Chapter 9 Nonparametric Statistics EPI 809 / Spring 2008

Chapter 9
Nonparametric Statistics
EPI 809 / Spring 2008
Learning Objectives
1.
Distinguish Parametric &
Nonparametric Test Procedures
2.
Explain commonly used
Nonparametric Test Procedures
3.
Perform Hypothesis Tests Using
Nonparametric Procedures
EPI 809 / Spring 2008
Hypothesis Testing Procedures
Hypothesis
Testing
Procedures
Parametric
Nonparametric
Wilcoxon
Rank Sum
Test
Z Test
t Test
One-Way
ANOVA
Kruskal-Wallis
H-Test
Many More Tests Exist!
EPI 809 / Spring 2008
Parametric Test Procedures
1.
Involve Population Parameters (Mean)
2.
Have Stringent Assumptions
(Normality)
3.
Examples: Z Test, t Test, 2 Test,
F test
EPI 809 / Spring 2008
Nonparametric Test
Procedures
1.
Do Not Involve Population Parameters
Example: Probability Distributions, Independence
2.
Data Measured on Any Scale (Ratio or
Interval, Ordinal or Nominal)
3.
Example: Wilcoxon Rank Sum Test
EPI 809 / Spring 2008
Advantages of
Nonparametric Tests
1.
Used With All Scales
2.
Easier to Compute
3.
Make Fewer Assumptions
4.
Need Not Involve
Population Parameters
5.
Results May Be as Exact
as Parametric Procedures
© 1984-1994 T/Maker Co.
EPI 809 / Spring 2008
Disadvantages of
Nonparametric Tests
1. May Waste Information
Parametric model more efficient
if data Permit
2. Difficult to Compute by
hand for Large Samples
3. Tables Not Widely Available
EPI 809 / Spring 2008
© 1984-1994 T/Maker Co.
Popular Nonparametric Tests
1. Sign Test
2. Wilcoxon Rank Sum Test
3. Wilcoxon Signed Rank Test
EPI 809 / Spring 2008
Sign Test
EPI 809 / Spring 2008
Sign Test
1.
Tests One Population Median, 
2. Corresponds to t-Test for 1 Mean
3. Assumes Population Is Continuous
4.
Small Sample Test Statistic: # Sample Values
Above (or Below) Median
5. Can Use Normal Approximation If n  10
EPI 809 / Spring 2008
Sign Test Concepts

Make null hypothesis about true median

Let S = number of values greater than median

Each sampled item is independent

If null hypothesis is true, S should have binomial
distribution with success probability .5
EPI 809 / Spring 2008
Sign Test Example

You’re an analyst for Chef-Boy-R-Dee. You’ve
asked 7 people to rate a new ravioli on a 5-point
scale (1 = terrible,…, 5 = excellent) The ratings
are: 2 5 3 4 1 4 5.

At the .05 level, is there evidence that the
median rating is at least 3?
EPI 809 / Spring 2008
Sign Test Solution

H0:
 Ha:
 =
 Test Statistic:
P-Value:
Decision:
Conclusion:
EPI 809 / Spring 2008
Sign Test Solution
H0:  = 3
 Ha:  < 3
 =
 Test Statistic:

P-Value:
Decision:
Conclusion:
EPI 809 / Spring 2008
Sign Test Solution
H0:  = 3
 Ha:  < 3
  = .05
 Test Statistic:

P-Value:
Decision:
Conclusion:
EPI 809 / Spring 2008
Sign Test Solution
H0:  = 3
 Ha:  < 3
  = .05
 Test Statistic:

P-Value:
S=2
(Ratings 1 & 2 Are
Less Than  = 3:
2, 5, 3, 4, 1, 4, 5)
Is observing 2 or
more a small prob
event?
Decision:
Conclusion:
EPI 809 / Spring 2008
Sign Test Solution
H0:  = 3
 Ha:  < 3
  = .05
 Test Statistic:

S=2
(Ratings 1 & 2 Are
Less Than  = 3:
2, 5, 3, 4, 1, 4, 5)
Is observing 2 or
more a small prob
event?
P-Value:
P(S  2) = 1 - P(S  1)
= .9297
(Binomial Table, n = 7,
p = 0.50)
Decision:
Conclusion:
EPI 809 / Spring 2008
Sign Test Solution
H0:  = 3
 Ha:  < 3
  = .05
 Test Statistic:

S=2
(Ratings 1 & 2 Are
Less Than  = 3:
2, 5, 3, 4, 1, 4, 5)
Is observing 2 or
more a small prob
event?
P-Value:
P(x  2) = 1 - P(x  1)
= . 9297
(Binomial Table, n = 7,
p = 0.50)
Decision:
Do Not Reject at  = .05
Conclusion:
EPI 809 / Spring 2008
Sign Test Solution
H0:  = 3
 Ha:  < 3
  = .05
 Test Statistic:

S=2
(Ratings 1 & 2 are
<  = 3:
2, 5, 3, 4, 1, 4, 5)
Is observing 2 or
more a small prob
event?
P-Value:
P(x  2) = 1 - P(x  1) =
= . 9297
(Binomial Table, n = 7,
p = 0.50)
Decision:
Do Not Reject at  = .05
Conclusion:
There is No evidence for
Median < 3
EPI 809 / Spring 2008
Wilcoxon Rank Sum
Test
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
1.Tests Two Independent Population
Probability Distributions
2.Corresponds to t-Test for 2 Independent
Means
3.Assumptions
Independent, Random Samples
Populations Are Continuous
4.Can Use Normal Approximation If ni  10
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Procedure
1. Assign Ranks, Ri, to the n1 + n2 Sample
Observations
If Unequal Sample Sizes, Let n1 Refer to Smaller-Sized Sample
Smallest Value = 1
2. Sum the Ranks, Ti, for Each Sample
Test Statistic Is TA (Smallest Sample)
Null hypothesis: both samples come from the same underlying
distribution
Distribution of T is not quite as simple as binomial, but it can be
computed
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Example

You’re a production planner. You want to see if
the operating rates for 2 factories is the same.
For factory 1, the rates (% of capacity) are 71,
82, 77, 92, 88. For factory 2, the rates are 85,
82, 94 & 97. Do the factory rates have the same
probability distributions at the .10 level?
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Solution





H0:
Ha:
=
n1 = n2 =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
 Ranks
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Solution





H0: Identical Distrib.
Ha: Shifted Left or
Right
=
n1 = n2 =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
 Ranks
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Solution





H0: Identical Distrib.
Ha: Shifted Left or
Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
 Ranks
EPI 809 / Spring 2008
Wilcoxon Rank Sum
Table 12 (Rosner) (Portion)
 = .05 two-tailed
n1
4
n2
4
5
6
:
TL
10
11
12
:
5
TU
26
29
32
:
TL
16
17
18
:
6
TU
34
38
42
:
EPI 809 / Spring 2008
TL
23
24
26
:
TU
43
48
52
:
..
..
..
..
..
:
Wilcoxon Rank Sum Test
Solution





H0: Identical Distrib.
Ha: Shifted Left or
Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Reject
Do Not
Reject
Reject
12
Test Statistic:
Decision:
Conclusion:
28  Ranks
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
Factory 2
Rate
Rank
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
82
77
92
88
Factory 2
Rate
Rank
85
82
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
77
92
88
Factory 2
Rate
Rank
85
82
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
77
2
92
88
Factory 2
Rate
Rank
85
82
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3
77
2
92
88
Factory 2
Rate
Rank
85
82
4
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
88
Factory 2
Rate
Rank
85
82
4 3.5
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
88
Factory 2
Rate
Rank
85
5
82
4 3.5
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
88
6
Factory 2
Rate
Rank
85
5
82
4 3.5
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
7
88
6
Factory 2
Rate
Rank
85
5
82
4 3.5
94
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
7
88
6
Factory 2
Rate
Rank
85
5
82
4 3.5
94
8
97
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
7
88
6
Factory 2
Rate
Rank
85
5
82
4 3.5
94
8
97
9
...
...
Rank Sum
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Computation Table
Factory 1
Rate
Rank
71
1
82
3 3.5
77
2
92
7
88
6
Rank Sum
Factory 2
Rate
Rank
85
5
82
4 3.5
94
8
97
9
...
...
19.5
EPI 809 / Spring 2008
25.5
Wilcoxon Rank Sum Test
Solution





H0: Identical Distrib.
Ha: Shifted Left or
Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Reject
Do Not
Reject
Reject
12
Test Statistic:
T2 = 5 + 3.5 + 8+ 9 = 25.5
(Smallest Sample)
Decision:
Conclusion:
28  Ranks
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Solution





H0: Identical Distrib.
Ha: Shifted Left or
Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Reject
Do Not
Reject
Reject
12
Test Statistic:
T2 = 5 + 3.5 + 8+ 9 = 25.5
(Smallest Sample)
Decision:
Do Not Reject at  = .10
Conclusion:
28  Ranks
EPI 809 / Spring 2008
Wilcoxon Rank Sum Test
Solution





H0: Identical Distrib.
Ha: Shifted Left or
Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Reject
Do Not
Reject
Reject
12
Test Statistic:
T2 = 5 + 3.5 + 8+ 9 = 25.5
(Smallest Sample)
Decision:
Do Not Reject at  = .10
Conclusion:
There is No evidence for
28  Ranks
unequal distrib
EPI 809 / Spring 2008