Introduction to Nonparametric...
Transcript of Introduction to Nonparametric...
Introduction toNonparametric Statistics
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Chapter Goals
After completing this chapter, you should beable to:Recognize when and how to use the WilcoxonWilcoxonsigned rank testsigned rank test for a population medianRecognize the situations for which the Wilcoxonsigned rank test applies and be able to use it fordecision-makingKnow when and how to perform a Manna Mann--WhitneyWhitneyUU--testtestPerform nonparametric analysis of variance usingthe KruskalKruskal--Wallis oneWallis one--way ANOVAway ANOVA
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Nonparametric Statistics
Nonparametric StatisticsFewer restrictive assumptions aboutdata levels and underlying probabilitydistributions
Population distributions may be skewedThe level of data measurement may onlybe ordinal or nominal
Data measured on any scaled (nominal,ordinal, ratio, interval)
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Advantages of Nonparametric TestsAdvantages of Nonparametric Tests
1. Used With All Scales2. Easier to Compute
Developed Originally Before WideComputer Use
3. Make Fewer Assumptions4. Need Not Involve
Population Parameters5. Results May Be as Exact
as Parametric Procedures © 1984-1994 T/Maker Co.
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Disadvantages ofNonparametric Tests
1. May Waste InformationIf Data Permit Using ParametricProceduresExample: Converting Data FromRatio to Ordinal Scale
2. Difficult to Compute byHand for Large Samples
3. Tables Not Widely Available
© 1984-1994 T/Maker Co.
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Frequently Used NonparametricTests
Sign Test
MannMann--Whitney UWhitney U--TestTest
WilcoxonWilcoxon MatchedMatched--PairsPairs Signed Rank TestSigned Rank Test
KruskalKruskal WallisWallis HH--TestTest
Spearman’s Rank Correlation Coefficient
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Nonparametric Tests for TwoPopulation Centers
NonparametricTests for Two
Population Centers
WilcoxonMatched-Pairs
Signed Rank Test
Mann-WhitneyU-test
LargeSamples
SmallSamples
LargeSamples
SmallSamples
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Mann-Whitney U-Test
Used to compare two samples from two populations
Assumptions:The two samples are independent and randomThe value measured is a continuous variableThe measurement scale used is at least ordinalIf they differ, the distributions of the two populations willdiffer only with respect to the central location
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Consider two samplescombine into a single list, but keep track ofwhich sample each value came fromrank the values in the combined list from lowto high
For ties, assign each the average rank of the tied values
separate back into two samples, each valuekeeping its assigned rankingsum the rankings for each sample
Mann-Whitney U-Test(continued)
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If the sum of rankings from one samplediffers enough from the sum of rankingsfrom the other sample, we conclude thereis a difference in the population medians
Mann-Whitney U-Test(continued)
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(continued)
Mann-Whitney U-Test
Mann-Whitney U-Statistics
111
211 2)1( TnnnnU
222
212 2)1( TnnnnU
where:n1 and n2 are the two sample sizes
T1 and T2 = sum of ranks for samples 1 and 2
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(continued)
Mann-Whitney U-Test
Claim: Median class size for Math is largerthan the median class size for English
A random sample of 9 Math and 9 Englishclasses is selected (samples do not have tobe of equal size), alfa = 0.05
Rank the combined values and then splitthem back into the separate samples
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Suppose the results are:
Class size (Math, M) Class size (English, E)
234534783466629581
304718344461542840
(continued)
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Size Rank18 123 228 330 434 634 634 640 844 9
Size Rank45 1047 1154 1261 1362 1466 1578 1681 1795 18
Ranking for combined samples
tied
(continued)
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Split back into the original samples:Class size(Math, M) Rank Class size
(English, E) Rank
234534783466629581
210616615141817
304718344461542840
411169131238
= 104 = 67
(continued)
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H0: MedianM = MedianE
HA: MedianM > MedianE
Claim: Median class size forMath is larger than themedian class size for English
221042
(9)(10)(9)(9)T2
1)(nnnnU 111
211
59672
(9)(10)(9)(9)T2
1)(nnnnU 222
212
Note: U1 + U2 = n1n2
(continued)
Mann-Whitney U-Test
Math:
English:
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• One side test
• For one-tailed tests like this one, check thealternative hypothesis to see if U1 or U2 should beused as the test statistic
• Since the alternative hypothesis indicates thatpopulation 1 (Math) has a higher median, use U1 asthe test statistic
• Use U1 as the test statistic: U = 22
• Compare U = 22 to the critical value U from theappropriate table
• If value for U is less than , where U=U1, Reject H0
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• n1 = 9, n2 = 9, U = 22
• So the table value is 0.0567 (> alfa 0.05)
• So the decision is do not reject Ho.
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Mann-Whitney U-Test forLarge Samples
The U statistic approaches a normal distributionas sample sizes increase
If samples (n1 and n2) are larger than 10, anormal approximation can be used
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Mann-Whitney U-Test forLarge Samples
The mean and standard deviation forMann-Whitney U Test Statistic:
(continued)
221
2
nnU
12)1)()(( 2121
2
nnnnU
Where n1 and n2 are sample sizes from populations 1 and 2
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Mann-Whitney U-Test forLarge Samples
Normal approximation for Mann-WhitneyU Test Statistic:
(continued)
12)1)()((
22121
212
nnnn
nnUz
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Large Sample Example
We wish to test
Suppose two samples are obtained:n1 = 40 , n2 = 50When rankings are completed, the sum of ranksfor sample 1 is T1 = 1475When rankings are completed, the sum of ranksfor sample 2 is T2 = 2620
H0: Median1 = Median2
HA: Median1 < Median2
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134514752
(40)(41)(40)(50)T2
1)(nnnnU 111
211
65526202
(50)(51)(40)(50)T2
1)(nnnnU 222
212
What’s the decision??
Reject Ho if Z < Z depends on the alternative hypothesis
Note: Normal approximation always use U2
Compute the U statistics:
Large Sample Example(continued)
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Since z = -2.80 < -1.645, we reject H0
645.1z
Reject H0
0MedianMedian:H0MedianMedian:H
21A
210
80.2
12)15040)(50)(40(
1000655
12)1nn)(n)(n(
2nnU
z2121
21
= .05
Do not reject H0
0
Large Sample Example(continued)
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Wilcoxon Matched-PairsSigned Rank Test
The Mann-Whitney U-Test is used whensamples from two populations are independent
If samples are paired, they are not independent
Use Wilcoxon Matched-Pairs Signed Rank Testwith paired samples
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The Wilcoxon T Test Statistic
Performing the Small-Sample WilcoxonMatched Pairs Test (for n < 15)
Calculate the test statistic T using these steps:
Step 1: collect sample data
Step 2: compute di = difference between thesample 1 value and its paired sample 2 value
Step 3: rank the differences, and give each rankthe same sign as the sign of the difference value
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The Wilcoxon T Test Statistic
Performing the Small-Sample WilcoxonMatched Pairs Test (for n < 15)
Step 4: The test statistic is the sum of the absolutevalues of the ranks for the group with the smallerexpected sum
Look at the alternative hypothesis to determinethe group with the smaller expected sumFor two tailed tests, just choose the smallersum
(continued)
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Small Sample Example
Paired samples, n = 9:
Value (before) Value (after)
384534583046425541
304718343431243840
baA
b0
MedianMedian:HMedianMedian:H a
Claim: Medianvalue is smallerafter than before
0MedianMedian:H0Median-Median:H
abA
a0 b
or can be stated as :
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Small Sample Example
Paired samples, n = 9:Value
(before)Value(after)
Differenced
Rankof d
Ranks with smallerexpected sum
364534583046425541
304718543831246240
6-2164-81518-71
4-283-679-51
2
6
5
= T = 13
(continued)
Dipilih jumlah T-
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The calculated T value is T = 13
Complete the test by comparing the calculatedT value to the critical T-value
For n = 9 and = .025 for a one-tailed test,T = 6
Since T > T , do not reject H0
T = 6
T = 13
do not reject H0reject H0
Small Sample Example(continued)
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Wilcoxon Matched Pairs Testfor Large Samples
The T statistic approaches a normaldistribution as sample size increases
If the number of paired values is larger than 15,a normal approximation can be used
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The mean and standard deviation forWilcoxon T :
(continued)
4)1(nn
T
24)12)(1)(( nnn
T
where n is the number of paired values
Wilcoxon Matched Pairs Testfor Large Samples
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Normal approximation for the Wilcoxon TTest Statistic:
(continued)
24)12)(1(
4)1(
nnn
nnTz
Wilcoxon Matched Pairs Testfor Large Samples
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Tests the equality of more than 2 populationmediansAssumptions:
variables have a continuous distribution.the data are at least ordinal.samples are independent.samples come from populations whose onlypossible difference is that at least one mayhave a different central location than theothers.
Kruskal-Wallis One-Way ANOVAClic
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Kruskal-Wallis Test Procedure
Obtain relative rankings for each value
In event of tie, each of the tied values getsthe average rank
Sum the rankings for data from each of the kgroups
Compute the H test statistic
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Kruskal-Wallis Test Procedure
The Kruskal-Wallis H test statistic:(with k – 1 degrees of freedom)
)1N(3nR
)1N(N12H
k
1i i
2i
where:N = Sum of sample sizes in all samplesk = Number of samplesRi = Sum of ranks in the ith sampleni = Size of the ith sample
(continued)
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Complete the test by comparing thecalculated H value to a critical 2 value fromthe chi-square distribution with k – 1degrees of freedom
(The chi-square distribution is Appendix G)Decision rule
Reject H0 if test statistic H > 2
Otherwise do not reject H0
(continued)
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Do different departments have different classsizes?
Kruskal-Wallis Example
Class size(Math, M)
Class size(English, E)
Class size(History, H)
2345547866
5560724570
3040183444
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Do different departments have different classsizes?
Kruskal-Wallis Example
Class size(Math, M) Ranking Class size
(English, E) Ranking Class size(History, H) Ranking
2341547866
2691512
5560724570
101114813
3040183444
35147
= 44 = 56 = 20
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The H statistic is
(continued)
Kruskal-Wallis Example
72.6)115(35
205
565
44)115(15
12
)1N(3nR
)1N(N12H
222
k
1i i
2i
equalareMedianspopulationallotN:H
MedianMedianMedian:H
A
HEM0
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Since H = 6.72 <do not reject H0
(continued)
Kruskal-Wallis Example
4877.9205.
Compare H = 6.72 to the critical value from thechi-square distribution for 5 – 1 = 4 degrees offreedom and = .05:
4877.9205.
There is not sufficient evidence to rejectthat the population medians are all equal
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Kruskal-Wallis Correction
If tied rankings occur, give each observation themean rank for which it is tiedThe H statistic is influenced by ties, and shouldbe corrected
Correction for tied rankings:NN
)tt(1 3
g
1ii
3i
where:g = Number of different groups of tiesti = Number of tied observations in the ith tied group of scoresN = Total number of observations
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H Statistic Corrected forTied Rankings
Corrected H statistic:
NN
)tt(1
)1N(3nR
)1N(N12
H
3
g
1ii
3i
k
1i i
2i
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