© 2011 Pearson Education, Inc

Statistics for Business and Economics Chapter 14 Nonparametric Statistics © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Content 14.1 Introduction: Distribution-Free Tests 14.2 Single Population Inferences 14.3 Comparing Two Populations: Independent Samples 14.4 Comparing Two Populations: Paired Difference Experiment 14.5 Comparing Three or More Populations: Completely Randomized Design As a result of this class, you will be able to... © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Content 14.6 Comparing Three or More Populations: Randomized Block Design 14.7 Rank Correlation As a result of this class, you will be able to... © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Learning Objectives Develop the need for inferential techniques that require fewer, or less stringent, assumptions than the methods of earlier chapters Introduce nonparametric tests that are based on ranks (i.e., on an ordering of the sample measurements according to their relative magnitudes) As a result of this class, you will be able to ... © 2011 Pearson Education, Inc

Introduction: Distribution-Free Tests 14.1 Introduction: Distribution-Free Tests :1, 1, 3 © 2011 Pearson Education, Inc

Parametric Test Procedures Involve population parameters Example: population mean Require interval scale or ratio scale Whole numbers or fractions Example: height in inches (72, 60.5, 54.7) Have stringent assumptions Example: normal distribution Examples: z-test, t-test, F-test, 2-test © 2011 Pearson Education, Inc

Nonparametric Test Procedures Do not involve population parameters Example: probability distributions, independence Data measured on any scale Ratio or interval Ordinal Example: good-better-best Nominal Example: male-female Example: Wilcoxon rank sum test © 2011 Pearson Education, Inc

Nonnormal Distributions - t-Statistic is Invalid © 2011 Pearson Education, Inc

Distribution-Free Tests Distribution-free tests are statistical tests that do not rely on any underlying assumptions about the probability distribution of the sampled population. The branch of inferential statistics devoted to distribution-free tests is called nonparametrics. Nonparametric statistics (or tests) based on the ranks of measurements are called rank statistics (or rank tests). © 2011 Pearson Education, Inc

Advantages of Nonparametric Tests Used with all scales Easier to compute Developed originally before wide computer use Make fewer assumptions Need not involve population parameters Results may be as exact as parametric procedures © 1984-1994 T/Maker Co. © 2011 Pearson Education, Inc

Disadvantages of Nonparametric Tests May waste information If data permit using parametric procedures Example: converting data from ratio to ordinal scale Difficult to compute by hand for large samples Tables not widely available © 1984-1994 T/Maker Co. © 2011 Pearson Education, Inc

Frequently Used Nonparametric Tests Sign Test Wilcoxon Rank Sum Test Wilcoxon Signed Rank Test Kruskal Wallis H-Test Spearman’s Rank Correlation Coefficient © 2011 Pearson Education, Inc

Single Population Inferences 14.2 Single Population Inferences :1, 1, 3 © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Sign Test Tests one population median,  (eta) Corresponds to t-test for one mean Assumes population is continuous Small sample test statistic: Number of sample values above (or below) median Alternative hypothesis determines Can use normal approximation if n  30 © 2011 Pearson Education, Inc

Sign Test Uses p-Value to Make Decision Binomial: n = 8 p = 0.5 Distribution has different shapes. Example 1: If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 0 defective item is about 0.6 (60%). If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 1 defective items is about .35 (35%). Example 2: If inspecting 5 items & the Probability of a defect is 0.5 (50%), the Probability of finding 1 defective items is about .18 (18%). Note: Could use formula or tables at end of text to get Probabilities. P-value is the probability of getting an observation at least as extreme as we got. If 7 of 8 observations ‘favor’ Ha, then p-value = p(x  7) = .031 + .004 = .035. If  = .05, then reject H0 since p-value  . © 2011 Pearson Education, Inc

Sign Test for a Population Median  One-Tailed Test H0:  = 0 Ha:  > 0 [or Ha:  < 0 ] Test statistic: S = Number of sample measurements greater than0 [or S = number of measurements less than 0] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Sign Test for a Population Median  Observed significance level: p-value = P(x ≥ S) where x has a binomial distribution with parameters n and p = .5 (Use Table II, Appendix B) Rejection region: Reject H0 if p-value ≤ .05 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Sign Test for a Population Median  Two-Tailed Test H0:  = 0 Ha:  ≠ 0 Test statistic: S = Larger of S1 and S2, where S1 is the number of sample measurements less than0 and S2 is the number of measurements greater than 0 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Sign Test for a Population Median  Observed significance level: p-value = 2P(x ≥ S) where x has a binomial distribution with parameters n and p = .5 (Use Table II, Appendix B) Rejection region: Reject H0 if p-value ≤ .05 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Conditions Required for Valid Application of the Sign Test The sample is selected randomly from a continuous probability distribution. [Note: No assumptions need to be made about the shape of the probability distribution.] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Large Sample Sign Test for a Population Median  One-Tailed Test H0:  = 0 Ha:  > 0 [or Ha:  < 0 ] Test statistic: S = Number of sample measurements greater than0 [or S = number of measurements less than 0] The “– .5” is the “correction for continuity.” Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Large Sample Sign Test for a Population Median  The null hypothesized mean value is np = .5n, and the standard deviation is Rejection region: z > z where tabulated z-values can be found in Table IV of Appendix B. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Large Sample Sign Test for a Population Median  Two-Tailed Test H0:  = 0 Ha:  ≠ 0 Test statistic: S = Larger of S1 and S2, where S1 is the number of sample measurements less than0 and S2 is the number of measurements greater than 0 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Large Sample Sign Test for a Population Median  The null hypothesized mean value is np = .5n, and the standard deviation is Rejection region: z > z where tabulated z-values can be found in Table IV of Appendix B. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Sign Test Example You’re an analyst for Chef-Boy-R-Dee. You’ve asked 7 people to rate a new ravioli on a 5-point Likert scale (1 = terrible to 5 = excellent). The ratings are: 2 5 3 4 1 4 5 At the .05 level of significance, is there evidence that the median rating is less than 3? Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Sign Test Solution H0: Ha:  = Test Statistic:  = 3  < 3 p-value: Decision: Conclusion: P(x  2) = 1 – P(x  1) = .937 (Binomial Table, n = 7, p = 0.50) .05 Note: More than 5 have been sold (6.4), but not enough to be significant. S = 2 (Ratings 1 & 2 are less than  = 3: 2, 5, 3, 4, 1, 4, 5) Do not reject at  = .05 There is no evidence median is less than 3 © 2011 Pearson Education, Inc

Comparing Two Populations: Independent Samples 14.3 Comparing Two Populations: Independent Samples :1, 1, 3 © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Wilcoxon Rank Sum Test Tests two independent population probability distributions Corresponds to t-test for two independent means Assumptions Independent, random samples Populations are continuous Can use normal approximation if ni  10 © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test: Independent Samples Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. One-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the right of D2 [or D1 is shifted to the left of D2] Test statistic: T1, if n1 < n2; T2, if n2 < n1 (Either rank sum can be used if n1 = n2.) Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test: Independent Samples Rejection region: T1: T1 ≥ TU [or T1 ≤ TL] T2: T2 ≤ TL [or T2 ≥ TU] where TL and TU are obtained from Table XIV of Appendix B Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test: Independent Samples Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. Two-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the left or to the right of D2 Test statistic: T1, if n1 < n2; T2, if n2 < n1 (Either rank sum can be used if n1 = n2.) We will denote this rank sum as T. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test: Independent Samples Rejection region: T ≤ TL or T ≥ TU where TL and TU are obtained from Table XIV of Appendix B Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Conditions Required for Valid Wilcoxon Rank Sum Test 1. The two samples are random and independent. 2. The two probability distributions from which the samples are drawn are continuous. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test Procedure Assign ranks, Ri, to the n1 + n2 sample observations If unequal sample sizes, let n1 refer to smaller- sized sample Smallest value = 1 Average ties Sum the ranks, Ti, for each sample Test statistic is Ti (smallest sample) © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test Example You’re a production planner. You want to see if the operating rates for two factories is the same. For factory 1, the rates (% of capacity) are 85, 82, 94, and 97. For factory 2, the rates are 71, 82, 77, 92, and 88 . Do the factory rates have the same probability distributions at the .10 level of significance? © 2011 Pearson Education, Inc 51

Wilcoxon Rank Sum Test Solution H0: Ha:  = n1 = n2 = Critical Value(s): Identical Distrib. Shifted Left or Right .10 4 5 © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Table (Portion)  = .05 one-tailed;  = .10 two-tailed n1 3 4 5 .. TL TU TL TU TL TU .. 3 6 15 7 17 7 20 .. n2 4 7 17 12 24 13 27 .. 5 7 20 13 27 19 36 .. : : : : : : : : © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test Solution H0: Identical Distrib. Ha: Shifted Left or Right  = .10 n1 = 4 n2 = 5 Critical Value(s): Reject H0 Do Not Reject H0 Reject H0 13 27  Ranks © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test Computation Table Factory 1 Factory 2 Rate Rank Rate Rank 85 5 71 1 82 3.5 3 82 4 94 8 77 2 97 9 92 7 ... 88 6 Rank Sum 19.5 25.5 © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test Solution H0: Identical Distrib. Ha: Shifted Left or Right  = .10 n1 = 4 n2 = 5 Critical Value(s): Test Statistic: Decision: Conclusion: T1 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest sample) Do not reject at  = .10 Reject H0 Do Not Reject H0 Reject H0 There is no evidence distrib. are not identical 13 27  Ranks © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10) Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. One-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the right of D2 [or D1 is shifted to the left of D2] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10) Test statistic: Rejection region: z > z (or z < –z) Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10) Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. Two-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the right or to the left of D2 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10) Test statistic: Rejection region: | z | > z/2 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Comparing Two Populations: Paired Differences Experiment 14.4 Comparing Two Populations: Paired Differences Experiment :1, 1, 3 © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test Tests probability distributions of two related populations Corresponds to t-test for dependent (paired) means Assumptions Random samples Both populations are continuous Can use normal approximation if n  25 n’ is the number of non-zero difference scores. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. One-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the right of D2 [or D1 is shifted to the left of D2] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Calculate the difference within each of the n matched pairs of observations. Then rank the absolute value of the n differences from the smallest (rank 1) to the highest (rank n) and calculate the rank sum T– of the negative differences and the rank sum T+ of the positive differences. [Note: Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly.] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Test statistic: T–, the rank sum of the negative differences [or T+, the rank sum of the positive differences] Rejection region: T– ≤ T0 [or T+ ≤ T0] where T0 is given in Table XV of Appendix B Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. Two-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the left or to the right of D2 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Calculate the difference within each of the n matched pairs of observations. Then rank the absolute value of the n differences from the smallest (rank 1) to the highest (rank n) and calculate the rank sum T– of the negative differences and the rank sum T+ of the positive differences. [Note: Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly.] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Test statistic: T–, the rank sum of the negative differences [or T+, the rank sum of the positive differences] Rejection region: T ≤ T0 where T0 is given in Table XV of Appendix B Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Conditions Required for a Valid Signed Rank Test 1. The sample of differences is randomly selected from the population of differences. 2. The probability distribution from which the sample of paired differences is drawn is continuous. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Signed Rank Test Procedure Obtain difference scores, Di = X1i – X2i Take absolute value of differences, |Di| Delete differences with 0 value Assign ranks, Ri, where smallest = 1 Assign ranks same signs as Di Sum ‘+’ ranks (T+) and ‘–’ ranks (T–) Test statistic is T– or T+ (one-tail test) Test statistic is T = smaller of T– or T+ (two-tail test) © 2011 Pearson Education, Inc

Signed Rank Test Computation Table X1i X2i Di = X1i - X2i |Di| Ri Sign Sign Ri X11 X21 D1 = X11 - X21 |D1| R1 ± ± R1 X12 X22 D2 = X12 - X22 |D2| R2 ± ± R2 X13 X23 D3 = X13 - X23 |D3| R3 ± ± R3 : : : : : : : X1n X2n Dn = X1n - X2n |Dn| Rn ± ± Rn Total T+ & T– © 2011 Pearson Education, Inc

Signed Rank Test Example You work in the finance department. Is the new financial package faster (α = .05)? You collect the following data entry times: User Current New Donna 9.98 9.88 Santosha 9.88 9.86 Sam 9.90 9.83 Tamika 9.99 9.80 Brian 9.94 9.87 Jorge 9.84 9.84 © 2011 Pearson Education, Inc © 1984-1994 T/Maker Co.

Signed Rank Test Solution H0: Ha:  = n' = Critical Value(s): Identical Distrib. Current Shifted Right .05 © 2011 Pearson Education, Inc

Signed Rank Test Computation Table X1i X2i Di |Di | Ri Sign Sign Ri 9.98 9.88 +0.10 +0.02 +0.07 +0.19 0.00 0.10 0.02 0.07 0.19 0.00 4 + ... +4 +1 +2.5 +5 Discard 9.88 9.86 1 9.90 9.83 2 2.5 9.99 9.80 5 9.94 9.87 3 2.5 9.84 9.84 ... Total T+= 15, T–= 0 © 2011 Pearson Education, Inc

Signed Rank Test Solution H0: Ha:  = n' = 5 (not 6; 1 elim.) Critical Value(s): Identical Distrib. Current Shifted Right .05 © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Table (Portion) One-Tailed Two-Tailed n = 5 n = 6 n = 7 .. a = .05 a = .10 1 2 4 .. a = .025 a = .05 1 2 .. a = .01 a = .02 .. a = .005 a = .01 .. n = 11 n = 12 n = 13 : : : : © 2011 Pearson Education, Inc

Signed Rank Test Solution H0: Ha:  = n' = 5 (not 6; 1 elim.) Critical Value(s): Identical Distrib. Current Shifted Right Test Statistic: Decision: Conclusion: T– = 0 .05 Reject at  = .05 Do Not Reject H0 Reject H0 There is evidence new package is faster 1 T0 © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for Large Samples (n ≥ 25) Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. One-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the right of D2 [or D1 is shifted to the left of D2] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Test statistic: Rejection region: z > z [or z < –z ] Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Assumptions: The sample size n is greater than or equal to 25. Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly. Tied absolute differences receive ranks equal to the average of the ranks they would have received had they not been tied. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for Large Samples (n ≥ 25) Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively. Two-Tailed Test H0: D1 and D2 are identical Ha: D1 is shifted to the left or to the right of D2 Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Test statistic: Rejection region: | z | > z Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Wilcoxon Signed Rank Test for a Paired Difference Experiment Assumptions: The sample size n is greater than or equal to 25. Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly. Tied absolute differences receive ranks equal to the average of the ranks they would have received had they not been tied. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Comparing Three or More Populations: Completely Randomized Design 14.5 Comparing Three or More Populations: Completely Randomized Design :1, 1, 3 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Tests the equality of more than two (p) population probability distributions Corresponds to ANOVA for more than two means Used to analyze completely randomized experimental designs Uses 2 distribution with p – 1 df if sample size nj ≥ 5 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test for Comparing k Probability Distributions H0: The k probability distributions are identical Ha: At least two of the k probability distributions differ in location. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. Test statistic: © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test for Comparing k Probability Distributions where nj = Number of measurements in sample j Rj = Rank sum for sample j, where the rank of each measurement is computed according to its relative magnitude in the totality of data for the k samples Rj = Rj/nj = Mean rank sum for jth sample R = Mean of all ranks = (n + 1)/2 n = Total sample size = n1 + n2 + . . . + nk Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test for Comparing k Probability Distributions Rejection region: H > with (k – 1) degrees of freedom Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Conditions Required for the Validity of the Kruskal-Wallis H-Test 1. The k samples are random and independent. 2. There are five or more measurements in each sample. 3. The k probability distributions from which the samples are drawn are continuous Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Procedure Assign ranks, Ri , to the n combined observations Smallest value = 1; largest value = n Average ties Sum ranks for each group Compute test statistic Squared total of each group © 2011 Pearson Education, Inc 111

Kruskal-Wallis H-Test Example As production manager, you want to see if three filling machines have different filling times. You assign 15 similarly trained and experienced workers, 5 per machine, to the machines. At the .05 level of significance, is there a difference in the distribution of filling times? Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 © 2011 Pearson Education, Inc 112

Kruskal-Wallis H-Test Solution Ha:  = df = Critical Value(s): Identical Distrib. At Least 2 Differ .05 p – 1 = 3 – 1 = 2 c 2 5.991  = .05 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution Raw Data Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 Ranks Mach1 Mach2 Mach3 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution Raw Data Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 Ranks Mach1 Mach2 Mach3 1 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution Raw Data Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 Ranks Mach1 Mach2 Mach3 2 1 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution Raw Data Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 Ranks Mach1 Mach2 Mach3 2 1 3 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution Raw Data Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 Ranks Mach1 Mach2 Mach3 14 9 2 15 6 7 12 10 1 11 8 4 13 5 3 © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution Raw Data Mach1 Mach2 Mach3 25.40 23.40 20.00 26.31 21.80 22.20 24.10 23.50 19.75 23.74 22.75 20.60 25.10 21.60 20.40 Ranks Mach1 Mach2 Mach3 14 9 2 15 6 7 12 10 1 11 8 4 13 5 3 65 38 17 Total © 2011 Pearson Education, Inc

Kruskal-Wallis H-Test Solution © 2011 Pearson Education, Inc 115

Kruskal-Wallis H-Test Solution Ha:  = df = Critical Value(s): Identical Distrib. At Least 2 Differ Test Statistic: Decision: Conclusion: H = 11.58 .05 p – 1 = 3 – 1 = 2 c 2 Reject at  = .05 5.991  = .05 There is evidence population distrib. are different © 2011 Pearson Education, Inc

Comparing Three or More Populations: Randomized Block Design 14.6 Comparing Three or More Populations: Randomized Block Design :1, 1, 3 © 2011 Pearson Education, Inc

Friedman Fr-statistic provides another method for testing to detect a shift in location of a set of k populations that have the same spread (or, scale) is based on the rank sums of the treatments, measures the extent to which the k samples differ with respect to their relative ranks within the blocks n’ is the number of non-zero difference scores. © 2011 Pearson Education, Inc

Friedman Fr-Test for a Randomized Block Design H0: The probability distributions for the k treatments are identical Ha: At least two of the probability distributions differ in location Test statistic: Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Friedman Fr-Test for a Randomized Block Design where b = Number of blocks k = Number of treatments Rj = Rank sum of the jth treatment, where the rank of each measurement is computed relative to its position within its own block Test statistic: Fr > with (k – 1) degrees of freedom Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Friedman Fr-Test for a Randomized Block Design Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Friedman Fr -Test Example Consider the data in the table. A pharmaceutical firm wants to compare the reaction times of subjects under the influence of three different drugs that it produces. Apply the Friedman Fr-test to the data. What conclusion can you draw? Test using  = .05. © 2011 Pearson Education, Inc

Friedman Fr -Test Example H0: The population distributions of reaction times are identical for the three drugs Ha: At least two of the three drugs have reaction time distributions that differ in location From the table: k = 3 treatments, b = 6 blocks, R1 = 7, R2 = 12, R3 = 17; so © 2011 Pearson Education, Inc

Friedman Fr -Test Example For  = .05,  2.05 = 5.99147, therefore Rejection region: H > 5.99147 © 2011 Pearson Education, Inc

Friedman Fr -Test Example Conclusion: Because H = 8.33 exceeds the critical value of 5.99, we reject the null hypothesis and conclude that at least two of the three drugs have distributions of reaction times that differ in location. That is, at least one of the drugs tends to yield reaction times that are faster than the others. © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc 14.7 Rank Correlation :1, 1, 3 © 2011 Pearson Education, Inc

Spearman’s Rank Correlation Coefficient Measures correlation between ranks Corresponds to Pearson product moment correlation coefficient Values range from –1 to +1 Formula (shortcut) di = ui – vi (difference in ranks of ith observation for samples 1 and 2) n = number of pairs of observations © 2011 Pearson Education, Inc

Spearman’s Rank Correlation Procedure Assign ranks, Ri , to the observations of each variable separately Calculate differences, di , between each pair of ranks Square differences, di2, between ranks Sum squared differences for each variable Use shortcut approximation formula © 2011 Pearson Education, Inc

Spearman’s Rank Correlation Example You’re a research assistant for the FBI. You’re investigating the relationship between a person’s attempts at deception and percent changes in their pupil size. You ask subjects a series of questions, some of which they must answer dishonestly. At the .05 level of significance, what is the correlation coefficient? Subj. Deception Pupil 1 87 10 2 63 6 3 95 11 4 50 7 5 43 0 © 2011 Pearson Education, Inc 112

Spearman’s Rank Correlation Table 2 Subj. Decep. R1i Pupil R2i di di 1 87 10 2 63 6 3 95 11 4 50 7 5 43 4 4 1 -1 1 2 3 2 5 5 2 3 1 1 Σdi2= © 2011 Pearson Education, Inc

Spearman’s Rank Correlation Solution © 2011 Pearson Education, Inc

Spearman’s Nonparametric Test for Rank Correlation One-Tailed Test H0:  = 0 Ha:  > 0 (or Ha:  < 0) Test statistic: Rejection region: rs > rs, or rs <–rs, when s < 0 where rs, is the value from Table XVI corresponding to the upper-tail area  and n pairs of observations Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Spearman’s Nonparametric Test for Rank Correlation Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Spearman’s Nonparametric Test for Rank Correlation Two-Tailed Test H0:  = 0 Ha:  ≠ 0 Test statistic: Rejection region: |rs | > rs, where rs, is the value from Table XVI corresponding to the upper-tail area  and n pairs of observations Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Spearman’s Nonparametric Test for Rank Correlation Ties: Assign tied measurements the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

Conditions Required for a Valid Spearman’s Test 1. The sample of experimental units on which the two variables are measured is randomly selected. 2. The probability distributions of the two variables are continuous. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Key Ideas Distribution-Free Tests Do not rely on assumptions about the probability distribution of the sampled population As a result of this class, you will be able to... © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Key Ideas Nonparametrics Distribution-free tests that are based on rank statistics One-sample nonparametric test for the population median – sign test Nonparametric test for two independent samples – Wilcoxon rank sum test Nonparametric test for matched pairs – Wilcoxon signed rank test As a result of this class, you will be able to... © 2011 Pearson Education, Inc

© 2011 Pearson Education, Inc Key Ideas Nonparametrics Nonparametric test for a completely randomized design – Kruskal-Wallis test Nonparametric test for a randomized block design – Friedman test Nonparametric test for rank correlation – Spearman’s test As a result of this class, you will be able to... © 2011 Pearson Education, Inc