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

2. Problem 6

3. X-bar Chart

6. A special cause occurred on the 3 rd and 4 th samples. Therefore, the X-bar chart is not in control. The process mean is unstable and, consequently, not predictable. Consequently, it is impossible to get a valid estimate of the mean. The operators are responsible for finding and correcting or institutionalizing special causes.

7. R Chart

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

11. Fraction Defective 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

12. Fraction Defective

13. Problem 8 (a)

16. 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.

17. 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.

18. 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.

19. Problem 10

20. c-Chart – Control Limits

22. Interpretation of Chart The process is in control. Each billing statement contains on average 4.4 errors, which is unacceptably high. Since the process is in control, management action is required to improve the process by reducing 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 bottles will be out of spec.

25. Problem 21 The index assumes that the process is on target. However, the process is not on target. Therefore, we should compute C T. z was computed using the NORMSDIST in Excel.. =2*NORMSDIST(G26) 26 G

26. .00003 0-4.17 z.00003 4.17 NORMSDIST(G26) = Area under curve to left of z Problem 21 USLLSL

27. Problem 21 Since the capability index is less than 1, the process is not capable. If process is center on target, the capability index would increase to 1.39, and the process would be capable.