1. Quality and Performance Chapter 5 05 - 01 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

2. What is Quality? Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05 - 02 Quality A term used by customers to describe their general satisfaction with a service or product.

3. Customer Satisfaction Conformance to Specifications Value Fitness for Use Support Psychological Impressions Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 03

4. Determinants of Service Quality ReliabilityConsistency of performance and dependability ResponsivenessWillingness or readiness of employees CompetenceRequired skills and knowledge AccessApproachability and ease of contact CourtesyPoliteness, respect, consideration, friendliness CommunicationKeeping customers informed CredibilityTrustworthiness, believability, honesty SecurityFreedom from danger, risk, or doubt Understanding/ knowing the customer Understand the customer’s needs TangiblesPhysical evidence of the service

5. International Quality Documentation Standards ISO 9001:2008 – Quality Standards ISO 14000:2004 – Environmental Management Standards ISO 26000:2010 – Social Responsibility Guidelines Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 5

6. Baldridge Performance Excellence Program Leadership Strategic Planning Customer Focus Measurement, Analysis, and Knowledge Management Workforce Focus Operations Focus Results Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 6

7. Total Quality Management Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 07 TQM A philosophy that stresses principles for achieving high levels of process performance and quality.

8. Continuous Improvement Kaizen Problem-solving tools PDSA Cycle Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 08

9. Costs of Quality Prevention Costs Appraisal Costs Internal Failure Costs External Failure Costs Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 09

10. 10 Failure (100) Correction (10) Prevention (1) $ $ $ $ $ $ $ $ $ “ An ounce of prevention is worth of a pound of cure”

11. 11 COST OF QUALITY “Defects are not free. Somebody makes them, and gets paid for making them”. W. Edwards Deming “Quality is free. It is not a gift, but it is free. What costs money are the unquality things-all the actions that involve not doing jobs right the first time.” Philip Crosby

12. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Six Sigma Approach 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 05- 12

13. Six Sigma Improvement Model Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 13 Six Sigma Certification Master Black Belts Black Belts Green Belts

14. 14 Quality Control Statistical Process Control Acceptance Sampling Variables Charts Attributes Charts Types of Quality Control 100% inspection

15. 15 Producer’s & Consumer’s Risk Producer's risk (  )- Type 1 Error – Probability of rejecting a good lot, probability that a good product will be rejected Consumer's risk (ß)- Type 2 Error – Probability of a non-corforming product will be available for sale, probability of accepting a bad lot

16.  Variability is inherent in every process  Natural or common causes  Special or assignable causes  Provides a statistical signal when assignable causes are present  Detect and eliminate assignable causes of variation Statistical Process Control (SPC)

17. SPC Applied to Services Fast-food restaurants – waiting time for service, customer complaints, cleanliness, food quality, order accuracy, employee courtesy Catalogue-order companies – order accuracy, operator knowledge and courtesy, packaging, delivery time, phone order waiting time Insurance companies – billing accuracy, timeliness of claims processing, agent availability and response time

18. SPC Applied to Services (cont.) Hospitals – timeliness and quickness of care, staff responses to requests, accuracy of lab tests, cleanliness, courtesy, accuracy of paperwork, speed of admittance and checkouts Grocery stores – waiting time to check out, frequency of out-of-stock items, quality of food items, cleanliness, customer complaints, checkout register errors Airlines – flight delays, lost luggage and luggage handling, waiting time at ticket counters and check-in, agent and flight attendant courtesy, accurate flight information, passenger cabin cleanliness and maintenance

19. Statistical Process Control (SPC) SPC The application of statistical techniques to determine whether a process is delivering what the customer wants. Performance Measurements – Variables - Characteristics that can be measured. – Attributes - Characteristics that can be counted. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 19

20. Sampling Sampling Plan – Size of the sample – Time between successive samples – Decision rules that determine when action should be taken Complete Inspection – Inspect each product at each stage Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 20

21. Sampling Statistics Sample Mean Sum of the observations divided by the total observations Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 21 where x i = observation of a quality characteristic (such as time) n = total number of observations = mean Sample Range Difference between the largest and smallest observation in a sample

22. Sampling Statistics Standard deviation– The square root of the variance of a distribution. An estimate of the process standard deviation based on a sample is given by: Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 22 where σ = standard deviation of a sample x i = observation of a quality characteristic (such as time) n = total number of observations = mean

23. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Sampling Statistics 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 05 - 23

24. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Sampling Statistics 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 05 - 24

25. 25 Properties of normal distribution Theoretical Basis of Control Charts x within fall x lal of 68.2% 

26. Sampling Distribution Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 26

27. Types of Variation Common cause – Variation that is random, unidentifiable and unavoidable Assignable cause – Variation that can be identified and eliminated Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 27

28. Effects of Assignable Cause Variation on the Process Distribution Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 28

29. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Time-ordered diagram used to determine whether observed variations are abnormal – 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 05 - 29

30. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts (a) Normal – No action 05- 30

31. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts (b) Run – Take action 05- 31

32. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts ( c) Sudden change – Monitor 05- 32

33. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts ( d) Exceeds control limits – Take action 05- 33

34. © 2011 Pearson Education, Inc. publishing as Prentice Hall Upper control limit Target Lower control limit Patterns in Control Charts One plot out above (or below). Investigate for cause. Process is “out of control.”

35. © 2011 Pearson Education, Inc. publishing as Prentice Hall Upper control limit Target Lower control limit Patterns in Control Charts Trends in either direction, 5 plots. Investigate for cause of progressive change.

36. © 2011 Pearson Education, Inc. publishing as Prentice Hall Upper control limit Target Lower control limit Patterns in Control Charts Two plots very near lower (or upper) control. Investigate for cause.

37. © 2011 Pearson Education, Inc. publishing as Prentice Hall Upper control limit Target Lower control limit Patterns in Control Charts Run of 5 above (or below) central line. Investigate for cause.

38. © 2011 Pearson Education, Inc. publishing as Prentice Hall Upper control limit Target Lower control limit Patterns in Control Charts Erratic behavior. Investigate.

39. Control Chart Types Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 39

40. Copyright © 2013 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 UCL c = c + z  c and LCL c = c – z  c 05 - 40

41. 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? 05 - 41Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

42. Example 5.4 a. The average number of defects per roll is 20. Therefore LCL c = c – z  c = 20 + 2(  20) = 28.94 = 20 – 2(  20) = 11.06 05- 42Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

43. Example 5.4 The process is technically out of control due to Sample 6. However, Sample 6 shows that the new supplier is a good one. 05- 43 b.

44. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 3 The Minnow County Highway Safety Department monitors accidents at the intersection of Routes 123 and 14. Accidents at the intersection have averaged three per month. a. Which type of control chart should be used? Construct a control chart with three sigma control limits. b.Last month, seven accidents occurred at the intersection. Is this sufficient evidence to justify a claim that something has changed at the intersection? 05 - 44

45. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 3 SOLUTION a.The safety department cannot determine the number of accidents that did not occur, so it has no way to compute a proportion defective at the intersection. Therefore, the administrators must use a c -chart for which There cannot be a negative number of accidents, so the LCL in this case is adjusted to zero. b.The number of accidents last month falls within the UCL and LCL of the chart. We conclude that no assignable causes are present and that the increase in accidents was due to chance. UCL c = c + z c LCL c = c – z c = 3 + 3 3 = 8.20 = 3 – 3 3 = –2.196 05 - 45

46. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 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 and 05 - 46

47. 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 Using three-sigma control limits, which will provide a Type I error of 0.26 percent, is the booking process out of statistical control? Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 47

48. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Example 5.3 Sample Number Wrong Account Numbers Sample Number Wrong Account Numbers 115724 21287 319910 42 17 5191115 64123 Total147 05 - 48

49. Example 5.3 05 - 49 147 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 Calculate the sample proportion defective and plot each sample proportion defective on the chart. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

50. Example 5.3 Fraction Defective Sample Mean UCL LCL.0091.0049.0007 |||||||||||| 123456789101112 X X X X X X X X X X X X The process is NOT in statistical control. 05 - 50

51. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 The data processing department of the Arizona Bank has five data entry clerks. Each working day their supervisor verifies the accuracy of a random sample of 250 records. A record containing one or more errors is considered defective and must be redone. The results of the last 30 samples are shown in the table. All were checked to make sure that none was out of control. 05 - 51

52. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 Sample Number of Defective Records Sample Number of Defective Records 17168 251712 319184 410196 5112011 682117 7122212 89236 96247 10132513 11182610 1252714 1316286 1442911 1511309 Total 300 05 - 52

53. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 a.Based on these historical data, set up a p -chart using z = 3. b.Samples for the next four days showed the following: SampleNumber of Defective Records Tues17 Wed15 Thurs22 Fri21 What is the supervisor’s assessment of the data- entry process likely to be? 05 - 53

54. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 SOLUTION a.From the table, the supervisor knows that the total number of defective records is 300 out of a total sample of 7,500 [or 30(250)]. Therefore, the central line of the chart is = 0.04 300 7,500 p = The control limits are 05 - 54

55. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 Sample Number of Defective Records Proportion Tues170.068 Wed150.060 Thurs220.088 Fri210.084 b.Samples for the next four days showed the following: Samples for Thursday and Friday are out of control. The supervisor should look for the problem and, upon identifying it, take corrective action. 05 - 55

56. Variable Control Charts Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 56 UCL x = x + A 2 R and LCL x = x – A 2 R

57. Variable Control Charts R-Chart Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 57 UCL R = D 4 R and LCL R = D 3 R

58. Calculating Control Chart Factors Table 5.1, page 171 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 58

59. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 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. 05- 59

60. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 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. 05- 60

61. 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? Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 61

62. Example 5.1 Compute the range for each sample and the control limits Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 62 UCL R = D 4 R =2.282(0.0021) = 0.00479 in. 0(0.0021) = 0 in. LCL R = D 3 R =

63. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Example 5.1 Process variability is in statistical control. 05- 63

64. Example 5.1 Compute the mean for each sample and the control limits. 0.5027 + 0.729(0.0021) = 0.5042 in. 0.5027 – 0.729(0.0021) = 0.5012 in. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 64

65. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Example 5.1 Process average is NOT in statistical control. 05 - 65

66. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 The Watson Electric Company produces incandescent light bulbs. The following data on the number of lumens for 40-watt light bulbs were collected when the process was in control. Observation Sample1234 1604612588600 2597601607603 3581570585592 4620605595588 5590614608604 05 - 66

67. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 Sample R 160124 260210 358222 460232 560424 Total2,991112 Average x = 598.2 R = 22.4 604 + 612 + 588 + 600 4 = 601 x = R =612 – 588 = 24 05 - 67

68. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 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 The x -chart control limits are 598.2 + 0.729(22.4) = 614.53 598.2 – 0.729(22.4) = 581.87 05 - 68

69. Process Capability Process Capability – The ability of the process to meet the design specification for a service or product – Nominal Value – Tolerance Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 69

70. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 202530 Minutes Upper specification Lower specification Nominal value (a) Process is capable Process distribution 05- 70

71. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 202530 Minutes Upper specification Lower specification Nominal value (b) Process is not capable Process distribution 05- 71

72. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability Lower specification Mean Upper specification Nominal value Six sigma Four sigma Two sigma 05- 72

73. Process Capability Index Measures how well a process is centered and whether the variability is acceptable Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 73 C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ where σ = standard deviation of the process distribution

74. 74 Process Capability Index

75. Process Capability Ratio A test to see if the process variability is capable of producing output within a product’s specifications. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 75 C p = Upper specification – Lower specification 6σ

76. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Quality Loss Function Loss (dollars) LowerNominalUpper specificationvaluespecification 05- 76

77. 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 + 5 minutes. Is the lab process capable of four sigma-level performance? Upper specification = 30 minutes Lower specification 20 minutes Average service 26.2 minutes  = 1.35 minutes Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05 - 77

78. Example 5.5 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 78 C pk = Minimum of 26.2 – 20, 30 – 26.2 3 ( 1.53) 3( 1.53) C pk = Minimum of, Process does not meets 4-sigma level of 1.33 C pk = Minimum of 1.53, 0.94 C pk = 0.94

79. Example 5.5 C p = Upper Specification – Lower Specification 6  C p = 30 – 20 = 1.23 6 (1.35) Process did not meet 4-sigma level of 1.33 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05 - 79

80. Example 5.5 C p = Upper - Lower 66 C p = 30 – 20 = 1.39 6 (1.20) Process meets 4-sigma level of 1.33 for variability New Data is collected: Upper specification = 30 minutes Lower specification 20 minutes Average service 26.1 minutes  = 1.20 minutes Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05 - 80

81. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 4 Pioneer Chicken advertises “lite” chicken with 30 percent fewer calories. (The pieces are 33 percent smaller.) The process average distribution for “lite” chicken breasts is 420 calories, with a standard deviation of the population of 25 calories. Pioneer randomly takes samples of six chicken breasts to measure calorie content. b.The product design calls for the average chicken breast to contain 400 ± 100 calories. Calculate the process capability index (target = 1.33) and the process capability ratio. Interpret the results. 05 - 81

82. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 4 Because the process capability ratio is 1.33, the process should be able to produce the product reliably within specifications. However, the process capability index is 1.07, so the current process is not centered properly for four-sigma performance. The mean of the process distribution is too close to the upper specification. The process capability ratio is b. The process capability index is C pk = Minimum of, = Minimum of = 1.60, = 1.07 420 – 300 3(25) 500 – 420 3(25) C p = Upper specification – Lower specification 6σ = = 1.33 500 – 300 6(25) 05 - 82