Question

Omm Lecture Exercise #11 TABLE 56.1 Factors for Computing Control Chart Limits (3 sigma) SAMPLE SIZE, MEAN FACTOR, UPPER RANGE, LOWER RANGE, n A2 D D 2 1.880 3.268 0 3 1.023 2.574 0 4 .729 2.282 0 5 .577 2.115 0 6 .483 2.004 0 7 .419 1.924 1.864 0.076 0.136 8 .373 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 R , the average of the sample ranges. 3. Construct the X. chart to check if the sample mean is in control 4. Construct the Range chart to check if process variation is in control.

Answer 1

B с D E j K 1 1 3 2 3 SAMPLE 1 SAMPLE 2 SAMPLE 3 SAMPLE 4 SAMPLE 5 2 0.5022 0.5041 0.5026 0.5034 0.5056 0.5014 0.5021 0.5018 0.5008 0.5041 4 0.5009 0.5032 0.5035 0.5024 0.5034 4 MEAN 0.5027 0.5018 0.502 0.50285 0.5023 0.50255 0.5015 0.502025 0.5039 0.50425 AVERAGE= 0.502695 MAXIMUM 0.5027 0.5041 0.5035 0.5034 0.5056 MINIMUM RANGE 0.5009 0.0018 0.502 0.0021 0.5018 0.0017 0.5008 0.0026 0.5034 0.0022 AVERAGE = 0.00208 5 6 7 8 9 10 11 X-DOUBLE BAR = AVERAGE X-DOUBLE BAR = 0.502695 NUMBER OF OBSERVATIONS IN A SAMPLE (N) = 4 UCL = X-DOUBLE BAR + (A2)x(R-bar) LCL = X-DOUBLE BAR - (A2)x(R-bar) #USE MEAN FACTOR TABLE TO FIND VALUE OF A2 ATN = 4 VALUE OF A2 = 0.729 12 R-BAR = AVERAGE R-BAR = 0.00208 NUMBER OF OBSERVATIONS IN A SAMPLE (N) = 4 LCL = (R-bar) x (D3) UCL = (R-bar) x (D4) #USE MEAN FACTOR TABLE TO FIND VALUE OF D3 and D4 AT N = 4 VALUE OF D3 = 0 VALUE OF D4 = 2.282 13 14 15 16 17 UCL = X-DOUBLE BAR + (A2)x(R-bar) UCL = 0.502695 + (0.729 X 0.00208) UCL = 0.50421 LCL = X-DOUBLE BAR - (A2)x(R-bar) LCL = 0.502695 - (0.729 X 0.00208) LCL = 0.501178 18 LCL = (R-bar) x (D3) LCL = 0.00208 X 0 = 0 LCL = 0 UCL = (R-bar) x (D4) UCL = 0.00208 X 2.282 UCL = 0.004746 19

Date Page The X-bar chart shows that last sample lies outside the control limits, so sample is not in control 0.5676 0-50441 UCL = 0.50421 0.56407 6.5038 036t 0.5032 0.5030 CL-6.502695 o.5661 0.5024 0.512a o. Sto 2014 05412 0.5olto LCL = 0.501178 2 3 4 5 Scanned by TapScanner

Page The R- chant shows that all the samples lie within the contool limits, therefore the process Variations are in control UCL -0.004746 0.005 0.0048 000046 0.0041 Orcoy27 0.001to 0.0381 0.4056 0.0034 Orap361 o.oabo 0.4648 Doopas 0.004 0.0020 ogo2o 0.0018 o apott CL=0.00208 3 ។ LCL=0 5 Scanned by TapScanner

All the parts have been done as shown.