# Nonparametric Statistics

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1 Nonparametric Statistics J. Lozano University of Goettingen Department of Genetic Epidemiology Interdisciplinary PhD Program in Applied Statistics & Empirical Methods Graduate Seminar in Applied Statistics July 3, 2006

2 2

3 Parametric vs. Nonparametric Hypothesis Testing Procedures Parametric Nonparametric z-test t-test One-Way ANOVA Wilcoxon Mann Whitney Test Kruskal-Wallis Test Many More Tests Exist! 3

4 Learning Objectives: 1. Differentiate nonparametric from parametric statistics 2. Discuss the advantages and disadvantages of nonparametric statistics 3. Enumerate and differentiate the different nonparametric tests 4. Apply some commonly used nonparametric test to hypothesis testing problems 4

5 Parametric Test Procedures 1. Involve Population Parameters Example: Population Mean t X μ 0 (S n) x 2. Have Stringent Assumptions Example: Normal Distribution 3. Require Interval Scale or Ratio Scale Whole Numbers or Fractions Example: Height in Inches (72, 60.5, 54.7) 4. Examples: z-test, t-test, ANOVA μ 5

6 Nonparametric Test Procedures 1. Do Not Involve Population Parameters 2. No Stringent Distribution Assumptions Distribution-free 3. Data Measured on Any Scale Ratio or Interval Ordinal Example: Good-Better-Best Nominal Example: Male-Female 4. Example: Wilcoxon-Mann-Whitney Test 6

7 Advantages of Nonparametric Tests 1. Used With All Scales 2. Easier to Compute Developed Originally Before Wide Computer Use 3. Make Fewer Assumptions 4. Need Not Involve Population Parameters 5. Results May Be as Exact as Parametric Procedures 7

8 Disadvantages of Nonparametric Tests 1. May Waste Information If Data Permit Using Parametric Procedures Example: Converting Data From Ratio to Ordinal Scale 2. Require a larger sample size than the corresponding parametric test in order to achieve the same power 3. Difficult to Compute by Hand for Large Samples 4. Stat tables are not readily available 8

9 Summary Table of Statistical Tests Level of Measurement Sample Characteristics Correlation 1 Sample 2 Sample K Sample (i.e., >2) Independent Dependent Independent Dependent Categorical or Nominal Rank or Ordinal Χ 2 Run s test Kolmogorov Smirnov Χ 2 Fisher Exact Median Test Binomial Wilcoxon- Mann- Whitney Mc Nemar s Χ 2 Cochran s Q Sign Test Wilcoxon Signed Ranks Kruskal-Wallis Friendman s ANOVA Kappa Agreement Test Spearman s rho Kendall Rank Parametric (Interval & Ratio) z-test or t-test t- test between groups Paired t-test 1 way ANOVA between groups 1 way ANOVA (within or repeated measure) Pearson s r Factorial ANOVA 9

10 Application of Commonly Used Nonparametric Statistics 10

11 Frequency x x2 => nonparametric statistics 11

12 Wilcoxon-Mann-Whitney Test

13 Wilcoxon-Mann-Whitney Test Also known as Wilcoxon-test, Wilcoxon rank sum test, U-test, Mann-Whitney-U-test Tests Two Independent Population - compare medians Corresponds to t-test for 2 Independent Means Assumptions Independent, Random Samples Populations Are Continuous Can Use Normal Approximation If n i 20 13

14 Wilcoxon-Mann-Whitney Test Procedure 1. Assign Ranks, r i, to the n 1 + n 2 Sample Observations If unequal sample sizes, let n 1 refer to smaller-sized sample Smallest Value = 1 Average Ties 2. Sum the Ranks, R i, for Each Sample 3. Test Statistic Null hypothesis: both samples come from the same underlying distribution n ( n 1) U n n R where R is the sum of ranks in sample1 Compute also (n 1 n 2 U). If this expression is larger than the above U, then this becomes the final U statistic. ** At N>=20, U begins to approximate t, so the test stat changes to a t-value. 14

15 Wilcoxon-Mann-Whitney Test Example You re an agriculturist. You want to determine if there was a difference in the biomass of male and female Juniper trees. Randomly select 6 trees of each gender from the field. Dry them to constant moisture and weigh in kg. Male trees Data: 71, 73, 78, 75, 72, 74 Female trees Data: 80, 73, 83, 84, 82, 79 15

16 Wilcoxon-Mann-Whitney Test Solution H0: Ha: = n1 = n2 = Critical Value: Test Statistic: Decision: Conclusion: 16

17 Wilcoxon-Mann-Whitney Test Computation Table Male Trees Female Trees Weight Rank Weight Rank Rank Sum

18 Wilcoxon-Mann-Whitney Test Solution H0: Median weights = Ha: Medians not = =0.05 n1 = 6 n2 = 6 Critical Value: U table = 31 * Depends on the U table you are using Test Statistic: U = (6)(6) + (6)(7)/ = 32.5 n1n2 - U = (6)(6) 32.5 = 3.5 *!!! final U Calc = 32.5 Decision: Reject H 0 at =.05 since U calc > U table Conclusion: There is evidence that the medians are not equal. 18

19 NPar Tests Mann-Whitney Test Wilcoxon-Mann-Whitney Test Solution (SPSS Output) Ranks Weight in kg Gender of Tree Male Female Total N Mean Rank Sum of Ranks Mann-Whitney U Wilcoxon W Z Test Statistics b Asymp. Sig. (2-tailed) Exact Sig. [2*(1-tailed Sig.)] a. Not corrected for ties. Weight in kg a b. Grouping Variable: Gender of Tree p < ; Reject Ho 19

20 Sign Test

21 Sign Test 1. Tests One Population Median, (eta) 2. Corresponds to t-test for 1 Mean 3. Assumes Population Is Continuous 4. Small Sample Test Statistic: # Sample Values Above (or Below) Median 5. Can Use Normal Approximation If n 10 21

22 Sign Test Example You re a marketing analyst for Chefs-R-Us. You ve asked 8 people to rate a new ravioli on a 5-point Likert scale 1 = terrible to 5 = excellent The ratings are: At the.05 level, is there evidence that the median rating is at least 3? 22

23 Sign Test Solution H0: Ha: = Test Statistic: P-Value: Decision: Conclusion: 23

24 30% 20% Sign Test Uses P-Value to Make Decision P(X) Binomial: n = 8 p = % 0% X P-Value is the probability of getting an observation at Least as extreme as we got. If 7 of 8 Observations Favor H a, Then P-Value P = P(x 7) = =.035. If =.05, Then Reject H 0 Since P-Value P. 24

25 Sign Test Solution H0: = 3 Ha: < 3 =.05 Test Statistic: S = 7 (Ratings 1 & 2 are less than = 3: ) P-Value: P(x 7) = =.035 (Binomial Table, n = 8, p = 0.50) n! x n X P( X ) (1 ) X!( n X )! Decision: Reject H 0 at =.05 Conclusion: There is evidence that the median is less than 3 25

26 Sign Test (R Output) R version (BSDA and ev10 package must be loaded) > x <- c(2,4,1,2,1,1,2,1) > sign.test(x,md=3, alternative="less") One-sample Sign-Test data: x s = 1, p-value = alternative hypothesis: true median is less than 3 95 percent confidence interval: -Inf 2 sample estimates: median of x 1.5 \$Confidence.Intervals Conf.Level L.E.pt U.E.pt Lower Achieved CI Inf 2 Interpolated CI Inf 2 Upper Achieved CI Inf 2 Warning message: multi-argument returns are deprecated in: return(rval, Confidence.Intervals) 26

27 Mc Nemar Change Test

28 McNemar Change Test (MCT) Typical examples: Testing the shift in the proportion of abnormal responses from before and after treatment in the same group of patients Comparing two ocular treatments when both are given to each patient, one in each eye 28

29 Data Layout: MCT Condition 1 No. of Responders No. of Non- Responders No. of Responders Condition 2 No. of Non- Responders TOTAL A B A + B C D C + D TOTAL A + C B + D N= A +B +C +D 29

30 McNemar Change Test (MCT) The hypothesis of interest is the equality of response proportions, p 1 and p 2, under conditions 1 and 2, respectively. The test statistic is based on the difference in the discordant cell frequencies (B, C). 30

31 Test Summary: MCT Null hypothesis: H 0 : p 1 = p 2 Alternative hypothesis: H a : p 1 p 2 Test Statistic: χ 2 χ 2 α 1 where 1 is the critical value from the chi-square table with significance level and 1 degree of freedom Decision Rule:Reject H 0 if χ 2 α 2 χ C B C B 2 31

32 Example: MCT Bilirubin Abnormalities Following Drug Treatment 86 patients were treated with an experimental drug for 3 months. Pre-post study clinical laboratory results showed abnormally high total bilirubin values. Is there evidence of a change in the preto post-treatment rates of abnormalities? 32

33 Solution: MCT Let p 1 and p 2 represent the proportions of patients with abnormally high bilirubin values ( Y ) before and after treatment, respectively. POST- Treatment N Y TOTAL PRE- N Treatment Y TOTAL Y = T.Bilirubin above upper limit of normal range 33

34 Test Summary: MCT Null hypothesis: H 0 : p 1 = p 2 Alternative hypothesis: H a : p 1 p 2 Test Statistic: B C χ 3.20 B C Decision: Do not reject H 0 since χ 2 < χ 2 α Conclusion: There is no sufficient evidence to conclude at 0.05 level of significance that a shift in abnormality rates occurs with treatment. 34

35 SAS Computer Output for MCT McNemar's Test Example: Bilirubin Abnormalities Following Drug Treatment TABLE OF PRE BY PST PRE(PRE) PST(PST) Frequency Percent Row Pct Col Pct N Y Total ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ N ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Y ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Total STATISTICS FOR TABLE OF PRE BY PST McNemar's Test Statistic = DF = 1 Prob =

36 References Dawson-Saunders, B. and Trapp, R.G. Basic and Clinical Biostatistics. Connecticut: Prentice Hall International, Inc. (1990). McGill, J.J. Chap. 14: Nonparametric Statistics Presentation. (May 2006). Siegel S. and Castellan N.J. Nonparametric Statistics for the Behavioral Sciences (2nd edition). New York: McGraw Hill (1988). Walker G.A. Common Statistical Methods for Clinical Research with SAS Examples, SAS Institute, Inc, Cary NC (1997). 36

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