic machine is use to fill an seal 0 ml tube of meicine. The process is sample in samples of four an the values of an are compute for each sample. After 5 samples, 55 an 30. If the specification limits are 0 ± 1.5, what are the conclusions regaring the ability of the process to prouce meicine tubes conforming to specifications? It is given that =.059 for n = 4. We en this unit by giving a summary of what we have covere in it..8 SUMMAY 1. Pareto iagram is a bar iagram, use for selection of quality characteristic for control in statistical quality control.. If the characteristic to be controlle is measureable such as weight, length, iameter, etc., control chart for variables can be use.