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by: Kristin Koelewyn

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Kristin Koelewyn
UA
GPA 3.4

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Notes for chapter 13b
COURSE
PROF.
Dr. S. Umashankar
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
KARMA
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This 5 page Class Notes was uploaded by Kristin Koelewyn on Tuesday April 12, 2016. The Class Notes belongs to BNAD277 at University of Arizona taught by Dr. S. Umashankar in Spring 2016. Since its upload, it has received 13 views. For similar materials see Business Statistics in Business at University of Arizona.

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Date Created: 04/12/16
Bnad277: Chapter 13b Notes Experimental Design and Analysis of Variance - Randomized Block Design o Experimental units are the objects of interest in the experiment. o A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. o If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design. o 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. ▯ 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. ▯ Example: 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. ▯ Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide. ▯ Mean Square Due to Treatments: • The overall sample mean is 29. Thus, SSTR = 5[(29.8 2 2 2 - 29) + (28.8 - 29) + (28.4 - 29) ] = 5.2 • MSTR = 5.2/(3 - 1) = 2.6 ▯ Mean Square Due to Blocks: • SSBL = 3[(30.333 - 29) + . . . + (25.667 - 29) ] = 51.33 • MSBL = 51.33/(5 - 1) = 12.8 ▯ Mean Square Due to Error: • SSE = 62 - 5.2 - 51.33 = 5.47 • MSE = 5.47/[(3 - 1)(5 - 1)] = .68 ▯ ANOVA Table: • ▯ Rejection Rule: • p-Value Approach: Reject H if0p-value < .05 • Critical Value Approach: Reject H i0 F > 4.46 ▯ Test Statistic: • F = MSTR/MSE = 2.6/.68 = 3.82 ▯ Conclusion: • 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 • There is insufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends. - Factorial Experiment o In some experiments we want to draw conclusions about more than one variable or factor. o Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required. o The term factorial is used because the experimental conditions include all possible combinations of the factors. o For example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations. - Two-Factor Factorial Experiement: o 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. ▯ 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. o Step 1: Compute the total sum of squares: ▯ o Step 2: Compute the sum of squares for factor A: o Step 3: Compute the sum of squares for factor B: o Step 4: Compute the sum of squares for interaction: o Step 5: Compute the sum of squares due to error: - Example: 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. o Factors: ▯ Factor A: Industry Type (2 levels) ▯ Factor B: Location (3 levels) o Replications: ▯ Each experimental condition is repeated 3 times o ANOVA Table: o Conclusions Using the Critical Value Approach: ▯ Industries: • F = 4.19 < F a 4.75 ▯ Locations: • F = 4.69 > F = 3.89 a ▯ Interaction: • F = 1.55 < F a 3.89

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