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. Costs of Quality Prevention Costs Appraisal Costs Internal Failure Costs External Failure Costs Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 03

4. Ethics and Quality Balancing the traditional measures of quality performance and the overall benefits to society. Identifying deceptive business practices. Developing a culture around ethics. Training employees to understand how ethics interfaces with their jobs. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 -04

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

6. Customer Satisfaction Conformance to Specifications Value Fitness for Use Support Psychological Impressions Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 06

7. Employee Involvement Cultural Change Teams – Employee Empowerment – Problem-solving teams – Special-purpose teams – Self-managed teams Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 07

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

9. What is Six Sigma? Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05 - 09 Six Sigma A comprehensive and flexible system for achieving, sustaining, and maximizing business success by minimizing defects and variability in processes.

10. 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- 10

11. Six Sigma Improvement Model Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 11 Six Sigma Certification Master Black Belts Black Belts Green Belts

12. Acceptance Sampling – The application of statistical techniques to determine if a quantity of material from a supplier should be accepted or rejected based on the inspection or test of one or more samples. Acceptable Quality Level – A statement of the proportion of defective items that the buyer will accept in a shipment. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 12

13. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Interface 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 05- 13

14. 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 - 14

15. 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 - 15

16. Sampling Statistics Sample Mean Sum of the observations divided by the total observations Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 16 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

17. 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 - 17 where σ = standard deviation of a sample x i = observation of a quality characteristic (such as time) n = total number of observations = mean

18. 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 - 18

19. 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 - 19

20. Sampling Distribution Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 20

21. 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 - 21

22. Effects of Assignable Cause Variation on the Process Distribution Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 22

23. 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 - 23

24. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Control Limits and Sampling Distribution Samples Assignable causes likely 1 2 3 UCL Nominal LCL 05- 24

25. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts (a) Normal – No action 05- 25

26. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts (b) Run – Take action 05- 26

27. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts ( c) Sudden change – Monitor 05- 27

28. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts ( d) Exceeds control limits – Take action 05- 28

29. Control Chart Errors Type I error – Concluding that a process is out of control when it is in control Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 29 Type II error – Concluding that a process is in control when it is out of control

30. Control Chart Types Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 30

31. Variable Control Charts R-Chart Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 31 UCL R = D 4 R and LCL R = D 3 R

32. Variable Control Charts Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 32 UCL x = x + A 2 R and LCL x = x – A 2 R

33. Calculating Control Chart Factors Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. 05 - 33

34. 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- 34

35. 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- 35

36. 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 - 36

37. Example 5.1 Compute the range for each sample and the control limits Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 37 UCL R = D 4 R =2.282(0.0021) = 0.00479 in. 0(0.0021) = 0 in. LCL R = D 3 R =

38. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Example 5.1 Process variability is in statistical control. 05- 38

39. 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- 39

40. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Example 5.1 Process average is NOT in statistical control. 05 - 40

41. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. An Alternate Form If the standard deviation of the process distribution is known, another form of the x - chart may be used: 05 - 41

42. 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. 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? 05 - 42Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

43. 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. 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? 05 - 43

44. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall. Example 5.2 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 The new process is an improvement. 05 - 44

45. 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 05 - 45Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

46. Application 5.1 Assuming that taking only 6 samples is sufficient, is the process in statistical control? 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. 05- 46

47. 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? 05- 47Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

48. Application 5.1 1.864(0.45) = 0.839 0.136(0.45) = 0.061 8.034 + 0.373(0.45) = 8.202 8.034 – 0.373(0.45) = 7.866 The resulting control charts indicate that the process is actually in control. 05- 48

49. 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 - 49

50. 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 - 50

51. 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 - 51

52. Example 5.3 05 - 52 147 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.

53. 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 - 53

54. 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 05- 54Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

55. 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. 05- 55Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

56. 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 - 56

57. 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 - 57Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

58. 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- 58Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

59. 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- 59 b.

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

61. Application Problem 5.3 05- 61Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

62. 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 - 62

63. 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- 63

64. 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- 64

65. 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- 65

66. 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 - 66 C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ where σ = standard deviation of the process distribution

67. 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 - 67 C p = Upper specification – Lower specification 6σ

68. 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 - 68

69. Example 5.5 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 69 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

70. 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 - 70

71. 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 - 71

72. Example 5.5 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 72 C pk = Minimum of 26.1 – 20, 30 – 26.1 3 ( 1.20) 3 ( 1.20) C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ C p = 1.08 Process does not meets 4-sigma level of 1.33

73. 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. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 73

74. Application 5.4 The value of 0.948 is far below the target of 1.33. Therefore, we can conclude that the process is not capable. 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: Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 74

75. Application 5.4 b.Process capability ratio: C p = Upper specification – Lower specification 6σ = = 1.0417 8.60 – 7.40 6(0.192) 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. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.05- 75

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

77. 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 - 77

78. 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 - 78

79. 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 - 79

80. 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 - 80

81. 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 - 81

82. 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 - 82

83. 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 - 83

84. 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 - 84

85. 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 - 85

86. 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 - 86

87. 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 - 87

88. 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 - 88

89. 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. a.Design an x -chart using the process standard deviation. 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 - 89

90. Solved Problem 4 SOLUTION a.For the process standard deviation of 25 calories, the standard deviation of the sample mean is 420 + 3(10.2) = 450.6 calories 420 – 3(10.2) = 389.4 calories 05 - 90Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

91. 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 - 91

92. 05 - 92 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.