Analysis of Variance ANOVA q A singlefactor ANOVA

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Analysis of Variance (ANOVA) q A single-factor ANOVA can be used to compare more

Analysis of Variance (ANOVA) q A single-factor ANOVA can be used to compare more than two means. For example, suppose a manufacturer of paper used for grocery bags is concerned about the tensile strength of the paper. Product engineers believe that tensile strength is a function of the hardwood concentration and want to test several concentrations for the effect on tensile strength. If there are 2 different hardwood concentrations (say, 5% and 15%), then a z-test or t-test is appropriate: H 0: μ 1 = μ 2 H 1: μ 1 ≠ μ 2 EGR 252 Spring 2009 - Ch. 13 Part 1 1

Comparing More Than Two Means q What if there are 3 different hardwood concentrations

Comparing More Than Two Means q What if there are 3 different hardwood concentrations (say, 5%, 10%, and 15%)? H 0: μ 1 = μ 2 H 0: μ 1 = μ 3 H 0: μ 2 = μ 3 and H 1: μ 1 ≠ μ 2 H 1: μ 1 ≠ μ 3 H 1: μ 2 ≠ μ 3 q How about 4 different concentrations (say, 5%, 10%, 15%, and 20%)? All of the above, PLUS H 0: μ 1 = μ 4 H 0: μ 2 = μ 4 H 0: μ 3 = μ 4 and H 1: μ 1 ≠ μ 4 H 1: μ 2 ≠ μ 4 H 1: μ 3 ≠ μ 4 q What about 5 concentrations? 10? EGR 252 Spring 2009 - Ch. 13 Part 1 2

Comparing Multiple Means - Type I Error q Suppose α = 0. 05 P(Type

Comparing Multiple Means - Type I Error q Suppose α = 0. 05 P(Type 1 error) = 0. 05 (1 – α) = P (accept H 0 | H 0 is true) = 0. 95 q Conducting multiple t-tests increases the probability of a Type 1 error q The greater the number of t-tests, the greater the error probability § 4 concentrations: (0. 95)4 = 0. 814 § 5 concentrations: (0. 95)5 = 0. 774 § 10 concentrations: (0. 95)10 = 0. 599 q Making the comparisons simultaneously (as in an ANOVA) reduces the error back to 0. 05 EGR 252 Spring 2009 - Ch. 13 Part 1 3

Analysis of Variance (ANOVA) Terms q Independent variable: that which is varied Ø Treatment

Analysis of Variance (ANOVA) Terms q Independent variable: that which is varied Ø Treatment Ø Factor q Level: the selected categories of the factor Ø In a single–factor experiment there a levels q Dependent variable: the measured result Ø Observations Ø Replicates Ø (N observations in the total experiment) q Randomization: performing experimental runs in random order so that other factors don’t influence results. EGR 252 Spring 2009 - Ch. 13 Part 1 4

The Experimental Design q Suppose a manufacturer is concerned about the tensile strength of

The Experimental Design q Suppose a manufacturer is concerned about the tensile strength of the paper used to produce grocery bags. Product engineers believe that tensile strength is a function of the hardwood concentration and want to test several concentrations for the effect on tensile strength. Six specimens were made at each of the 4 hardwood concentrations (5%, 10%, 15%, and 20%). The 24 specimens were tested in random order on a tensile test machine. q Terms Ø Ø Factor: Hardwood Concentration Levels: 5%, 10%, 15%, 20% a=4 N = 24 EGR 252 Spring 2009 - Ch. 13 Part 1 5

The Results and Partial Analysis q The experimental results consist of 6 observations at

The Results and Partial Analysis q The experimental results consist of 6 observations at each of 4 levels for a total of N = 24 items. To begin the analysis, we calculate the average and total for each level. Hardwood Observations Concentration 1 2 3 4 5% 7 8 15 11 9 10 60 10. 00 10% 12 17 13 18 19 15 94 15. 67 15% 14 18 19 17 16 18 102 17. 00 20% 19 25 22 23 18 20 127 21. 17 383 15. 96 EGR 252 Spring 2009 - Ch. 13 Part 1 5 6 Totals Averages 6

To determine if there is a difference in the response at the 4 levels

To determine if there is a difference in the response at the 4 levels … 1. 2. 3. 4. 5. 6. Calculate sums of squares Calculate degrees of freedom Calculate mean squares Calculate the F statistic Organize the results in the ANOVA table Conduct the hypothesis test EGR 252 Spring 2009 - Ch. 13 Part 1 7

Calculate the sums of squares EGR 252 Spring 2009 - Ch. 13 Part 1

Calculate the sums of squares EGR 252 Spring 2009 - Ch. 13 Part 1 8

Additional Calculations Calculate Degrees of Freedom dftreat = a – 1 = 3 df

Additional Calculations Calculate Degrees of Freedom dftreat = a – 1 = 3 df error = a(n – 1) = 20 dftotal = an – 1 = 23 Mean Square, MS = SS/df MStreat = 382. 7917/3 = 127. 5972 MSE = 130. 1667 /20 = 6. 508333 Calculate F = MStreat / MSError = 127. 58 / 6. 51 = 19. 61 EGR 252 Spring 2009 - Ch. 13 Part 1 9

Organizing the Results Build the ANOVA table ANOVA Source of Variation SS df Treatment

Organizing the Results Build the ANOVA table ANOVA Source of Variation SS df Treatment 382. 79 3 Error 130. 17 20 Total 512. 96 23 MS F 127. 6 19. 6 P-value F crit 3. 6 E-06 3. 1 6. 5083 Determine significance Ø fixed α-level compare to Fα, a-1, a(n-1) Ø p – value find p associated with this F with degrees of freedom a-1, a(n-1) EGR 252 Spring 2009 - Ch. 13 Part 1 10

Conduct the Hypothesis Test Null Hypothesis: The mean tensile strength is the same for

Conduct the Hypothesis Test Null Hypothesis: The mean tensile strength is the same for each hardwood concentration. Alternate Hypothesis: The mean tensile strength differs for at least one hardwood concentration Compare Fcrit to Fcalc Draw the graphic State your decision with respect to the null hypothesis State your conclusion based on the problem statement EGR 252 Spring 2009 - Ch. 13 Part 1 11

Hypothesis Test Results Null Hypothesis: The mean tensile strength is the same for each

Hypothesis Test Results Null Hypothesis: The mean tensile strength is the same for each hardwood concentration. Alternate Hypothesis: The mean tensile strength differs for at least one hardwood concentration Fcrit less than Fcalc Draw the graphic Reject the null hypothesis Conclusion: The mean tensile strength differs for at least one hardwood concentration. EGR 252 Spring 2009 - Ch. 13 Part 1 12

Hypothesis Test Results Null Hypothesis: The mean tensile strength is the same for each

Hypothesis Test Results Null Hypothesis: The mean tensile strength is the same for each hardwood concentration. Alternate Hypothesis: The mean tensile strength differs for at least one hardwood concentration Fcrit less than Fcalc Draw the graphic Reject the null hypothesis Conclusion: The mean tensile strength differs for at least one hardwood concentration. EGR 252 Spring 2009 - Ch. 13 Part 1 13

Post-hoc Analysis: “Hand Calculations” 1. Calculate and check residuals, eij = Oi - Ei

Post-hoc Analysis: “Hand Calculations” 1. Calculate and check residuals, eij = Oi - Ei Ø plot residuals vs treatments Ø normal probability plot 2. Perform ANOVA and determine if there is a difference in the means 3. If the decision is to reject the null hypothesis, identify which means are different using Tukey’s procedure: 4. Model: yij = μ + αi + εij EGR 252 Spring 2009 - Ch. 13 Part 1 14

Graphical Methods - Computer Level 5% 10% 15% 20% Individual 95% CIs For Mean

Graphical Methods - Computer Level 5% 10% 15% 20% Individual 95% CIs For Mean Based on Pooled St. Dev N Mean St. Dev +---------+-----+----6 10. 000 2. 828 (----*----) 6 15. 667 2. 805 (----*-----) 6 17. 000 1. 789 (----*-----) 6 21. 167 2. 639 (-----*----) +---------+-----+----8. 0 12. 0 16. 0 EGR 252 Spring 2009 - Ch. 13 Part 1 20. 0 15

Numerical Methods - Computer Ø Tukey’s test Ø Duncan’s Multiple Range test Ø Easily

Numerical Methods - Computer Ø Tukey’s test Ø Duncan’s Multiple Range test Ø Easily performed in Minitab § Tukey 95% Simultaneous Confidence Intervals (partial results) 10% subtracted from: Lower Center Upper ----+---------+-----+----15% -2. 791 1. 333 5. 458 (-----*-----) 20% 1. 376 5. 500 9. 624 (-----*-----) ----+---------+-----7. 0 0. 0 7. 0 14. 0 EGR 252 Spring 2009 - Ch. 13 Part 1 16

Blocking q Creating a group of one or more people, machines, processes, etc. in

Blocking q Creating a group of one or more people, machines, processes, etc. in such a manner that the entities within the block are more similar to each other than to entities outside the block. q Balanced design: n = 1 for each treatment/block category q Model: yij = μ + α i + βj + εij EGR 252 Spring 2009 - Ch. 13 Part 1 17

Example: Robins Air Force Base uses CO 2 to strip paint from F-15’s. You

Example: Robins Air Force Base uses CO 2 to strip paint from F-15’s. You have been asked to design a test to determine the optimal pressure for spraying the CO 2. You realize that there are five machines that are being used in the paint stripping operation. Therefore, you have designed an experiment that uses the machines as blocking variables. You emphasized the importance of balanced design and a random order of testing. The test has been run with these results (values are minutes to strip one fighter): EGR 252 Spring 2009 - Ch. 13 Part 1 18

ANOVA: One-Way with Blocking 1. Construct the ANOVA table Where, EGR 252 Spring 2009

ANOVA: One-Way with Blocking 1. Construct the ANOVA table Where, EGR 252 Spring 2009 - Ch. 13 Part 1 19

Blocking Example Your turn: fill in the blanks in the following ANOVA table (from

Blocking Example Your turn: fill in the blanks in the following ANOVA table (from Excel): ANOVA Source of Variation SS df MS P-value F crit 44. 867 8. 492 0. 0105 4. 458968 0. 0553 _______ Rows 89. 733 2 Columns 77. 733 _____ Error 42. 267 8 5. 2833 Total 209. 73 ___ F ____ 2. Make decision and draw conclusions: EGR 252 Spring 2009 - Ch. 13 Part 1 20

Two-Way ANOVA q Blocking is used to keep extraneous factors from masking the effects

Two-Way ANOVA q Blocking is used to keep extraneous factors from masking the effects of the one treatment you are interested in studying. q A two-way ANOVA is used when you are interested in determining the effect of two treatments. q Model: yijk = μ + α i + βj + (α β)ijk + εij Ø α is the main effect of Treatment A Ø β is the main effect of Treatment B Ø The α β component is the interaction effect EGR 252 Spring 2009 - Ch. 13 Part 1 21

Two-Way ANOVA w/ Replication q Your fame as an experimental design expert grows. You

Two-Way ANOVA w/ Replication q Your fame as an experimental design expert grows. You have been called in as a consultant to help the Pratt and Whitney plant in Columbus determine the best method of applying the reflective stripe that is used to guide the Automated Guided Vehicles (AGVs) along their path. There are two ways of applying the stripe (paint and coated adhesive tape) and three types of flooring (linoleum and two types of concrete) in the facilities using the AGVs. You have set up two identical “test tracks” on each type of flooring and applied the stripe using the two methods under study. You run 3 replications in random order and count the number of tracking errors per 1000 ft of track. The results are as follows: EGR 252 Spring 2009 - Ch. 13 Part 1 22

Two-Way ANOVA Example q Analysis is the similar to the one-way ANOVA; however we

Two-Way ANOVA Example q Analysis is the similar to the one-way ANOVA; however we are now concerned with interaction effects q The two-way ANOVA table displays three calculated F values EGR 252 Spring 2009 - Ch. 13 Part 1 23

Two-Way ANOVA EGR 252 Spring 2009 - Ch. 13 Part 1 24

Two-Way ANOVA EGR 252 Spring 2009 - Ch. 13 Part 1 24

Your Turn q Fill in the blanks … ANOVA Source of Variation SS df

Your Turn q Fill in the blanks … ANOVA Source of Variation SS df MS F ____ 2. 39 _ 76 Sample 0. 4356 1 Columns 4. 48 2 Interaction 0. 9644 ___ 0. 4822 Within 2. 18 ___ 0. 1817 Total 8. 06 17 2. 24 12. 33 _____ P-value F crit 0. 14748 4. 74722 0. 00123 3. 88529 0. 11104 3. 88529 q What does this mean? EGR 252 Spring 2009 - Ch. 13 Part 1 25

What if Interaction Effects are Significant? q For example, suppose a new test was

What if Interaction Effects are Significant? q For example, suppose a new test was run using different types of paint and adhesive, with the following results: ANOVA Source of Variation SS Sample 0. 109 Columns 1. 96 Interaction 2. 831 df MS P-value F crit 1 0. 1089 1. 071 0. 3211 4. 7472 2 0. 98 9. 639 0. 0032 3. 8853 2 1. 4156 13. 92 0. 0007 3. 8853 Within 1. 22 12 0. 1017 Total 6. 12 17 EGR 252 Spring 2009 - Ch. 13 Part 1 F 26

Understanding Interaction Effects q Graphical methods: Ø graph means vs factors Ø identify where

Understanding Interaction Effects q Graphical methods: Ø graph means vs factors Ø identify where the effect will change the result for one factor based on the value of the other. EGR 252 Spring 2009 - Ch. 13 Part 1 27