1. 5 – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 Total Quality Management Course No. 403434 Power Point Slides by Osama Aljarrah Six Sigma

2. 5 – 2 Costs of Quality Defect can be defined as any failure lead to customer dissatisfaction  Prevention costs  Appraisal costs  Internal failure costs  External failure costs  Ethics and quality

3. 5 – 3 Total Quality Management Customer satisfaction

4. 5 – 4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Total Quality Management Customer satisfaction  Conformance to specifications  Value  Fitness for use  Support  Psychological impressions Employee involvement  Cultural change  Teams

5. 5 – 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Total Quality Management Continuous improvement  Kaizen  A philosophy  Not unique to quality  Problem solving process

6. 5 – 6 The Deming Wheel Plan Do Study Act

7. 5 – 7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Six Sigma X X X X X X X X X X X X X X X X X Process average OK; too much variation Process variability OK; process off target Process on target with low variability Reduce spread Center process X X X X X X X X X Figure 5.3 – Six-Sigma Approach Focuses on Reducing Spread and Centering the Process

8. 5 – 8 8 WHAT IS SIX SIGMA? Six Sigma - A highly disciplined process that enables organizations deliver nearly perfect products and services. The figure of six arrived statistically from current average maturity of most business enterprises A philosophy and a goal: as perfect as practically possible. A methodology and a symbol of quality. Contd…

9. 5 – 9 WHAT IS SIX SIGMA? A comprehensive and flexible project for achieving, sustaining, and maximizing business success by minimizing system process muda, mura, and muri Six Sigma utilizes many established quality-management statistical tools But, it is much more!

10. 5 – 10 Six Sigma Improvement Model Control Improve Analyze Measure Define

11. 5 – 11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Application of statistical techniques Acceptable quality level (AQL) Linked through supply chains

12. 5 – 12 Acceptance Sampling Firm A uses TQM or Six Sigma to achieve internal process performance Supplier uses TQM or Six Sigma to achieve internal process performance YesNo YesNo fan motors fan blades Accept blades? Supplier Manufactures fan blades TARGET: Firm A’s specs Accept motors? Motor inspection Blade inspection Firm A Manufacturers furnace fan motors TARGET: Buyer’s specs Buyer Manufactures furnaces

13. 5 – 13 Statistical Process Control Used to detect process change Variation of outputs Performance measurement – variables Performance measurement – attributes Sampling Sampling distributions

14. 5 – 14 Sampling Distributions 1.The sample mean is the sum of the observations divided by the total number of observations where x i = observation of a quality characteristic (such as time) n = total number of observations x = mean

15. 5 – 15 Sampling Distributions 2.The range is the difference between the largest observation in a sample and the smallest. The standard deviation is the square root of the variance of a distribution. An estimate of the process standard deviation based on a sample is given by where σ = standard deviation of a sample

16. 5 – 16 Sample and Process Distributions Distribution of sample means 25Time Mean Process distribution Figure 5.6 –Relationship Between the Distribution of Sample Means and the Process Distribution

17. 5 – 17 Causes of Variation Common causes  Random, unavoidable sources of variation  Location  Spread  Shape Assignable causes  Can be identified and eliminated  Change in the mean, spread, or shape  Used after a process is in statistical control

18. 5 – 18 Assignable Causes (a) Location Time Average Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

19. 5 – 19 Assignable Causes (b) Spread Time Average Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

20. 5 – 20 Assignable Causes (c) Shape Time Average Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

21. 5 – 21 Control Charts Time-ordered diagram of process performance  Mean  Upper control limit  Lower control limit Steps for a control chart 1.Random sample 2.Plot statistics 3.Eliminate the cause, incorporate improvements 4.Repeat the procedure

22. 5 – 22 Control Charts Samples Assignable causes likely 1 2 3 How Control Limits Relate to the Sampling Distribution: Observations from Three Samples UCL Nominal LCL

23. 5 – 23 Nominal UCL LCL Variations Sample number Control Charts Control Chart Examples (a) Normal – No action

24. 5 – 24 Nominal UCL LCL Variations Sample number Control Charts Control Chart Examples (b) Run – Take action

25. 5 – 25 Nominal UCL LCL Variations Sample number Control Charts Control Chart Examples (c) Sudden change – Monitor

26. 5 – 26 Nominal UCL LCL Variations Sample number Control Charts Control Chart Examples (d) Exceeds control limits – Take action

27. 5 – 27 Control Charts Two types of error are possible with control charts A type I error occurs when a process is thought to be out of control when in fact it is not A type II error occurs when a process is thought to be in control when it is actually out of statistical control These errors can be controlled by the choice of control limits

28. 5 – 28 SPC Methods Control charts for variables  R -Chart UCL R = D 4 R and LCL R = D 3 R where R =average of several past R values and the central line of the control chart D 3, D 4 =constants that provide three standard deviation (three-sigma) limits for the given sample size

29. 5 – 29 Control Chart Factors TABLE 5.1|FACTORS FOR CALCULATING THREE-SIGMA LIMITS FOR |THE x -CHART AND R -CHART Size of Sample ( n ) Factor for UCL and LCL for x -Chart ( A 2 ) Factor for LCL for R -Chart ( D 3 ) Factor for UCL for R -Chart ( D 4 ) 21.88003.267 31.02302.575 40.72902.282 50.57702.115 60.48302.004 70.4190.0761.924 80.3730.1361.864 90.3370.1841.816 100.3080.2231.777

30. 5 – 30 SPC Methods UCL x = x + A 2 R and LCL x = x – A 2 R Control charts for variables  x -Chart where x =central line of the chart, which can be either the average of past sample means or a target value set for the process A 2 =constant to provide three-sigma limits for the sample mean

31. 5 – 31 Steps for x - and R -Charts 1.Collect data 2.Compute the range 3.Use Table 5.1 to determine R -chart control limits 4.Plot the sample ranges. If all are in control, proceed to step 5. Otherwise, find the assignable causes, correct them, and return to step 1. 5.Calculate x for each sample

32. 5 – 32 Steps for x - and R -Charts 6.Use Table 5.1 to determine x -chart control limits 7.Plot the sample means. If all are in control, the process is in statistical control. Continue to take samples and monitor the process. If any are out of control, find the assignable causes, correct them, and return to step 1. If no assignable causes are found, assume out-of-control points represent common causes of variation and continue to monitor the process.

33. 5 – 33 Using x - and R -Charts EXAMPLE 5.1 The management of West Allis Industries is concerned about the production of a special metal screw used by several of the company’s largest customers. The diameter of the screw is critical to the customers. Data from five samples appear in the accompanying table. The sample size is 4. Is the process in statistical control? SOLUTION Step 1:For simplicity, we use only 5 samples. In practice, more than 20 samples would be desirable. The data are shown in the following table.

34. 5 – 34 Using x - and R -Charts Data for the x - and R -Charts: Observation of Screw Diameter (in.) Observation Sample Number 1234 Rx 10.50140.50220.50090.50270.00180.5018 20.50210.50410.50240.50200.00210.5027 30.50180.50260.50350.50230.00170.5026 40.50080.50340.50240.50150.00260.5020 50.50410.50560.50340.50470.00220.5045 Average0.00210.5027 Step 2:Compute the range for each sample by subtracting the lowest value from the highest value. For example, in sample 1 the range is 0.5027 – 0.5009 = 0.0018 in. Similarly, the ranges for samples 2, 3, 4, and 5 are 0.0021, 0.0017, 0.0026, and 0.0022 in., respectively. As shown in the table, R = 0.0021.

35. 5 – 35 Using x - and R -Charts Step 3:To construct the R -chart, select the appropriate constants from Table 5.1 for a sample size of 4. The control limits are UCL R = D 4 R =2.282(0.0021) = 0.00479 in. 0(0.0021) = 0 in. Step 4:Plot the ranges on the R -chart, as shown in Figure 5.10. None of the sample ranges falls outside the control limits so the process variability is in statistical control. If any of the sample ranges fall outside of the limits, or an unusual pattern appears, we would search for the causes of the excessive variability, correct them, and repeat step 1. LCL R = D 3 R =

36. 5 – 36 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Figure 5.10 –Range Chart from the OM Explorer x and R-Chart Solver for the Metal Screw, Showing That the Process Variability Is in Control

37. 5 – 37 Using x - and R -Charts Step 5:Compute the mean for each sample. For example, the mean for sample 1 is 0.5014 + 0.5022 + 0.5009 + 0.5027 4 = 0.5018 in. Similarly, the means of samples 2, 3, 4, and 5 are 0.5027, 0.5026, 0.5020, and 0.5045 in., respectively. As shown in the table, x = 0.5027.

38. 5 – 38 Using x - and R -Charts Step 7:Plot the sample means on the control chart, as shown in Figure 5.11. The mean of sample 5 falls above the UCL, indicating that the process average is out of statistical control and that assignable causes must be explored, perhaps using a cause-and-effect diagram. LCL x = x – A 2 R = 0.5027 + 0.729(0.0021) = 0.5042 in. 0.5027 – 0.729(0.0021) = 0.5012 in. UCL x = x + A 2 R = Step 6:Now construct the x -chart for the process average. The average screw diameter is 0.5027 in., and the average range is 0.0021 in., so use x = 0.5027, R = 0.0021, and A 2 from Table 5.1 for a sample size of 4 to construct the control limits:

39. 5 – 39 Using x - and R -Charts The x -Chart from the OM Explorer x and R -Chart Solver for the Metal Screw, Showing That Sample 5 is out of Control

40. 5 – 40 An Alternate Form UCL x = x + z σ x and LCL x = x – zσ x If the standard deviation of the process distribution is known, another form of the x -chart may be used: where σ x = σ / n σ = standard deviation of the process distribution n = sample size x = central line of the chart z = normal deviate number

41. 5 – 41 Using Process Standard Deviation EXAMPLE 5.2 For Sunny Dale Bank the time required to serve customers at the drive-by window is an important quality factor in competing with other banks in the city. Mean time to process a customer at the peak demand period is 5 minutes Standard deviation of 1.5 minutes Sample size of six customers Design an x -chart that has a type I error of 5 percent After several weeks of sampling, two successive samples came in at 3.70 and 3.68 minutes, respectively. Is the customer service process in statistical control?

42. 5 – 42 Using Process Standard Deviation SOLUTION x = 5 minutes σ = 1.5 minutes n = 6 customers z = 1.96 UCL x = x + z σ/  n = LCL x = x – z σ/  n = 5.0 + 1.96(1.5)/  6 = 6.20 minutes 5.0 – 1.96(1.5)/  6 = 3.80 minutes The process variability is in statistical control, so we proceed directly to the x -chart. The control limits are

43. 5 – 43 Using Process Standard Deviation Obtain the value for z in the following way For a type I error of 5 percent, 2.5 percent of the curve will be above the UCL and 2.5 percent below the LCL From the normal distribution table (see Appendix 1) we find the z value that leaves only 2.5 percent in the upper portion of the normal curve (or 0.9750 in the table) So z = 1.96 The two new samples are below the LCL of the chart, implying that the average time to serve a customer has dropped Assignable causes should be explored to see what caused the improvement

44. 5 – 44 Application 5.1 Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped. Webster is concerned whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes are taken and each tube is weighed in ounces. Tube Number Sample12345678AvgRange 17.988.348.027.948.447.687.818.118.0400.76 28.238.127.988.418.318.187.998.068.1600.43 37.897.777.918.048.007.897.938.097.9400.32 48.248.187.838.057.908.167.978.078.0500.41 57.878.137.927.998.107.818.147.887.9800.33 68.138.148.118.138.148.128.138.148.1300.03 Avgs8.0500.38

45. 5 – 45 Application 5.1 Assuming that taking only 6 samples is sufficient, is the process in statistical control? UCL R = D 4 R = LCL R = D 3 R = 1.864(0.38) = 0.708 0.136(0.38) = 0.052 The range chart is out of control since sample 1 falls outside the UCL and sample 6 falls outside the LCL. This makes the x calculation moot. Conclusion on process variability given R = 0.38 and n = 8:

46. 5 – 46 Application 5.1 Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures. Tube Number Sample12345678AvgRange 17.988.348.027.948.447.687.818.118.0400.76 28.238.127.988.418.318.187.998.068.1600.43 37.897.777.918.048.007.897.938.097.9400.32 48.248.187.838.057.908.167.978.078.0500.41 57.878.137.927.998.107.818.147.887.9800.33 Avgs8.0340.45 What is the conclusion on process variability and process average?

47. 5 – 47 Application 5.1 UCL R = D 4 R = LCL R = D 3 R = 1.864(0.45) = 0.839 0.136(0.45) = 0.061 UCL x = x + A 2 R = LCL x = x – A 2 R = 8.034 + 0.373(0.45) = 8.202 8.034 – 0.373(0.45) = 7.832 Now R = 0.45, x = 8.034, and n = 8 The resulting control charts indicate that the process is actually in control.

48. 5 – 48 Control Charts for Attributes p -charts are used to control the proportion defective Sampling involves yes/no decisions so the underlying distribution is the binomial distribution The standard deviation is p = the center line on the chart UCL p = p + zσ p and LCL p = p – zσ p and

49. 5 – 49 Using p -Charts Periodically a random sample of size n is taken The number of defectives is counted The proportion defective p is calculated If the proportion defective falls outside the UCL, it is assumed the process has changed and assignable causes are identified and eliminated If the proportion defective falls outside the LCL, the process may have improved and assignable causes are identified and incorporated

50. 5 – 50 Using a p -Chart EXAMPLE 5.3 Hometown Bank is concerned about the number of wrong customer account numbers recorded Each week a random sample of 2,500 deposits is taken and the number of incorrect account numbers is recorded The results for the past 12 weeks are shown in the following table Is the booking process out of statistical control? Use three-sigma control limits, which will provide a Type I error of 0.26 percent.

51. 5 – 51 Using a p -Chart Sample Number Wrong Account Numbers Sample Number Wrong Account Numbers 115724 21287 319910 42 17 5191115 64123 Total147

52. 5 – 52 Using a p -Chart 147 12(2,500) = = 0.0049 p = Total defectives Total number of observations σ p =  p (1 – p )/ n =  0.0049(1 – 0.0049)/2,500 = 0.0014 UCL p = p + zσ p LCL p = p – zσ p = 0.0049 + 3(0.0014) = 0.0091 = 0.0049 – 3(0.0014) = 0.0007 Step 1:Using this sample data to calculate p

53. 5 – 53 Using a p -Chart Step 2:Calculate the sample proportion defective. For sample 1, the proportion of defectives is 15/2,500 = 0.0060. Step 3:Plot each sample proportion defective on the chart, as shown in Figure 5.12. The p -Chart from POM for Windows for Wrong Account Numbers, Showing That Sample 7 is Out of Control Fraction Defective Sample Mean UCL LCL.0091.0049.0007 |||||||||||| 123456789101112 X X X X X X X X X X X X

54. 5 – 54 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.2 A sticky scale brings Webster’s attention to whether caulking tubes are being properly capped. If a significant proportion of the tubes aren’t being sealed, Webster is placing their customers in a messy situation. Tubes are packaged in large boxes of 144. Several boxes are inspected and the following numbers of leaking tubes are found: Sample TubesSample TubesSample Tubes 13 86155 25 94160 33109172 44112186 52126192 64135201 72141Total =72

55. 5 – 55 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.2 Calculate the p -chart three-sigma control limits to assess whether the capping process is in statistical control. The process is in control as the p values for the samples all fall within the control limits.

56. 5 – 56 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts for Attributes c -charts count the number of defects per unit of service encounter The underlying distribution is the Poisson distribution The mean of the distribution is c and the standard deviation is  c UCL c = c + z  c and LCL c = c – z  c

57. 5 – 57 Using a c -Chart EXAMPLE 5.4 The Woodland Paper Company produces paper for the newspaper industry. As a final step in the process, the paper passes through a machine that measures various product quality characteristics. When the paper production process is in control, it averages 20 defects per roll. a.Set up a control chart for the number of defects per roll. For this example, use two-sigma control limits. b.Five rolls had the following number of defects: 16, 21, 17, 22, and 24, respectively. The sixth roll, using pulp from a different supplier, had 5 defects. Is the paper production process in control?

58. 5 – 58 Using a c -Chart SOLUTION a.The average number of defects per roll is 20. Therefore UCL c = c + z  c LCL c = c – z  c = 20 + 2(  20) = 28.94 = 20 – 2(  20) = 11.06 The control chart is shown in Figure 5.13

59. 5 – 59 Using a c -Chart Figure 5.13 –The c -Chart from POM for Windows for Defects per Roll of Paper b.Because the first five rolls had defects that fell within the control limits, the process is still in control. Five defects, however, is less than the LCL, and therefore, the process is technically “out of control.” The control chart indicates that something good has happened.

60. 5 – 60 Application 5.3 At Webster Chemical, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Even when the process is in control, there will still be an average of 4 lumps per tube of caulk. Testing for the presence of lumps destroys the product, so Webster takes random samples. The following are results of the study: Tube #LumpsTube #LumpsTube #Lumps 1656 95 2564100 3071119 4486122 Determine the c -chart two-sigma upper and lower control limits for this process.

61. 5 – 61 Application 5.3

62. 5 – 62 Process Capability Process capability refers to the ability of the process to meet the design specification for the product or service Design specifications are often expressed as a nominal value and a tolerance

63. 5 – 63 Process Capability 202530 Minutes Upper specification Lower specification Nominal value (a) Process is capable Process distribution The Relationship Between a Process Distribution and Upper and Lower Specifications

64. 5 – 64 Process Capability 202530 Minutes Upper specification Lower specification Nominal value (b) Process is not capable The Relationship Between a Process Distribution and Upper and Lower Specifications Process distribution

65. 5 – 65 Process Capability Figure 5.15 – Effects of Reducing Variability on Process Capability Lower specification Mean Upper specification Nominal value Six sigma Four sigma Two sigma

66. 5 – 66 Process Capability The process capability index measures how well a process is centered and whether the variability is acceptable C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ where σ = standard deviation of the process distribution

67. 5 – 67 Process Capability The process capability ratio tests whether process variability is the cause of problems C p = Upper specification – Lower specification 6σ

68. 5 – 68 Determining Process Capability Step 1.Collect data on the process output, and calculate the mean and the standard deviation of the process output distribution. Step 2.Use the data from the process distribution to compute process control charts, such as an x - and an R -chart.

69. 5 – 69 Determining Process Capability Step 3.Take a series of at least 20 consecutive random samples from the process and plot the results on the control charts. If the sample statistics are within the control limits of the charts, the process is in statistical control. If the process is not in statistical control, look for assignable causes and eliminate them. Recalculate the mean and standard deviation of the process distribution and the control limits for the charts. Continue until the process is in statistical control.

70. 5 – 70 Determining Process Capability Step 4.Calculate the process capability index. If the results are acceptable, the process is capable and document any changes made to the process; continue to monitor the output by using the control charts. If the results are unacceptable, calculate the process capability ratio. If the results are acceptable, the process variability is fine and management should focus on centering the process. If not, management should focus on reducing the variability in the process until it passes the test. As changes are made, recalculate the mean and standard deviation of the process distribution and the control limits for the charts and return to step 3.

71. 5 – 71 Assessing Process Capability EXAMPLE 5.5 The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.35 minutes The nominal value for this service is 25 minutes with an upper specification limit of 30 minutes and a lower specification limit of 20 minutes The administrator of the lab wants to have four-sigma performance for her lab Is the lab process capable of this level of performance?

72. 5 – 72 Assessing Process Capability SOLUTION The administrator began by taking a quick check to see if the process is capable by applying the process capability index: Lower specification calculation = = 1.53 26.2 – 20.0 3(1.35) Upper specification calculation = = 0.94 30.0 – 26.2 3(1.35) C pk = Minimum of [1.53, 0.94] = 0.94 Since the target value for four-sigma performance is 1.33, the process capability index told her that the process was not capable. However, she did not know whether the problem was the variability of the process, the centering of the process, or both. The options available to improve the process depended on what is wrong.

73. 5 – 73 Assessing Process Capability She next checked the process variability with the process capability ratio: The process variability did not meet the four-sigma target of 1.33. Consequently, she initiated a study to see where variability was introduced into the process. Two activities, report preparation and specimen slide preparation, were identified as having inconsistent procedures. These procedures were modified to provide consistent performance. New data were collected and the average turnaround was now 26.1 minutes with a standard deviation of 1.20 minutes. C p = = 1.23 30.0 – 20.0 6(1.35)

74. 5 – 74 Assessing Process Capability She now had the process variability at the four-sigma level of performance, as indicated by the process capability ratio: However, the process capability index indicated additional problems to resolve: C p = = 1.39 30.0 – 20.0 6(1.20) C pk =, = 1.08 (26.1 – 20.0) 3(1.20) (30.0 – 26.1) 3(1.20) Minimum of

75. 5 – 75 Application 5.4 Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces ± 0.60 ounces. The target process capability ratio is 1.33, signifying that management wants 4-sigma performance. The current distribution of the filling process is centered on 8.054 ounces with a standard deviation of 0.192 ounces. Compute the process capability index and process capability ratio to assess whether the filling process is capable and set properly.

76. 5 – 76 Application 5.4 Recall that a capability index value of 1.0 implies that the firm is producing three-sigma quality (0.26% defects) and that the process is consistently producing outputs within specifications even though some defects are generated. The value of 0.948 is far below the target of 1.33. Therefore, we can conclude that the process is not capable. Furthermore, we do not know if the problem is centering or variability. C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ = Minimum of = 1.135, = 0.948 8.054 – 7.400 3(0.192) 8.600 – 8.054 3(0.192) a.Process capability index:

77. 5 – 77 Application 5.4 b.Process capability ratio: C p = Upper specification – Lower specification 6σ = = 1.0417 8.60 – 7.40 6(0.192) Recall that if the C pk is greater than the critical value (1.33 for four-sigma quality) we can conclude that the process is capable. Since the C pk is less than the critical value, either the process average is close to one of the tolerance limits and is generating defective output, or the process variability is too large. The value of C p is less than the target for four-sigma quality. Therefore we conclude that the process variability must be addressed first, and then the process should be retested.

78. 5 – 78 Quality Engineering Quality engineering is an approach originated by Genichi Taguchi that involves combining engineering and statistical methods to reduce costs and improve quality by optimizing product design and manufacturing processes. The quality loss function is based on the concept that a service or product that barely conforms to the specifications is more like a defective service or product than a perfect one.

79. 5 – 79 Quality Engineering Loss (dollars) LowerNominalUpper specificationvaluespecification Taguchi’s Quality Loss Function

80. 5 – 80 Solved Problem 1 The Watson Electric Company produces incandescent lightbulbs. The following data on the number of lumens for 40- watt lightbulbs were collected when the process was in control. Observation Sample1234 1604612588600 2597601607603 3581570585592 4620605595588 5590614608604 a.Calculate control limits for an R -chart and an x -chart. b.Since these data were collected, some new employees were hired. A new sample obtained the following readings: 570, 603, 623, and 583. Is the process still in control?

81. 5 – 81 Solved Problem 1 Sample R 160124 260210 358222 460232 560424 Total2,991112 Average x = 598.2 R = 22.4 x 604 + 612 + 588 + 600 4 = 601 x = R = 612 – 588 = 24 SOLUTION a.To calculate x, compute the mean for each sample. To calculate R, subtract the lowest value in the sample from the highest value in the sample. For example, for sample 1,

82. 5 – 82 Solved Problem 1 The R -chart control limits are b.First check to see whether the variability is still in control based on the new data. The range is 53 (or 623 – 570), which is outside the UCL for the R -chart. Since the process variability is out of control, it is meaningless to test for the process average using the current estimate for R. A search for assignable causes inducing excessive variability must be conducted. 2.282(22.4) = 51.12 0(22.4) = 0 UCL R = D 4 R = LCL R = D 3 R = The x -chart control limits are UCL x = x + A 2 R = LCL x = x – A 2 R = 598.2 + 0.729(22.4) = 614.53 598.2 – 0.729(22.4) = 581.87