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

Learning Objectives Describe Analysis of Variance (ANOVA) Explain the Rationale of ANOVA Compare Experimental Designs Test the Equality of 2 or More Means Completely Randomized Design Factorial Design As a result of this class, you will be able to ...

Experiments

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

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

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

Completely Randomized Design

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

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

Randomized Design Example Factor (Training Method) Factor levels Level 1 Level 2 Level 3 (Treatments) Are the mean training times the same for 3 different methods? 9 subjects 3 methods (factor levels) Experimental    units 21 hrs. 17 hrs. 31 hrs. Dependent variable 27 hrs. 25 hrs. 28 hrs. (Response) 29 hrs. 20 hrs. 22 hrs. 78

One-Way ANOVA F-Test

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

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 Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales?

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

One-Way ANOVA F-Test Hypotheses H0: 1 = 2 = 3 = ... = k All population means are equal No treatment effect Ha: 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 1 2 3 X f(X)  =

Variances AMONG differ Variances WITHIN differ Why Variances? Same treatment variation Different random variation Pop 1 Pop 2 Pop 3 Pop 4 Pop 6 Pop 5 Variances AMONG differ B Different treatment variation Same random variation Pop 1 Pop 2 Pop 3 Pop 4 Pop 6 Pop 5 Variances WITHIN differ A In all cases, the population means are DIFFERENT! Should reject Ho. Panel A: Same treatment effect - means are equal. Different random variation - standard deviations are different. As variances WITHIN get bigger, we are more likely to conclude equal means. In Populations 4, 5, & 6, it is possible to draw a sample and falsely conclude population means are equal. Panel B: Different treatment effect - means are different. Same random variation - standard deviations are equal. As variances AMONG get bigger, we are more likely to conclude population means are different. In Populations 4,5, & 6, it is possible to draw a sample and falsely conclude population means are equal. Possible to conclude means are equal! 82

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

One-Way ANOVA Partitions Total Variation Variation due to treatment Total variation Variation due to random sampling Variation due to Random Sampling are due to Individual Differences Within Groups. 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 85

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

Treatment Variation Response, X X3 X X2 X1 Group 1 Group 2 Group 3

Random (Error) Variation Response, X X3 X2 X1 Group 1 Group 2 Group 3

One-Way ANOVA F-Test Test Statistic F = MST / MSE MST is Mean Square for Treatment MSE is Mean Square for Error Degrees of Freedom 1 = k - 1 2 = n - k k = Number of groups n = Total sample size

One-Way ANOVA Summary Table Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment k - 1 SST MST = SST/(k - 1) MST MSE n = sum of sample sizes of all populations. k = number of factor levels All values are positive. Why? (squared terms) Degrees of Freedom & Sum of Squares are additive; Mean Square is NOT. Error n - k SSE MSE = SSE/(n - k) Total n - 1 SS(Total) = SST+SSE

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

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? 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

One-Way ANOVA F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 Not all equal Test Statistic: Decision: Conclusion: .05 2 12 F 3.89  = .05

Summary Table Solution From Computer Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment (Machines) 3 - 1 = 2 47.1640 23.5820 25.60 Error 15 - 3 = 12 11.0532 .9211 Total 15 - 1 = 14 58.2172

One-Way ANOVA F-Test Solution H0: 1 = 2 = 3 Ha: Not All Equal  = .05 1 = 2 2 = 12 Critical Value(s): Test Statistic: Decision: Conclusion: F MST MSE  23 5820 9211 25.6 . Reject at  = .05  = .05 There is evidence population means are different F 3.89

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)? M1 M2 M3 M4 10 11 13 18 9 16 8 23 5 9 9 25 Use the following table. You assign randomly 3 people to each method, making sure that they are similar in intelligence etc. © 1984-1994 T/Maker Co.

Summary Table (Partially Completed) Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment 348 (Methods) Error 80 Total

One-Way ANOVA F-Test Solution* H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 = 4 Not all equal Test Statistic: Decision: Conclusion: .05 3 8 F 4.07  = .05

Summary Table Solution* Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment (Methods) 4 - 1 = 3 348 116 11.6 Error 12 - 4 = 8 80 10 Total 12 - 1 = 11 428

One-Way ANOVA F-Test Solution* H0: 1 = 2 = 3 = 4 Ha: Not All Equal  = .05 1 = 3 2 = 8 Critical Value(s): Test Statistic: Decision: Conclusion: F MST MSE  116 10 11 6 . Reject at  = .05  = .05 There is evidence population means are different F 4.07

Tukey Procedure Tells which population means are significantly different Example: μ1 = μ 2 ≠ μ 3 Post hoc procedure Done after rejection of equal means in ANOVA Output from many statistical computer programs X f(X) m 1 = 2 3 2 Groupings 105

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

Randomized Block Design Total Variation Partitioning Variation Due to Random Sampling SS(Total) SSB Total Variation Variation Due to Blocks Variation Due to Treatment SSE SST

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

Randomized Block Design F-Test Test Statistic F = MST / MSE MST is Mean Square for Treatment MSE is Mean Square for Error Degrees of Freedom 1 = k – 1 2 = n – k – b + 1 k = Number of groups n = Total sample size b = Number of blocks

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 1 5.4 3.6 4.0 2 4.1 3.8 2.9 3 6.1 5.6 4.3 4 3.6 2.3 2.6 5 5.3 4.7 3.4

Random Block Design F-Test Solution* H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 Not all equal Test Statistic: Decision: Conclusion: .05 2 8 F 4.46  = .05

Summary Table Solution* Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment (Methods) 3 - 1 = 2 5.43 2.71 12.9 Block (Employee) 5 - 1 = 4 10.69 2.67 12.7 Error 15 - 3 - 5 + 1 = 8 1.68 .21 Total 15 - 1 = 14 17.8

Random Block Design F-Test Solution* H0: 1 = 2 = 3 Ha: Not all equal  = .05 1 = 2 2 = 8 Critical Value(s): Test Statistic: Decision: Conclusion: F MST MSE  2.71 .21 12.9 Reject at  = .05  = .05 There is evidence population means are different F 4.46

Factorial Experiments

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

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

Two-Way ANOVA Data Table Factor Factor B A 1 2 ... b Observation k 1 X111 X121 ... X1b1 Xijk X112 X122 ... X1b2 2 X211 X221 ... X2b1 X212 X222 ... X2b2 Level i Factor A Level j Factor B : : : : : Treatment a Xa11 Xa21 ... Xab1 Xa12 Xa22 ... Xab2

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

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

Two-Way ANOVA

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

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

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

Graphs of Interaction Effects of motivation (high or low) and training method (A, B, C) on mean learning time Interaction No Interaction Average Average Response High Response High Low Low A B C A B C

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

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

Two-Way ANOVA Hypotheses Test for Treatment Means H0: The ab treatment means are equal Ha: At least two of the treatment means differ Test Statistic F = MST / MSE Degrees of Freedom 1 = ab – 1 2 = n – ab

Two-Way ANOVA Hypotheses Test for Factor Interaction H0: The factors do not interact Ha: The factors do interact Test Statistic F = MS(AB) / MSE Degrees of Freedom 1 = (a – 1)(b – 1) 2 = n – ab

Two-Way ANOVA Hypotheses Test for Main Effect of Factor A H0: No difference among mean levels of factor A Ha: At least two factor A mean levels differ Test Statistic F = MS(A) / MSE Degrees of Freedom 1 = (a – 1) 2 = n – ab

Two-Way ANOVA Hypotheses Test for Main Effect of Factor B H0: No difference among mean levels of factor B Ha: At least two factor B mean levels differ Test Statistic F = MS(B) / MSE Degrees of Freedom 1 = (b – 1) 2 = n – ab

Two-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) a - 1 SS(A) MS(A) MS(A) MSE B (Column) b - 1 SS(B) MS(B) MS(B) MSE AB (Interaction) (a - 1)(b - 1) SS(AB) MS(AB) MS(AB) MSE Error n - ab SSE MSE Same as other designs Total n - 1 SS(Total)

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 Classroom Computer 15 hr. 10 hr. 22 hr. Motivation High 11 hr. 12 hr. 17 hr. 27 hr. 31 hr. Low 29 hr. 49 hr.

Treatment Means F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): The 6 treatment means are equal At least 2 differ Test Statistic: Decision: Conclusion: .05 5 6 F 4.39  = .05

Two-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Square F Model 5 546.75 240.35 7.65 Error 6 188.5 31.42 Corrected Total 11 735.25

Treatment Means F-Test Solution H0: The 6 treatment means are equal Ha: At least 2 differ  = .05 1 = 5 2 = 6 Critical Value(s): Test Statistic: Decision: Conclusion: F 4.39  = .05 Reject at  = .05 There is evidence population means are different

Interaction F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): The factors do not interact The factors interact Test Statistic: Decision: Conclusion: .05 2 6 F 5.14  = .05

Two-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1 546.75 546.75 17.40 B (Column) 2 531.5 265.75 8.46 AB (Interaction) 2 123.5 61.76 1.97 Error 6 188.5 31.42 Same as other designs Total 11 SS(Total)

Interaction F-Test Solution H0: The factors do not interact Ha: The factors interact  = .05 1 = 2 2 = 6 Critical Value(s): Test Statistic: Decision: Conclusion: F 5.14  = .05 Do not reject at  = .05 There is no evidence the factors interact

Main Factor A F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): No difference between motivation levels Motivation levels differ Test Statistic: Decision: Conclusion: .05 1 6 F 5.99  = .05

Two-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1 546.75 546.75 17.40 B (Column) 2 531.5 265.75 8.46 AB (Interaction) 2 123.5 61.76 1.97 Error 6 188.5 31.42 Same as other designs Total 11 SS(Total)

Main Factor A F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): No difference between motivation levels Motivation levels differ Test Statistic: Decision: Conclusion: .05 1 6 Reject at  = .05 F 5.99  = .05 There is evidence motivation levels differ

Main Factor B F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): No difference between training methods Training methods differ Test Statistic: Decision: Conclusion: .05 2 6 F 5.14  = .05

Two-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1 546.75 546.75 17.40 B (Column) 2 531.5 265.75 8.46 AB (Interaction) 2 123.5 61.76 1.97 Error 6 188.5 31.42 Same as other designs Total 11 SS(Total)

Main Factor B F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): No difference between training methods Training methods differ Test Statistic: Decision: Conclusion: .05 2 6 Reject at  = .05 F 5.14  = .05 There is evidence training methods differ

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