Submitted by: Submitted by chocolatecandi83
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Category: Business and Industry
Date Submitted: 08/21/2011 12:58 PM
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...