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

© 2003 Prentice-Hall, Inc. Chap 11-2 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 The Randomized Block Design F Test for the Difference in c Means The Tukey Procedure Chapter Topics

© 2003 Prentice-Hall, Inc. Chap 11-3 Chapter Topics The Factorial Design: Two-Way Analysis of Variance Examine Effects of Factors and Interaction Kruskal-Wallis Rank Test for Differences in c Medians Friedman Rank Test for Differences in c Medians (continued)

© 2003 Prentice-Hall, Inc. Chap 11-4 General Experimental Setting Investigator Controls One or More Independent Variables Called treatment variables or factors Each treatment factor contains two or more groups (or levels) Observe Effects on Dependent Variable Response to groups (or levels) of independent variable Experimental Design: The Plan Used to Test Hypothesis

© 2003 Prentice-Hall, Inc. Chap 11-5 Completely Randomized Design Experimental Units (Subjects) are Assigned Randomly to Groups Subjects are assumed to be homogeneous Only One Factor or Independent Variable With 2 or more groups (or levels) Analyzed by One-Way Analysis of Variance (ANOVA)

© 2003 Prentice-Hall, Inc. Chap 11-6 Factor (Training Method) Factor Levels (Groups) Randomly Assigned Units Dependent Variable (Response) 21 hrs17 hrs31 hrs 27 hrs25 hrs28 hrs 29 hrs20 hrs22 hrs Randomized Design Example   

© 2003 Prentice-Hall, Inc. Chap 11-7 One-Way Analysis of Variance F Test Evaluate the Difference Among the Mean Responses of 2 or More (c ) Populations E.g., Several types of tires, oven temperature settings Assumptions Samples are randomly and independently drawn This condition must be met Populations are normally distributed F Test is robust to moderate departure from normality Populations have equal variances Less sensitive to this requirement when samples are of equal size from each population

© 2003 Prentice-Hall, Inc. Chap 11-8 Why ANOVA? Could Compare the Means One by One using Z or t Tests for Difference of Means Each Z or t Test Contains Type I Error The Total Type I Error with k Pairs of Means is 1- (1 -  ) k E.g., If there are 5 means and use  =.05 Must perform 10 comparisons Type I Error is 1 – (.95) 10 =.40 40% of the time you will reject the null hypothesis of equal means in favor of the alternative when the null is true!

© 2003 Prentice-Hall, Inc. Chap 11-9 Hypotheses of One-Way ANOVA All population means are equal No treatment effect (no variation in means among groups) At least one population mean is different (others may be the same!) There is a treatment effect Does not mean that all population means are different

© 2003 Prentice-Hall, Inc. Chap 11-12 One-Way ANOVA (Partition of Total Variation) Variation Due to Group SSA Variation Due to Random Sampling SSW Total Variation SST Commonly referred to as:  Within Group Variation  Sum of Squares Within  Sum of Squares Error  Sum of Squares Unexplained Commonly referred to as:  Among Group Variation  Sum of Squares Among  Sum of Squares Between  Sum of Squares Model  Sum of Squares Explained  Sum of Squares Treatment = +

© 2003 Prentice-Hall, Inc. Chap 11-19 Within-Group Variation (continued) For c = 2, this is the pooled-variance in the t test. If more than 2 groups, use F Test. For 2 groups, use t test. F Test more limited.

© 2003 Prentice-Hall, Inc. Chap 11-21 One-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Squares (Variance) F Statistic Among (Factor) c – 1SSA MSA = SSA/(c – 1 ) MSA/MSW Within (Error) n – cSSW MSW = SSW/(n – c ) Totaln – 1 SST = SSA + SSW

© 2003 Prentice-Hall, Inc. Chap 11-22 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 = c -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

© 2003 Prentice-Hall, Inc. Chap 11-23 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)

© 2003 Prentice-Hall, Inc. Chap 11-24 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

© 2003 Prentice-Hall, Inc. Chap 11-25 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

© 2003 Prentice-Hall, Inc. Chap 11-27 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

© 2003 Prentice-Hall, Inc. Chap 11-28 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...

© 2003 Prentice-Hall, Inc. Chap 11-30 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

© 2003 Prentice-Hall, Inc. Chap 11-31 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.

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

© 2003 Prentice-Hall, Inc. Chap 11-35 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.

© 2003 Prentice-Hall, Inc. Chap 11-36 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

© 2003 Prentice-Hall, Inc. Chap 11-39 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

© 2003 Prentice-Hall, Inc. Chap 11-40 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

© 2003 Prentice-Hall, Inc. Chap 11-41 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

© 2003 Prentice-Hall, Inc. Chap 11-42 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

© 2003 Prentice-Hall, Inc. Chap 11-48 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

© 2003 Prentice-Hall, Inc. Chap 11-49 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

© 2003 Prentice-Hall, Inc. Chap 11-51 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

© 2003 Prentice-Hall, Inc. Chap 11-52 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.

© 2003 Prentice-Hall, Inc. Chap 11-55 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.

© 2003 Prentice-Hall, Inc. Chap 11-57 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?

© 2003 Prentice-Hall, Inc. Chap 11-58 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)

© 2003 Prentice-Hall, Inc. Chap 11-59 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)

© 2003 Prentice-Hall, Inc. Chap 11-66 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

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

© 2003 Prentice-Hall, Inc. Chap 11-68 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

© 2003 Prentice-Hall, Inc. Chap 11-69 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

© 2003 Prentice-Hall, Inc. Chap 11-70 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

© 2003 Prentice-Hall, Inc. Chap 11-71 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

© 2003 Prentice-Hall, Inc. Chap 11-72 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)

© 2003 Prentice-Hall, Inc. Chap 11-73 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)

© 2003 Prentice-Hall, Inc. Chap 11-74 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?

© 2003 Prentice-Hall, Inc. Chap 11-75 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

© 2003 Prentice-Hall, Inc. Chap 11-77 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

© 2003 Prentice-Hall, Inc. Chap 11-79 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

© 2003 Prentice-Hall, Inc. Chap 11-80 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

© 2003 Prentice-Hall, Inc. Chap 11-81 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

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

© 2003 Prentice-Hall, Inc. Chap 11-84 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

© 2003 Prentice-Hall, Inc. Chap 11-85 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

© 2003 Prentice-Hall, Inc. Chap 11-86 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)

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

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

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