Fu_ch9_Nonpara (5)

Chapter 9Nonparametric Statistics

EPI 809 / Spring 2008

Learning Objectives1.Distinguish Parametric & Nonparametric Test Procedures 2.Explain commonly used Nonparametric Test Procedures3. Perform Hypothesis Tests Using Nonparametric Procedures


Hypothesis Testing ProceduresMany More Tests Exist!


Parametric Test Procedures1.Involve Population Parameters (Mean)

2.Have Stringent Assumptions (Normality)

3.Examples: Z Test, t Test, 2 Test, F test


Nonparametric Test Procedures1.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


Advantages of Nonparametric Tests1.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.


Disadvantages of Nonparametric Tests1.May Waste Information Parametric model more efficient if data Permit2.Difficult to Compute by hand for Large Samples3.Tables Not Widely Available 1984-1994 T/Maker Co.


Popular Nonparametric Tests1.Sign Test

2.Wilcoxon Rank Sum Test

3.Wilcoxon Signed Rank Test


Sign Test


Sign Test Tests One Population Median,

2.Corresponds to t-Test for 1 Mean

3.Assumes Population Is Continuous

Small Sample Test Statistic: # Sample Values Above (or Below) Median

5. Can Use Normal Approximation If n 10


Sign Test ConceptsMake 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


Sign Test ExampleYoure an analyst for Chef-Boy-R-Dee. Youve 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?


Sign Test SolutionH0: Ha: = Test Statistic:P-Value: Decision:

Conclusion:


Sign Test SolutionH0: = 3Ha: < 3 = Test Statistic:P-Value: Decision:

Conclusion:


Sign Test SolutionH0: = 3Ha: < 3 = .05Test Statistic:P-Value: Decision:

Conclusion:



Conclusion:

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?



Conclusion:

P(S 2) = 1 - P(S 1) = .9297 (Binomial Table, n = 7, p = 0.50)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?



Conclusion:

Do Not Reject at = .05P(x 2) = 1 - P(x 1) = . 9297 (Binomial Table, n = 7, p = 0.50)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?



Conclusion:

Do Not Reject at = .05There is No evidence for Median < 3P(x 2) = 1 - P(x 1) == . 9297 (Binomial Table, n = 7, p = 0.50)S = 2 (Ratings 1 & 2 are < = 3: 2, 5, 3, 4, 1, 4, 5)Is observing 2 or more a small prob event?


Wilcoxon Rank Sum Test


Wilcoxon Rank Sum Test 1.Tests Two Independent Population Probability Distributions2.Corresponds to t-Test for 2 Independent Means3.AssumptionsIndependent, Random SamplesPopulations Are Continuous4.Can Use Normal Approximation If ni 10


Wilcoxon Rank Sum Test Procedure1.Assign Ranks, Ri, to the n1 + n2 Sample ObservationsIf 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 distributionDistribution of T is not quite as simple as binomial, but it can be computed


Wilcoxon Rank Sum Test Example

Youre 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?


Wilcoxon Rank Sum Test SolutionH0:Ha: =n1 =n2 = Critical Value(s):Test Statistic: Decision:

Conclusion:

Ranks


Wilcoxon Rank Sum Test SolutionH0: Identical Distrib.Ha: Shifted Left or Right =n1 =n2 = Critical Value(s):Test Statistic: Decision:

Conclusion:

Ranks


Wilcoxon Rank Sum Test SolutionH0: Identical Distrib.Ha: Shifted Left or Right = .10n1 = 4 n2 = 5 Critical Value(s):Test Statistic: Decision:

Conclusion:

Ranks


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


n1

4

5

6

..

TL

TU

TL

TU

TL

TU

..

4

10

26

16

34

23

43

..

n2

5

11

29

17

38

24

48

..

6

12

32

18

42

26

52

..

:

:

:

:

:

:

:

:


Conclusion:

RejectRejectDo Not Reject1228 Ranks


Wilcoxon Rank Sum Test Computation TableFactory 1Factory 2RateRankRateRankRank Sum


Wilcoxon Rank Sum Test Computation TableFactory 1Factory 2RateRankRateRank718582827794929788......Rank Sum


Wilcoxon Rank Sum Test Computation TableFactory 1Factory 2RateRankRateRank71185823 3.5824 3.577294929788......Rank Sum


Wilcoxon Rank Sum Test Computation TableFactory 1Factory 2RateRankRateRank711855823 3.5824 3.5772948927979886......Rank Sum19.525.5



Conclusion:

RejectRejectDo Not Reject1228 RanksT2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)



Conclusion:

Do Not Reject at = .10RejectRejectDo Not Reject1228 RanksT2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)



Conclusion:

Do Not Reject at = .10There is No evidence for unequal distribRejectRejectDo Not Reject1228 RanksT2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)


As a result of this class, you will be able to ...12479Assume that the population is normally distributed.Allow students about 10 minutes to solve this.Note: More than 5 have been sold (6.4), but not enough to be significant.Note: More than 5 have been sold (6.4), but not enough to be significant.Note: More than 5 have been sold (6.4), but not enough to be significant.Note: More than 5 have been sold (6.4), but not enough to be significant.Note: More than 5 have been sold (6.4), but not enough to be significant.Note: More than 5 have been sold (6.4), but not enough to be significant.Note: More than 5 have been sold (6.4), but not enough to be significant.47951

Documents

Fu_ch9_Nonpara (5)