OPM assignment 4 part 2

OPM assignment 4 - 0.1 0.04 30 4 0.04 0.08 0.04 0.1 0.04 Average Given Data n 100 Assume Z 1.96 Output Total Late 120 P-Bar 0.04 σp =

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Ian Schechter OPM Assignment 4 Alabama Airline case Sample Week Late flights Fraction Deviation UCL LCL P-BAR UCL LCL 1 2 0.02 0.08 0 0.04 0.1 0.04 2 4 0.04 0.08 0 0.04 0.1 0.04 3 10 0.1 0.08 0 0.04 0.1 0.04 4 4 0.04 0.08 0 0.04 0.1 0.04 5 1 0.01 0.08 0 0.04 0.1 0.04 6 1 0.01 0.08 0 0.04 0.1 0.04 7 13 0.13 0.08 0 0.04 0.1 0.04 8 9 0.09 0.08 0 0.04 0.1 0.04 9 11 0.11 0.08 0 0.04 0.1 0.04 10 0 0 0.08 0 0.04 0.1 0.04 11 3 0.03 0.08 0 0.04 0.1 0.04 12 4 0.04 0.08 0 0.04 0.1 0.04 13 2 0.02 0.08 0 0.04 0.1 0.04 14 2 0.02 0.08 0 0.04 0.1 0.04 15 8 0.08 0.08 0 0.04 0.1 0.04 16 2 0.02 0.08 0 0.04 0.1 0.04 17 3 0.03 0.08 0 0.04 0.1 0.04 18 7 0.07 0.08 0 0.04 0.1 0.04 19 3 0.03 0.08 0 0.04 0.1 0.04 20 2 0.02 0.08 0 0.04 0.1 0.04 21 3 0.03 0.08 0 0.04 0.1 0.04 22 7 0.07 0.08 0 0.04 0.1 0.04 23 4 0.04 0.08 0 0.04 0.1 0.04 24 3 0.03 0.08 0 0.04 0.1 0.04 25 2 0.02 0.08 0 0.04 0.1 0.04 26 2 0.02 0.08 0 0.04 0.1 0.04 27 0 0 0.08 0 0.04 0.1 0.04 28 1 0.01 0.08 0 0.04 0.1 0.04 29 3 0.03 0.08 0 0.04
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Unformatted text preview: 0.1 0.04 30 4 0.04 0.08 0.04 0.1 0.04 Average Given Data n 100 Assume Z 1.96 Output Total Late 120 P-Bar 0.04 σp = sqrt[pbar*(1-pbar)/n] 0.02 UCLp = pbar + zσp 0.08 LCLp = pbar – zσp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.2 Column C Column D Column E Column F 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.05 0.1 0.15 Column C Column F Column G Column H...
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This note was uploaded on 04/07/2008 for the course OPM 311 taught by Professor Rui during the Spring '08 term at Binghamton University.

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