HW05_key-Sp11[1]

# HW05_key-Sp11[1] - MGMT 36100: HW 05 40 Points Due: April...

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MGMT 36100: HW 05 40 Points Due: April 14, 2011 Question 1 (Points: 4) (a) UTL = 100 + 10 = 110; LTL = 100 – 10 = 90 = (110 – 90) / (6 * 4) = 0.833. Whether this is acceptable or not depends upon the target process capability value. Probability of acceptance is (b) C PK = min {( μ – LTL) / (3 σ ), (UTL – μ ) / (3 σ )} = ( μ – LTL) / (3 σ ) = (92 – 90) / (3 * 4) = 0.167 Probability of acceptance is , which is very low. Question 2 (Points: 8) Item Child Level Final 1 3,7 0 √ 0 2 3,10 0 √ 0 3 4,6 0 1 √ 1 4 7,8 0 1 2 2 5 6,10 0 √ 0 6 9 0 1 2 √ 2 7 - 0 1 2 3 3 8 - 0 1 2 3 3 9 - 0 1 2 3 3 10 8 0 1 √ 1 Determine the item levels using the procedure discussed in class and draw neat product structure diagrams for data in the table on right. Page 1

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Question 3 (Points: 16, 16) Part A Page 2
Q2 Part A Period 0 1 2 3 4 5 6 7 8 9 W LT = 2 L4L SS=10 GR 30 30 100 65 100 80 50 65 40 SR 100 OH 135 205 175 75 10 10 10 10 10 10 PR 100 80 50 65 40 POR 0 0 100 80 50 65 40 0 0 T LT = 1 L4L SS=50

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## This note was uploaded on 03/03/2012 for the course MGMT 361 taught by Professor Panwalker during the Spring '10 term at Purdue.

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HW05_key-Sp11[1] - MGMT 36100: HW 05 40 Points Due: April...

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