4QA3 F12 Week 3 Lecture Notes

R ucl x z d2 r lcl x z d2 n n z is

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Unformatted text preview: 96 4.99 5.08 5.07 4.96 4.96 4.99 4.89 5.01 5.03 4.99 5.08 5.09 4.99 4.98 5.00 4.97 4.96 4.99 5.01 5.02 5.05 5.08 5.03 0.08 0.12 0.08 0.14 0.13 0.10 0.14 0.11 0.15 0.10 A. Gandomi 50.09 1.15 24 ∑ R = 1.15 = 0.115 R= k 10 LCL = d3 R = 0 × 0.115 = 0 UCL = d 4 R = 2.11 × 0.115 = 0.243 0.28 – 0.24 – Range 0.20 – 0.16 – UCL = 0.243 R = 0.115 0.12 – 0.08 – 0.04 – 0 – 4QA3 F12 LCL = 0 | | | | | | 1 2 3 4 5 6 Sample number | 7 | 8 | 9 | 10 25 To test if the mean of a process is stable. ●  ●  The method relies on the fact that averages of observations tend towards the normal distribution by the Central Limit Theorem. 4QA3 F12 A. Gandomi 26 1.  2.  3.  4.  5.  4QA3 F12 Form Subgroups of data of size 4 or 5 typically. Average observations within a subgroup. Maintain a running graph on the subgroup averages. De3ine UCL and LCL as the upper and lower control limits. They are of the form: !R$ UCL = X + z # & " d2 % !R$ LCL = X − z # & " d2 % n n z is typically 3. A. Gandomi 27 5.10 – 5.08 – 5.06 – ∑ X = 50.1 = 5.01 X= 5.04 – 10 Mean k UCL = 5.08 UCL = 5.01 + (3) ( 0.115 / 2.326 ) 5 = 5.08 LCL = 5.01 − (3) ( 0.115 / 2.326 ) 5 = 4.94 = x = 5.01 5.02 – 5.00 – 4.98 – 4.96 – LCL = 4.94 4.94 – 4.92 – | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | 9 10 Sample number 4QA3 F12 A. Gandomi 28 Average Transaction Time (4 samples each) (minutes) 14.0 UCL =13.1 13.0 12.0 X = 11.2 11.0 10.0 LCL = 9.3 9.0 8.0 Range within subgroup (m...
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