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1 Nonparametric Statistical Techniques Chapter 17

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Nonparametric Statistical Techniques. Chapter 17. 17.1 Introduction. The statistical techniques introduced in this chapter deal with ordinal data. We test to determine whether the population locations differ. - PowerPoint PPT Presentation

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  • Nonparametric StatisticalTechniquesChapter 17

  • 17.1 IntroductionThe statistical techniques introduced in this chapter deal with ordinal data.We test to determine whether the population locations differ. In testing the locations we will not refer to any parameter, thus the procedures name.

  • 17.1 IntroductionWhen comparing two populations the hypotheses generally are:H0: The population locations are the sameH1: (i) The locations differ, or (ii) Population 1 is located to the right (left) of population 2 The random variable X1 is generally larger (smaller) than X2.

  • 17.2 Wilcoxon Rank Sum TestThe problem characteristics of this test are:The problem objective is to compare two populations.The data are either ordinal or interval (but not normal).The samples are independent.

  • Wilcoxon Rank Sum Test ExampleExample 17.1Based on the two samples shown below, can we infer at 5% significance level that the location of population 1 is to the left of the location of population 2? Sample 1: 22, 23, 20; Sample 2: 18, 27, 26; The hypotheses are: H0: The two population locations are the same.H1: The location of population 1 is to the left of the location of population 2.

  • Graphical DemonstrationWhy use the sum of ranks to test locations? If the locations of the two populations are about the same, (the null hypothesis is true)we would expect the ranks to be evenly spread between the samples.In this case the sum of ranks for the two samples will be close to one another.Two hypothetical populations and their corresponding samples are presented, the GREEN population and the PURPLE population.PopulationsLet us rank the observations of the two samples together

  • Graphical DemonstrationWhy use the sum of ranks to test locations? Allow the GREEN population to shift to the left of the PURPLE population.

  • Graphical DemonstrationWhy use the sum of ranks to test locations? The green sample is expected to shift to the left too.As a result, several observations exchange location.What happens to the sum of ranks? Click.

  • Graphical DemonstrationWhy use the sum of ranks to test locations? 671345810111292The green sum decreases , and the purple sum increases.Changing the relative location of two populations affect the sum of ranks of the two samples combined.

  • Wilcoxon Rank Sum Test ExampleExample 17.1 continuedTest statistic 1. Rank all the six observations (1 for the smallest). 3. Let T = 9 be the test statistic (We arbitrarily define the test statistic as the rank sum of sample 1.

  • Wilcoxon Rank Sum Test RationaleExample 17.1 - continuedIf T is sufficiently small then most of the smaller observations are located in population 1. Reject the null hypothesis.Question: How small is sufficiently small?We need to look at the distribution of T.

  • 1,2,36 7 8 9 10 11 12 13 14 151,2,41,2,51,2,61,3,41,3,61,3,51,4,51,4,61,5,62,3,42,3,52,3,62,4,52,4,62,5,63,4,53,4,63,5,64,5,6T.05.10.15T is the rank sum of a sample of size 3.This sample received the ranks 3, 4, 5If H0 is true (the two populations have the same location), each ranking is equally likely, and each possible value of T has the same probability = 1/20This sample received the ranks 1, 2, 3The distribution of T under H0 for two samples of size 3

  • The distribution of T under H0 for two samples of size 31,2,36 7 8 9 10 11 12 13 14 151,2,41,2,51,2,61,3,41,3,61,3,51,4,51,4,61,5,62,3,42,3,52,3,62,4,52,4,62,5,63,4,53,4,63,5,64,5,6T.05.10.15 The significance level is 5%,and under H0 P(T 6) = .05.Thus, the critical value of T is 6.

  • Wilcoxon Rank Sum Test ExampleExample 17.1 - continuedConclusionH0 is rejected if T6. Since T = 9, there is insufficient evidence to conclude that population 1 is located to the left of population 2, at the 5% significance level.

  • Critical values of the Wilcoxon Rank Sum Testa = .025 for two tail test, or a = .05 for one tail testUsing the table: For given two samples of sizes n1 and n2, P(TTU)= a.For a two tail test: P(T25) = .025 if n1=4 and n2=4.For a one tail test: P(T25) = .05 if n1=4 and n2=4.11 25A similar table exists for a = .05 (one tail test) and a = .10 (two tail test)TL TU TL TU TL TU TL TU

    Sheet1

    n2n1

    345...10

    46 1811 2517 33...61 89

    56 2112 2818 37...64 96

    .

    .

    .

    109 3316 4424 56...79 131

  • Wilcoxon rank sum test for samples where n > 10The test statistic is approximately normally distributed with the following parameters: Therefore,

    Z =T - E(T)sT

  • Wilcoxon rank sum test for samples where n > 10, ExampleA pharmaceutical company is planning to introduce a new painkiller.To determine the effectiveness of the drug, 30 people were randomly selected.15 were given the tested drug (Sample 1).15 were given aspirin (Sample 2).Each participant was asked to indicate which one of five statements best represented the effectiveness of the drug they took. Example 17.2 (using Wilcoxon rank sum test with ordinal data)

  • Wilcoxon test for samples where n > 10, ExampleExample 17.2 continuedSummary of the experiment results.SolutionThe objective is to compare two populations of ordinal data.The two samples are independent.Wilcoxon rank test is the appropriate technique to apply.

    Sheet1

    The drug taken wasPainkillerAspirin

    extremely effective (5)61

    quite effective (4)35

    somewhat effective (3)43

    slightly effective (2)14

    not at all effective (1)12

  • Wilcoxon rank sum test for samples where n > 10, ExampleThe hypothesesH0: The locations of population 1 and 2 are the sameH1: The location of population 1 is to the right of the location of population 2.Received the new painkillerReceived AspirinSolving by handTo reject the null hypothesis, we need to show that z is large enough.First we rank the observations, Secondly, we run a z-test, with rejection region of Z > Za.

  • Wilcoxon rank sum test for samples where n > 10, ExampleRanking the raw dataThere are three observationswith an effectiveness score of 1.The original ranks for these observations are 1, 2 , and 3.This tie is broken by giving eachobservation the average rank of 2.Sum of ranks:T1=276.5T2=188.5

    Sheet1

    PainkillerRankAspirinRank

    1212

    2612

    31226

    31226

    31226

    31226

    419.5312

    419.5312

    419.5312

    527419.5

    527419.5

    527419.5

    527419.5

    527419.5

    527527

    &A

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  • Wilcoxon rank sum test for samples where n > 10, ExampleTo standardize the test statistic we need:E(T) = n1(n1+n2+1)/2= (15)(31)/2=232.5

  • Wilcoxon rank sum test for samples where n > 10, ExampleFor 5% significance level z=1.645.Since z = 1.83 > 1.645, there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.At 5% significance level, the new drugs is perceived as more effective than Aspirin.

  • Wilcoxon rank sum test for samples where n > 10, ExampleExcel solution (Xm17-02)

    WILCOXON1

    Wilcoxon Rank Sum Test

    Rank sumObservations

    New276.515

    Aspirin188.515

    z Stat1.83

    P(Z

  • Wilcoxon rank sum test for non-normal interval data, ExampleThe human resource manager of a large company wanted to compare how long business and non-business graduates worked for the company before quitting.Two samples of 25 business graduates and 20 non-business graduates were randomly selected.The data representing their time with the company were recorded. Retaining Workers

  • Wilcoxon rank sum test for non-normal interval data, ExampleCan the personnel manager conclude at 5% significance level that a difference in duration of employment exists between business and non-business graduates? Retaining workers - continued

    Sheet1

    BusinessNon-Bus

    6025

    1160

    1822

    1924

    523

    2536

    ..

    ..

    ..

    1716

    3728

    49

    860

    2829

    2716

    1122

    6060

    2517

    560

    1332

    22

    11

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  • Wilcoxon rank sum test for non-normal interval data, ExampleSolution The problem objective is to compare two populations of interval data.The samples are independent.The non-normality of the two populations is apparent from the sample histograms:

  • Wilcoxon rank sum test for non-normal interval data, Example Solution continuedThe Wilcoxon rank test is the correct procedure to run.H0: The two population locations are the sameH1: The location of population 1(business graduates) is different from the location of population 2 (non- business graduates).

  • Wilcoxon rank sum test for non-normal interval data, ExampleSolution continuedSolving by hand The rejection region is After the ranking process is completed, we have: T = Tbusiness graduates = 463. E(T) = n1(n1+n2+1)/2=575; sT=[n1n2(n1+n2+1)/12]1/2=43.8

  • Wilcoxon rank sum test for non-normal interval data, ExampleExcel solution (Workers.xls)

    There is a strong evidence to infer that the duration of employment is different for business and non-business graduates

    WILCOXON2

    Wilcoxon Rank Sum Test

    Rank SumObservations

    Business46325

    Non-Bus57220

    z Stat-2.56

    P(Z

  • Required conditions for nonparametric testsA rejection of the null hypothesis when performing a nonparametric test can occur due to:different locationdifferent spread (variance)different shape (distribution).Since we are interested in the location, we require that the two distributions are identical, except for location.

  • 17.3 Sign Test and Wilcoxon Signed Rank Sum TestTwo techniques for matched pairs experiment are introduced. the objective is to compare two populations.the data are either ordinal or interval (but not normal).The samples are matched by pairs.

  • The Sign TestThis test is employed when:The problem objective is to compare two populations, andThe data are ordinal, andThe experimental design is matched pairs.The hypotheses H0: The two population locations are the same H1: The two population locations differ or population 1 is right (left) of population 2

  • The Sign Test Statistic and Sampling DistributionA matched pair experiment calls for a test of matched pair differences. The test statistic and sampling distributionRecord the sign of all the matched-pair-differences.The number of positive (or negative) differences is the test statistic.

  • The Sign Test - RationaleThe number of positive or negative differences is binomial, with:n = the number of non-zero differencesp = the probability that a difference is positive (negative) If the two populations have the same locations (H0 is true), it is expected that

    Thus, under H0: p = 0.5Number of positive differences = Number of negative differences

  • The Sign Test - RationaleThe test statistic and sampling distributionThe hypotheses:H0: The two population locations are the sameH1: The two population locations are different

    H0: p = .5H1: p .5

  • The Sign Test Statistic and Sampling DistributionThe Test continuedThe hypotheses testedH0: p = .5H1: p .5The binomial variable can be approximated by a normal variable if np and n(1-p) > 5.The Z- statistic becomes

  • The Sign Test ExampleExample 17.3 (Xm17-03)In an experiment to determine which car is perceived to have the more comfortable ride, 25 people took two rides:One ride in a European model.One ride in a North American car.Each person ranked the cars on a scale of 1 (ride is very uncomfortable) to 5 (ride is very comfortable).

  • The Sign Test ExampleDo these data allow us to conclude at 5% significance level that the European car is perceived to be more comfortable?

    Sheet1

    RespondentEuropeanAmericanDifference

    145-1

    2211

    3541

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    5211

    6532

    713-2

    8422

    9422

    ...0

    ...1

    ...1

    13211

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  • The Sign Test ExampleSolution We compare two populations The data are ordinal A matched pair experiment

    Sheet1

    RespondentEuropeanAmericanDifference

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    713-2

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    ...0

    ...1

    ...1

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    Sheet1

    RespondentEuropeanAmericanDifference

    145-1

    2211

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    5211

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    713-2

    8422

    9422

    1022.

    11321

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  • The Sign Test ExampleSolutionThe hypotheses are:

    H0: The two population location are the same. H1: The European car population is located to the right of the American car population.The test.There were 18 positive, 5 negatives, and 2 zero differences. Thus, X = 18, n = 23(!).Z = [x-np]/[np(1-p)].5 = [18-.5(23)]/[.5{23}.5] = 2.71The rejection region is z > za. For a = .05 we have z > 1.645. The p-value = P(Z > 2.71) = .0034

  • The Sign Test ExampleUsing the computer: Tools > Data Analysis Plus > Sign TestExcel Solution (Xm17-03)

    SIGNTEST2

    Sign Test

    DifferenceEuropean - American

    Positive Differences18

    Negative Differences5

    Zero Differences2

    z Stat2.71

    P(Z

  • The Sign Test ExampleConclusion: Since the p-value < a we reject the null hypothesis.At 5% significance level there is sufficient evidence to infer that the European car is perceived as more comfortable than the American car.

  • The Sign Test ExampleChecking the required conditionsObserve the sample histograms (Xm17-03)

    The populations are similar in shape and spread

    Chart1

    1

    7

    6

    7

    4

    0

    Euro

    European cars

    Sheet1

    EuropeanAmericanBin

    451

    212

    543

    324

    215

    53

    13BinEuro

    4211

    4227

    2236

    3247

    4354

    21More0

    34

    21BinFrequency

    4316

    2127

    4337

    5444

    3151

    42More0

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    23

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    0

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    Frequency

    American cars

    0

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    Euro

    European cars

    Chart2

    6

    7

    7

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    Frequency

    American cars

    Sheet1

    EuropeanAmericanBin

    451

    212

    543

    324

    215

    53

    13BinEuro

    4211

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    2236

    3247

    4354

    21More0

    34

    21BinFrequency

    4316

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    42More0

    33

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    Sheet1

    Frequency

    American cars

    Euro

    European cars

  • Wilcoxon Signed Rank Sum TestThis test is used whenthe problem objective is to compare two populations,the data are interval but not normal,the samples are matched pairs.The test statistic and sampling distributionT is based on rank sum of the absolute values of the positive and negative differencesWhen n TU or T 30, T is approximately normally distributed. Use a Z-test.

  • Wilcoxon Signed Rank Sum Test,ExampleExample 17.4Does flextime work-schedule help reduce the travel time of workers to work?A random sample of 32 workers was selected, and workers recorded their travel time before and after the program was implemented.The hypotheses test areThe two population locations are the same.The two population locations are different.

  • Wilcoxon Signed Rank Sum Test, ExampleExample 17.4Does flextime work-schedule help reduce the travel time of workers to work?A random sample of 32 workers was selected, and workers recorded their travel time before and after the program was implemented.The hypotheses areH0: The two population locations are the same.H1: The two population locations are different.The rejection region:|z| > za/2

  • This data were sorted by the absolute valueof the differences.12345678Ties were broken by assigning the average rank to the tied observations

    Sheet1

    Worker8:00-ArrFlextimeDifferenceABS(Diff.)Ranks

    34344-114.5

    51615114.5This data was sorted by the absolute value

    83839-114.5of the differences.

    121312114.5

    161819-114.5The ties were broken by assigning the average

    231918114.5of all the ranks that belong to a certain tie, to

    275150114.5all the tie members.

    302019114.5

    446442213

    62628-2213

    96163-2213

    105254-2213

    136971-2213

    155355-2213

    1825232213

    2840382213

    311921-2213

    134313321

    1168653321

    1741383321

    1917143321

    223033-3321

    244851-3321

    2624213321

    235314427

    2144404427

    252933-4427

    2926224427

    3242384427

    768635531

    1418135531

    2026215531

    Sheet2

    Sheet3

  • T is the rank sum of the positive differences. T = T+ = 367.5E(T) = n(n+1)/4 = 32(33)/4 = 264 sT = [n(n+1)(2n+1)/24].5 = 53.48The test statistic is:

    Z =

    T= 1.94367.5 -

    Sheet1

    Worker8:00-ArrFlextimeDiffertenceABS(Diff.)Ranks

    34344-114.5

    51615114.5This data was sorted by the absolute value

    83839-114.5of the differences.

    121312114.5

    161819-114.5The ties were broken by assigning the average

    231918114.5of all the ranks that belong to a certain tie, to

    275150114.5all the tie members.

    302019114.5

    446442213

    62628-2213

    96163-2213

    105254-2213

    136971-2213

    155355-2213

    1825232213

    2840382213

    311921-2213

    134313321

    1168653321

    1741383321

    1917143321

    223033-3321

    244851-3321

    2624213321

    235314427

    2144404427

    252933-4427

    2926224427

    3242384427

    768635531

    1418135531

    2026215531

    Sheet2

    Sheet3

  • Wilcoxon Signed Rank Sum Test,ExampleExcel solution (Xm17-04)

    Sheet1

    8:00-ArrFlextime

    3431

    3531

    4344

    4644

    1615

    2628

    6863

    3839

    6163

    5254

    6865

    1312

    6971

    1813

    5355

    1819

    4138

    2523Wilcoxon Signed Rank Sum Test

    1714

    2621Difference8:00-Arr - Flextime

    4440

    3033T+367.5

    1918T-160.5

    4851Observations (for test)32

    2933z Stat1.94

    2421P(Z

  • Wilcoxon Signed Rank Sum Test,ExampleSolution continued The rejection region for a = .05 is |z| > z.025 = 1.96Conclusion: Since |1.94| < 1.96, There is insufficient evidence to infer that the flextime program was effective at 5% significance level.

  • 17.4 Kruskal-Wallis TestThe problem characteristics for this test are:The problem objective is to compare two or more populations.The data are either ordinal or interval but not normal. The samples are independent.The hypotheses areH0: The location of all the k populations are the same.H1: At least two population locations differ.

  • Kruskal-Wallis Test StatisticRank the data from 1(smallest) to n (largest).Calculate the rank sums T1, T2,Tk for all the k samples.Calculate the statistic H as follows:

  • Test Rationale and Rejection regionIf all the populations have the same location (H0 is true)The ranks should be evenly distributed among the k samples. The statistic H will be small.

  • Test Rationale and Rejection RegionSampling distributionWhen the sample sizes 5, H is approximately chi-squared distributed with k-1 degrees of freedom.The rejection region: Since a large value of H justifies the rejection of H0, we have:

  • The Kruskal-Wallis Test ExampleExample 17.5How do customers rate three shifts with respect to speed of service in a certain restaurant?Three samples of 10 customer response-cards were randomly selected, one sample from each shift.Customer ratings were recorded.

  • The Kruskal-Wallis Test ExampleCan we conclude that customers perceive the speed of service to be different among the three shifts at 5% significance level?

    Sheet1

    4:00-midMid-8:008:00-4:00

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  • The Kruskal-Wallis Test ExampleSolutionThe problem objective is to compare three populations.The data are ordinal.The hypotheses:H0: The locations of all three populations are the same.H1: At least two population locations differ.

  • The Kruskal-Wallis Test ExampleSolution - continuedTest statistic:

    T1 = 186.5 T2 = 156.0 T3 = 122.5n = n1 + n2 + n3 = 10+10+10 = 30Ranking

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  • The Kruskal-Wallis Test ExampleFor a = .05, c2a,k-1 = c2.05,2 = 5.99147Solution - continuedThe critical value

  • The Kruskal-Wallis Test ExampleSolution Excel (Xm17-05)

    WALLIS1

    Kruskal-Wallis Test

    GroupRank SumObservations

    4:00-mid186.510

    Mid-8:0015610

    8:00-4:00122.510

    H Stat2.64

    df2

    p-value0.2665

    chi-squared Critical5.9915

    Sheet1

    4:00-midMid-8:008:00-4:00

    433

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  • The Kruskal-Wallis Test ExampleConclusion: Since H=2.64 < 5.99147, do not reject the null hypothesis. There is insufficient evidence to conclude at 5% significance level, that there is a difference in customers perception regarding service speed among the three shifts.

  • 17.5 Friedman TestThe problem characteristics of this test are:The problem objective is to compare two or more populations.The data are either ordinal or interval but not normal. (For normal populations we use ANOVA).The data are generated from a blocked experiment (samples are not independent).The hypotheses areThe location of all the k populations are the same.At least two population locations differ.

  • Test Statistic and Rejection RegionThe test statistic is

    The rejection region is b = the number of blocksK = the number of treatments

  • The Friedman Test ExampleExample 17.6Four managers evaluate applicants for a job in an accounting firm on several dimensions.Eight applicants were randomly selected, and their evaluations by the four managers recorded.Can we conclude at 5% significance level thatthere are differences inthe way managersevaluate candidates?

    Sheet1

    Manager

    Applicant1234

    12122

    24232

    32223

    43132

    53235

    62234

    74155

    83253

    &A

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  • The Friedman Test ExampleSolutionThe problem objective is to compare four populationsData are ordinal.This is a randomized block design experiment because each applicant (block) was ranked four times.The appropriate procedure is the Friedman test

  • The Friedman Test ExampleSolutionThe hypotheses areH0: The locations of all four populations are the same.H1: At least two population locations differ.The data

    Sheet1

    Manager

    Applicant1234

    12122

    24232

    32223

    43132

    53235

    62234

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  • The Friedman Test ExampleT1 = 21 T2 = 10 T3 = 24.5 T4 = 24.53423.52.51.522.511.5211.51113323.52.533.5431.542443.52.5How to rank, block by block.

    Applicant 1:Scores: 2 1 2 2Actual ranks: 2 1 3 4Averaged ranks: 3 1 3 3

    Sheet1

    Manager

    Applicant1234

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  • The Friedman Test ExampleSolutionIn our problem:b = 8 (number of blocks) k = 4 (number of treatments, populations)

  • The Friedman Test ExampleSolutionWe have : Fr = 10.61; Let a = .05, then c2.05, 4-1 = 7.8147

  • The Friedman Test ExampleSolution Excel (Xm17-06)

    FRIEDMAN1

    Friedman Test

    GroupRank Sum

    Manager121

    Manager210

    Manager324.5

    Manager424.5

    Fr Stat10.61

    df3

    p-value0.0140

    chi-squared Critical7.8147

    Sheet1

    Manager1Manager2Manager3Manager4

    2122

    4232

    2223

    3132

    3235

    2234

    4155

    3253

  • The Friedman Test ExampleConclusion: Since Fr =10.61> 7.8147, reject the null hypothesis. There is sufficient evidence to conclude at 5% significance level, that the managers evaluations differ.