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Delta Plastics, Inc. (B)

Terry Lamb

Strayer University

1. Prepare a 3-sigma control chart for both production processes, using the new and standard material (use of quality report in “Delta Plastics, Inc, Case A,” Chapter 5).

The quality report for 4 weeks is given in this problem. Now the objective is to find out the process which is producing higher or lower defects. Hence the defects are summed up and used to draw the control charts. The total number of defects for the standard material, for 20 days (5 working days multiplied by 4 weeks) is given below.

The mean of the total number of defects is calculated as

Process mean = (Total defects) / (Number of samples)

Process mean = 193 / 20 = 9.65 defects

Total defects = Uneven edges + Cracks + Scratches + Air bubbles + Thickness variation

Week 1 Week 2 Week 3 Week 4

Day-1 11 Day-6 12 Day-11 12 Day-16 11

Day-2 8 Day-7 8 Day-12 7 Day-17 8

Day-3 12 Day-8 6 Day-13 8 Day-18 6

Day-4 12 Day-9 11 Day-14 6 Day-19 12

Day-5 12 Day-10 9 Day-15 13 Day-20 9

Total defects for 20 days = 193

Hence S.D = SQRT (108.55 / 19) = 2.39

Given that z = 3,

UCL = 9.65 + 3* (2.39 / SQRT (20)) = 11.25

LCL = 9.65 - 3* (2.39 / SQRT (20)) = 8.05

For super plastic:

Total defects = Uneven edges + Cracks + Scratches + Air bubbles + Thickness variation

Week 1 Week 2 Week 3 Week 4

Day-1 11 Day-6 11 Day-11 10 Day-16 15

Day-2 9 Day-7 10 Day-12 14 Day-17 16

Day-3 11 Day-8 9 Day-13 17 Day-18 12

Day-4 8 Day-9 13 Day-14 11 Day-19 15

Day-5 12 Day-10 11 Day-15 10 Day-20 15

Total defects for 20 days = 240

The mean of the total number of defects is calculated as

Process mean = (Total defects) / (Number of samples)

Process mean = 240 / 20 = 12 defects

Hence S.D = SQRT (124/ 19) = 2.55

Given that z = 3,

UCL = 12 + 3* (2.55 / SQRT (20)) = 13.71

LCL =12 - 3* (2.55 / SQRT (20)) = 10.29

2. Discuss whether or not both...