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286 Part IV: Quality Assuranceˆ c + c + c + ... + c1 2 3l = c =mm.(19.45)Then the three-sigma control limits for the c control chart are defined as follows:UCL = c +3 c(19.46)CL = c (19.47)LCL = c 3 c(19.48)Note that for small values of c – ( 5) the Poisson distribution is asymmetric; thevalue of a type I error (a) above the UCL and below the LCL is usually notthe same. Thus, for small values of c – it may be more prudent to use probabilitycontrol limits rather than the three-sigma control limits. The probability controllimits can be found by using Poisson distribution tables.To illustrate the construction of a c control chart using three-sigma controllimits we consider the data in Table 19.7 of Example 19.9.Part IV.B.4EXAMPLE 19.9A paper mill has detected that almost 90 percent of rejected paper rolls are due to nonconformitiesof two types, holes and wrinkles in the paper. The Six Sigma Green Beltteam in the mill decided to set up control charts to reduce the number of or eliminatethese nonconformities. To set up control charts, the team decided to collect some databy taking random samples of five rolls each day for 30 days and counting the number ofnonconformities (holes and wrinkles) in each sample. The data are shown in Table 19.7.Set up a c control chart using these data.Solution:Using the data in Table 19.7, the estimate of the population parameter is given by30c iˆ i=1222l = c = = = 74 . .30 30Therefore, using equations (19.46) through (19.48), the three-sigma control limits of thephase I c control chart are given byUCL = 74 . + 3 74 . = 1556 .CL = 74 .LCL = 74 . 3 74 . = – 076 . = 0.Note that if LCL turns out to be negative, as in this example, then we set LCL at zerosince the number of nonconformities can not be negative. The c control chart for thedata in Table 19.7 is shown in Figure 19.13.Continued
Chapter 19: B. Statistical Process Control 287ContinuedTable 19.7 Total number of nonconformities in samples of five rolls of paper.Total number of Total number of Total number ofDay nonconformities Day nonconformities Day nonconformities1 8 11 7 21 92 6 12 6 22 63 7 13 6 23 84 7 14 8 24 75 8 15 6 25 66 7 16 6 26 97 8 17 8 27 98 7 18 9 28 79 6 19 8 29 710 9 20 9 30 8c Chart of Nonconformities1614UCL = 15.5612Sample count108642c – = 7.4Part IV.B.403 6 9 12 15 18 21 24 27 30SampleLCL = 0Figure 19.13 c control chart of nonconformities for the data in Table 19.7.From Figure 19.13, it is quite clear that the process is stable. In other words, thereare no special causes present and the only causes that are affecting the process arecommon causes. Thus, to eliminate the imperfections in the paper, the managementmust take action on the system, such as examining the quality of wood chips and pulp,changing old equipment, or providing more training for the workers. Also, to enhancethe process further and eliminate the nonconformities, the quality engineers shoulduse the techniques available in design of experiments.
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Continued
Table 19.7 Total number of nonconformities in samples of five rolls of paper.
Total number of Total number of Total number of
Day nonconformities Day nonconformities Day nonconformities
1 8 11 7 21 9
2 6 12 6 22 6
3 7 13 6 23 8
4 7 14 8 24 7
5 8 15 6 25 6
6 7 16 6 26 9
7 8 17 8 27 9
8 7 18 9 28 7
9 6 19 8 29 7
10 9 20 9 30 8
c Chart of Nonconformities
16
14
UCL = 15.56
12
Sample count
10
8
6
4
2
c – = 7.4
Part IV.B.4
0
3 6 9 12 15 18 21 24 27 30
Sample
LCL = 0
Figure 19.13 c control chart of nonconformities for the data in Table 19.7.
From Figure 19.13, it is quite clear that the process is stable. In other words, there
are no special causes present and the only causes that are affecting the process are
common causes. Thus, to eliminate the imperfections in the paper, the management
must take action on the system, such as examining the quality of wood chips and pulp,
changing old equipment, or providing more training for the workers. Also, to enhance
the process further and eliminate the nonconformities, the quality engineers should
use the techniques available in design of experiments.