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: solution
A light-aircraft engine repair shop switched the payment method it used
from hourly wage to hourly wage plus a bonus computed on the time
Change= before - after. +ve value indicates improvement
required to disassemble, repair, and reassemble an engine. The
following are data collected for 25 engines before the change and (and H0: median(change) = 0 vs H1: median(change) > 0
the same) 25 engines after the change. At a 0.10 significance level, did
the 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
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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 the
mnew 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. 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 - µU
Test 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 U
and has N(0,1) distribution
σU So Reject H0 at 10% level and conclude that the change has reduced
the average output level
3