ime261a12s - IME 2610 Fall 2010 Assignment 12 SOLUTIONS Due Monday December 6 2010 Material Covered Chapter 10 11 Problem A Problem 10.12 Problem 10.14

# ime261a12s - IME 2610 Fall 2010 Assignment 12 SOLUTIONS Due...

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IME 2610 - Fall 2010 Assignment 12 SOLUTIONS Due Monday December 6, 2010 Material Covered: Chapter 10 & 11 Problem A
Problem 10.12
Problem 10.14 a) Test for a difference in the median lead concentration Kruskal-Wallis H o : M 1 = M 2 = M 3 H 1 : Not all M are equal
Kruskal-Wallis Test: Lead versus Treatment Kruskal-Wallis Test on Lead Treatment N Median Ave Rank Z Magnetic 8 35.50 10.0 -1.22 Softener 8 76.00 14.9 1.19 Untreated 8 37.50 12.6 0.03 Overall 24 12.5 H = 1.95 DF = 2 P = 0.377 H = 1.96 DF = 2 P = 0.376 (adjusted for ties) At a 5% significance level, there is no evidence of a difference in the median lead concentration for the three types of water treatments. b) Descriptive Statistics: Lead Variable Treatment Mean StDev Median Lead Magnetic 41.7 34.2 35.5 Softener 127.3 138.4 76.0 Untreated 90.4 97.7 37.5 The standard deviations of each population do not appear to be equal which would be in violation of the assumptions of an ANOVA test. A Kruskal-Wallis test would still be valid even without assuming equal variances (textbook p 502). Problem 11.4 General Linear Model: Density versus Strength, Time