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# Session 19

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### Session 19

1. 1. 03-09-2012 About Non-parametric (NP) Methods • Valid without restrictive assumptions about the population Nonparametric methods • Rank based (detailed data not available or used) • More robust – against outlier – arbitrary population distribution Session XIX • Less efficient A List of NP Methods Tests for central tendency One-sample Two-sample K-sample • Kolmogorov-Smirnov test • Sign test • Mann-Whitney U test/Rank-sum test One-sample Two-sample T-/Z- test for T-/Z- test for ANOVA • Run test: A test for independence mean Equality of mean • Kruskal-Wallis test Sign-test Kruskal- Non-Parametric Rank sum test What are their respective parametric counter-parts? For median Wallis test (Mann-Whitney) Problem 1 Problem 1: solutionA light-aircraft engine repair shop switched the payment method it usedfrom hourly wage to hourly wage plus a bonus computed on the time Change= before - after. +ve value indicates improvementrequired to disassemble, repair, and reassemble an engine. Thefollowing are data collected for 25 engines before the change and (and H0: median(change) = 0 vs H1: median(change) > 0the same) 25 engines after the change. At a 0.10 significance level, didthe new plan increase productivity? H0: π=0.5 vs. H1: π > 0.5, where π =P(change > 0) Hours Required Hours Required Before After Before After ------------------- ------------------- 29 32 25 34 32 19 42 27 32 22 20 26 19 21 25 25 31 20 33 31 22 24 34 19 28 25 20 22 31 31 21 32 32 18 22 31 44 22 45 30 41 24 43 29 23 26 31 20 34 41 1
2. 2. 03-09-2012 Problem 1: solution Sign test (Assuming ‘the same’ 25 engines) Change= before - after. +ve value indicates improvement• For paired comparison of ‘center’(median) H0: median(change) = 0 vs H1: median(change) > 0 – One-sample test for median H0: π=0.5 vs. H1: π > 0.5, where π =P(change > 0)• Similarly test for percentiles – H0: 1st quartile of score out of 60 is 20 p - 0.5 Test Statistic Z = – T.S. is Y= no. of scores < 20 has B(n, .25) under the 0.5 × 0.5 n null hypothesis C.R. at level α = 0.1 is Z > 1.28 Conclusion at 10% level: no• Take p-value approach if n is small 13 significant increase in - 0.5 Observed value of the T.S. is 23 = .63 productivity. 0.5 × 0.5 23 Statistics and Sleeping Problem 2 A manufacturer of toys changed the type of plastic molding machines it was using because a new one gave evidence of being more economical. As the Christmas season began, however, productivity seemed somewhat lower than last year. Because production records of the past years were readily available, the production manager decided to compare the monthly output for the 15 months when the old machines were used and the 11 months of production so far this year. Records show these output amounts with the old and new machines. Monthly output in Units ------------------------------------------------------------------------------------------------- Old Machines New Machines ------------------------------------------------- ----------------------------------- 992 966 965 956 945 889 1054 900 938 972 912 938 1027 940 850 892 873 796 983 1016 911 1014 897 877 1258 902 Problem 2: solution Mann-Whitney U-test • For comparing averages of two populations • Combine all data and rank them. H 0 : m old = m new vs H1 : m old > m new – Smallest observation is assigned rank 1; second smallest observation is ranked 2 …Highest observation is ranked n – Give mid- rank in case of tie.mold average (median) output with old machine • If the average of the first population is smaller, then themnew average (median) output with new machine combined rank of observations from first populations would be small • Need sample sizes to be 10 or more for normal approximation (after standardizing). Tables are available for smaller sample. • Also known as Wilcoxon rank sum test 2
3. 3. 03-09-2012 Calculation Problem 2 R1: Total rank of observations drawn from population 1 H 0 : m old = m new vs H1 : m old > m new R2: Total rank of observations drawn from population 2 n2 (n2 + 1) 11× 12 Test Statistic U = n1n2 + − R2 = 15 ×11 + − R2 = 231 − R2 2 2 n1 ( n1 + 1) U - µUTest Statistic U = n1n2 + − R1 At α = 0.1, the C.R is > 1.28 2 σU= No. of (X i , Y j ) pairs with X i < Y j ( R1 = rank sum of all X obs) where, µU = n1n2 165 = n n (n + n + 1) , σU = 1 2 1 2 = 19.27 2 2 12 n1n2 n1n2 (n1 + n2 + 1)Under H 0 , µU = , σU = 2 12 (231 − 115.5) − 82.5 R2=115.5, Observed value of the T.S. is 19.27 = 1.71 U-µ Uand has N(0,1) distribution σU So Reject H0 at 10% level and conclude that the change has reduced the average output level 3