Data are sometimes not compatible with the assumptions of parametric statistical tests (i.e. t-test, regression, ANOVA)

Transcription

1 BSTT523 Pagano & Gauvreau Chapter 13 1 Nonparametric Statistics Data are sometimes not compatible with the assumptions of parametric statistical tests (i.e. t-test, regression, ANOVA) In particular, data may not be normally distributed with homogeneous variance, even after data transformation In these cases, alternatives to parametric methods are needed These alternatives are called nonparametric techniques since the null and alternative hypotheses are not stated in terms of specific parameters (μ) of an assumed distribution (normal) For example, H0: Two groups are from populations with the same median H1: The medians of the two populations are different The median, to be sensible, only assumes that the variable of interest in on at least an ordinal scale no assumption about a parametric distribution is made

2 BSTT523 Pagano & Gauvreau Chapter 13 2 Advantages of nonparametric methods: In most cases, they are easy to apply Useful for analysis of ranks or ordinal data (i.e. no need to assume interval or ratio-scaled data) Fewer and less stringent assumptions are made than for parametric methods Depending on the procedure, they may be almost as powerful as the corresponding parametric procedure when the assumptions of the latter are met More reliable when parametric assumptions are not reasonable Disadvantages of nonparametric methods: If the assumptions of the parametric methods can be met, they are more efficient (i.e. greater statistical power) Parametric methods are fairly robust to certain types of departure from the normal assumptions Difficult to construct elaborate models (e.g. models with many independent variables and/or interactions) with nonparametric methods

3 BSTT523 Pagano & Gauvreau Chapter 13 3 Nonparametric (NP) Test Procedures Association NP counterpart to Pearson correlation Spearman correlation, based on ranks Paired samples NP counterparts to paired t-test Sign Test Wilcoxn Signed-Rank Test Two independent samples NP counterparts to independent-samples t-test Median Test Wilcoxon Rank Sum Test (Mann-Whitney U Test) Multiple samples NP counterpart to one-way ANOVA Kruskal-Wallis Test Blocked designs NP counterpart to Randomized Blocks ANOVA Friedman s Test

4 BSTT523 Pagano & Gauvreau Chapter 13 4 Sign Test random sample of pairs of observations H0: the median difference is zero Data: N pairs (X i, Y i ) i = 1,, N Procedure: for each pair assign the sign if X i < Y i then sign = + if X i > Y i then sign = - if X i = Y i then sign = 0 (tied observations excluded) Test statistic: n + = number of + pairs If H0 is true, distribution of n + is binomial(n,1/2) (where n is the number of untied pairs) compare n + with binomial table Large-sample approximation (for n>20): Z = n + n 2 n 4 ~N(0,1) under H0 note: E(n + H 0 ) = n 2, V(n + H 0 ) = n 4

5 BSTT523 Pagano & Gauvreau Chapter 13 5 Example: Matched-pairs design involving change scores in self-perception of health (Y) among hyptertensives Pair Group Total Mean Treatment Control Total Difference n + = 11 From Binomial table, n + needs to be at least 12 to reject with 2-sided α=.05 Use Table A.1 in P&G: P(n + 12 n = 15, p =.5) = P(n + 3 n = 15, p =.5) = =.0176 two-tailed p-value = 2(.0176)=.0352 P(n + 11 n = 15, p =.5) = P(n + 4 n = 15, p =.5) = =.0593 two-tailed p-value = 2(.0593)=.1186 Normal approximation: Z = = = 1.81 p=2(.035)=.07 cannot reject H0 that the median difference is zero (paired t-test yields T=3.109, df=14, p<.008)

6 BSTT523 Pagano & Gauvreau Chapter 13 6 Wilcoxon Signed Rank Test Sign test discards any scale information (of the differences) Wilcoxon, based on ranks, is more powerful than sign test Data must be ordinally scaled (as in sign test) H0 is same as for sign test n pairs of data, compute D i = Y i X i for each pair Procedure: 1. Discard pairs where D i = 0 2. Order D i s in ascending order and assign ranks (assign average ranks to equal D i s) 3. Affix sign of difference (Y i X i ) to rank; signed ranks denoted by R i 4. Compute T + = Ri >0 R i sum of positive signed ranks 5. Compare T + to Table A.6 Large-sample approximation (for n>20): Z = T + [n(n+1) 4] ~N(0,1) under H0 n(n+1)(2n+1) 24 note: E(T + H 0 ) = n(n + 1) 4, V(T + H 0 ) = n(n + 1)(2n + 1) 24

7 BSTT523 Pagano & Gauvreau Chapter 13 7 Assumptions regarding ranks: must be able to rank within a pair e.g. pre>post for subject i must be able to rank difference scores between pairs e.g. (pre-post) for subject i > (pre-post) for subject j Latter assumption is used for signed-rank test but not for sign test, and allows for a more powerful test.

8 BSTT523 Pagano & Gauvreau Chapter 13 8 Back to example: Pair Group Total Mean Treatment Control Total Difference pair sign D i R i T + = (Table A.6: n=15, critical values are 25,95 for 2-sided α=.05) Z = (15)(6) 4 (15)(16)(31) 24 = = 2.41 (p<.016) reject H0 that median difference is zero (paired t-test yields T=3.109, df=14, p<.008)

9 BSTT523 Pagano & Gauvreau Chapter 13 9 SAS program for previous example: data one; input pair treat diff=treat-control; datalines; ; proc univariate; var diff; id pair; run; From SAS output: Tests for Location: Mu0=0 Test -Statistic p Value Student's t t Pr > t Sign M 3.5 Pr >= M Signed Rank S 42.5 Pr >= S Notes: Student s t shows paired t-test (parametric test) Sign shows M = n + n 2 Signed Rank shows S = T + n(n + 1) 4 p-values are 2-sided

10 BSTT523 Pagano & Gauvreau Chapter Two independent samples: Wilcoxon Rank Sum Test (a.k.a. Mann-Whitney U Test) Assumptions: random samples from each of two independent populations H0: the two populations have the same distribution Data: n 1 and n 2 observations from populations 1 and 2 N = n 1 + n 2 = total sample size Procedure: 1. Rank all observations from 1 to N (average ranks to tied values) 2. Sum the ranks for each sample T 1 and T 2 = sums of sample 1 and 2 ranks, respectively T 1 + T 2 = N(N + 1) 2 3. Compare T 1 with Table A.7 4. Normal approximation: Z = T 1 n 1 (N+1) 2 ~N(0,1) under H0 n 1 (n 2 )(N+1) 12 E(T 1 H 0 ) = n 1 (N + 1) 2, V(T 1 H 0 ) = n 1 (n 2 )(N + 1) 12 Mann-Whitney U Test Alternative formulation of the Wilcoxon Rank Sum Test, based on U = T 1 n 1 (n 1 + 1) 2

11 BSTT523 Pagano & Gauvreau Chapter Example: Distributions of height in two diet treatment groups Protein-Rich Diet (n=13) Height (in) Rank Protein-Poor Diet (n=14) Height (in) Rank T 1 = 224 and T 2 = 154 Z = 224 (13x28) 2 (13x14x28) 12 = = (p=.042) reject H0 that the distributions are the same (t-test yields T=2.45, df=25, p=.022) SAS program: data; input group datalines; ; proc npar1way wilcoxon; class group; var height; run;

12 BSTT523 Pagano & Gauvreau Chapter SAS output: The NPAR1WAY Procedure Wilcoxon Scores (Rank Sums) for Variable height Classified by Variable group Sum of Expected Std Dev Mean group N Scores Under H0 Under H0 Score ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ Average scores were used for ties. Wilcoxon Two-Sample Test Statistic Normal Approximation Z One-Sided Pr > Z Two-Sided Pr > Z t Approximation One-Sided Pr > Z Two-Sided Pr > Z Z includes a continuity correction of 0.5. Kruskal-Wallis Test Chi-Square DF 1 Pr > Chi-Square Notes: sum of scores listed for each group expected sum of scores under Ho is n i (N + 1) 2 for group i standard deviation under H 0 correction for continuity n 1 n 2 (N + 1.5) 12.5 Z based on correction for continuity Z = T 1 n 1 (N+1) 2.5 n 1 n 2 (N+1.5) 12.5 t approximation takes sample size into account (df=n-2) Kruskal_wallis, NP ANOVA analogue, test statistic = Z 2

13 BSTT523 Pagano & Gauvreau Chapter k Independent Samples the Kruskal Wallis test Assumption: random samples from each of k independent populations H 0 : the k populations have the same distribution Data: n 1, n 2,, n k observations from populations 1,,k Total sample size is N = k i=1 n i Procedure: Rank observations from 1 to N R ii = rank of jth observation in ith sample R i = sum of ith sample s ranks R = sum of all ranks = N(N + 1) 2 Test statistic T = 12 k n N(N+1) i=1 i(r i n i R N) 2 = 12 k n N(N+1) i=1 i(r i R ) 2 = 12 k R N(N+1) i=1 i 2 n i 2 3(N + 1)~χ k 1 under H 0 Notes: adjustments made if there are a large number of ties For k=2, Kruskal-Wallis T = square of large=sample approximation for Wilcoxon rank sum test (2-sided, without correction for continuity)

14 BSTT523 Pagano & Gauvreau Chapter Example: Five treatments for fever blisters, including a placebo, were randomly assigned to 30 patients. The data in the table identify, for each treatment, the number of days from initial appearance of the blisters until healing is complete. placebo group 2 group 3 group 4 group 5 value R ii value R ii value R ii value R ii value R ii R 1 = 145 R 1 = R 2 = 75.5 R 2 = R 3 = 53.5 R 3 = 8.92 R 4 = 81.5 R 4 = R 5 = R 5 = R 2 i = (145) 2 + (75.5) 2 + (53.5) 2 + (81.5) 2 + (109.5) 2 = T = 12 k N(N+1) R i 2 n i i=1 3(N + 1) = = at α=.05, critical χ 4,.94 = 9.49 < reject H 0 the five group distributions are not all the same

15 BSTT523 Pagano & Gauvreau Chapter Example using SAS: data one; input group numdays datalines; ; proc npar1way wilcoxon; class group; var numdays; SAS output: The NPAR1WAY Procedure Wilcoxon Scores (Rank Sums) for Variable numdays Classified by Variable group Sum of Expected Std Dev Mean group N Scores Under H0 Under H0 Score ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ Average scores were used for ties. Kruskal-Wallis Test Chi-Square DF 4 Pr > Chi-Square Notes: WILCOXON option gives Kruskal-Wallis when k>2 (always gives K-W, just no Wilcoxon when k>2) T = 11.01, corrected for ties, p=.0265 Analogous 1-way ANOVA yields F=3.9, p=.014

16 BSTT523 Pagano & Gauvreau Chapter k Matched Samples = Friedman s test Assumptions: ordinal data from k matched samples H 0 : the k samples were drawn from the same population Data: blocks: groups: block 1 block 2... block b group 1 Y 11 Y Y 1b group k Y k1 Y k2... Y kk one observation per cell each matched set of units serves as a block Procedure: Rank observations separately within each block (R ii ) Test statistic T = 12b k n k(k+1) i=1 i(r i b (k + 1) 2) 2 = 12 k R 2 bb(k+1) i=1 i 2 3b(k + 1)~χ k 1 under H 0 Notes: adjustments made if there are a large number of ties for k=2, Friedman T = square of large-sample approximation for sign test (2-sided)

17 BSTT523 Pagano & Gauvreau Chapter Example: A private research corporation conducted an experiment to investigate the toxic effects of three chemicals used in the tire-manufacturing industry. In this experiment, three adjacent 1-inch squares were marked on the back of each of eight rats, and the three chemicals were applied separately to the three squares on each rat. The squares were then rated from 0 to 10, depending on the degree of irritation. chemical 1 chemical 2 chemical 3 rat rating R ii rating R ii rating R ii R 1 = 14.5 R 2 = 22 R 3 = 11.5 R 2 i = (14.5) 2 + (22) 2 + (11.5) 2 = T = 12 k bb(k+1) R 2 i i=1 3b(k + 1) = = at α=.05, critical χ 2,.95 = 5.99 reject H 0 the chemicals are not all the same

18 BSTT523 Pagano & Gauvreau Chapter Example using SAS: data one; input rat chemical irrit datalines; ; proc freq; tables rat*chemical*irrit /cmh2 scores=rank noprint; SAS output: The FREQ Procedure Summary Statistics for chemical by irrit Controlling for rat Cochran-Mantel-Haenszel Statistics (Based on Rank Scores) Statistic Alternative Hypothesis DF Value Prob ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ 1 Nonzero Correlation Row Mean Scores Differ Total Sample Size = 24 Notes: Friedman s T= 8.36, corrected for ties, df=2, p=.0153 (2 nd CMH test statistic) Analogous RB ANOVA yields F=7.00, df=2,14, p=.0078