Question

In: Operations Management

A finishing process packages assemblies into boxes. You have noticed variability in the boxes and desire...

A finishing process packages assemblies into boxes. You have noticed variability in the boxes and desire to improve the process to fix the problem because some products fit too tightly into the boxes and others fit too closely. Following are width measurements of the box. Using x and R charts, plot and interpret the process.

							MEAN	CI x-bar (Grand Mean)	LCL x-bar (CI-A2*R-bar)	UCL x-bar (CI+A2*R-bar)	Sample Range (Max-Min)	R-bar (Mean)	UCL R (R-bar*D4)	Spring 2020
1	68.51	68.46	68.54	68.34	68.46	68.46
2	68.94	68.2	68.54	68.56	68.7	68.7
3	68.66	68.44	68.55	68.77	68.7	68.64
4	68.49	68.94	68.56	68.62	68.69	68.56
5	68.64	68.63	68.62	68.32	68.34	68.24
6	68.34	68.42	68.99	68.02	68.03	68.47
7	68.99	68.94	68.95	68.95	68.94	68.97
8	68.92	68.91	68.97	68.93	68.96	68.95
						GRAND MEAN
		n=
		A2=
		D4=
			info from chart

Expert Solution

n= 6
A2= 0.483
D4=2.004
D3 = 0

Formula

R-chart

CL (Center line) = R-bar = 0.4

UCL = D4*R-bar = 2.004*0.4 = 0.8016

LCL = D3*R-bar = 0

We can see the process is not in control for R chart

Graph Data

Sample Range	Center line	UCL	LCL
0.2	0.4	0.8016	0
0.74	0.4	0.8016	0
0.33	0.4	0.8016	0
0.45	0.4	0.8016	0
0.4	0.4	0.8016	0
0.97	0.4	0.8016	0
0.05	0.4	0.8016	0
0.06	0.4	0.8016	0

x-bar chart,

CL (Center line) = x-bar-bar = Grand mean = 68.63479167

UCL = x-bar-bar + A2*R-bar = 68.63479167+0.483*0.4 = 68.82799167

LCL = x-bar-bar - A2*R-bar = 68.63479167-0.483*0.4 = 68.44159167

We can see the process is not in control for x-bar chart

Graph Data

Sample	Sample mean	Center line	UCL	LCL
1	68.461667	68.635	68.828	68.442
2	68.606667	68.635	68.828	68.442
3	68.626667	68.635	68.828	68.442
4	68.643333	68.635	68.828	68.442
5	68.465	68.635	68.828	68.442
6	68.378333	68.635	68.828	68.442
7	68.956667	68.635	68.828	68.442
8	68.94	68.635	68.828	68.442

keosha answered 8 months ago

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