Langston Dunlap_9.docx - Langston Dunlap Dr Wickramage MGMT 3016 26 November 2019 Assignment 9 1 A a flow chart would be generally valuable in this

Langston Dunlap_9.docx - Langston Dunlap Dr Wickramage MGMT...

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Langston Dunlap Dr. Wickramage MGMT 3016 26 November 2019 Assignment 9 1. A) a flow chart would be generally valuable in this circumstance. By distinguishing the reason they can discover an answer. It is valuable when discovering quality issues. b) A histogram is the most valuable since it would best speak to this data. c) A cause and effect chart would work. This is on the grounds that they need to discover the reason for the changes. 2. A) Total defects = 9 9/1800= .005 *1000000= 5000 EMPO= 5000/2= 2500 The empo is 2500.
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b) The sigma level is 4
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4. 5. a) mean=8.05, average range=R-bar=0.38 x-bar chart LCL= _bar-A2* =8.05-0.373*0.38=7.9083 UCL= _bar+A2* =8.05+0.373*0.38=8.1917 R-bar chart LCL=D3* =0.136*0.38=0.0517 UCL=D4* =0.373*0.38=0.1417
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sample 1 2 3 4 5 6 7 8 mean range(R) 1 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.04 0.76 2 8.23 8.12 7.98 8.41 8.31 8.18 7.99 8.06 8.16 0.43 3 7.89 7.77 7.91 8.04 8 7.89 7.93 8.09 7.94 0.32 4 8.24 8.18 7.83 8.05 7.9 8.16 7.97 8.07 8.05 0.41 5 7.87 8.13 7.92 7.99 8.1 7.81 8.14 7.88 7.98 0.33 6 8.13 8.14 8.11 8.13 8.14 8.12 8.13 8.14 8.13 0.03 mean= 8.05 0.38 B) X-bar charts R-bar charts
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C) but based on R-charts the process is not in statistical control as most of the points out side the
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Unformatted text preview: LCL and UCL 6. Total number of defects is 72. Total number sampled is 144*20 = 2880. This makes p_bar = 72/2880 = 0.025 This makes q_bar = 1 – 0.025 = 0.975 The standard deviation is calculated using the formula This makes SD = 2.21 The control limits are calculated using UCL = p_bar + 3*sqrt(p_bar*q_bar/n_bar) LCL = p_bar - 3*sqrt(p_bar*q_bar/n_bar) Here n_bar = 144 Using this we have UCL = 0.064 LCL = -0.014 Since proportions of defects cannot be negative we shall take LCL as 0. P-Chart Based on the chart seen, the process is within control. None of the points have breached the UCL or LCL line. 7. Nmumber of days = 1 week = 7 days Sum of errors = 30 Average number of errors per day(C-bar) = Sum of errors / Number of days = 30/7= 4.29 UCL = C-bar + [3(Sqrt of C-bar)] = 4.29 + [3(Sqrt of 4.29)] = 4.29 + (3X2.071) = 4.29 + 6.213 = 10.503 LCL = C-bar - [3(Sqrt of C-bar)] = 4.29 - [3(Sqrt of 4.29)] = 4.29 - (3X2.071) = 4.29 - 6.213 = - 1.923 = 0 (as when the LCL is less than zero,it is considered as zero) 8....
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