1. 1 1 Slide Slides by JOHN LOUCKS St. Edward’s University

2. 2 2 Slide Chapter 13, Part B Experimental Design and Analysis of Variance n Factorial Experiments( 因子试验 ) n Randomized Block Design( 随机区组设计）

3. 3 3 Slide n Experimental units are the objects of interest in the experiment. n A completely randomized design （纯随机设计） is an experimental design in which the treatments are randomly assigned to the experimental units. n If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design （随机区组设计）. Randomized Block Design

4. 4 4 Slide For a randomized block design the sum of squares total (SST) is partitioned into three groups: sum of squares due to treatments, sum of squares due to blocks, and sum of squares due to error. For a randomized block design the sum of squares total (SST) is partitioned into three groups: sum of squares due to treatments, sum of squares due to blocks, and sum of squares due to error. n ANOVA Procedure Randomized Block Design SST = SSTR + SSBL + SSE The total degrees of freedom, n T - 1, are partitioned such that k - 1 degrees of freedom go to treatments, b - 1 go to blocks, and ( k - 1)( b - 1) go to the error term. The total degrees of freedom, n T - 1, are partitioned such that k - 1 degrees of freedom go to treatments, b - 1 go to blocks, and ( k - 1)( b - 1) go to the error term.

5. 5 5 Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquare F Treatments Error Total k - 1 n T - 1 SSTR SSE SST Randomized Block Design n ANOVA Table BlocksSSBL b - 1 ( k – 1)( b – 1) p - Value

6. 6 6 Slide

7. 7 7 Randomized Block Design n Example: Crescent Oil Co. Crescent Oil has developed three Crescent Oil has developed three new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the miles per gallon ratings of the three blends is being conducted to determine if the mean ratings are the same for the three blends.

8. 8 8 Slide Randomized Block Design n Example: Crescent Oil Co. Five automobiles have been Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide. Factor... Gasoline blend Treatments... Blend X, Blend Y, Blend Z Blocks... Automobiles Response variable... Miles per gallon

9. 9 9 Slide Randomized Block Design 29.8 28.8 28.4 TreatmentMeans 12345 31302933263029293125302928292630.33329.33328.66731.00025.667 Type of Gasoline (Treatment) BlockMeans Blend X Blend Y Blend Z Automobile(Block)

10. 10 Slide n Mean Square Due to Error Randomized Block Design MSE = 5.47/[(3 - 1)(5 - 1)] =.68 SSE = 62 - 5.2 - 51.33 = 5.47 MSBL = 51.33/(5 - 1) = 12.8 SSBL = 3[(30.333 - 29) 2 +... + (25.667 - 29) 2 ] = 51.33 MSTR = 5.2/(3 - 1) = 2.6 SSTR = 5[(29.8 - 29) 2 + (28.8 - 29) 2 + (28.4 - 29) 2 ] = 5.2 The overall sample mean is 29. Thus, n Mean Square Due to Treatments n Mean Square Due to Blocks

11. 11 Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquare F Treatments Error Total 2 14 5.20 5.47 62.00 8 2.60.68 3.82 n ANOVA Table Randomized Block Design Blocks51.3312.804 p -Value.07

12. 12 Slide n Rejection Rule Randomized Block Design For  =.05, F.05 = 4.46 For  =.05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator) (2 d.f. numerator and 8 d.f. denominator) p -Value Approach: Reject H 0 if p -value <.05 Critical Value Approach: Reject H 0 if F > 4.46

13. 13 Slide n Conclusion Randomized Block Design There is insufficient evidence to conclude that There is insufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends. The p -value is greater than.05 (where F = 4.46) and less than.10 (where F = 3.11). (Excel provides a p -value of.07). Therefore, we cannot reject H 0. The p -value is greater than.05 (where F = 4.46) and less than.10 (where F = 3.11). (Excel provides a p -value of.07). Therefore, we cannot reject H 0. F = MSTR/MSE = 2.6/.68 = 3.82 n Test Statistic

14. 14 Slide Factorial Experiments （因子试验） n In some experiments we want to draw conclusions about more than one variable or factor. n Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required. n For example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations. n The term factorial is used because the experimental conditions include all possible combinations of the factors.

15. 15 Slide The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment. The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment. n ANOVA Procedure SST = SSA + SSB + SSAB + SSE The total degrees of freedom, n T - 1, are partitioned such that ( a – 1) d.f go to Factor A, ( b – 1) d.f go to Factor B, ( a – 1)( b – 1) d.f. go to Interaction, and ab ( r – 1) go to Error. The total degrees of freedom, n T - 1, are partitioned such that ( a – 1) d.f go to Factor A, ( b – 1) d.f go to Factor B, ( a – 1)( b – 1) d.f. go to Interaction, and ab ( r – 1) go to Error. Two-Factor Factorial Experiment We again partition the sum of squares total (SST) into its sources. We again partition the sum of squares total (SST) into its sources.

16. 16 Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquare F Factor A Error Total a - 1 n T - 1 SSA SSE SST Factor B SSB b - 1 ab ( r – 1) Two-Factor Factorial Experiment Interaction SSAB ( a – 1)( b – 1) p - Value

17. 17 Slide n Step 3 Compute the sum of squares for factor B Two-Factor Factorial Experiment n Step 1 Compute the total sum of squares n Step 2 Compute the sum of squares for factor A

18. 18 Slide n Step 4 Compute the sum of squares for interaction Two-Factor Factorial Experiment SSE = SST – SSA – SSB - SSAB n Step 5 Compute the sum of squares due to error

19. 19 Slide A survey was conducted of hourly wages A survey was conducted of hourly wages for a sample of workers in two industries for a sample of workers in two industries at three locations in Ohio. Part of the purpose of the survey was to determine if differences exist determine if differences exist in both industry type and in both industry type and location. The sample data are shown on the next slide. n Example: State of Ohio Wage Survey Two-Factor Factorial Experiment

20. 20 Slide n Example: State of Ohio Wage Survey Two-Factor Factorial Experiment IndustryCincinnatiClevelandColumbusI$12.10$11.80$12.90 I 11.80 11.80 11.20 11.20 12.70 12.70 I 12.10 12.10 12.00 12.00 12.20 12.20 II 12.40 12.40 12.60 12.60 13.00 13.00 II 12.50 12.50 12.00 12.00 12.10 12.10 II 12.00 12.00 12.50 12.50 12.70 12.70

21. 21 Slide n Factors Two-Factor Factorial Experiment Each experimental condition is repeated 3 times Each experimental condition is repeated 3 times Factor B: Location (3 levels) Factor B: Location (3 levels) Factor A: Industry Type (2 levels) Factor A: Industry Type (2 levels) n Replications

22. 22 Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquare F Factor A Error Total 1 17.50 1.43 3.42 12.50.12 4.19 n ANOVA Table Factor B 1.12.562 Two-Factor Factorial Experiment Interaction.37.192 4.69 1.55 p -Value.06.03.25

23. 23 Slide n Conclusions Using the Critical Value Approach Two-Factor Factorial Experiment Interaction is not significant. Interaction: F = 1.55 < F  = 3.89 Interaction: F = 1.55 < F  = 3.89 Mean wages differ by location. Locations: F = 4.69 > F  = 3.89 Locations: F = 4.69 > F  = 3.89 Mean wages do not differ by industry type. Industries: F = 4.19 < F  = 4.75 Industries: F = 4.19 < F  = 4.75

24. 24 Slide n Conclusions Using the p -Value Approach Two-Factor Factorial Experiment Interaction is not significant. Interaction: p -value =.25 >  =.05 Interaction: p -value =.25 >  =.05 Mean wages differ by location. Locations: p -value =.03 <  =.05 Locations: p -value =.03 <  =.05 Mean wages do not differ by industry type. Industries: p -value =.06 >  =.05 Industries: p -value =.06 >  =.05

25. 25 Slide End of Chapter 13, Part B