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 50. Results are given for the last 10 samples. Sample Defectives 1 4 2 7 3 2 4 6 5 12 6 5 7 4 8 4 9 2 10 12 The type of control chart that is best to monitor this process is a) Establish the control limits to include 99.73% of the random variation in defectives. UCLA = (enter your response as a number between 0 and 1, rounded to three decimal places). LCL, = (enter your response as a number between 0 and 1, rounded to three decimal places). b) Has the process been in control? Based on the developed control limits, the number of defectives has been c) If the sample size were 25 instead, how would your limits and conclusions change? UCL = (enter your response as a number between 0 and 1, rounded to three decimal places). LCL, = (enter your response as a number between 0 and 1, rounded to three decimal places). Based on the developed control limits, the number of defectives has been

Answer 1

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Answer: P chart will be used to monitor the proportion of defects in given sample

Answer a: 99.73% control limits corresponds to 3 sigma limits

UCLp= 0.252

LCLp= 0.000

Yes, The process is in control: As we can see that NO sample is beyond control limits in charts

Explanation:

Day	Defectives	Observations
1	4	50
2	7	50
3	2	50
4	6	50
5	12	50
6	5	50
7	4	50
8	4	50
9	2	50
10	12	50
total	58	500
steps and formulas, for z=3
Proportion of defects=Pbar	total defectives/total observations	0.116
Q=	1-P	0.884
N=	average sample size	50
Standard deviation	squareroot(P*Q/N)	0.045
UCL=	P + z*squareroot(P*Q/N)	0.252
LCL=	P - z*squareroot(P*Q/N)	-0.020
	defects cannot be negative, therefore negative LCL is taken as '0'	0.000

P Chart 0.300 0.250 0.200 0.150 0.100 0.050 0.000 8 1 2 3 4 5 6 proportion of defect=defectives/observations 7 - 9 10 UCL ---LCL

Use the table below for the chart

Day	Defectives	Observations	proportion of defect=defectives/observations	P	UCL	LCL
1	4	50	0.080	0.116	0.252	0.000
2	7	50	0.140	0.116	0.252	0.000
3	2	50	0.040	0.116	0.252	0.000
4	6	50	0.120	0.116	0.252	0.000
5	12	50	0.240	0.116	0.252	0.000
6	5	50	0.100	0.116	0.252	0.000
7	4	50	0.080	0.116	0.252	0.000
8	4	50	0.080	0.116	0.252	0.000
9	2	50	0.040	0.116	0.252	0.000
10	12	50	0.240	0.116	0.252	0.000

Answer b: if sample size is 25

UCLp= 0.485

LCLp= 0.000

Yes, The process is in control: As we can see that NO sample is beyond control limits in charts

Explanation:

Day	Defectives	Observations
1	4	25
2	7	25
3	2	25
4	6	25
5	12	25
6	5	25
7	4	25
8	4	25
9	2	25
10	12	25
total	58	250
steps and formulas, for z=3
Proportion of defects=Pbar	total defectives/total observations	0.232
Q=	1-P	0.768
N=	average sample size	25
Standard deviation	squareroot(P*Q/N)	0.084
UCL=	P + z*squareroot(P*Q/N)	0.485
LCL=	P - z*squareroot(P*Q/N)	-0.021
	defects cannot be negative, therefore negative LCL is taken as '0'	0.000

P Chart 0.600 0.500 0.400 0.300 0.200 0.100 0.000- 1 2 3 4 5 6 7 8 9 10 proportion of defect=defectives/observations --P - UCL L CL

Day	Defectives	Observations	proportion of defect=defectives/observations	P	UCL	LCL
1	4	25	0.160	0.232	0.485	0.000
2	7	25	0.280	0.232	0.485	0.000
3	2	25	0.080	0.232	0.485	0.000
4	6	25	0.240	0.232	0.485	0.000
5	12	25	0.480	0.232	0.485	0.000
6	5	25	0.200	0.232	0.485	0.000
7	4	25	0.160	0.232	0.485	0.000
8	4	25	0.160	0.232	0.485	0.000
9	2	25	0.080	0.232	0.485	0.000
10	12	25	0.480	0.232	0.485	0.000