Chapter 14Nonparametric StatisticsIn all of the inference procedures we have seen INTHE PREVIOUS CHAPTERS there are someassumptions that must be satisfied and we must checkthem before proceeding with the procedure.ResponseInferences for AssumptionsCategorical Proportions (p or p1 – p2) Contingency tables Logistic RegressionSRSLarge Sample(s) QuantitativeMeans (µ, µd, or µ1 – µ2)ANOVASRL, MLRSRSNormality Constant 2 In general we prefer to use tests based on normalityassumptions since they perform “better” even whenthere is a slight deviation from a normal distribution. Chapter 14, Fall 2007 Page 1 of 19However, we should not use them when There is some evidence in the sample data toindicate that the population is highly skewed The sample sizes are too smallIn such cases we prefer to use non-parametric methodswhere there are fewer and less restrictive assumptions.In fact, the methods of nonparametric statistics are alsocalled “distribution free” methods.Nonparametric Procedures The response variable is quantitative but does nothave to satisfy the assumptions of the methodsbased on normal distribution of response. Use these methods with o Small samples and/or o Skewed distributions and/oro Groups have very different variances Assumptionso Random sampleso Quantitative Data Inferences are about the population medians, NOTabout mean(s)Chapter 14, Fall 2007 Page 2 of 19We will see three of these tests:Type of inferenceRank Tests Normal Tests Two independentsamples Wilcoxon Rank Sum Test (Mann-Whitney Test)CIs and Significancetests for µ1 – µ2(t-test)Matched pairsWilcoxon signed rank test CIs and Significancetests for µd (t-test)Several independentgroupsKruskal Wallis TestANOVAChapter 14, Fall 2007 Page 3 of 1914.1 Wilcoxon Rank Sum Test (For 2 independent samples)Two Examples: Suppose we want to compare thegrades of two students in each of the followingexamples. Example 1Example 2Bob Hope65 (3)78 (7)68 (4)61 (1)63 (2)69 (5)70 (6)Jill Jack70 (6.5)68 (3.5)72 (8.5)72 (8.5)70 (6.5)68 (3.5)63 (1)69 (5)65 (2)Rank all observations(for the 2 students together) from 1 to 7Rank all observations (for the 2 students together) from 1 to 9. Break tiesSum Ranks of each Bob: 3+7+4+1 = 15Hope: 2+5+6 = 13Sum Ranks of each Jill: 6.5+3.5+8.5+8.5+6.5 = 33.5Jack: 3.5+1+5+2 = 11.5Find Average of RanksBob: 15/4 = 3.75Hope: 13/3 = 4.33Find Average of RanksJill: 33.5/5 = 6.7Jack: 11.5/4 = 2.875Chapter 14, Fall 2007 Page 4 of 19Using the Minitab Output, we can state the following:Example 1 Example 294.8% CI for 1 – 2 is (–9, 15). Since 0 is in CI we fail to reject Ho when we use α = 0.052 (=1 – 0.948)96.3% CI for 1 – 2 is (– 0.001, 8.999). The CI (barely) includes zero, so we fail to reject Ho when α = 0.037 = (1 – 0.963)p-value = 0.8597 (2-sided)We fail to reject Ho for any level of significance.p-value = 0.0250 (2-sided) so we reject Ho for α = 0.05 or 0.10[for any α > 0.0250].Conclusion: The observed data do not indicate any significant difference between the performance of Bob and Hope.Conclusion: The observed data give some evidence of significant difference in the performance of Jack and Jill.The Minitab output appears on the next pageChapter 14, Fall 2007 Page 5 of 19Example 1 - Student grades - are they different? Bob Hope 65 63 78 69 68 70 61 Mann-Whitney Test and CI: Bob, Hope N Median Bob 4 66.50 Hope 3 69.00 Point estimate for ETA1-ETA2 is -1.50 94.8 Percent CI for ETA1-ETA2 is (-9.00, 15.00) W = 15.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.8597 Example 2 - Student grades - is student 1 better? Jill Jack 70 68 68 63 72 69 72 65 70 Mann-Whitney Test and CI: Jill, Jack N Median Jill 5 70.000 Jack 4 66.500 Point estimate for ETA1-ETA2 is 4.000 96.3 Percent CI for ETA1-ETA2 is (-0.001, 8.999) W = 33.5 Test of ETA1 = ETA2 vs ETA1 > ETA2 is significant at 0.0250 The test is significant at 0.0236 (adjusted for ties) Note: When you have few observations you may use a worksheet as shown below for easy ranking (especially when there are ties):Jill 63(1) 65 (2) 68 (3.5) 70 (6.5) 72 (8.5)Jack 68 (3.5) 69 (5) 70 (6.5) 72 (8.5)Chapter 14, Fall 2007 Page 6 of 19Alternatively you may order all observations and put inone row and the corresponding ranks underneath each observed value,identified by the groupObserva-tions162171193196207211224241243250271288484Sum of ranksMean rankA 1 2 3 4 7 8 2525/6= 4.17B 5 6 9 10 11 12 13 6666/7= 9.43Example: Compare the reaction times (in milliseconds)of subjects under 2 drugs (different subjects for each drug).Drug A Drug B1.96 (4)2.24 (7)1.71 (2)2.41 (8)1.62 (1)1.93 (3)Sum of Ranks = 25Mean rank = 25/6 = 4.172.11 (6)2.43 (9)2.07 (5)2.71 (11)2.50 (10)4.84 (13)2.88 (12)Sum of Ranks = 66Mean rank = 66/7 = 9.43Chapter 14, Fall 2007 Page 7 of 19What type of a problem do we have here? Treatments: 2 treatments (Drug A and B) Groups: 2 independent groups Response: Reaction time (Quantitative)Comparing the mean (or median) of a quantitativevariable in 2 populations (groups) using independentsamples. Which test to use: Should we use the t-test or theWilcoxon Rank sum test? It seems that we could useeither. However, first we must check the assumptions.Assumptions for Wilcoxon rank sum test: SRS and  Quantitative response (both satisfied)] Assumptions for the t-test:  SRS (OK) and Normal populations (Not OK, there is an outlier)So we should not use the t-test. Chapter 14, Fall 2007 Page 8 of 19Steps for Wilcoxon Test: 1) Rank the ALL observations together2) Sum the ranks of each group3) Find the average of ranks for each group4) Look at the computer output: 96.2% CI for 1 – 2 is (– 0.9497, – 0.1099) 2-sided p-value = 0.0184 Conclusion: The observed data indicatesignificant differences between the median(mean?) reaction times of the two drugs at 5%and 10% levels of significance.Note: Minitab reports Mann-Whitney test which is“equivalent” to the Wilcoxon Rank-Sum test. In sometexts you may see this as the Wilcoxon-Mann-Whitneytest.Example - Reaction Time under Drugs A and B A B 1.96 2.11 2.24 2.43 1.71 2.07 2.41 2.71 1.62 2.50 1.93 2.84 2.88