1 1 Slide © 2005 Thomson/South-Western AK/ECON 3480 M & N WINTER 2006 n Power Point Presentation n Professor Ying Kong School of Analytic Studies and Information Technology Atkinson Faculty of Liberal and Professional Studies York University

2 2 Slide © 2005 Thomson/South-Western Chapter 13, Part B Analysis of Variance and Experimental Design n Factorial Experiments n An Introduction to Experimental Design n Completely Randomized Designs n Randomized Block Design

3 3 Slide © 2005 Thomson/South-Western An Introduction to Experimental Design n Statistical studies can be classified as being either experimental or observational. n In 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 In an observational study, no attempt is made to control the factors. n Cause-and-effect relationships are easier to establish in experimental studies than in observational studies.

4 4 Slide © 2005 Thomson/South-Western An Introduction to Experimental Design n A factor is a variable that the experimenter has selected for investigation. n A treatment is a level of a factor. n Experimental units are the objects of interest in the experiment. n A completely randomized design 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.

5 5 Slide © 2005 Thomson/South-Western The between-samples estimate of  2 is referred The between-samples estimate of  2 is referred to as the mean square due to treatments (MSTR). n Between-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 6 Slide © 2005 Thomson/South-Western The second estimate of  2, the within-samples estimate, is referred to as the mean square due to error (MSE). The second estimate of  2, the within-samples estimate, is referred to as the mean square due to error (MSE). n Within-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 7 Slide © 2005 Thomson/South-Western n ANOVA 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 8 Slide © 2005 Thomson/South-Western AutoShine, Inc. is considering marketing a long- 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 Example: AutoShine, Inc. In order to test the durability 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 9 Slide © 2005 Thomson/South-Western Completely Randomized Design The number of times each car went through the 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 Example: AutoShine, Inc.

10 Slide © 2005 Thomson/South-Western Sample Mean Sample Variance Observation Wax Type 1 Wax Type 2 Wax Type Completely Randomized Design

11 Slide © 2005 Thomson/South-Western n Hypotheses Completely Randomized Design where:  1 = mean number of washes for Type 1 wax  2 = mean number of washes for Type 2 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 Slide © 2005 Thomson/South-Western Because the sample sizes are all equal: 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 Mean Square Error n Mean Square Between Treatments = ( )/3 = 29.8

13 Slide © 2005 Thomson/South-Western n Rejection 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 Slide © 2005 Thomson/South-Western n Test 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 Conclusion F = MSTR/MSE = 2.6/2.77 =.939 The p -value is greater than.10, where F = (Excel provides a p -value of.42.) (Excel provides a p -value of.42.) Therefore, we cannot reject H 0. Therefore, we cannot reject H 0.

15 Slide © 2005 Thomson/South-Western Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total Completely Randomized Design n ANOVA Table

16 Slide © 2005 Thomson/South-Western 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.

17 Slide © 2005 Thomson/South-Western 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 ANOVA Table BlocksSSBL b - 1 ( k – 1)( b – 1)

18 Slide © 2005 Thomson/South-Western 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.

19 Slide © 2005 Thomson/South-Western 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.

20 Slide © 2005 Thomson/South-Western Randomized Block Design TreatmentMeans Type of Gasoline (Treatment) BlockMeans Blend X Blend Y Blend Z Automobile(Block)

21 Slide © 2005 Thomson/South-Western n Mean Square Due to Error Randomized Block Design MSE = 5.47/[(3 - 1)(5 - 1)] =.68 SSE = = 5.47 MSBL = 51.33/(5 - 1) = 12.8 SSBL = 3[( ) ( ) 2 ] = MSTR = 5.2/(3 - 1) = 2.6 SSTR = 5[( ) 2 + ( ) 2 + ( ) 2 ] = 5.2 The overall sample mean is 29. Thus, n Mean Square Due to Treatments n Mean Square Due to Blocks

22 Slide © 2005 Thomson/South-Western Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total n ANOVA Table Randomized Block Design Blocks

23 Slide © 2005 Thomson/South-Western 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

24 Slide © 2005 Thomson/South-Western 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

25 Slide © 2005 Thomson/South-Western Factorial Experiments 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.

26 Slide © 2005 Thomson/South-Western 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.

27 Slide © 2005 Thomson/South-Western 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 Slide © 2005 Thomson/South-Western 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

29 Slide © 2005 Thomson/South-Western 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

30 Slide © 2005 Thomson/South-Western 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

31 Slide © 2005 Thomson/South-Western n Example: State of Ohio Wage Survey Two-Factor Factorial Experiment IndustryCincinnatiClevelandColumbusI I I II II II

32 Slide © 2005 Thomson/South-Western 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

33 Slide © 2005 Thomson/South-Western Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Factor A Error Total n ANOVA Table Factor B Two-Factor Factorial Experiment Interaction

34 Slide © 2005 Thomson/South-Western n Conclusions Using the p -Value Approach Two-Factor Factorial Experiment ( p -values were found using Excel) 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

35 Slide © 2005 Thomson/South-Western 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

36 Slide © 2005 Thomson/South-Western End of Chapter 13, Part B