Question

Management at Webster Chemical Company is concerned as to whether caulking tubes are being properly capped. If a significant proportion of the tubes are not being​ sealed, Webster is placing its customers in a messy situation. Tubes are packaged in large boxes of

145145.

Several boxes are​ inspected, and the following numbers of leaking tubes are​ found:

                                                                                                                        

Sample	Tubes	Sample	Tubes	Sample	Tubes
1	11	8	22	15	55
2	99	9	88	16	44
3	44	10	22	17	11
4	99	11	11	18	44
5	88	12	55	19	55
6	22	13	66	20	22
7	66	14	22	Total	8686

Calculate​ p-chart

threethree​-sigma

control limits to assess whether the capping process is in statistical control.The

UCL Subscript pUCLp

equals

nothing

and the

LCL Subscript pLCLp

equals

nothing.

​(Enter

your responses rounded to three decimal

places.

If your answer for

LCL Subscript pLCLp

is​ negative, enter this value as

0.​)

Is the capping process in​ control?

A.

All of the sample proportions are within the control​ limits, so the capping process is not in control.

B.

All of the sample proportions are within the control limitsAll of the sample proportions are within the control limits​,

so the capping process

isis

in control.

C.

At least one of the sample proportions is below the lower control limit and at least one of the sample proportions is above the upper control​ limit, so the capping process is not in control.

D.

At least one of the sample proportions is above the upper control limitAt least one of the sample proportions is above the upper control limit​,

so the capping process

is notis not

in control.

E.

At least one of the sample proportions is below the lower control​ limit, so the capping process is not in control.

Answer 1

Answer: UCLp= 0.072, LCLp= 0.000

Explanation:

Sample	defectives	observation
1	1	145
2	9	145
3	4	145
4	9	145
5	8	145
6	2	145
7	6	145
8	2	145
9	8	145
10	2	145
11	1	145
12	5	145
13	6	145
14	2	145
15	5	145
16	4	145
17	1	145
18	4	145
19	5	145
20	2	145
total	86	2900
steps and formulas,z= 3 sigma process
Proportion of defects=P=	total defective/total observations	0.030
Q=	1-P	0.970
N=	average sample size	145
Standard deviation	squareroot(P*Q/N)	0.014
UCL=	P + z*squareroot(P*Q/N)	0.0719
LCL=	P - z*squareroot(P*Q/N)	-0.0126
	defects cannot be negative, therefore negative LCL is taken as '0'	0.00

Answer b: All of the sample proportions are within the control​ limits, so the capping process is not in control.

Chart Pbar 0.080 0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 1 17 2 3 4 5 6 7 8 9 10 11 12 proportion-defectives/observations 13 14 UCL 15 16 ---LCL 18 - 19 20 P

Use the table below for the charts

Sample	defectives	observation	proportion= defectives/observations	UCL	LCL	P
1	1	145	0.007	0.072	0.000	0.030
2	9	145	0.062	0.072	0.000	0.030
3	4	145	0.028	0.072	0.000	0.030
4	9	145	0.062	0.072	0.000	0.030
5	8	145	0.055	0.072	0.000	0.030
6	2	145	0.014	0.072	0.000	0.030
7	6	145	0.041	0.072	0.000	0.030
8	2	145	0.014	0.072	0.000	0.030
9	8	145	0.055	0.072	0.000	0.030
10	2	145	0.014	0.072	0.000	0.030
11	1	145	0.007	0.072	0.000	0.030
12	5	145	0.034	0.072	0.000	0.030
13	6	145	0.041	0.072	0.000	0.030
14	2	145	0.014	0.072	0.000	0.030
15	5	145	0.034	0.072	0.000	0.030
16	4	145	0.028	0.072	0.000	0.030
17	1	145	0.007	0.072	0.000	0.030
18	4	145	0.028	0.072	0.000	0.030
19	5	145	0.034	0.072	0.000	0.030
20	2	145	0.014	0.072	0.000	0.030

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