Question

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

Determine the p−p− , S_p, 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 bar Sp UCL LCL

Also what comments can you make about the process?

a.Process is in statistical control.

b.Process is out of statistical control.

Answer 1

Sample n number of defectives fraction defectives

1 15 0 0

2 15 2 0.133

3 15 0 0

4 15 3 0.200

5 15 1 0.067

6 15 3 0.200

7 15 1 0.067

8 15 0 0

9 15 0 0

10 15 0 0

(Note : fraction defectives = number of defectives/n)

Sample size(n) = 15

Number of samples = 10

Total number of observation (n ) = Sample size x number of observation = 15 x 10 = 150

Sum of number of defectives(np)= 0+2+0+3+1+3+1+0+0+0 = 10

At 95% confidence value of Z = 1.96

P-bar = np/ n = 10/150 = 0.067

Sp = √{[P-bar(1-P-bar)] / n}

= √ {[0.067(1-0.067)] / 15}

= √ [(0.067 x 0.933) /15]

= √ (0.062511/15)

= √0.0041674

= 0.065

UCL = P-bar + Z(Sp) = 0.067 + (1.96x0.065) = 0.067 + 0.1274 = 0.194

LCL = P-bar - Z(Sp) = 0.067 - (1.96x0.065) = 0.067 - 0.1274 = -0.060 = 0 (when the LCL value is negative it is taken as 0)

b) The process is not in statistical control because the fraction defectives for sample 4 and 6 are 0.200 which is beyond the upper control limit of 0.194