1. Statistics for Business and Economics Chapter 8 Design of Experiments and Analysis of Variance

2. Learning Objectives 1.Describe Analysis of Variance (ANOVA) 2.Explain the Rationale of ANOVA 3.Compare Experimental Designs 4.Test the Equality of 2 or More Means Completely Randomized Design Factorial Design

3. Experiments

4. Experiment Investigator controls one or more independent variables –Called treatment variables or factors –Contain two or more levels (subcategories) Observes effect on dependent variable –Response to levels of independent variable Experimental design: plan used to test hypotheses

5. Examples of Experiments 1.Thirty stores are randomly assigned 1 of 4 (levels) store displays (independent variable) to see the effect on sales (dependent variable). 2.Two hundred consumers are randomly assigned 1 of 3 (levels) brands of juice (independent variable) to study reaction (dependent variable).

6. Experimental Designs Factorial One-Way ANOVA Experimental Designs Completely Randomized Two-Way ANOVA

7. Completely Randomized Design

8. Experimental Designs Factorial One-Way ANOVA Experimental Designs Completely Randomized Two-Way ANOVA

9. Completely Randomized Design Experimental units (subjects) are assigned randomly to treatments –Subjects are assumed homogeneous One factor or independent variable –Two or more treatment levels or classifications Analyzed by one-way ANOVA

10. Factor (Training Method) Factor levels (Treatments) Level 1 2 3 Experimental units Dependent 21 hrs.17 hrs.31 hrs. variable 27 hrs.25 hrs.28 hrs. (Response) 29 hrs.20 hrs.22 hrs. Randomized Design Example 

11. One-Way ANOVA F-Test

12. Experimental Designs Factorial One-Way ANOVA Experimental Designs Completely Randomized Two-Way ANOVA

13. One-Way ANOVA F-Test Tests the equality of two or more (k) population means Variables –One nominal scaled independent variable  Two or more (k) treatment levels or classifications –One interval or ratio scaled dependent variable Used to analyze completely randomized experimental designs

14. Conditions Required for a Valid ANOVA F-test: Completely Randomized Design 1.Randomness and independence of errors Independent random samples are drawn 2.Normality Populations are approximately normally distributed 3.Homogeneity of variance Populations have equal variances

15. One-Way ANOVA F-Test Hypotheses H 0 :  1 =  2 =  3 =... =  k — All population means are equal — No treatment effect H a : Not All  i Are Equal — At least 2 pop. means are different — Treatment effect —  1   2 ...   k is Wrong X f(X)  1 =  2 =  3 123 X  = 

16. Why Variances? Same treatment variation Different random variation Possible to conclude means are equal! Pop 1Pop 2Pop 3 Pop 4Pop 6 Pop 5 Variances WITHIN differ A Pop 1Pop 2Pop 3 Pop 4Pop 6 Pop 5 Variances AMONG differ B Different treatment variation Same random variation

17. 1.Compares two types of variation to test equality of means 2.Comparison basis is ratio of variances 3.If treatment variation is significantly greater than random variation then means are not equal 4.Variation measures are obtained by ‘partitioning’ total variation One-Way ANOVA Basic Idea

18. One-Way ANOVA Partitions Total Variation Total variation Variation due to treatment Variation due to random sampling Sum of Squares Among Sum of Squares Between Sum of Squares Treatment Among Groups Variation Sum of Squares Within Sum of Squares Error Within Groups Variation

19. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-19 Partitioning the Variation Total variation can be split into two parts: SST = Total Sum of Squares (Total variation) SSA = Sum of Squares Among Groups (Treatment or Among-group variation) SSW = Sum of Squares Within Groups (Random or Within-group variation) SST = SSA + SSW

20. Total Variation X Group 1Group 2Group 3 Response, X

21. Among Group Variation X X3X3 X2X2 X1X1 Group 1Group 2Group 3 Response, X

22. With Group (Random) Variation X2X2 X1X1 X3X3 Group 1Group 2Group 3 Response, X

26. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-26 Obtaining the Mean Squares The Mean Squares are obtained by dividing the various sum of squares by their associated degrees of freedom Mean Square Among (d.f. = c-1) Mean Square Within (d.f. = n-c) Mean Square Total (d.f. = n-1)

27. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-27 One-Way ANOVA Table Source of Variation Sum Of Squares Degrees of Freedom Mean Square (Variance) Among Groups c - 1MSA = Within Groups SSWn - cMSW = TotalSSTn – 1 SSA MSA MSW F c = number of groups n = sum of the sample sizes from all groups df = degrees of freedom SSA c - 1 SSW n - c F STAT =

28. One-Way ANOVA F-Test Critical Value  If means are equal, F = MSA / MSW  1. Only reject large F! Always One-Tail! F (α; c – 1, n – c) 0 Reject H 0 Do Not Reject H 0 F © 1984-1994 T/Maker Co.

29. One-Way ANOVA F-Test Example As production manager, you want to see if three filling machines have different mean 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 mean filling times? Mach1Mach2Mach3 25.4023.4020.00 26.3121.8022.20 24.1023.5019.75 23.7422.7520.60 25.1021.6020.40

30. One-Way ANOVA F-Test Solution H 0 : H a :  = 1 = 2 = Critical Value(s): F 03.89  =.05  1 =  2 =  3 Not All Equal.05 2 12

31. Summary Table Solution From Computer Treatment (Machines) 3 - 1 = 247.164023.582025.60 Error15 - 3 = 1211.0532.9211 Total15 - 1 = 1458.2172 Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) F

32. One-Way ANOVA F-Test Solution Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence population means are different F MSA MSW  235820 9211 25.6..

33. 表 14.1 A 、 B 、 C 廠牌汽車耗油試驗

34. 表 14.5 變異數分析表

35. 圖 14.6 汽車耗油的檢定

36. One-Way ANOVA F-Test Thinking Challenge You’re a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods (  =.05)? M1M2M3M4 10111318 916823 59925 Use the following table. © 1984-1994 T/Maker Co.

37. Summary Table Solution* Treatment (Methods) 4 - 1 = 334811611.6 Error12 - 4 = 88010 Total12 - 1 = 11428 Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) F

38. One-Way ANOVA F-Test Solution* H 0 : H a :  = 1 = 2 = Critical Value(s): 04.07  =.05  1 =  2 =  3 =  4 Not All Equal  

39. One-Way ANOVA F-Test Solution* Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence population means are different F MSA MSW  116 10 116.

40. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-40 ANOVA Assumptions Randomness and Independence –Select random samples from the c groups (or randomly assign the levels) Normality –The sample values for each group are from a normal population Homogeneity of Variance –All populations sampled from have the same variance –Can be tested with Levene’s Test

41. Randomized Block Design Reduces sampling variability (MSE) Matched sets of experimental units (blocks) One experimental unit from each block is randomly assigned to each treatment

42. Randomized Block Design Total Variation Partitioning Variation Due to Random Sampling SST SSB Total Variation Variation Due to Blocks Variation Due to Treatment SSW SSA

43. Conditions Required for a Valid ANOVA F-test: Randomized Block Design 1.The blocks are randomly selected, and all treatments are applied (in random order) to each block 2.The distributions of observations corresponding to all block-treatment combinations are approximately normal 3.All block-treatment distributions have equal variances

44. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-44 Partitioning the Variation Total variation can now be split into three parts: SST = Total variation SSA = Among-Group variation SSBL = Among-Block variation SSE = Random variation SST = SSA + SSBL + SSE

45. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-45 Sum of Squares for Blocks Where: c = number of groups r = number of blocks X i. = mean of all values in block i X = grand mean (mean of all data values) SST = SSA + SSBL + SSE

46. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-46 Partitioning the Variation Total variation can now be split into three parts: SST and SSA are computed as they were in One-Way ANOVA SST = SSA + SSBL + SSE SSE = SST – (SSA + SSBL)

47. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-47 Mean Squares

48. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-48 Randomized Block ANOVA Table Source of Variation dfSSMS Among Groups SSAMSA Error n-r-c+1 SSEMSE Totaln - 1SST c - 1 MSA MSE F c = number of populationsn=rc = total number of observations r = number of blocksdf = degrees of freedom Among Blocks SSBr - 1MSB MSE

49. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-49 Main Factor test: df 1 = c – 1 df 2 = (r – 1)(c – 1) MSA MSE F STAT = Reject H 0 if F STAT > F α Testing For Factor Effect

50. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-50 Test For Block Effect Blocking test: df 1 = r – 1 df 2 = (r – 1)(c – 1) MSB MSE F STAT = Reject H 0 if F STAT > F α

51. Randomized Block Design Example A production manager wants to see if three assembly methods have different mean assembly times (in minutes). Five employees were selected at random and assigned to use each assembly method. At the.05 level of significance, is there a difference in mean assembly times? Employee Method 1 Method 2 Method 3 15.43.64.0 24.13.82.9 36.15.64.3 43.62.32.6 55.34.73.4

52. Random Block Design F-Test Solution* H 0 : H a :  = 1 = 2 = Critical Value(s): F 04.46  =.05  1 = ., 2 = ., 3 Not all equal.05 2  8

53. Summary Table Solution* Treatment (Methods) 3 - 1 = 25.432.7112.9 Error15 - 3 - 5 + 1 = 8 = 2*4 1.68.21 Total15 - 1 = 1417.8 Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) F Block (Employee) 5 - 1 = 410.692.6712.7

54. Random Block Design F-Test Solution* Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence population means are different F MST MSE  2.71.21 12.9

55. Random Block Design F-Test Solution* H 0 : H a :  = 1 = 2 = Critical Value(s): F 04.46  =.05  1,. =  2,. =  3,. Not all equal.05 4  8

56. Random Block Design F-Test Solution* Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence block means are different F MSBL MSE  2.67.21 12.7

57. Factorial Experiments

58. Experimental Designs Factorial One-Way ANOVA Experimental Designs Completely Randomized Two-Way ANOVA

59. Factorial Design Experimental units (subjects) are assigned randomly to treatments –Subjects are assumed homogeneous Two or more factors or independent variables – Each has two or more treatments (levels) Analyzed by two-way ANOVA

60. Two-Way ANOVA Data Table XijkXijk Level i Factor A Level j Factor B Observation k FactorFactor B A12...b 1 X 111 X 121...X 1b1 X 112 X 122...X 1b2 2 X 211 X 221...X 2b1 X 212 X 222...X 2b2 : :::: a X a11 X a21...X ab1 X a12 X a22...X ab2 Treatment

61. Factorial Design Example Factor 2 (Training Method) Factor Levels Level 1Level 2Level 3 Level 1 15 hr. 10 hr. 22 hr. Factor 1 (Motivation) (High) 11 hr. 12 hr. 17 hr. Level 2 27 hr.  15 hr.  31 hr.  (Low) 29 hr.  17 hr.  49 hr.  Treatment

62. Advantages of Factorial Designs Saves time and effort –e.g., Could use separate completely randomized designs for each variable Controls confounding effects by putting other variables into model Can explore interaction between variables

63. Graphs of Interaction Effects of motivation (high or low) and training method (A, B, C) on mean learning time InteractionNo Interaction Average Response ABC High Low Average Response ABC High Low

64. Two-Way ANOVA

65. Experimental Designs Factorial One-Way ANOVA Experimental Designs Completely Randomized Two-Way ANOVA

66. Two-Way ANOVA Tests the equality of two or more population means when several independent variables are used Same results as separate one-way ANOVA on each variable –No interaction can be tested Used to analyze factorial designs

67. Interaction Occurs when effects of one factor vary according to levels of other factor When significant, interpretation of main effects (A and B) is complicated Can be detected –In data table, pattern of cell means in one row differs from another row –In graph of cell means, lines cross

68. Graphs of Interaction Effects of motivation (high or low) and training method (A, B, C) on mean learning time InteractionNo Interaction Average Response ABC High Low Average Response ABC High Low

69. Two-Way ANOVA Total Variation Partitioning Variation Due to Random Sampling Variation Due to Interaction SS(AB) SST Total Variation Variation Due to Treatment A Variation Due to Treatment B SSA SSB SSE

70. Conditions Required for Valid F-Tests in Factorial Experiments 1.Normality Populations are approximately normally distributed 2.Homogeneity of variance Populations have equal variances 3.Independence of errors Independent random samples are drawn

71. Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) r - 1SS(A)MS(A) MSE B (Column) c - 1SS(B)MS(B) MSE AB (Interaction) (r - 1)(c - 1)SS(AB)MS(AB) MSE Errorn - rcSSEMSE Totaln - 1SS(Total) Two-Way ANOVA Summary Table Same as other designs

72. Two-Way ANOVA Hypotheses Test for Main Effect of Factor A H 0 : No difference among mean levels of factor A H a : At least two factor A mean levels differ Test Statistic F = MS(A) / MSE Degrees of Freedom 1 = (r – 1) 2 = n – rc

73. Two-Way ANOVA Hypotheses Test for Main Effect of Factor B H 0 : No difference among mean levels of factor B H a : At least two factor B mean levels differ Test Statistic F = MS(B) / MSE Degrees of Freedom 1 = (c – 1) 2 = n – rc

74. Two-Way ANOVA Hypotheses Test for Factor Interaction H 0 : The factors do not interact H a : The factors do interact Test Statistic F = MS(AB) / MSE Degrees of Freedom 1 = (r – 1)(c – 1) 2 = n – rc

75. Two-Way ANOVA Hypotheses Test for Treatment Means H 0 : The ab treatment means are equal H a : At least two of the treatment means differ Test Statistic F = MST / MSE Degrees of Freedom 1 = rc – 1 2 = n – rc

76. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-76 Two-Way ANOVA Equations Total Variation: Factor A Variation: Factor B Variation:

77. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-77 Two-Way ANOVA Equations Interaction Variation: Sum of Squares Error: (continued)

78. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-78 Two-Way ANOVA Equations where: r = number of levels of factor A c = number of levels of factor B n ’ = number of replications in each cell (continued)

79. Basic Business Statistics, 11e © 2009 Prentice-Hall, Inc.. Chap 11-79 Two-Way ANOVA Equations where: r = number of levels of factor A c = number of levels of factor B n ’ = number of replications in each cell (continued)

80. Factorial Design Example Human Resources wants to determine if training time is different based on motivation level and training method. Conduct the appropriate ANOVA tests. Use α =.05 for each test. Training Method Factor Levels Self– paced ClassroomComputer 15 hr.10 hr.22 hr. Motivation High 11 hr.12 hr.17 hr. 27 hr. 15 hr. 31 hr. Low 29 hr.17 hr.49 hr.

81. Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1546.75 B (Column) 2531.5265.75 AB (Interaction) 2123.561.76 Error 6188.531.42 Total11SS(Total) Two-Way ANOVA Summary Table Same as other designs 17.40 8.46 1.97

82. Main Factor A F-Test Solution H 0 : H a :  = 1 = 2 = Critical Value(s): F 05.99  =.05 No difference between motivation levels Motivation levels differ.05 1  6

83. Main Factor A F-Test Solution Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence motivation levels differ

84. Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1546.75 B (Column) 2531.5265.75 AB (Interaction) 2123.561.76 Error 6188.531.42 Total11SS(Total) Two-Way ANOVA Summary Table Same as other designs 17.40 8.46 1.97

85. Main Factor B F-Test Solution H 0 : H a :  = 1 = 2 = Critical Value(s): F 05.14  =.05 No difference between training methods Training methods differ.05 2  6

86. Main Factor B F-Test Solution Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence training methods differ

87. Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1546.75 B (Column) 2531.5265.75 AB (Interaction) 2123.561.76 Error 6188.531.42 Total11SS(Total) Two-Way ANOVA Summary Table Same as other designs 17.40 8.46 1.97

88. Interaction F-Test Solution H 0 : H a :  = 1 = 2 = Critical Value(s): F 05.14  =.05 The factors do not interact The factors interact.05 2 6

89. Interaction F-Test Solution Test Statistic: Decision: Conclusion: Do not reject at  =.05 There is no evidence the factors interact

90. Treatment Means F-Test Solution H 0 : H a :  = 1 = 2 = Critical Value(s): F 04.39  =.05 The 6 treatment means are equal At least 2 differ.05 5 6

91. Source of Variation Degrees of Freedom Sum of Squares Mean Square F Model5546.75240.35 Error6 188.5 31.42 Corrected Total Two-Way ANOVA Summary Table 7.65 11 735.25

92. Treatment Means F-Test Solution Test Statistic: Decision: Conclusion: Reject at  =.05 There is evidence population means are different

93. Conclusion 1.Described Analysis of Variance (ANOVA) 2.Explained the Rationale of ANOVA 3.Compared Experimental Designs 4.Tested the Equality of 2 or More Means Completely Randomized Design Factorial Design