non parametric tests
DESCRIPTION
Presentation on Non Parametric TestsTRANSCRIPT
-
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