1 / 30

Chapter 6 Revised

Chapter 6 Revised. Homework Problems 6, 8, 10, 21. Problem 6 (a). Given Target = 12 ounces If process in control, R-bar = .6 Is process in control at these levels? n = 6 k = 5. Problem 6 (a) – X-bar Chart. Problem 6 (a) – X-bar Chart. Problem 6 (a) – X-bar Chart. X-bar Chart.

les
Download Presentation

Chapter 6 Revised

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Revised Homework Problems 6, 8, 10, 21

  2. Problem 6 (a) • Given • Target = 12 ounces • If process in control, R-bar = .6 • Is process in control at these levels? • n = 6 • k = 5

  3. Problem 6 (a) – X-bar Chart

  4. Problem 6 (a) – X-bar Chart

  5. Problem 6 (a) – X-bar Chart

  6. X-bar Chart • The means of the last three samples fall outside of the control limits. • Therefore, the X-bar chart is not in control. • The process mean is unstable and, consequently, not predictable. • Consequently, the mean is not on target. • The operators are responsible for identifying and removing the special causes responsible for the instability.

  7. X-bar Chart • Once the special causes are removed and the mean is in control, the mean will be stable but it may or may not be on target. • If it is not, management action is required to correct the mean to the target value. • The interpretation of the X-bar chart assumes that R chart is in control.

  8. Problem 6 (a) – R Chart

  9. R Chart

  10. R Chart

  11. The R chart is in control The process variance is therefore stable and predictable. The variance can be estimated as R Chart

  12. Fraction Defective (Extra) • Assume that USL = 12.5 ounces and LSL = 11.5 ounces. • If the mean was in control and centered on the target of 12 ounces, the fraction defective would be

  13. Fraction Defective (Extra)

  14. Problem 8 (a)

  15. Problem 8 (a)

  16. Problem 8 (a)

  17. Problem 8 (a) • All the sample fraction defectives fall within the control limits and form a random pattern. • The process appears in control. • We can therefore estimate the process fraction defective. • Our best estimate is the p-bar of .09.

  18. Problem 8 (a) • Thus, 9% of the tires produced are defective. • Since the process is stable, management action is required to improve the process by reducing the fraction defective. • On Day 8, no defective tires were found. • Since this point is on the LCL, it should be investigate for a special cause, which may have a favorable impact on the fraction defective.

  19. Problem 8 (b) The sample fraction defective on Day 11 falls above the UCL. The process fraction defective is Therefore out of control. The underlying special cause has an unfavorable affect on the process because it shifted the process fraction defective upward.

  20. Problem 10

  21. u-Chart – Control Limits

  22. Interpretation of Chart • The process is in control. • Each billing statement contains on average .044 errors, or less than an average of one billing error per statement. • Since there should be no billing errors, management action is required to achieve further reductions in the average number of errors.

  23. Problem 21 Since the capability index is greater than 1, the process is capable.

  24. Problem 21 Process fraction defective = .00003, or 3 out of every 100,000 packages will be outside of the specification limits.

  25. .000015 .000015 z -4.17 0 4.17 Problem 21 NORMSDIST(-4.17) = Area under curve to left of z LSL USL

  26. Problem 21 • The Cp index assumes that the process is on target. • However, the process is not on target. • The mean is 9.8 and the target is 10. • The fraction defection will therefore be greater than 3 out of 100,000. • Therefore, we should compute CT.

  27. Problem 21

  28. Problem 21 • Since the capability index is less than 1, the process is not capable. • The actual process fraction defective is .006, or 6 out of 1,000 packages. • If process is center on target, the capability index would increase to 1.39, and the process fraction defective would decrease to 3 out of 100,000 packages.

  29. Problem 21

More Related