Bus 515 - Operations Management - Delta Plastics

DELTA (B) 1

Delta Plastics, Inc. (B)

Veronica Garcia

Professor: Diana Echols

BUS 515 Operations Management

Strayer University

November 13, 2011

DELTA (B) 2

DELTA PLASTICS, INC (B)

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).

Delta Plastics has already created a quality report that will be used to create the 3-sigma control chart for both production processes using the standard material and the new material (super plastic). The quality report is done for 4 weeks, totaling 20 days; 5 days per week. Both materials are tested for 5 different defects including uneven edges, cracks, scratches, air bubbles, and thickness variation. Now the objective is to find out which process has the highest and lowest level of defects. These defects are summed up and used to create the 3-sigma control chart.

First is the standard material. Taking the quality report previously provided, the total number of defects for the standard material, for 20 days, is measured by using an X-bar chart. The mean of the total number of defects is calculated as process mean = (Total defects) / (Number of samples), and for this material the process mean = 193 / 20 = 9.65 defects. The total number of defects for the 20 days is 193. The rest of the chart information is as follows:

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

Now for the new material, the super plastic, the same 5 defects are tested and the same amounts of samples were tested; 20 days total 5 days per week for 4 weeks. The mean of the total number of defects is calculated as process mean = (Total defects) / (Number of samples)

DELTA (B) 3

and the process mean for this material = 240 / 20 = 12 defects. The total number of defects for 20 days is 240 defects. The rest of the chart information for the super material is as follows:

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