Question

Jamison Kovach Supply Company manufactures paper clips and other office products. Although​ inexpensive, paper clips have provided the firm with a high margin of profitability. Sample size is 40. Results are given for the last 10 samples Sample 1 2 3 4 5 6 7 8 9 10 Defectives 4 7 3 6 6 5 4 4 2 12 Establish the control limits to include​ 99.73% of the random variation in defectives. UCL= LCL= Has the process been in​ control? Based on the developed control​ limits, the number of defectives has been?

Answer 1

Answer: 99.73% explains the requirement of 3 sigma control limits

Day	Defective	Observations	proportion of defect= defectives/observations
1	4	40	0.100
2	7	40	0.175
3	3	40	0.075
4	6	40	0.150
5	6	40	0.150
6	5	40	0.125
7	4	40	0.100
8	4	40	0.100
9	2	40	0.050
10	12	40	0.300
total	53	400

steps and formulas
Probability of defects=P=	total defectives /total observations			0.1325
Q=	1-P			0.8675
N=	average sample size			40
UCL=	P + 3*squareroot(P*Q/N)			0.293
LCL=	P - 3*squareroot(P*Q/N)			-0.0283
	defects cannot be negative, therefore negative LCL is taken as '0'			0.000
Standard deviation		squareroot(P*Q/N)		0.0536

• 10th sample is out of control. As the proportion of defectives is 0.300 in 10th sample and UCL is 0.293. so, 10th sample is beyond UCL
• Total no. of defectives have been 53.

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