Ten samples of 15 parts each were taken from an on going pro

Ten samples of 15 parts each were taken from an on going process to establish a p-chart for control. The samples and the number of defectives in each are shown in the following table:

Solution

Sample   Number  Number of Defective    Proportion Defective

1          15           3                       0.30

2          15           1                       0.10

3          15           0                        0

4          15           0                        0

5          15           0                        0

6          15           2                       0.20     

7          15           0                         0

8          15           3                       0.30

9          15           1                       0.10

10         15           0                         0

Number of Defective is 10

10 samples each contain 15 parts

P=  Total defectives / Total samples observations

10 / 10 *15

=0.07

Sp =   p * (1- p)/n

= 0.07 * 1-0.07/ 15

=0.0659

The control limits are computed as follows:

UCL= p+  z (p(1-p)/ n)

=0.07 +1.96 (0.07(1-0.07/15)

=0.07 +1.96 *0.0659

=0.07+0.129

=0.199

LCL =  p - z (p(1-p)/ n)

=0.07  - 1.96 (0.07(1-0.07/15)

=0.07  - 1.96 *0.0659

=0.07 - 0.129

=0.059

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b) Process is in statistical control

Process is in statistical control because  lower limit  showing  5 out of 10 samples are 0 in number

Even Higher limit showing only 3 are defective.

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