1. © 2003 Prentice-Hall, Inc.Chap 11-1 Analysis of Variance & Post-ANOVA ANALYSIS IE 340/440 PROCESS IMPROVEMENT THROUGH PLANNED EXPERIMENTATION Dr. Xueping Li University of Tennessee

2. © 2003 Prentice-Hall, Inc. Chap 11-2 What If There Are More Than Two Factor Levels? The t-test does not directly apply There are lots of practical situations where there are either more than two levels of interest, or there are several factors of simultaneous interest The analysis of variance (ANOVA) is the appropriate analysis “engine” for these types of experiments – Chapter 3, textbook The ANOVA was developed by Fisher in the early 1920s, and initially applied to agricultural experiments Used extensively today for industrial experiments

3. © 2003 Prentice-Hall, Inc. Chap 11-3 Figure 3.1 (p. 61) A single-wafer plasma etching tool.

4. © 2003 Prentice-Hall, Inc. Chap 11-4 Table 3.1 (p. 62) Etch Rate Data (in Å/min) from the Plasma Etching Experiment)

5. © 2003 Prentice-Hall, Inc. Chap 11-5 The Analysis of Variance (Sec. 3-3, pg. 65) In general, there will be a levels of the factor, or a treatments, and n replicates of the experiment, run in random order…a completely randomized design (CRD) N = an total runs We consider the fixed effects case…the random effects case will be discussed later Objective is to test hypotheses about the equality of the a treatment means

6. © 2003 Prentice-Hall, Inc. Chap 11-6 Models for the Data There are several ways to write a model for the data:

7. © 2003 Prentice-Hall, Inc. Chap 11-7 The Analysis of Variance The name “analysis of variance” stems from a partitioning of the total variability in the response variable into components that are consistent with a model for the experiment The basic single-factor ANOVA model is

8. © 2003 Prentice-Hall, Inc. Chap 11-8 The Analysis of Variance Total variability is measured by the total sum of squares: The basic ANOVA partitioning is:

9. © 2003 Prentice-Hall, Inc. Chap 11-9 The Analysis of Variance A large value of SS Treatments reflects large differences in treatment means A small value of SS Treatments likely indicates no differences in treatment means Formal statistical hypotheses are:

10. Chap 11-10 The Analysis of Variance While sums of squares cannot be directly compared to test the hypothesis of equal means, mean squares can be compared. A mean square is a sum of squares divided by its degrees of freedom: If the treatment means are equal, the treatment and error mean squares will be (theoretically) equal. If treatment means differ, the treatment mean square will be larger than the error mean square.

11. © 2003 Prentice-Hall, Inc. Chap 11-11 The Analysis of Variance is Summarized in a Table Computing…see text, pp 70 – 73 The reference distribution for F 0 is the F a-1, a(n-1) distribution Reject the null hypothesis (equal treatment means) if

12. © 2003 Prentice-Hall, Inc. Chap 11-12 Features of One-Way ANOVA F Statistic The F Statistic is the Ratio of the Among Estimate of Variance and the Within Estimate of Variance The ratio must always be positive df 1 = a -1 will typically be small df 2 = N - c will typically be large The Ratio Should Be Close to 1 if the Null is True

13. © 2003 Prentice-Hall, Inc. Chap 11-13 Features of One-Way ANOVA F Statistic If the Null Hypothesis is False The numerator should be greater than the denominator The ratio should be larger than 1 (continued)

14. © 2003 Prentice-Hall, Inc. Chap 11-14 The Reference Distribution:

15. © 2003 Prentice-Hall, Inc. Chap 11-15 Table 3.1 (p. 62) Etch Rate Data (in Å/min) from the Plasma Etching Experiment)

16. © 2003 Prentice-Hall, Inc. Chap 11-16 Table 3.4 (p. 71) ANOVA for the Plasma Etching Experiment

17. © 2003 Prentice-Hall, Inc. Chap 11-17 Table 3.5 (p. 72) Coded Etch Rate Data for Example 3.2 Coding the observations More about manual calculation p.70-71

18. Chap 11-18 Graphical View of the Results

19. © 2003 Prentice-Hall, Inc. Chap 11-19 Model Adequacy Checking in the ANOVA Text reference, Section 3-4, pg. 76 Checking assumptions is important Normality Constant variance Independence Have we fit the right model? Later we will talk about what to do if some of these assumptions are violated

20. © 2003 Prentice-Hall, Inc. Chap 11-20 Model Adequacy Checking in the ANOVA Examination of residuals (see text, Sec. 3-4, pg. 76) Design-Expert generates the residuals Residual plots are very useful Normal probability plot of residuals

21. © 2003 Prentice-Hall, Inc. Chap 11-21 Table 3.6 (p. 76) Etch Rate Data and Residuals from Example 3.1 a.

22. © 2003 Prentice-Hall, Inc. Chap 11-22 Figure 3.4 (p. 77) Normal probability plot of residuals for Example 3-1.

23. © 2003 Prentice-Hall, Inc. Chap 11-23 Figure 3.5 (p. 78) Plot of residuals versus run order or time.

24. © 2003 Prentice-Hall, Inc. Chap 11-24 Figure 3.6 (p. 79) Plot of residuals versus fitted values.

25. © 2003 Prentice-Hall, Inc. Chap 11-25 Other Important Residual Plots

26. © 2003 Prentice-Hall, Inc. Chap 11-26 Post-ANOVA Comparison of Means The analysis of variance tests the hypothesis of equal treatment means Assume that residual analysis is satisfactory If that hypothesis is rejected, we don’t know which specific means are different Determining which specific means differ following an ANOVA is called the multiple comparisons problem There are lots of ways to do this…see text, Section 3-5, pg. 86 We will use pairwise t-tests on means…sometimes called Fisher’s Least Significant Difference (or Fisher’s LSD) Method

27. © 2003 Prentice-Hall, Inc. Chap 11-27 Tukey’s Test H0: Mu_i = Mu_j ; H1: Mu_i <> Mu_j T statistic Whether Where f is the DF of MSE a is the number of groups

28. © 2003 Prentice-Hall, Inc. Chap 11-28 The Tukey-Kramer Procedure Tells which Population Means are Significantly Different E.g.,  1 =  2   3 2 groups whose means may be significantly different Post Hoc (A Posteriori) Procedure Done after rejection of equal means in ANOVA Pairwise Comparisons Compare absolute mean differences with critical range X f(X)  1 =  2  3

29. © 2003 Prentice-Hall, Inc. Chap 11-29 The Tukey-Kramer Procedure: Example 1. Compute absolute mean differences: Machine1 Machine2 Machine3 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 2. Compute critical range: 3. All of the absolute mean differences are greater than the critical range. There is a significant difference between each pair of means at the 5% level of significance.

30. © 2003 Prentice-Hall, Inc. Chap 11-30 Fisher’s LSD H0: Mu_i = Mu_j Least Significant Difference Whether Where

31. © 2003 Prentice-Hall, Inc. Chap 11-31 Design-Expert Output Treatment Means (Adjusted, If Necessary) EstimatedStandard MeanError 1-159.801.27 2-2015.401.27 3-2517.601.27 4-3021.601.27 5-3510.801.27 MeanStandardt for H0 TreatmentDifferenceDFErrorCoeff=0Prob > |t| 1 vs 2-5.6011.80-3.120.0054 1 vs 3-7.8011.80-4.340.0003 1 vs 4-11.8011.80-6.57< 0.0001 1 vs 5-1.0011.80-0.560.5838 2 vs 3-2.2011.80-1.230.2347 2 vs 4-6.2011.80-3.450.0025 2 vs 54.6011.802.560.0186 3 vs 4-4.0011.80-2.230.0375 3 vs 56.8011.803.790.0012 4 vs 510.8011.806.01< 0.0001

32. © 2003 Prentice-Hall, Inc. Chap 11-32 Figure 3.12 (p. 99) Design-Expert computer output for Example 3-1.

33. © 2003 Prentice-Hall, Inc. Chap 11-33 Figure 3.13 (p. 100) Minitab computer output for Example 3-1.

34. © 2003 Prentice-Hall, Inc. Chap 11-34

35. © 2003 Prentice-Hall, Inc. Chap 11-35 Graphical Comparison of Means Text, pg. 89

36. © 2003 Prentice-Hall, Inc. Chap 11-36 For the Case of Quantitative Factors, a Regression Model is often Useful Response:Strength ANOVA for Response Surface Cubic Model Analysis of variance table [Partial sum of squares] Sum ofMeanF SourceSquaresDFSquareValue Prob > F Model441.813147.2715.85 < 0.0001 A90.84190.849.78 0.0051 A2343.211343.2136.93 < 0.0001 A364.98164.986.99 0.0152 Residual195.15219.29 Lack of Fit33.95133.954.210.0535 Pure Error161.20208.06 Cor Total636.9624 CoefficientStandard 95% CI 95% CI FactorEstimateDFErrorLowHighVIF Intercept19.4710.9517.4921.44 A-Cotton %8.1012.592.7113.499.03 A2-8.8611.46-11.89-5.831.00 A3-7.6012.87-13.58-1.629.03

37. Chap 11-37 The Regression Model Final Equation in Terms of Actual Factors: Strength = +62.61143 -9.01143* Cotton Weight % +0.48143 * Cotton Weight %^2 -7.60000E- 003 * Cotton Weight %^3 This is an empirical model of the experimental results

38. © 2003 Prentice-Hall, Inc. Chap 11-38 Figure 3.7 (p. 83) Plot of residuals versus ŷ ij for Example 3-5.

39. © 2003 Prentice-Hall, Inc. Chap 11-39 Table 3.9 (p. 83) Variance-Stabilizing Transformations

40. © 2003 Prentice-Hall, Inc. Chap 11-40 Figure 3.8 (p. 84) Plot of log S i versus log for the peak discharge data from Example 3.5.

41. © 2003 Prentice-Hall, Inc. Chap 11-41 Figure 3.12 (p. 99) Design-Expert computer output for Example 3-1.

42. © 2003 Prentice-Hall, Inc. Chap 11-42 Figure 3.13 (p. 100) Minitab computer output for Example 3-1.

43. © 2003 Prentice-Hall, Inc. Chap 11-43 Display on page 103

44. © 2003 Prentice-Hall, Inc. Chap 11-44 Example 3-1 p70 EX3-1 p112

45. © 2003 Prentice-Hall, Inc. Chap 11-45

46. © 2003 Prentice-Hall, Inc. Chap 11-46 One-Way ANOVA F Test Example As production manager, you want to see if 3 filling machines have different mean filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the.05 significance level, is there a difference in mean filling times? Machine1 Machine2 Machine3 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

47. © 2003 Prentice-Hall, Inc. Chap 11-47 One-Way ANOVA Example: Scatter Diagram 27 26 25 24 23 22 21 20 19 Time in Seconds Machine1 Machine2 Machine3 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

48. © 2003 Prentice-Hall, Inc. Chap 11-48 One-Way ANOVA Example Computations Machine1 Machine2 Machine3 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

49. © 2003 Prentice-Hall, Inc. Chap 11-49 Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Squares (Variance) F Statistic Among (Factor) 3-1=247.164023.5820 MSA/MSW =25.60 Within (Error) 15-3=1211.0532.9211 Total15-1=1458.2172

50. © 2003 Prentice-Hall, Inc. Chap 11-50 One-Way ANOVA Example Solution F 03.89 H 0 :  1 =  2 =  3 H 1 : Not All Equal  =.05 df 1 = 2 df 2 = 12 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = 0.05. There is evidence that at least one  i differs from the rest.  = 0.05 F MSA MSW   235820 9211 256...

51. © 2003 Prentice-Hall, Inc. Chap 11-51 Solution in Excel Use Tools | Data Analysis | ANOVA: Single Factor Excel Worksheet that Performs the One-Factor ANOVA of the Example

52. © 2003 Prentice-Hall, Inc. Chap 11-52 The Tukey-Kramer Procedure Tells which Population Means are Significantly Different E.g.,  1 =  2   3 2 groups whose means may be significantly different Post Hoc (A Posteriori) Procedure Done after rejection of equal means in ANOVA Pairwise Comparisons Compare absolute mean differences with critical range X f(X)  1 =  2  3

53. © 2003 Prentice-Hall, Inc. Chap 11-53 The Tukey-Kramer Procedure: Example 1. Compute absolute mean differences: Machine1 Machine2 Machine3 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 2. Compute critical range: 3. All of the absolute mean differences are greater than the critical range. There is a significant difference between each pair of means at the 5% level of significance.

54. © 2003 Prentice-Hall, Inc. Chap 11-54 Solution in PHStat Use PHStat | c-Sample Tests | Tukey-Kramer Procedure … Excel Worksheet that Performs the Tukey- Kramer Procedure for the Previous Example

55. © 2003 Prentice-Hall, Inc. Chap 11-55 Levene’s Test for Homogeneity of Variance The Null Hypothesis The c population variances are all equal The Alternative Hypothesis Not all the c population variances are equal

56. © 2003 Prentice-Hall, Inc. Chap 11-56 Levene’s Test for Homogeneity of Variance: Procedure 1.For each observation in each group, obtain the absolute value of the difference between each observation and the median of the group. 2.Perform a one-way analysis of variance on these absolute differences.

57. © 2003 Prentice-Hall, Inc. Chap 11-57 Levene’s Test for Homogeneity of Variances: Example As production manager, you want to see if 3 filling machines have different variance in filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the.05 significance level, is there a difference in the variance in filling times? Machine1 Machine2 Machine3 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

58. © 2003 Prentice-Hall, Inc. Chap 11-58 Levene’s Test: Absolute Difference from the Median

59. © 2003 Prentice-Hall, Inc. Chap 11-59 Summary Table

60. © 2003 Prentice-Hall, Inc. Chap 11-60 Levene’s Test Example: Solution F 03.89 H 0 : H 1 : Not All Equal  =.05 df 1 = 2 df 2 = 12 Critical Value(s): Test Statistic: Decision: Conclusion: Do not reject at  = 0.05. There is no evidence that at least one differs from the rest.  = 0.05

61. © 2003 Prentice-Hall, Inc. Chap 11-61 Randomized Blocked Design Items are Divided into Blocks Individual items in different samples are matched, or repeated measurements are taken Reduced within group variation (i.e., remove the effect of block before testing) Response of Each Treatment Group is Obtained Assumptions Same as completely randomized design No interaction effect between treatments and blocks

62. © 2003 Prentice-Hall, Inc. Chap 11-62 Randomized Blocked Design (Example)          Factor (Training Method) Factor Levels (Groups) Blocked Experiment Units Dependent Variable (Response) 21 hrs17 hrs31 hrs 27 hrs25 hrs28 hrs 29 hrs20 hrs22 hrs

63. © 2003 Prentice-Hall, Inc. Chap 11-63 Randomized Block Design (Partition of Total Variation) Variation Due to Group SSA Variation Among Blocks SSBL Variation Among All Observations SST Commonly referred to as:  Sum of Squares Error  Sum of Squares Unexplained Commonly referred to as:  Sum of Squares Among  Among Groups Variation = + + Variation Due to Random Sampling SSW Commonly referred to as:  Sum of Squares Among Block

64. © 2003 Prentice-Hall, Inc. Chap 11-64 Total Variation

65. © 2003 Prentice-Hall, Inc. Chap 11-65 Among-Group Variation

66. © 2003 Prentice-Hall, Inc. Chap 11-66 Among-Block Variation

67. © 2003 Prentice-Hall, Inc. Chap 11-67 Random Error

68. © 2003 Prentice-Hall, Inc. Chap 11-68 Randomized Block F Test for Differences in c Means No treatment effect Test Statistic Degrees of Freedom 0  Reject

69. © 2003 Prentice-Hall, Inc. Chap 11-69 Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Squares F Statistic Among Group c – 1SSA MSA = SSA/(c – 1) MSA/ MSE Among Block r – 1SSBL MSBL = SSBL/(r – 1) MSBL/ MSE Error (r – 1)  c – 1) SSE MSE = SSE/[(r – 1)  (c– 1)] Total rc – 1SST

70. © 2003 Prentice-Hall, Inc. Chap 11-70 Randomized Block Design: Example As production manager, you want to see if 3 filling machines have different mean filling times. You assign 15 workers with varied experience into 5 groups of 3 based on similarity of their experience, and assigned each group of 3 workers with similar experience to the machines. At the.05 significance level, is there a difference in mean filling times? Machine1 Machine2 Machine3 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

71. © 2003 Prentice-Hall, Inc. Chap 11-71 Randomized Block Design Example Computation Machine1 Machine2 Machine3 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

72. © 2003 Prentice-Hall, Inc. Chap 11-72 Randomized Block Design Example: Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Squares F Statistic Among Group 2 SSA= 47.164 MSA = 23.582 23.582/1.0503 =22.452 Among Block 4 SSBL= 2.6507 MSBL =.6627.6627/1.0503 =.6039 Error  SSE= 8.4025 MSE = 1.0503 Total 14 SST= 58.2172

73. © 2003 Prentice-Hall, Inc. Chap 11-73 Randomized Block Design Example: Solution F 04.46 H 0 :  1 =  2 =  3 H 1 : Not All Equal  =.05 df 1 = 2 df 2 = 8 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = 0.05. There is evidence that at least one  i differs from the rest.  = 0.05 F MSA MSE   23582 1.0503 22.45.

74. © 2003 Prentice-Hall, Inc. Chap 11-74 Randomized Block Design in Excel Tools | Data Analysis | ANOVA: Two Factor Without Replication Example Solution in Excel Spreadsheet

75. © 2003 Prentice-Hall, Inc. Chap 11-75 The Tukey-Kramer Procedure Similar to the Tukey-Kramer Procedure for the Completely Randomized Design Case Critical Range

76. © 2003 Prentice-Hall, Inc. Chap 11-76 The Tukey-Kramer Procedure: Example 1. Compute absolute mean differences: Machine1 Machine2 Machine3 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 2. Compute critical range: 3. All of the absolute mean differences are greater. There is a significance difference between each pair of means at 5% level of significance.

77. © 2003 Prentice-Hall, Inc. Chap 11-77 The Tukey-Kramer Procedure in PHStat PHStat | c-Sample Tests | Tukey-Kramer Procedure … Example in Excel Spreadsheet

78. © 2003 Prentice-Hall, Inc. Chap 11-78 Two-Way ANOVA Examines the Effect of: Two factors on the dependent variable E.g., Percent carbonation and line speed on soft drink bottling process Interaction between the different levels of these two factors E.g., Does the effect of one particular percentage of carbonation depend on which level the line speed is set?

79. © 2003 Prentice-Hall, Inc. Chap 11-79 Two-Way ANOVA Assumptions Normality Populations are normally distributed Homogeneity of Variance Populations have equal variances Independence of Errors Independent random samples are drawn (continued)

80. © 2003 Prentice-Hall, Inc. Chap 11-80 SSE Two-Way ANOVA Total Variation Partitioning Variation Due to Factor A Variation Due to Random Sampling Variation Due to Interaction SSA SSAB SST Variation Due to Factor B SSB Total Variation d.f.= n-1 d.f.= r-1 = + + d.f.= c-1 + d.f.= (r-1)(c-1) d.f.= rc(n’-1)

81. © 2003 Prentice-Hall, Inc. Chap 11-81 Two-Way ANOVA Total Variation Partitioning

82. © 2003 Prentice-Hall, Inc. Chap 11-82 Total Variation

83. © 2003 Prentice-Hall, Inc. Chap 11-83 Factor A Variation Sum of Squares Due to Factor A = the difference among the various levels of factor A and the grand mean

84. © 2003 Prentice-Hall, Inc. Chap 11-84 Factor B Variation Sum of Squares Due to Factor B = the difference among the various levels of factor B and the grand mean

85. © 2003 Prentice-Hall, Inc. Chap 11-85 Interaction Variation Sum of Squares Due to Interaction between A and B = the effect of the combinations of factor A and factor B

86. © 2003 Prentice-Hall, Inc. Chap 11-86 Random Error Sum of Squares Error = the differences among the observations within each cell and the corresponding cell means

87. © 2003 Prentice-Hall, Inc. Chap 11-87 Two-Way ANOVA: The F Test Statistic F Test for Factor B Main Effect F Test for Interaction Effect H 0 :  1. =  2. = =  r. H 1 : Not all  i. are equal H 0 :  ij = 0 (for all i and j) H 1 :  ij  0 H 0 :    1 = . 2 = =   c H 1 : Not all . j are equal Reject if F > F U F Test for Factor A Main Effect

88. © 2003 Prentice-Hall, Inc. Chap 11-88 Two-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Squares F Statistic Factor A (Row) r – 1SSA MSA = SSA/(r – 1) MSA/ MSE Factor B (Column) c – 1SSB MSB = SSB/(c – 1) MSB/ MSE AB (Interaction) (r – 1)(c – 1)SSAB MSAB = SSAB/ [(r – 1)(c – 1)] MSAB/ MSE Error r  c  n ’ – 1) SSE MSE = SSE/[r  c  n ’ – 1)] Total r  c  n ’ – 1 SST

89. © 2003 Prentice-Hall, Inc. Chap 11-89 Features of Two-Way ANOVA F Test Degrees of Freedom Always Add Up rcn’-1=rc(n’-1)+(c-1)+(r-1)+(c-1)(r-1) Total=Error+Column+Row+Interaction The Denominator of the F Test is Always the Same but the Numerator is Different The Sums of Squares Always Add Up Total=Error+Column+Row+Interaction

90. © 2003 Prentice-Hall, Inc. Chap 11-90 Kruskal-Wallis Rank Test for c Medians Extension of Wilcoxon Rank Sum Test Tests the equality of more than 2 (c) population medians Distribution-Free Test Procedure Used to Analyze Completely Randomized Experimental Designs Use  2 Distribution to Approximate if Each Sample Group Size n j > 5 df = c – 1

91. © 2003 Prentice-Hall, Inc. Chap 11-91 Kruskal-Wallis Rank Test Assumptions Independent random samples are drawn Continuous dependent variable Data may be ranked both within and among samples Populations have same variability Populations have same shape Robust with Regard to Last 2 Conditions Use F test in completely randomized designs and when the more stringent assumptions hold

92. © 2003 Prentice-Hall, Inc. Chap 11-92 Kruskal-Wallis Rank Test Procedure Obtain Ranks In event of tie, each of the tied values gets their average rank Add the Ranks for Data from Each of the c Groups Square to obtain T j 2

93. © 2003 Prentice-Hall, Inc. Chap 11-93 Kruskal-Wallis Rank Test Procedure Compute Test Statistic # of observation in j –th sample H may be approximated by chi-square distribution with df = c –1 when each n j >5 (continued)

94. © 2003 Prentice-Hall, Inc. Chap 11-94 Kruskal-Wallis Rank Test Procedure Critical Value for a Given  Upper tail Decision Rule Reject H 0 : M 1 = M 2 = = M c if test statistic H > Otherwise, do not reject H 0 (continued)

95. © 2003 Prentice-Hall, Inc. Chap 11-95 Machine1 Machine2 Machine3 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 Kruskal-Wallis Rank Test: Example As production manager, you want to see if 3 filling machines have different median filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the.05 significance level, is there a difference in median filling times?

96. © 2003 Prentice-Hall, Inc. Chap 11-96 Machine1 Machine2 Machine3 14 9 2 15 6 7 12 10 1 11 8 4 13 5 3 Example Solution: Step 1 Obtaining a Ranking Raw DataRanks 6538 17 Machine1 Machine2 Machine3 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

97. © 2003 Prentice-Hall, Inc. Chap 11-97 Example Solution: Step 2 Test Statistic Computation

98. © 2003 Prentice-Hall, Inc. Chap 11-98 Kruskal-Wallis Test Example Solution H 0 : M 1 = M 2 = M 3 H 1 : Not all equal  =.05 df = c - 1 = 3 - 1 = 2 Critical Value(s): Reject at Test Statistic: Decision: Conclusion: There is evidence that population medians are not all equal.  =.05  =.05. H = 11.58

99. © 2003 Prentice-Hall, Inc. Chap 11-99 Kruskal-Wallis Test in PHStat PHStat | c-Sample Tests | Kruskal-Wallis Rank Sum Test … Example Solution in Excel Spreadsheet

100. © 2003 Prentice-Hall, Inc. Chap 11-100 Friedman Rank Test for Differences in c Medians Tests the equality of more than 2 (c) population medians Distribution-Free Test Procedure Used to Analyze Randomized Block Experimental Designs Use  2 Distribution to Approximate if the Number of Blocks r > 5 df = c – 1

101. © 2003 Prentice-Hall, Inc. Chap 11-101 Friedman Rank Test Assumptions The r blocks are independent The random variable is continuous The data constitute at least an ordinal scale of measurement No interaction between the r blocks and the c treatment levels The c populations have the same variability The c populations have the same shape

102. © 2003 Prentice-Hall, Inc. Chap 11-102 Friedman Rank Test: Procedure  Replace the c observations by their ranks in each of the r blocks; assign average rank for ties  Test statistic:  R.j 2 is the square of the rank total for group j  F R can be approximated by a chi-square distribution with (c –1) degrees of freedom  The rejection region is in the right tail

103. © 2003 Prentice-Hall, Inc. Chap 11-103 Friedman Rank Test: Example As production manager, you want to see if 3 filling machines have different median filling times. You assign 15 workers with varied experience into 5 groups of 3 based on similarity of their experience, and assigned each group of 3 workers with similar experience to the machines. At the.05 significance level, is there a difference in median filling times? Machine1 Machine2 Machine3 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

104. © 2003 Prentice-Hall, Inc. Chap 11-104 Friedman Rank Test: Computation Table

105. © 2003 Prentice-Hall, Inc. Chap 11-105 Friedman Rank Test Example Solution H 0 : M 1 = M 2 = M 3 H 1 : Not all equal  =.05 df = c - 1 = 3 - 1 = 2 Critical Value: Reject at Test Statistic: Decision: Conclusion: There is evidence that population medians are not all equal.  =.05 F R = 8.4

106. © 2003 Prentice-Hall, Inc. Chap 11-106 Chapter Summary Described the Completely Randomized Design: One-Way Analysis of Variance ANOVA Assumptions F Test for Difference in c Means The Tukey-Kramer Procedure Levene’s Test for Homogeneity of Variance Discussed the Randomized Block Design F Test for the Difference in c Means The Tukey Procedure

107. © 2003 Prentice-Hall, Inc. Chap 11-107 Chapter Summary Described the Factorial Design: Two-Way Analysis of Variance Examine effects of factors and interaction Discussed Kruskal-Wallis Rank Test for Differences in c Medians Illustrated Friedman Rank Test for Differences in c Medians (continued)