1. Statistics for clinicians Biostatistics course by Kevin E. Kip, Ph.D., FAHA Professor and Executive Director, Research Center University of South Florida, College of Nursing Professor, College of Public Health Department of Epidemiology and Biostatistics Associate Member, Byrd Alzheimer’s Institute Morsani College of Medicine Tampa, FL, USA 1

2. SECTION 4.1 Module Overview and Introduction Hypothesis testing for 2 or more independent groups and non- parametric methods.

3. SECTION 4.6 More than two-samples: independent groups – categorical outcome

4. 4. More Than Two Independent Groups-Categorical Outcome  Parameter:Difference in proportions for >2 groups (χ 2 test)  H 0 :Distribution (proportions) of the outcome is independent of the groups  H 1 :H 0 is false  Test statistic: χ 2 value Find critical value in Table 3 (df = (r – 1)(c – 1) r = rows, c = columns  By convention, the grouping variable (e.g. predictor) is shown in the rows, and the outcome variable is shown in the columns.  “Expected” frequencies are based on the null hypothesis of independence between the grouping variable and outcome.

5. 4. More Than Two Independent Groups-Categorical Outcome 1)Set up the hypothesis and determine level of statistical significance H 0 : Living arrangement and exercise are independent H 1 : H 0 is false α = 0.05 2)Select the appropriate test statistic Example: Living arrangement and level of exercise (row %) Living Arrangement No Regular Exercise Sporadic Exercise Regular ExerciseTotal Dormitory32 (0.356)30 (0.333)28 (0.311)90 (1.0) On-campus apartment74 (0.411)64 (0.356)42 (0.233)180 (1.0) Off-campus apartment110 (0.733)25 (0.167)15 (0.100)150 (1.0) At home39 (0.780)6 (0.120)5 (0.100)50 (1.0) Total255 (0.543)125 (0.266)90 (0.191)470

6. 4. More Than Two Independent Groups-Categorical Outcome 3)Set up the decision rule (df = (r – 1)(c – 1) (df = (4 – 1)(3 – 1) = 6; α = 0.05 From Table 3, critical value = 12.59 Reject H 0 if χ 2 > 12.59 Example: Living arrangement and level of exercise (row %) Living Arrangement No Regular Exercise Sporadic Exercise Regular ExerciseTotal Dormitory32 (0.356)30 (0.333)28 (0.311)90 (1.0) On-campus apartment74 (0.411)64 (0.356)42 (0.233)180 (1.0) Off-campus apartment110 (0.733)25 (0.167)15 (0.100)150 (1.0) At home39 (0.780)6 (0.120)5 (0.100)50 (1.0) Total255 (0.543)125 (0.266)90 (0.191)470

7. 4. More Than Two Independent Groups-Categorical Outcome Living Arrangement No Regular Exercise Sporadic Exercise Regular ExerciseTotal Dormitory32 (0.356)30 (0.333)28 (0.311)90 (1.0) On-campus apartment74 (0.411)64 (0.356)42 (0.233)180 (1.0) Off-campus apartment110 (0.733)25 (0.167)15 (0.100)150 (1.0) At home39 (0.780)6 (0.120)5 (0.100)50 (1.0) Total255 (0.543)125 (0.266)90 (0.191)470 4)Compute the test statistic: Row total x column total Expected cell frequency-------------------------------- N Living Arrangement No Regular Exercise Sporadic Exercise Regular ExerciseTotal Dormitory48.8 (0.543)23.9 (0.266)17.2 (0.191)90 (1.0) On-campus apartment97.7 (0.543)47.9 (0.266)34.5 (0.191)180 (1.0) Off-campus apartment81.4 (0.543)39.9 (0.266)28.7 (0.191)150 (1.0) At home27.1 (0.543)13.3 (0.266)9.6 (0.191)50 (1.0) Total255 (0.543)125 (0.266)90 (0.191)470 OBSERVED (O) EXPECTED (E)

8. 4. More Than Two Independent Groups-Categorical Outcome Living Arrangement No Regular Exercise Sporadic Exercise Regular ExerciseTotal Dormitory32 (0.356)30 (0.333)28 (0.311)90 (1.0) On-campus apartment74 (0.411)64 (0.356)42 (0.233)180 (1.0) Off-campus apartment110 (0.733)25 (0.167)15 (0.100)150 (1.0) At home39 (0.780)6 (0.120)5 (0.100)50 (1.0) 4)Compute the test statistic: Living Arrangement No Regular Exercise Sporadic Exercise Regular ExerciseTotal Dormitory48.8 (0.543)23.9 (0.266)17.2 (0.191)90 (1.0) On-campus apartment97.7 (0.543)47.9 (0.266)34.5 (0.191)180 (1.0) Off-campus apartment81.4 (0.543)39.9 (0.266)28.7 (0.191)150 (1.0) At home27.1 (0.543)13.3 (0.266)9.6 (0.191)50 (1.0) OBSERVED (O) EXPECTED (E) χ 2 = ((32 – 48.8) 2 / 48.8) + ((30 – 23.9) 2 / 23.9) + ((28 – 17.2) 2 / 17.2) + ((74 – 97.7) 2 / 97.7) + ((64 – 47.9) 2 / 47.9) + ((42 – 34.5) 2 / 34.5) + ((110 – 81.4) 2 / 81.4) + ((25 – 39.9) 2 / 39.9) + ((15 – 28.7) 2 / 28.7) + ((39 – 27.1) 2 / 27.1) + ((6 – 13.3) 2 / 13.3) + ((5 – 9.6) 2 / 9.6) χ 2 = 60.5 5) Conclusion: Reject H 0 ; 60.5 > 12.59

9. 4. More Than Two Independent Groups-Categorical Outcome (Practice) 1)Set up the hypothesis and determine level of statistical significance H 0 :__________________________ H 1 : __________________________ α = 0.05 2)Select the appropriate test statistic Example: Smoking status and exercise >= 3 times/week (row %) Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (_____)20 (_____)44 (1.0) Former smoker72 (_____)153 (_____)225 (1.0) Never smoker99 (_____)128 (_____)227 (1.0) Total195 (_____)301 (_____)496

10. 4. More Than Two Independent Groups-Categorical Outcome (Practice) 1)Set up the hypothesis and determine level of statistical significance H 0 : Smoking status and exercise are independent H 1 : H 0 is false α = 0.05 2)Select the appropriate test statistic Example: Smoking status and exercise >= 3 times/week (row %) Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (0.545)20 (0.455)44 (1.0) Former smoker72 (0.320)153 (0.680)225 (1.0) Never smoker99 (0.436)128 (0.564)227 (1.0) Total195 (0.393)301 (0.607)496

11. 4. More Than Two Independent Groups-Categorical Outcome (Practice) Example: Smoking status and exercise >= 3 times/week (row %) Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (_____)20 (_____)44 (1.0) Former smoker72 (_____)153 (_____)225 (1.0) Never smoker99 (_____)128 (_____)227 (1.0) Total195 (_____)301 (_____)496 3)Set up the decision rule (df = (r – 1)(c – 1) (df = ________ From Table 3, critical value = ___________ Reject H 0 if: ____________

12. 4. More Than Two Independent Groups-Categorical Outcome (Practice) Example: Smoking status and exercise >= 3 times/week (row %) Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (0.545)20 (0.455)44 (1.0) Former smoker72 (0.320)153 (0.680)225 (1.0) Never smoker99 (0.436)128 (0.564)227 (1.0) Total195 (0.393)301 (0.607)496 3)Set up the decision rule (df = (r – 1)(c – 1) (df = (3 – 1)(2 – 1) = 2; α = 0.05 From Table 3, critical value = 5.99 Reject H 0 if χ 2 > 5.99

13. 4. More Than Two Independent Groups-Categorical Outcome (Practice) 4)Compute the test statistic: OBSERVED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (0.545)20 (0.455)44 (1.0) Former smoker72 (0.320)153 (0.680)225 (1.0) Never smoker99 (0.436)128 (0.564)227 (1.0) Total195 (0.393)301 (0.607)496 EXPECTED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker Former smoker Never smoker Total195 (0.393)301 (0.607)496 Row total x column total Expected cell frequency-------------------------------- N

14. 4. More Than Two Independent Groups-Categorical Outcome (Practice) 4)Compute the test statistic: OBSERVED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (0.545)20 (0.455)44 (1.0) Former smoker72 (0.320)153 (0.680)225 (1.0) Never smoker99 (0.436)128 (0.564)227 (1.0) Total195 (0.393)301 (0.607)496 EXPECTED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker17.21 (0.393)26.70 (0.607)44 (1.0) Former smoker88.46 (0.393)136.54 (0.607)225 (1.0) Never smoker89.24 (0.393137.76 (0.607)227 (1.0) Total195 (0.393)301 (0.607)496 Row total x column total Expected cell frequency-------------------------------- N

15. 4. More Than Two Independent Groups-Categorical Outcome (Practice) 4)Compute the test statistic: OBSERVED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (0.545)20 (0.455)44 (1.0) Former smoker72 (0.320)153 (0.680)225 (1.0) Never smoker99 (0.436)128 (0.564)227 (1.0) EXPECTED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker17.21 (0.393)26.70 (0.607)44 (1.0) Former smoker88.46 (0.393)136.54 (0.607)225 (1.0) Never smoker89.24 (0.393137.76 (0.607)227 (1.0) χ 2 = 5.Conclusion: ___________________________

16. 4. More Than Two Independent Groups-Categorical Outcome (Practice) 4)Compute the test statistic: OBSERVED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker24 (0.545)20 (0.455)44 (1.0) Former smoker72 (0.320)153 (0.680)225 (1.0) Never smoker99 (0.436)128 (0.564)227 (1.0) EXPECTED Smoking Status Exercise <3 Times/Week Exercise >3 Times/WeekTotal Current smoker17.21 (0.393)26.70 (0.607)44 (1.0) Former smoker88.46 (0.393)136.54 (0.607)225 (1.0) Never smoker89.24 (0.393137.76 (0.607)227 (1.0) χ 2 = ((24 – 17.21) 2 / 17.21) + ((20 – 26.7) 2 / 26.7) + ((72 – 88.46) 2 / 88.46) + ((153 – 136.54) 2 / 136.43) + ((99 – 89.24) 2 / 89.24) + ((128 – 137.76) 2 / 137.76) = 2.68 + 1.68 + 3.06 + 1.98 + 1.07 + 0.69 χ 2 = 11.17 5.Conclusion: Reject H 0 ; 11.17 > 5.99

17. 3. More Than Two Independent Groups-Categorical Outcome (Practice) Example: Smoking status and exercise >= 3 times/week (α = 0.05) SPSS Analyze Descriptive Statistics Crosstabs Row Variable: Smoking status Column Variable: Exercise > 3 times/week Statistics: Chi-square Cells: Observed Expected Row percentages

18. SECTION 4.7 Non-parametric statistical methods: Introduction

19. Overview: (Non-Parametric Tests):  Most statistical tests are “parametric” tests that are based on assumptions for appropriate use, such as the outcome variable follows a normal distribution (e.g. comparing means between groups by use of student t test).  Parametric tests rely on probability distributions: Non-parametric tests are “distribution free” - do not rely on normality.  Non-parametric tests are particularly useful for outcome variables that are highly skewed.  In general, non-parametric are less powerful (in terms of hypothesis testing) compared to analogous parametric tests.

20. Overview: (Non-Parametric Tests):  The parametric assumption of normality is particularly worrisome for small sample sizes (n < 30). Nonparametric tests are often a good option for these data.  In parametric tests, hypotheses are not about population parameters, such as µ 1 = µ 2. Instead, hypotheses are more general, such as two populations are equal in terms of their central tendency.

21. CharacteristicParametricNon-Parametric Assumed distributionNormalAny Assumed varianceHomogeneousAny Typical dataRatio or intervalOrdinal or nominal Dataset relationshipsIndependentAny Usual central measureMeanMedian BenefitsCan draw more conclusions Simplicity: less affected by outliers

22. TestsParametricNon-Parametric CorrelationPearsonSpearman Independent, 2-groupsStudent t testMann-Whitney (Wilcoxon Rank Sum Test) Independent, > 2-groupsOne-way ANOVAKruskal-Wallis Test Dependent, 2-groupsPaired t testWilcoxon Signed Rank Test

23. Positively skewed

24. Non-Hispanic Positively skewed

25. Hispanic Approximately normally distributed

26. SAS code: options ls=80; libname test 'G:\NGR 7848 2012\Lectures\Unit 8'; data test; input id group score; cards; 1116 2112 3113 4112 5114 6118 7119 8111 9116 10194 11222 12226 13221 14220 15222 16225 17220 18226 19224 20223 ; run; ID Group Score

27. proc univariate; class group; var score; run; Moments N10Sum Weights10 Mean22.5Sum Observations225 Std Deviation25.2641599Variance638.277778 Skewness3.09666404Kurtosis9.6937323 Uncorrected SS10807Corrected SS5744.5 Coeff Variation112.285155Std Error Mean7.98922886 Basic Statistical Measures LocationVariability Mean22.50000Std Deviation25.26416 Median15.00000Variance638.27778 Mode12.00000Range83.00000 Interquartile Range6.00000 Group 1

28. Moments N10Sum Weights10 Mean22.9Sum Observations229 Std Deviation2.28278582Variance5.21111111 Skewness0.15271431Kurtosis-1.434527 Uncorrected SS5291Corrected SS46.9 Coeff Variation9.96849704Std Error Mean0.72188026 Basic Statistical Measures LocationVariability Mean22.90000Std Deviation2.28279 Median22.50000Variance5.21111 Mode20.00000Range6.00000 Interquartile Range4.00000 Group 2

29. proc ttest; class group; var score; run; groupMethodMean95% CL MeanStd Dev95% CL Std Dev 122.50004.427140.572925.264217.377646.1225 222.900021.267024.53302.28281.57024.1675 Diff (1-2)Pooled-0.4000-17.253116.453117.937213.553626.5260 Diff (1-2)Satterth waite -0.4000-18.502217.7022 MethodVariancesDFt ValuePr > |t| PooledEqual18-0.050.9608 SatterthwaiteUnequal9.1469-0.050.9613 Equality of Variances MethodNum DFDen DFF ValuePr > F Folded F99122.48<.0001 t-Test (inappropriate)

30. proc npar1way wilcoxon; class group; var score; run; Wilcoxon Scores (Rank Sums) for Variable score Classified by Variable group groupNSum of Scores Expected Under H0 Std Dev Under H0 Mean Score 11065.0105.013.2038676.50 210145.0105.013.20386714.50 Average scores were used for ties. Wilcoxon Two-Sample Test Statistic65.0000 Normal Approximation Z-2.9915 One-Sided Pr < Z0.0014 Two-Sided Pr > |Z|0.0028 t Approximation One-Sided Pr < Z0.0038 Two-Sided Pr > |Z|0.0075 Wilcoxon-Test (appropriate)

32. SECTION 4.8 Two independent samples: Wilcoxon Rank Sum Test

33. 1. Two Independent Samples: Wilcoxon Rank Sum Test  Also called the Mann-Whitney “U” Test.  Tests whether 2 samples are likely to derive from the same population (i.e. the 2 samples have the same shape).  Analogous to comparing the median between the 2 samples.  H 0 : The 2 populations are equal; H 1 : The 2 populations are not equal.  Almost always tested as 2-sided.

34. Group Ordered Ranks From our previous example: Sum the ranks for each group, denoted as R 1 and R 2. If H 0 is true, then we expect that R 1 is approximately equal to R 2.

35. From the Previous Table: n 1 = 10; n 2 = 10; R 1 = 65; R 2 = 145 n 1 (n 1 +1)n 2 (n 2 +1) U 1 = n 1 n 2 + ----------- - R 1 U 2 = n 1 n 2 + ----------- - R 2 2 2 10(10+1) U 1 = 10(10) + ----------- - 65= 90.0 2 10(10+1) U 2 = 10(10) + ----------- - 145= 10.0 2 The test statistic, U, is evaluated by the smaller of U 1 and U 2 1. Two Independent Samples: Wilcoxon Rank Sum Test

36. 1.Set up the hypothesis and level of significance: H 0 : The two populations are equal H 1 : The two populations are not equalα = 0.05 2.Select the appropriate test statistic: Mann-Whitney U-test 3.Set up the decision rule: Reject H 0 if U < 23 (see Table 5) 4.Compute the test statistic: From previous slide: U = min(R 1, R 2 )U = 10.0 5.Conclusion: Reject H 0 because 10.0 < 23 (populations are not equal) 1. Two Independent Samples: Wilcoxon Rank Sum Test

37. Two Independent Samples: Wilcoxon Rank Sum Test (Practice) GroupOrdered ValuesRanks PlaceboDrugPlaceboDrugPlaceboDrug 73 56 64 42 121 R1R2 Sum of Ranks  ________ Example: Consider a trial of a new drug to reduce symptoms of asthma in children (episodes of shortness of breath); one-sided; α=0.05 n 1 = ___; n 2 = ___; R 1 = ___; R 2 = ___ n 1 (n 1 +1) n 2 (n 2 +1) U 1 = n 1 n 2 + ----------- - R 1 U 2 = n 1 n 2 + ----------- - R 2 2 2 U 1 = U 2 =

38. Two Independent Samples: Wilcoxon Rank Sum Test (Practice) GroupOrdered ValuesRanks PlaceboDrugPlaceboDrugPlaceboDrug 73 1 1 56 2 2 64 3 3 42444.5 1215 6 667.5 7 9 12 10 R1R2 Sum of Ranks  3718 Example: Consider a trial of a new drug to reduce symptoms of asthma in children (episodes of shortness of breath); one-sided; α=0.05 n 1 = 5; n 2 = 5; R 1 = 37; R 2 = 18 n 1 (n 1 +1) n 2 (n 2 +1) U 1 = n 1 n 2 + ----------- - R 1 U 2 = n 1 n 2 + ----------- - R 2 2 2 5(6) U 1 = 5(5) + ------- - 37= 3 2 5(6) U 2 = 5(5) + ------- - 18= 22 2

39. 1.Set up the hypothesis and level of significance: H 0 : The two populations are equal H 1 : The two populations are not equalα = 0.05 2.Select the appropriate test statistic: Mann-Whitney U-test 3.Set up the decision rule: Reject H 0 if U < _____ (see Table 5) 4.Compute the test statistic: From previous slide: U = min(R 1, R 2 )U = ______ 5.Conclusion: _________________________________________ Two Independent Samples: Wilcoxon Rank Sum Test

40. 1.Set up the hypothesis and level of significance: H 0 : The two populations are equal H 1 : The two populations are not equalα = 0.05 2.Select the appropriate test statistic: Mann-Whitney U-test 3.Set up the decision rule: Reject H 0 if U < 4 (see Table 5) 4.Compute the test statistic: From previous slide: U = min(R 1, R 2 )U = 3 5.Conclusion: Reject H 0 because 3(U) < 4(critical value)(populations are not equal) Two Independent Samples: Wilcoxon Rank Sum Test

41. SECTION 4.9 Matched samples: Wilcoxon Signed Rank Test

42. 2. Matched Samples: Wilcoxon Signed Rank Test  Similar to the paired t test, the Wilcoxon Signed Rank test considers the difference scores between the pairs, including positive or negative, and the magnitude of difference.  Ranks are based on absolute values of the difference scores.  H 0 : The median difference is zero;  H 1 : The median difference is positive.  Reject H 0 if test statistic (W) is < critical value (table 7)  Research hypothesis can be 1-sided or 2-sided.

43. 2. Matched Samples: Wilcoxon Signed Rank Test Child Before Treatment After Treatment Difference (Before – After) Ordered Absolute Diff.Ranks Signed Ranks 1857510-51 27050201033 34050-10 3-3 46540251033 58020601555 67565102066 75540152577 82025-56088 Example: Consider examination of the effectiveness a new drug to reduce (i.e. one-sided) repetitive behaviors in children with autism. Repetitive behavior before and after treatment  H 0 : The median difference is zero;  H 1 : The median difference is positiveone-sided, α = 0.05

44. 2. Matched Samples: Wilcoxon Signed Rank Test Child Before Treatment After Treatment Difference (Before – After) Ordered Absolute Diff.Ranks Signed Ranks 1857510-51 27050201033 34050-10 3-3 46540251033 58020601555 67565102066 75540152577 82025-56088 Repetitive behavior before and after treatment The test statistic, W, is the smaller of W+ and W-, which are the sums of the positive and negative ranks (using absolute values). W+ = (3+3+5+6+7+8) = 32 W- = (1+3) = 4 W = min(W+, W-) = 4refer to table 7n=8, critical value = 6 Reject H 0 : W = 4 < 6

45. 2. Matched Samples: Wilcoxon Signed Rank Test (Practice) Child Before Treatment After Treatment Difference (Before – After) Ordered Absolute Diff.Ranks Signed Ranks 16375 24830 37248 45228 59054 66440 75258 Example: Consider examination of the effectiveness a new drug to reduce (i.e. one-sided) repetitive behaviors in children with autism. Repetitive behavior before and after treatment  H 0 : ___________________________  H 1 : ___________________________one-sided, α = 0.05

46. 2. Matched Samples: Wilcoxon Signed Rank Test (Practice) Child Before Treatment After Treatment Difference (Before – After) Ordered Absolute Diff.Ranks Signed Ranks 16375-12-61 2483018-122-2 37248241833 4522824 55 59054362455 6644024 55 75258-63677 Example: Consider examination of the effectiveness a new drug to reduce (i.e. one-sided) repetitive behaviors in children with autism. Repetitive behavior before and after treatment  H 0 : The median difference is zero;  H 1 : The median difference is positiveone-sided, α = 0.05

47. 2. Matched Samples: Wilcoxon Signed Rank Test (Practice) Child Before Treatment After Treatment Difference (Before – After) Ordered Absolute Diff.Ranks Signed Ranks 16375 24830 37248 45228 59054 66440 75258 Repetitive behavior before and after treatment The test statistic, W, is the smaller of W+ and W-, which are the sums of the positive and negative ranks (using absolute values). W+ = _____________ W- = ____________ W = min(W+, W-) = ____refer to table 7n=____, critical value = _____ Conclusion:__________________________

48. 2. Matched Samples: Wilcoxon Signed Rank Test (Practice) Child Before Treatment After Treatment Difference (Before – After) Ordered Absolute Diff.Ranks Signed Ranks 16375-12-61 2483018-122-2 37248241833 4522824 55 59054362455 6644024 55 75258-63677 Repetitive behavior before and after treatment The test statistic, W, is the smaller of W+ and W-, which are the sums of the positive and negative ranks (using absolute values). W+ = (3+5+5+5+7) = 25 W- = (1+2) = 3 W = min(W+, W-) = 3refer to table 7n=7, critical value = 4 Conclusion: Reject H 0 : W = 3 < 4

49. SECTION 4.10 More than 2 Independent Groups: Kruskal-Wallis Test

50. 3. Tests with >2 Independent Groups: Kruskal-Wallis Test  Conceptually similar to ANOVA for comparison of means when the outcome is approximately normally distributed.  Uses ranks like the Wilcoxon Rank Sum Test, but for > 2 groups (k).  H 0 : The k population medians are equal.  H 1 : The k population medians are not equal.  Reject H 0 if test statistic (H) is > critical value (table 3)  The test is usually conducted as 2-sided.

51. Tests with >2 Independent Groups: Kruskal-Wallis Test Albumin Levels (g/dL) Based on Protein Diet (%) Ordered Sample Ranked Sample 5%10%15%5%10%15%5%10%15% 3.13.84.02.6 1 4.15.52.9 2.5 2.9 5.03.1 4 3.44.8 3.4 5 4.2 3.8 6 4.0 7 4.1 8 4.2 9 4.8 10 5.0 11 5.5 12 Example: Evaluate difference in albumin levels based on 3-different protein diets (2-sided, α = 0.05).

52. Tests with >2 Independent Groups: Kruskal-Wallis Test Albumin Levels (g/dL) Based on Protein Diet (%) Ordered Sample Ranked Sample 5%10%15%5%10%15%5%10%15% 3.13.84.02.6 1 4.15.52.9 2.5 2.9 5.03.1 4 3.44.8 3.4 5 4.2 3.8 6 4.0 7 4.1 8 4.2 9 4.8 10 5.0 11 5.5 12 H = - 3 (13) = 7.52 12 12(13) 7.5 2 3 + 30.5 2 5 + 40 2 4 See Table 8; n 1 = 3; n 2 = 5; n 3 =4; α= 0.05. Reject H 0 if H is > 5.656 7.52 > 5.656; Reject H 0.

53. Tests with >2 Independent Groups: Kruskal-Wallis Test (Practice) Example: Evaluate difference in albumin levels based on 3-different protein diets (2-sided, α = 0.05). Order and Rank the 3 Groups Below Albumin Levels (g/dL) Based on Protein Diet (%) Ordered Sample Ranked Sample 5%10%15%5%10%15%5%10%15% 2.83.33.2 4.34.57.1 2.73.93.4 2.1 5.5 6.2 6.8 Sum of Ranks - 

54. Tests with >2 Independent Groups: Kruskal-Wallis Test (Practice) Albumin Levels (g/dL) Based on Protein Diet (%) Ordered Sample Ranked Sample 5%10%15%5%10%15%5%10%15% 2.83.33.2 4.34.57.1 2.73.93.4 2.1 5.5 6.2 6.8 Sum of Ranks -  Critical Value = _______ (Table 8) Conclusion

55. Tests with >2 Independent Groups: Kruskal-Wallis Test (Practice) Albumin Levels (g/dL) Based on Protein Diet (%) Ordered Sample Ranked Sample 5%10%15%5%10%15%5%10%15% 2.83.33.22.1 1 4.34.57.12.7 2 3.93.42.8 3 2.1 5.5 3.2 4 6.2 3.3 5 6.8 3.4 6 3.9 7 4.3 8 4.5 9 5.5 10 6.2 11 6.8 12 7.1 13 Sum of Ranks -  142156 H = - 3 (14) = 9.33 13 13(14) 14 2 4 + 21 2 3 + 56 2 6 Critical Value = 5.61 (Table 8) H = 9.33 > 5.61, Reject H 0