Statistics for Business and EconomicsNonparametric Statistics Chapter 14

Learning Objectives1.Distinguish Parametric & Nonparametric Test Procedures 2.Explain a Variety of Nonparametric Test Procedures3.Solve Hypothesis Testing Problems Using Nonparametric Tests4.Compute Spearmans Rank Correlation

Hypothesis Testing ProceduresMany More Tests Exist!

Parametric Test Procedures1.Involve Population ParametersExample: Population Mean2.Require Interval Scale or Ratio ScaleWhole Numbers or FractionsExample: Height in Inches (72, 60.5, 54.7)3.Have Stringent AssumptionsExample: Normal Distribution4.Examples: Z Test, t Test, 2 Test

Nonparametric Test Procedures1.Do Not Involve Population ParametersExample: Probability Distributions, Independence2.Data Measured on Any ScaleRatio or IntervalOrdinalExample: Good-Better-BestNominalExample: Male-Female3.Example: Wilcoxon Rank Sum Test

Advantages of Nonparametric Tests1.Used With All Scales2.Easier to ComputeDeveloped Originally Before Wide Computer Use3.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 If Data Permit Using Parametric ProceduresExample: Converting Data From Ratio to Ordinal Scale2.Difficult to Compute by Hand for Large Samples3.Tables Not Widely Available 1984-1994 T/Maker Co.

Frequently Used Nonparametric Tests1.Sign Test 2.Wilcoxon Rank Sum Test3.Wilcoxon Signed Rank Test4.Kruskal Wallis H-Test5. Friedmans Fr-Test6. Spearmans Rank Correlation Coefficient

Sign Test

Sign Test 1.Tests One Population Median, (eta)2.Corresponds to t-Test for 1 Mean3.Assumes Population Is Continuous4.Small Sample Test Statistic: # Sample Values Above (or Below) MedianAlternative Hypothesis Determines5.Can Use Normal Approximation If n 10

Sign Test Uses P-Value to Make DecisionBinomial: n = 8 p = 0.5P-Value Is the Probability of Getting an Observation At Least as Extreme as We Got. If 7 of 8 Observations Favor Ha, Then P-Value = P(x 7) = .031 + .004 = .035. If = .05, Then Reject H0 Since P-Value .

Sign Test ExampleYoure an analyst for Chef-Boy-R-Dee. Youve asked 7 people to rate a new ravioli on a 5-point Likert scale (1 = terrible to 5 = excellent. The ratings are: 2 5 3 4 1 4 5. At the .05 level, is there evidence that the median rating is less than 3?

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

Conclusion:

Do Not Reject at = .05There Is No Evidence Median Is Less Than 3P(x 2) = 1 - P(x 1) = .937 (Binomial Table, n = 7, p = 0.50)S = 2 (Ratings 1 & 2 Are Less Than = 3: 2, 5, 3, 4, 1, 4, 5)

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 = 1Average Ties2.Sum the Ranks, Ti, for Each Sample3.Test Statistic Is TA (Smallest Sample)

Wilcoxon Rank Sum Test ExampleYoure 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 Table (Portion) = .05 one-tailed; = .10 two-tailed

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

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

Conclusion:

Do Not Reject at = .10There Is No Evidence Distrib. Are Not EqualRejectRejectDo Not Reject1327 RanksT2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)

Wilcoxon Signed Rank Test

Wilcoxon Signed Rank Test 1.Tests Probability Distributions of 2 Related Populations2.Corresponds to t-test for Dependent (Paired) Means3.AssumptionsRandom SamplesBoth Populations Are Continuous4.Can Use Normal Approximation If n 25

Signed Rank Test Procedure1.Obtain Difference Scores, Di = X1i - X2i 2.Take Absolute Value of Differences, Di3.Delete Differences With 0 Value4.Assign Ranks, Ri, Where Smallest = 15.Assign Ranks Same Signs as Di6.Sum + Ranks (T+) & - Ranks (T-)Test Statistic Is T- (One-Tailed Test)Test Statistic Is Smaller of T- or T+ (2-Tail)

Signed Rank Test Computation Table

X1i

X2i

Di = X1i - X2i

|Di|

Ri

Sign

Sign Ri

X11

X21

D1 = X11 - X21

|D1|

R1

R1

X12

X22

D2 = X12 - X22

|D2|

R2

R2

X13

X23

D3 = X13 - X23

|D3|

R3

R3

:

:

:

:

:

:

:

X1n

X2n

Dn = X1n - X2n

|Dn|

Rn

Rn

Total

T+ & T-

Signed Rank TestExampleYou work in the finance department. Is the new financial package faster (.05 level)? You collect the following data entry times:UserCurrentNewDonna9.989.88Santosha9.889.86Sam9.909.83Tamika9.999.80Brian9.949.87Jorge9.849.84 1984-1994 T/Maker Co.

Signed Rank Test Computation Table

Wilcoxon Signed Rank Table (Portion)

One-Tailed

Two-Tailed

n = 5

n = 6

n = 7

..

= .05

= .10

1

2

4

..

= .025

= .05

1

2

..

= .01

= .02

0

..

= .005

= .01

..

n = 11

n = 12

n = 13

:

:

:

:

Signed Rank Test SolutionH0: Identical Distrib.Ha: Current Shifted Right = .05n = 5 (not 6; 1 elim.)Critical Value(s):Test Statistic: Decision:

Conclusion:

Reject at = .05There Is Evidence New Package Is FasterRejectDo Not Reject1T0Since One-Tailed Test & Current Shifted Right, Use T-: T- = 0

Kruskal-Wallis H-Test

Kruskal-Wallis H-Test1.Tests the Equality of More Than 2 (p) Population Probability Distributions2.Corresponds to ANOVA for More Than 2 Means3.Used to Analyze Completely Randomized Experimental Designs 4.Uses 2 Distribution with p - 1 df If At Least 1 Sample Size nj > 5

Kruskal-Wallis H-Test Assumptions1.Independent, Random Samples 2.At Least 5 Observations Per Sample3.Continuous Population Probability Distributions

Kruskal-Wallis H-Test Procedure1.Assign Ranks, Ri , to the n Combined ObservationsSmallest Value = 1; Largest Value = nAverage Ties2.Sum Ranks for Each Group

Kruskal-Wallis H-Test Procedure1.Assign Ranks, Ri , to the n Combined ObservationsSmallest Value = 1; Largest Value = nAverage Ties2.Sum Ranks for Each Group3.Compute Test StatisticSquared total of each group

Kruskal-Wallis H-Test ExampleAs production manager, you want to see if 3 filling machines have different filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the .05 level, is there a difference in the distribution of filling times?Mach1Mach2Mach3 25.4023.4020.00 26.3121.8022.20 24.1023.5019.75 23.7422.7520.60 25.1021.6020.40

Kruskal-Wallis H-Test SolutionRaw DataMach1Mach2Mach3 25.4023.4020.00 26.3121.8022.20 24.1023.5019.75 23.7422.7520.60 25.1021.6020.40RanksMach1Mach2Mach3 1492 1567 12101 1184 1353 653817Total

Kruskal-Wallis H-Test Solution

Kruskal-Wallis H-Test SolutionH0: Identical Distrib.Ha: At Least 2 Differ = .05df = p - 1 = 3 - 1 = 2Critical Value(s):

205.991Test Statistic: Decision:

Conclusion:

Reject at = .05There Is Evidence Pop. Distrib. Are Different = .05H = 11.58

Friedman Fr-Test for a Randomized Block Design

Friedman Fr-Test1.Tests the Equality of More Than 2 (p) Population Probability Distributions2.Corresponds to ANOVA for More Than 2 Means3.Used to Analyze Randomized Block Experimental Designs 4.Uses 2 Distribution with p - 1 df If either p, the number of treatments, or b, the number of blocks, exceeds 5

Friedman Fr-Test Assumptions1. The p treatments are randomly assigned to experimental units within the b blocks Samples 2. The measurements can be ranked within the blocks3. Continuous population probability distributions

Friedman Fr-Test Procedure1.Assign Ranks, Ri = 1 p, to the p treatments in each of the b blocksSmallest Value = 1; Largest Value = pAverage Ties2.Sum Ranks for Each Treatment

Friedman Fr-Test Procedure1.Assign Ranks, Ri = 1 p, to the p treatments in each of the b blocksSmallest Value = 1; Largest Value = pAverage TiesSum Ranks for Each TreatmentCompute Test Statistic

Squared total of each treatment

Friedman Fr-Test ExampleThree new traps were tested to compare their ability to trap mosquitoes. Each of the traps, A, B, and C were placed side-by-side at each five different locations. The number of mosquitoes in each trap was recorded. At the .05 level, is there a difference in the distribution of number of mosquitoes caught by the three traps?TrapA TrapBTrapC 35 0 231715 1157 842 19115

Friedman Fr-Test SolutionRaw DataTrapA TrapBTrapC 35 0 231715 1157 842 19115

RanksTrapA TrapBTrapC 231 32131232132114106Total

Friedman Fr-Test Solution

Friedman Fr-Test SolutionH0: Identical Distrib.Ha: At Least 2 Differ = .05df = p - 1 = 3 - 1 = 2Critical Value(s):

205.991Test Statistic: Decision:

Conclusion:

Reject at = .05There Is Evidence Pop. Distrib. Are Different = .05Fr = 6.64

Spearmans Rank Correlation Coefficient

Spearmans Rank Correlation Coefficient1.Measures Correlation Between Ranks2.Corresponds to Pearson Product Moment Correlation Coefficient3.Values Range from -1 to +1

Spearmans Rank Correlation Coefficient1.Measures Correlation Between Ranks2.Corresponds to Pearson Product Moment Correlation Coefficient3.Values Range from -1 to +14.Equation (Shortcut)

Spearmans Rank Correlation Procedure1.Assign Ranks, Ri , to the Observations of Each Variable Separately2.Calculate Differences, di , Between Each Pair of Ranks3.Square Differences, di 2, Between Ranks4.Sum Squared Differences for Each Variable5.Use Shortcut Approximation Formula

Spearmans Rank Correlation ExampleYoure a research assistant for the FBI. Youre investigating the relationship between a persons attempts at deception & % changes in their pupil size. You ask subjects a series of questions, some of which they must answer dishonestly. At the .05 level, what is the correlation coefficient?Subj.DeceptionPupil 18710 2636 39511 4507 5430

Spearmans Rank Correlation Table

Subj.

Decep.

R1i

Pupil

R2i

di

di2

1

87

4

10

4

0

0

2

63

3

6

2

1

1

3

95

5

11

5

0

0

4

50

2

7

3

-1

1

5

43

1

0

1

0

0

Total

2

Spearmans Rank Correlation Solution

As a result of this class, you will be able to ...12479Distribution has different shapes.1st Graph:If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 0 defective item is about 0.6 (60%).If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 1 defective items is about .35 (35%).2nd Graph:If inspecting 5 items & the Probability of a defect is 0.5 (50%), the Probability of finding 1 defective items is about .18 (18%).Note:Could use formula or tables at end of text to get Probabilities.

Assume 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.47951479n is the number of non-zero difference scores.Look up n = 5 in table; not 6!Rejection region does not include cut-off.106911111111211510691111111121151069112