Question

Question 4 [20 marks] By utilising Annexure A, answer the following questions: (a) 15 samples of n 8 have been taken from a cleaning operation. The average sample range for the 20 samples was 0.016 minute, and the average mean was 3 minutes. Determine the three-sigma control limits for this process. (4 marks) (b) 15 samples of n 10 observations have been taken from a milling process. The average sample range is 0.01 centimetres. Determine upper and lower control limits for sample ranges. (c) Determine which of these three processes are capable: (4 marks) (12 marks) Upper Process Mean Standard Lower Deviation Specification Specification 1 5.5 6.7 6.0 0.14 2 7.5 0.10 8.0 7.0 3 4.9 4.6 0.12 4.3 [TURN OVER 7 ANNEXURE-A: Table 1: Control Chart parameters FACTORS FOR R Number of Factor for X-bar FACTORS FORR CHARTS CHARTS Observations in Chart Upper Control Limit Subgroup A2 Lower Control Limit D4 D3 2 1.88 0 3.27 3 1.02 C 2.57 4 0.73 C 2.28 5 0.58 0 2.11 6 0.48 0 2.00 0.42 1.92 7 0.08 0.37 0.14 1.86 9 0.34 0.18 1.82 1.78 0.31 10 0.22 1.74 11 0.29 0.26 12 0.27 0.28 1.72 13 0.25 0.31 1.69 14 0.24 0.33 1.67 15 0.22 0.35 165 o ae

Answer 1

Solution (a) Alven data, Sample range as R = 0.016 and acbage mean I= 3. from the given table when the sample size n=8 then tabulated A2 = 0.37 Now we have to calculate a segna lints le UCL : I+AZR -> 370-37(0.016) -> 3.00592 UCL : 3.0059. LCL : 1 - APR >> 320-37(0.016) -> 2.99008 : LCL = 2.1 u| Recured control limites: LLCL = 2.9941 and UCL = 3.0059. (b) given dato, - n=10, awerage sample range R = 0.01 from the given table when the sample size nalo then tabulated 03= 0.22 & Du = 1.18. Now UCL = Du(R) => 1.78 (0.01) > 0.0178 .. UCL -0.0178 LCL = D3(R] => 0.22(0.00 => 0.0022. :. LCL - 6.00-2. .. Required control limit = 0.0022 (c) which of these 3 process are capable Cpk = men(mean-lower speciprcation /3.5D, Upper Specpication -- mean/3-so). ie

(2 Cpkı Geo men (6.0 - 5.5/Bro.lu), 6.7-6.0/3(0-1)) = men (1.19,167) cpk2 = min (7.5-7.0|63(0-0), 8-0-.5360.10)) = men (1.67,1.67) cpk3 a min (.6-0.31660.12)), .9-4.61360.12)) = min (0.83,0·83) Now we have to calculate the minimum of cpk values ie -> CPK l minimum value = y**.1.19 Yes it as per than 1.33 so the epk I process as capable. -> cpk 2 minimum value = 1.67. Yes it is greater than 1.33 so the cpk 2 process as capable. -> cplc 3 minimum value = 0.83. 1. No it as less than 1.33 so the cPk 3 process is not capable. Conclusion: cpk 2 is greater than 1.33 then process 2 as capable remaining two process ie cpkl and cpk 3 are not capable producing withen the specifications. and o