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Chapter 10

Chapter 10. Quality Control. Phases of Quality Assurance. Figure 10-1. Inspection before/after Production e.g raw materials. Corrective action during Production “Machine in Control”. Quality built into the process. Acceptance sampling. Process control. Continuous improvement.

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Chapter 10

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  1. Chapter 10 Quality Control

  2. Phases of Quality Assurance Figure 10-1 Inspection before/after Production e.g raw materials Corrective action during Production “Machine in Control” Quality built into the process Acceptance sampling Process control Continuous improvement The least progressive The most progressive

  3. Inputs Transformation Outputs Acceptance sampling Acceptance sampling Process control Inspection

  4. Inspection Costs Figure 10-3 Cost Total Cost Cost of inspection Cost of passing defectives Optimal Amount of Inspection

  5. Where to Inspect in the Process • Raw materials and purchased parts • Finished products • Before a costly operation • Before an irreversible process • Before a covering process

  6. Examples of Inspection Points Table 10-1

  7. Statistical Process Control • The Control Process • Define • Measure • Compare to a standard • Evaluate • Take corrective action • Evaluate corrective action

  8. Statistical Process Control • Variations and Control • Random variation: Natural variations in the output of process, created by countless minor factors • Assignable variation: A variation whose source can be identified (and hopefully corrected).

  9. Mean Sampling Distribution Samplingdistribution (Plot the average of points (sample) from the process distribution) Processdistribution

  10. Normal Distribution Figure 10-5 Standard deviation     Mean 95.5% 99.7%

  11. Mean Type I Error Figure 10-7 /2 /2 LCL UCL Probabilityof Type I error Type I error: Concluding a process is not in control when it actually is. Type II error: Concluding a process is in control when it is not.

  12. Control Limits Samplingdistribution Processdistribution Mean Lowercontrollimit Uppercontrollimit Notice: the control limits are of the sampling distribution and not the process.

  13. Abnormal variationdue to assignable sources Out ofcontrol UCL Mean Normal variationdue to chance LCL Abnormal variationdue to assignable sources 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sample number Control Chart Figure 10-8

  14. 1 2 3 4 Observations from Sample Distribution Figure 10-9 UCL LCL Sample number

  15. UCL x-Chart LCL UCL LCL Mean and Range Charts (process mean is shifting upward) Sampling Distribution Detects shift Does notdetect shift R-chart Mean charts allow you to check to see if the mean of a process changes.

  16. Upper control limit (UCL): = x + zsx Lower control limit (LCL): = x - zsx = where sx = s/ n sx = Standard deviation of distribution of sample means s = Process standard deviation n = Sample size z = Standard normal deviate x = Average of sample means = First of two ways to calculate mean charts Equation 10-1 = X-chart UCL LCL Use this formula when you have a good idea of the standard deviation

  17. Example 1 Sample 123 45 1 2 3 4 12.11 12.10 12.11 12.08 12.15 12.12 12.10 12.11 12.09 12.09 12.11 12.15 12.12 12.10 12.08 12.10 12.09 12.14 12.13 12.12 Observation Mean 12.10 12.12 12.11 12.10 12.12 A quality inspector took five samples, each with four observations, of the length of time to process a loan application at a credit union. The analyst computed the mean of each sample and then computed the grand mean. All values are in minutes. Use this information to obtain three-sigma (i.e., z = 3) control limits for means of future times. It is known from previous experience that the standard deviation of the process is .02 minutes. Solution 12.10 + 12.12 + 12.11 + 12.10 + 12.12 = 12.11 5 Using Formula 10-1, with z = 3, n = 4 observations per sample, and s = .02, we find

  18. In Class Exercise Pg. 469 #2 • Mean chart using standard deviation

  19. Solution

  20. = UCL = x + A2R = LCL = x - A2R where R = Average of sample ranges UCL LCL Second way to find mean charts (using range because we don’t know standard deviation) Equation 10-2 X-chart

  21. Table 10-2Factors for three-sigma control limits for x and R charts Number of FACTORS FOR R CHARTS Observations Factor for Lower Upper in Subgroup, x Chart, Control Limit, Control Limit, n A2D3D4 2 1.88 0 3.27 3 1.02 0 2.57 4 0.73 0 2.28 5 0.58 0 2.11 6 0.48 0 2.00 7 0.42 0.08 1.92 8 0.37 0.14 1.86 9 0.34 0.18 1.82 10 0.31 0.22 1.78 11 0.29 0.26 1.74 12 0.27 0.28 1.72 13 0.25 0.31 1.69 14 0.24 0.33 1.67 15 0.22 0.35 1.65 16 0.21 0.36 1.64 17 0.20 0.38 1.62 18 0.19 0.39 1.61 19 0.19 0.40 1.60 20 0.18 0.41 1.59 Source: Adapted from Eugene Grant and Richard Leavenworth, Statistical Quality Control, 5th ed. (New York: McGraw-Hill, 1980).

  22. Example 2 Twenty samples of n = 8 have been taken from a cleaning operation. The average sample range for the 20 samples was 0.016 minutes, and the average mean was 3 minutes. Determine three-sigma control limits for this process. Solution x = 3 cm R = 0.016 A2 = 0.37 for n = 8 (from Table 10-2) UCL = x + A2R= 3 + 0.37(0.016) = 3.006 minutes = = = LCL = x-A2R= 3 + 0.37(0.016) = 2.994 minutes Note that this approach assumes that the range is in control at the start.

  23. UCL Does notreveal increase x-Chart LCL UCL Range Charts (Variability) Sampling Distribution (process variability is increasing) R-chart Reveals increase LCL Range charts allow you to check to see if the variability of a process changes.

  24. UCLR = D4R LCLR = D3R UCL LCL Range Charts (Variability) Equation 10-3 where values of D3 and D4 are obtained from Table 10-2 R-chart

  25. UCLR = D4R LCLR = D3R Example 3 Twenty-five samples of n = 10 observations have been taken from a milling process. The average sample range was 0.01 centimeter. Determine upper and lower control limits for sample ranges. Solution R = 0.01 cm, n = 10 From Table 10-2, for n = 10, D4 = 1.78 and D3 = 0.22. UCLR = 1.78(0.01) = 0.0178 or 0.018 LCLR = 0.22(0.01) = 0.0022 or 0.002

  26. In Class Exercise pg 470 #4Mean and range charts

  27. solution

  28. Control Chart for Attributes • p-Chart - Control chart used to monitor the proportion of defectives in a process (count the defects). E.g., number of defects in a sample of 200. • c-Chart - Control chart used to monitor the number of defects per unit (count the defects). E.g., number of accidents per day

  29. Use of p-Charts • When observations can be placed into two categories. • Good or bad • Pass or fail • Operate or don’t operate • When the data consists of multiple samples of several observations each

  30. If p is unknown, it can beestimated from samples. That estimate, , replaces p in the preceding formulas, as illustrated in Example 4. Note: Because the formula is an approximation, it sometimes happens that the computed LCL is negative. In those instances, zero is used as the lower limit. p-Chart equations UCLp = p + zsp LCLp = p-zsp

  31. Number of Number of Sample Defectives Sample Defectives 1 14 11 8 2 10 12 12 3 12 13 9 4 13 14 10 5 9 15 11 6 11 16 10 7 10 17 8 8 12 18 12 9 13 19 10 10 10 20 16 220 Example 4 An inspector counted the number of defective computer chips in each of 20 samples. Using the following information, construct a control chart that will describe 99.74 percent of the chance variation in the process when the process is in control. Each sample contained 100 chips. Continued

  32. Solution z for 99.74 percent is 3.00 (from Appendix Table A). Total number of defectives Total number of observations 220 20(100) p = = = .11 p(1 -p) n .11(1 - .11) 100 ^ = sp = = .03 LCLp = p - z(sp) = .11 - 3.00(.03) = .02 ^ ^ Example 4(Continued) Control limits are: UCLp = p + z(sp) = .11 + 3.00(.03) = .20 Plotting the control limits and the sample fraction defective, you can see that the process is initially in control. Continued

  33. 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0 0 2 4 6 8 10 12 14 16 18 20 Example 4(Continued) UCL = .20 Fraction defective LCL = .02 Sample number

  34. In Class Exercise Pg. 470 #6p-Chart

  35. Solution

  36. Use of c-Charts • Use only when the number of occurrences per unit of measure can be counted; non-occurrences cannot be counted. • Scratches, chips, dents, or errors per item • Cracks or faults per unit of distance • Breaks or Tears per unit of area

  37. Number of Number of Sample Defects Sample Defects 1 3 10 1 2 2 11 3 3 4 12 4 4 5 13 2 5 1 14 4 6 2 15 2 7 4 16 1 8 1 17 3 9 2 18 1 45 Example 5 c-Chart Rolls of coiled wire are monitored using a c-chart. Eighteen rolls have been examined, and the number of defects per roll has been recorded in the following table. Is the process in control? Plot the values on a control chart using three standard deviation control limits.

  38. Solution c = 45/18 = 2.5 UCLc = c + 3 c = 2.5 + 3 2.5 = 7.24 LCLc = c - 3 c = 2.5 - 3 2.5 = -0.66 0 8 UCL = 7.24 6 4 Defects per unit c = 2.5 2 LCL = 0 0 0 2 4 6 8 10 12 14 16 18 Sample number Example 5 (Continued)

  39. In Class Exercise Pg. 470 #7c-Chart

  40. solution

  41. Counting Above/Below Median Runs (7 runs) B A A B A B B B A A B Counting Up/Down Runs (8 runs) U U D U D U D U U D Counting Runs

  42. Median: Up/down: OR where N = Total number of observations r = Observed number of runs with either As and Bs or Us and Ds, depending on which is involved.

  43. Sample A/B Mean U/D Sample A/B Mean U/D 1 B 10.0 -- 11 B 10.7 D 2 B 10.4 U 12 A 11.3 U 3 B 10.2 D 13 B 10.8 D 4 A 11.5 U 14 A 11.8 U 5 B 10.8 D 15 A 11.2 D 6 A 11.6 U 16 A 11.6 U 7 A 11.1 D 17 A 11.2 D 8 A 11.2 U 18 B 10.6 D 9 B 10.6 D 19 B 10.7 U 10 B 10.9 U 20 A 10.9 U A/B: 10 runs U/D: 17 runs Twenty sample means have been taken from a process. The means are shown in the following table. Use median and up/down run tests with z = 2 to determine if assignable causes of variation are present. Assume the median is 11.0. Solution The means are marked according to above/below the median and up/down. The solid lines represent the runs.

  44. Example 6 (Continued) N 20 E ( r ) = + 1 = + 1 = 11 med 2 2 2 N - 1 2 ( 20 ) - 1 E ( r ) = = = 13 u / d 3 3 The expected number of runs for each test is: The standard deviations are Although the median does not reveal any pattern, since its ztest value is less than +2, the up/down test does; its value exceeds +2. Consequently, nonrandom variations are probably present in the data and, hence, the process is not in control. The ztest values are

  45. In Class Exercise pg 472 #13

  46. LowerSpecification UpperSpecification Process variability matches specifications LowerSpecification UpperSpecification Process variability well within specifications LowerSpecification UpperSpecification Process variability exceeds specifications Process Capability Process  * 3 or 6

  47. 3 Sigma and 6 Sigma Quality Upperspecification Lowerspecification 1350 ppm 1350 ppm 1.7 ppm 1.7 ppm Processmean +/- 3 Sigma +/- 6 Sigma

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