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EPI 809 / Spring 2008 Chapter 9 Nonparametric Statistics

Chap. 9: Nonparametric Statistics - SRM University 809 / Spring 2008 Advantages of Nonparametric Tests 1. Used With All Scales 2. Easier to Compute 3. Make Fewer Assumptions 4. Need

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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

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, χ2 Test, F test

EPI 809 / Spring 2008

Nonparametric Test Procedures

1. Do Not Involve Population ParametersExample: 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 Scales2. Easier to Compute3. Make Fewer Assumptions4. Need Not Involve

Population Parameters5. 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

© 1984-1994 T/Maker Co.

EPI 809 / Spring 2008

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, η

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: α = Test Statistic:

P-Value:

Decision:

Conclusion:

EPI 809 / Spring 2008

Sign Test Solution

H0: η = 3Ha: η < 3α = Test Statistic:

P-Value:

Decision:

Conclusion:

EPI 809 / Spring 2008

Sign Test Solution

H0: η = 3Ha: η < 3α = .05Test Statistic:

P-Value:

Decision:

Conclusion:

EPI 809 / Spring 2008

Sign Test Solution

H0: η = 3Ha: η < 3α = .05Test Statistic:

P-Value:

Decision:

Conclusion:

S = 2 (Ratings 1 & 2 Are Less Than η = 3:2, 5, 3, 4, 1, 4, 5)Is observing 2 or more a small prob event?

EPI 809 / Spring 2008

Sign Test Solution

H0: η = 3Ha: η < 3α = .05Test Statistic:

P-Value:

Decision:

Conclusion:

P(S ≥ 2) = 1 - P(S ≤ 1) = .9297

(Binomial Table, n = 7, p = 0.50)

S = 2 (Ratings 1 & 2 Are Less Than η = 3:2, 5, 3, 4, 1, 4, 5)Is observing 2 or more a small prob event?

EPI 809 / Spring 2008

Sign Test Solution

H0: η = 3Ha: η < 3α = .05Test Statistic:

P-Value:

Decision:

Conclusion:Do Not Reject at α = .05

P(x ≥ 2) = 1 - P(x ≤ 1) = . 9297

(Binomial Table, n = 7, p = 0.50)

S = 2 (Ratings 1 & 2 Are Less Than η = 3:2, 5, 3, 4, 1, 4, 5)Is observing 2 or more a small prob event?

EPI 809 / Spring 2008

Sign Test Solution

H0: η = 3Ha: η < 3α = .05Test Statistic:

P-Value:

Decision:

Conclusion:Do Not Reject at α = .05

There is No evidence for Median < 3

P(x ≥ 2) = 1 - P(x ≤ 1) == . 9297

(Binomial Table, n = 7, p = 0.50)

S = 2 (Ratings 1 & 2 are < η = 3:2, 5, 3, 4, 1, 4, 5)Is observing 2 or more a small prob event?

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.AssumptionsIndependent, Random SamplesPopulations 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 SampleSmallest Value = 1

2. Sum the Ranks, Ti, for Each SampleTest 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:α =n1 = n2 = Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Σ Ranks

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Solution

H0: Identical Distrib.Ha: Shifted Left or Rightα =n1 = n2 = Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Σ Ranks

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Solution

H0: Identical Distrib.Ha: Shifted Left or Rightα = .10n1 = 4 n2 = 5 Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Σ Ranks

EPI 809 / Spring 2008

Wilcoxon Rank Sum Table 12 (Rosner) (Portion)α = .05 two-tailed

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Solution

H0: Identical Distrib.Ha: Shifted Left or Rightα = .10n1 = 4 n2 = 5 Critical Value(s):

Test Statistic:

Decision:

Conclusion:Reject RejectDo Not Reject

12 28 Σ Ranks

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 8582 8277 9492 9788 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 8582 8277 9492 9788 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 8582 8277 2 9492 9788 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 8582 3 82 477 2 9492 9788 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 8582 3 3.5 82 4 3.577 2 9492 9788 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 85 582 3 3.5 82 4 3.577 2 9492 9788 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 85 582 3 3.5 82 4 3.577 2 9492 9788 6 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 85 582 3 3.5 82 4 3.577 2 9492 7 9788 6 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 85 582 3 3.5 82 4 3.577 2 94 892 7 9788 6 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 85 582 3 3.5 82 4 3.577 2 94 892 7 97 988 6 ... ...

Rank Sum

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Computation Table

Factory 1 Factory 2Rate Rank Rate Rank71 1 85 582 3 3.5 82 4 3.577 2 94 892 7 97 988 6 ... ...

Rank Sum 19.5 25.5

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Solution

H0: Identical Distrib.Ha: Shifted Left or Rightα = .10n1 = 4 n2 = 5 Critical Value(s):

Test Statistic:

Decision:

Conclusion:Reject RejectDo Not Reject

12 28 Σ Ranks

T2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Solution

H0: Identical Distrib.Ha: Shifted Left or Rightα = .10n1 = 4 n2 = 5 Critical Value(s):

Test Statistic:

Decision:

Conclusion:Do Not Reject at α = .10

Reject RejectDo Not Reject

12 28 Σ Ranks

T2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)

EPI 809 / Spring 2008

Wilcoxon Rank Sum Test Solution

H0: Identical Distrib.Ha: Shifted Left or Rightα = .10n1 = 4 n2 = 5 Critical Value(s):

Test Statistic:

Decision:

Conclusion:Do Not Reject at α = .10

There is No evidence for unequal distrib

Reject RejectDo Not Reject

12 28 Σ Ranks

T2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)