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Slides by John Loucks St. Edward’s University

Slides by John Loucks St. Edward’s University. Chapter 13, Part B Experimental Design and Analysis of Variance. Factorial Experiment. Randomized Block Design. Randomized Block Design. Experimental units are the objects of interest in the experiment.

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Slides by John Loucks St. Edward’s University

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  1. Slides by John Loucks St. Edward’s University

  2. Chapter 13, Part B Experimental Design and Analysis of Variance • Factorial Experiment • Randomized Block Design

  3. Randomized Block Design • Experimental unitsare the objects of interest in the experiment. • A completely randomized designis an experimental design in which the treatments are randomly assigned to the experimental units. • If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.

  4. Randomized Block Design • ANOVA Procedure • 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. SST = SSTR + SSBL + SSE • The total degrees of freedom, nT - 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. Randomized Block Design • ANOVA Table p- Value Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments k - 1 SSTR Blocks SSBL b - 1 Error SSE (k – 1)(b – 1) Total SST nT - 1

  6. Randomized Block Design • Example: Crescent Oil Co. 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.

  7. Randomized Block Design • Example: Crescent Oil Co. 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

  8. Randomized Block Design Type of Gasoline (Treatment) Automobile (Block) Block Means Blend X Blend Y Blend Z 1 2 3 4 5 31 30 29 33 26 30 29 29 31 25 30 29 28 29 26 30.333 29.333 28.667 31.000 25.667 Treatment Means 29.8 28.8 28.4

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

  10. Randomized Block Design • ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square p-Value F .07 Treatments 5.20 2.60 3.82 2 Blocks 51.33 4 12.80 Error 5.47 8 .68 Total 62.00 14

  11. Randomized Block Design • Rejection Rule p-Value Approach:Reject H0 if p-value < .05 Critical Value Approach:Reject H0 if F> 4.46 For  = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator)

  12. Randomized Block Design • Test Statistic • Conclusion F = MSTR/MSE = 2.6/.68 = 3.82 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 H0. There is insufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends.

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

  14. Two-Factor Factorial Experiment • ANOVA Procedure • The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment. • We again partition the sum of squares total (SST) into its sources. SST = SSA + SSB + SSAB + SSE • The total degrees of freedom, nT - 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.

  15. Two-Factor Factorial Experiment p- Value Source of Variation Sum of Squares Degrees of Freedom Mean Square F SSA Factor A a - 1 Factor B SSB b - 1 Interaction SSAB (a – 1)(b – 1) Error SSE ab (r – 1) Total SST nT - 1

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

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

  18. Two-Factor Factorial Experiment • Example: State of Ohio Wage Survey A survey was conducted of hourly wages 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 in both industry type and location. The sample data are shown on the next slide.

  19. Two-Factor Factorial Experiment • Example: State of Ohio Wage Survey

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

  21. Two-Factor Factorial Experiment • ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square p-Value F Factor A .50 .50 4.19 .06 1 Factor B 1.12 2 .56 4.69 .03 Interaction .37 2 .19 1.55 .25 Error 1.43 12 .12 Total 3.42 17

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

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

  24. End of Chapter 13, Part B

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