Ten samples of 15 parts each were taken from an ongoing process to establish a p-chart for control. The samples and the number of defectives in each are shown in the following table: |
SAMPLE | n | NUMBER OF DEFECTIVE ITEMS IN THE SAMPLE | |||
1 | 15 | 0 | |||
2 | 15 | 2 | |||
3 | 15 | 0 | |||
4 | 15 | 3 | |||
5 | 15 | 1 | |||
6 | 15 | 3 | |||
7 | 15 | 1 | |||
8 | 15 | 0 | |||
9 | 15 | 0 | |||
10 | 15 | 0 | |||
a. |
Determine the p−p−, Sp, UCL and LCL for a p-chart of 95 percent confidence (1.96 standard deviations). (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 3 decimal places.) |
p−p− | |
Sp | |
UCL | |
LCL | |
b. | What comments can you make about the process? | ||||
|
Solution:
Z = 1.96
Number of samples, N = 10
Sample size, n = 15
(a) P = Total number of defective items in the sample / (Number of samples x Sample size)
P = (2 + 3 + 1 + 3 + 1) / (10 x 15)
P = 0.067
Sp = SQRT [P x (1 - P)] / n
Sp = SQRT [0.067 x (1 - 0.067)] / 15
Sp = 0.065
UCL = P + [Z x Sp]
UCL = 0.067 + [1.96 x 0.065]
UCL = 0.194
LCL = P - [Z x Sp]
LCL = 0.067 - [1.96 x 0.065]
LCL = - 0.060 or 0 (Number of defects cannot be negative)
LCL = 0
(b) Since, the number of defective items in few of the samples is more than the Upper Control Limit (UCL = 0.194), the process is out of statistical control.
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