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2 votes
You are attempting to develop a quality monitoring

system for some parts purchased from Charles Sox Manufacturing
Co. These parts are either good or defective. You have decided
to take a sample of 100 units. Develop a table of the appropriate upper and lower control chart limits for various values of the
average fraction defective in the samples taken. The values for p
in this table should range from 0.02 to 0.10 in increments of 0.02.
Develop the upper and lower control limits for a 99.73% confidence level. p = 0.2; 0.4; 0.6; 0.8; 0.10

1 Answer

3 votes

To develop the upper and lower control limits for a 99.73% confidence level, we can use the formulas for the control limits in a p-chart. The control limits are typically calculated based on the average fraction defective (p) and the sample size (n).

The formulas for the upper control limit (UCL) and lower control limit (LCL) in a p-chart are as follows:

UCL = p + 3 * sqrt((p*(1-p))/n)

LCL = p - 3 * sqrt((p*(1-p))/n)

We will calculate the UCL and LCL for each value of p ranging from 0.02 to 0.10 in increments of 0.02, with a sample size of 100.

For p = 0.02:

UCL = 0.02 + 3 * sqrt((0.02*(1-0.02))/100) = 0.0274

LCL = 0.02 - 3 * sqrt((0.02*(1-0.02))/100) = 0.0126

For p = 0.04:

UCL = 0.04 + 3 * sqrt((0.04*(1-0.04))/100) = 0.0484

LCL = 0.04 - 3 * sqrt((0.04*(1-0.04))/100) = 0.0316

For p = 0.06:

UCL = 0.06 + 3 * sqrt((0.06*(1-0.06))/100) = 0.0674

LCL = 0.06 - 3 * sqrt((0.06*(1-0.06))/100) = 0.0526

For p = 0.08:

UCL = 0.08 + 3 * sqrt((0.08*(1-0.08))/100) = 0.0864

LCL = 0.08 - 3 * sqrt((0.08*(1-0.08))/100) = 0.0736

For p = 0.10:

UCL = 0.10 + 3 * sqrt((0.10*(1-0.10))/100) = 0.1044

LCL = 0.10 - 3 * sqrt((0.10*(1-0.10))/100) = 0.0956

The table of upper and lower control chart limits for the given values of p is as follows:

p UCL LCL

0.02 0.0274 0.0126

0.04 0.0484 0.0316

0.06 0.0674 0.0526

0.08 0.0864 0.0736

0.10 0.1044 0.0956

These values represent the upper and lower control limits for a p-chart with a sample size of 100 and a 99.73% confidence level.

answered
User SuhasD
by
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