Question

TABLE 56.1 SAMPLE SIZE, n Lecture Exercise #11 2 3 Factors for Computing Control Chart Limits (3 sigma) MEAN FACTOR, UPPER RANGE, LOWER RANGE, A DA D 1.880 3.268 0 1.023 2.574 0 .729 2.282 0 .577 2.115 0 .483 2.004 0 .419 1.924 0.076 .373 1.864 0.136 4 5 6 7 8 9 .337 1.816 0.184 10 .308 1.777 0.223 12 .266 1.716 0.284 We wish to determine if screw production is in statistical control. We have no prior information, other than the 5 samples (where each sample has 4 observations) Number 2 3 4 Sample 1 0.5014 0.5022 0.5009 0.5027 Sample 2 0.5021 0.5041 0.5032 0.5020 Sample 3 0.5018 0.5026 0.5035 0.5023 Sample 4 0.5008 0.5034 0.5024 0.5015 Sample 5 0.5041 0.5056 0.5034 0.5039 1 1. Compute the sample range and mean for each of the 5 samples. 2. Compute x , the average of the sample means, and the average of the sample ranges. 3. Construct theX chart to check if the sample mean is in control 4. Construct the Range chart to check if process variation is in

Answer 1

Question – 1:

Number	1	2	3	4	Mean	Range
Sample 1	0.5014	0.5022	0.5009	0.5027	0.5018	0.0018
Sample 2	0.5021	0.5041	0.5032	0.502	0.50285	0.0021
Sample 3	0.5018	0.5026	0.5035	0.5023	0.50255	0.0017
Sample 4	0.5008	0.5034	0.5024	0.5015	0.502025	0.0026
Sample 5	0.5041	0.5056	0.5034	0.5039	0.50425	0.0022

Question – 2:

x-bar-bar = average of all sample means = (0.5018 + 0.50285 + 0.50255 + 0.502025 + 0.50425)/5 = 0.502695

R-bar = (0.0018 + 0.0021 + 0.0017 + 0.0026 + 0.0022)/5 = 0.00208

Question – 3:

Here,

n = 4

A2 = 0.729

UCLx = x-bar-bar + A2*R-bar = 0.502695 + 0.729*0.00208 = 0.50421132

LCLx = x-bar-bar – A2*R-bar = 0.502695 – 0.729*0.00208 = 0.501179

As per mean chart, the process is not in control as observed data are not bounded with UCL and LCL

Question – 4:

Here,

n = 4

D3 = 0

D4 = 2.282

UCL-R = D4*R-bar = 2.282*0.00208 = 0.004747

LCL-R = D3*R-bar = 0

As per Range chart, the process is in control as observed data are not bounded with UCL and LCL

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