1. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 1 Slide Slide Slides Prepared by Juei-Chao Chen Fu Jen Catholic University Slides Prepared by Juei-Chao Chen Fu Jen Catholic University

2. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 2 Slide Slide Chapter 13, Part B Analysis of Variance and Experimental Design nFactorial Experiments n nAn Introduction to Experimental Design n nCompletely Randomized Designs n nRandomized Block Design

3. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 3 Slide Slide An Introduction to Experimental Design n nStatistical studies can be classified as being either experimental or observational. n nIn an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. n nIn an observational study, no attempt is made to control the factors. n nCause-and-effect relationships are easier to establish in experimental studies than in observational studies.

4. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 4 Slide Slide An Introduction to Experimental Design n nA factor is a variable that the experimenter has selected for investigation. n nA treatment is a level of a factor. n nExperimental units are the objects of interest in the experiment. n nA completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. n nIf the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.

5. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 5 Slide Slide The between-samples estimate of  2 is referred to as the mean square due to treatments (MSTR). n nBetween-Treatments Estimate of Population Variance Completely Randomized Design denominator is the degrees of freedom associated with SSTR numerator is called the sum of squares due to treatments (SSTR)

6. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 6 Slide Slide The second estimate of  2, the within-samples estimate, is referred to as the mean square due to error (MSE). n nWithin-Treatments Estimate of Population Variance Completely Randomized Design denominator is the degrees of freedom associated with SSE numerator is called the sum of squares due to error (SSE)

7. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 7 Slide Slide n nANOVA Table Completely Randomized Design Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total k - 1 n T - 1 SSTR SSE SST n T - k

8. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 8 Slide Slide AutoShine, Inc. is considering marketing a long- lasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. Completely Randomized Design n n Example: AutoShine, Inc. In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration.

9. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 9 Slide Slide Completely Randomized Design The number of times each car went through the carwash is shown on the next slide. AutoShine, Inc. must decide which wax to market. Are the three waxes equally effective? n n Example: AutoShine, Inc.

10. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 10 Slide Slide 12345273029283133283130302928303231 Sample Mean Sample Variance Observation Wax Type 1 Wax Type 2 Wax Type 3 2.5 3.3 2.5 2.5 3.3 2.5 29.0 30.430.0 Completely Randomized Design

11. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 11 Slide Slide nHypotheses Completely Randomized Design where:  1 = mean number of washes for Type 1 wax  2 = mean number of washes for Type 2 wax  3 = mean number of washes for Type 3 wax H 0 :  1  =  2  =  3  H a : Not all the means are equal

12. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 12 Slide Slide Because the sample sizes are all equal: Completely Randomized Design MSE = 33.2/(15 - 3) = 2.77 MSTR = 5.2/(3 - 1) = 2.6 SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2 SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2 n nMean Square Error n nMean Square Between Treatments = (29 + 30.4 + 30)/3 = 29.8

13. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 13 Slide Slide nRejection Rule Completely Randomized Design where F.05 = 3.89 is based on an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom p-Value Approach: Reject H 0 if p-value <.05 Critical Value Approach: Reject H 0 if F > 3.89

14. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 14 Slide Slide nTest Statistic Completely Randomized Design There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same. n nConclusion F = MSTR/MSE = 2.6/2.77 =.939 The p-value is greater than.10, where F = 2.81. (Excel provides a p-value of.42.) Therefore, we cannot reject H 0.

15. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 15 Slide Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total 2 14 5.2 33.2 38.4 12 Completely Randomized Design 2.60 2.77.939 n nANOVA Table

16. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 16 Slide 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. n nANOVA 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.

17. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 17 Slide Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total k - 1 n T - 1 SSTR SSE SST Randomized Block Design n nANOVA Table BlocksSSBL b - 1 ( k – 1)( b – 1)

18. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 18 Slide Slide Randomized Block Design nExample: 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.

19. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 19 Slide Slide Randomized Block Design n nExample: 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.

20. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 20 Slide 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)

21. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 21 Slide Slide nMean 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 nMean Square Due to Treatments n nMean Square Due to Blocks

22. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 22 Slide Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total 2 14 5.20 5.47 62.00 8 2.60.68 3.82 n nANOVA Table Randomized Block Design Blocks51.3312.804

23. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 23 Slide Slide nRejection Rule Randomized Block Design For  =.05, F.05 = 4.46 (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

24. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 24 Slide Slide nConclusion Randomized Block Design 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. F = MSTR/MSE = 2.6/.68 = 3.82 n nTest Statistic

25. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 25 Slide Slide Factorial Experiments n nIn some experiments we want to draw conclusions about more than one variable or factor. n nFactorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required. n nFor example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations. n nThe term factorial is used because the experimental conditions include all possible combinations of the factors.

26. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 26 Slide Slide The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment. n nANOVA 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. Two-Factor Factorial Experiment We again partition the sum of squares total (SST) into its sources.

27. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 27 Slide Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquares 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)

28. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 28 Slide Slide nStep 3 Compute the sum of squares for factor B Two-Factor Factorial Experiment n nStep 1 Compute the total sum of squares n nStep 2 Compute the sum of squares for factor A

29. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 29 Slide Slide nStep 4 Compute the sum of squares for interaction Two-Factor Factorial Experiment SSE = SST – SSA – SSB - SSAB n nStep 5 Compute the sum of squares due to error

30. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 30 Slide Slide 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. n nExample: State of Ohio Wage Survey Two-Factor Factorial Experiment

31. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 31 Slide Slide n nExample: State of Ohio Wage Survey Two-Factor Factorial Experiment IndustryCincinnatiClevelandColumbus I12.1011.8012.90 I11.8011.2012.70 I12.1012.0012.20 II12.4012.6013.00 II12.5012.0012.10 II12.0012.5012.70

32. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 32 Slide Slide nFactors Two-Factor Factorial Experiment Each experimental condition is repeated 3 times Factor B: Location (3 levels) Factor A: Industry Type (2 levels) n nReplications

33. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 33 Slide Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Factor A Error Total 1 17.50 1.43 3.42 12.50.12 4.19 n nANOVA Table Factor B 1.12.562 Two-Factor Factorial Experiment Interaction.37.192 4.69 1.55

34. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 34 Slide Slide n nConclusions Using the p-Value Approach Two-Factor Factorial Experiment (p-values were found using Excel) Interaction is not significant. Interaction: p-value =.25 >  =.05 Mean wages differ by location. Locations: p-value =.03 <  =.05 Mean wages do not differ by industry type. Industries: p-value =.06 >  =.05

35. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 35 Slide Slide n nConclusions Using the Critical Value Approach Two-Factor Factorial Experiment Interaction is not significant. Interaction: F = 1.55 < F  = 3.89 Mean wages differ by location. Locations: F = 4.69 > F  = 3.89 Mean wages do not differ by industry type. Industries: F = 4.19 < F  = 4.75

36. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 36 Slide Slide End of Chapter 13, Part B