© 2011 Pearson Education, Inc

Statistics for Business and Economics
Chapter 8 Design of Experiments and Analysis of Variance © 2011 Pearson Education, Inc

Contents 8.1 Elements of a Designed Experiment 8.2 The Completely Randomized Design: Single Factor 8.3 Multiple Comparisons of Means 8.4 The Randomized Block Design 8.5 Factorial Experiments: Two Factors As a result of this class, you will be able to ... © 2011 Pearson Education, Inc

Learning Objectives Discuss critical elements in the design of a sampling experiment Learn how to set up three experimental designs for comparing more than two population means: completely randomized, randomized block, and factorial design Show how to analyze data collected from a designed experiment using a technique called an analysis of variance (ANOVA) As a result of this class, you will be able to ... © 2011 Pearson Education, Inc

Elements of a Designed Experiment
8.1 Elements of a Designed Experiment :1, 1, 3 © 2011 Pearson Education, Inc

Factors Factors are those variables whose effect on the response is of interest to the experimenter. Quantitative factors are measured on a numerical scale, whereas qualitative factors are those that are not (naturally) measured on a numerical scale. Factors are also referred to as independent variables. © 2011 Pearson Education, Inc

Designed and Observational Experiment
A designed experiment is one for which the analyst controls the specification of the treatments and the method of assigning the experimental units to each treatment. An observational experiment is one for which the analyst simply observes the treatments and the response on a sample of experimental units. © 2011 Pearson Education, Inc

Experiment Investigator controls one or more independent variables Called treatment variables or factors Contain two or more levels (subcategories) Observes effect on dependent variable Response to levels of independent variable Experimental design: plan used to test hypotheses © 2011 Pearson Education, Inc

Examples of Experiments
Thirty stores are randomly assigned 1 of 4 (levels) store displays (independent variable) to see the effect on sales (dependent variable). Two hundred consumers are randomly assigned 1 of 3 (levels) brands of juice (independent variable) to study reaction (dependent variable). © 2011 Pearson Education, Inc

The Completely Randomized Design: Single Factor
8.2 The Completely Randomized Design: Single Factor :1, 1, 3 © 2011 Pearson Education, Inc

Completely Randomized Design
A completely randomized design is a design for which independent random samples of experimental units are selected for each treatment. © 2011 Pearson Education, Inc

Completely Randomized Design
Experimental units (subjects) are assigned randomly to treatments Subjects are assumed homogeneous One factor or independent variable Two or more treatment levels or classifications Analyzed by one-way ANOVA © 2011 Pearson Education, Inc

Randomized Design Example
Factor (Training Method) Factor levels Level 1 Level 2 Level 3 (Treatments) Are the mean training times the same for 3 different methods? 9 subjects 3 methods (factor levels) Experimental    units 21 hrs. 17 hrs. 31 hrs. Dependent variable 27 hrs. 25 hrs. 28 hrs. (Response) 29 hrs. 20 hrs. 22 hrs. © 2011 Pearson Education, Inc 78

ANOVA F-Test Tests the equality of two or more (k) population means Variables One nominal scaled independent variable Two or more (k) treatment levels or classifications One interval or ratio scaled dependent variable Used to analyze completely randomized experimental designs Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

ANOVA F-Test to Compare k Treatment Means: Completely Randomized Design H0: µ1 = µ2 = … = µk Ha: At least two treatment means differ Rejection region: F > F, where F is based on (k – 1) numerator degrees of freedom (associated with MST) and (n – k) denominator degrees of freedom (associated with MSE). Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

Conditions Required for a Valid ANOVA F-test: Completely Randomized Design 1. The samples are randomly selected in an independent manner from the k treatment populations. (This can be accomplished by randomly assigning the experimental units to the treatments.) 2. All k sampled populations have distributions that are approximately normal. 3. The k population variances are equal (i.e., © 2011 Pearson Education, Inc

ANOVA F-Test Hypotheses
H0: 1 = 2 = 3 = ... = k All population means are equal No treatment effect Ha: Not All i Are Equal At least 2 pop. means are different Treatment effect 1  2  ...  k is Wrong x f(x)  1 = 2 3 1 2 3 x f(x)  = © 2011 Pearson Education, Inc

Variances AMONG differ Variances WITHIN differ
Why Variances? Same treatment variation Different random variation Pop 1 Pop 2 Pop 3 Pop 4 Pop 6 Pop 5 Variances AMONG differ B Different treatment variation Same random variation Pop 1 Pop 2 Pop 3 Pop 4 Pop 6 Pop 5 Variances WITHIN differ A In all cases, the population means are DIFFERENT! Should reject Ho. Panel A: Same treatment effect - means are equal. Different random variation - standard deviations are different. As variances WITHIN get bigger, we are more likely to conclude equal means. In Populations 4, 5, & 6, it is possible to draw a sample and falsely conclude population means are equal. Panel B: Different treatment effect - means are different. Same random variation - standard deviations are equal. As variances AMONG get bigger, we are more likely to conclude population means are different. In Populations 4,5, & 6, it is possible to draw a sample and falsely conclude population means are equal. Possible to conclude means are equal! © 2011 Pearson Education, Inc 82

ANOVA Basic Idea Compares two types of variation to test equality of means Comparison basis is ratio of variances If treatment variation is significantly greater than random variation then means are not equal Variation measures are obtained by ‘partitioning’ total variation © 2011 Pearson Education, Inc

What Do You Do When the Assumptions Are Not Satisfied for an ANOVA for a Completely Randomized Design? Answer: Use a nonparametric statistical method such as the Kruskal-Wallis H-test. © 2011 Pearson Education, Inc

Steps for Conducting an ANOVA for a Completely Randomized Design
1. Be sure the design is truly completely randomized, with independent random samples for each treatment. 2. Check the assumptions of normality and equal variances. © 2011 Pearson Education, Inc

3. Create an ANOVA summary table that specifies the variability attributable to treatments and error, making sure that it leads to the calculation of the F-statistic for testing the null hypothesis that the treatment means are equal in the population. Use a statistical software program to obtain the numerical results. If no such package is available, use the calculation formulas in Appendix C. © 2011 Pearson Education, Inc

4. If the F-test leads to the conclusion that the means differ, a. Conduct a multiple comparisons procedure for as many of the pairs of means as you wish to compare. Use the results to summarize the statistically significant differences among the treatment means. b. If desired, form confidence intervals for one or more individual treatment means. © 2011 Pearson Education, Inc

5. If the F-test leads to the nonrejection of the null hypothesis that the treatment means are equal, consider the following possibilities: a. The treatment means are equal–that is, the null hypothesis is true. © 2011 Pearson Education, Inc

b. The treatment means really differ, but other important factors affecting the response are not accounted for by the completely randomized design. These factors inflate the sampling variability, as measured by MSE, resulting in smaller values of the F-statistic. Either increase the sample size for each treatment or use a different experimental design that accounts for the other factors affecting the response. © 2011 Pearson Education, Inc

ANOVA Partitions Total Variation
Variation due to treatment Total variation Variation due to random sampling Variation due to Random Sampling are due to Individual Differences Within Groups. Sum of Squares Among Sum of Squares Between Sum of Squares Treatment Among Groups Variation Sum of Squares Within Sum of Squares Error Within Groups Variation © 2011 Pearson Education, Inc 85

Random (Error) Variation
Response, x x3 x2 x1 Group 1 Group 2 Group 3 © 2011 Pearson Education, Inc

ANOVA F-Test Test Statistic
F = MST / MSE MST is Mean Square for Treatment MSE is Mean Square for Error Degrees of Freedom 1 = k – 1 numerator degrees of freedom 2 = n – k denominator degrees of freedom k = Number of groups n = Total sample size © 2011 Pearson Education, Inc

Mean Square (Variance)
ANOVA Summary Table Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment k – 1 SST MST = SST/(k – 1) MST MSE n = sum of sample sizes of all populations. k = number of factor levels All values are positive. Why? (squared terms) Degrees of Freedom & Sum of Squares are additive; Mean Square is NOT. Error n – k SSE MSE = SSE/(n – k) Total n – 1 SS(Total) = SST+SSE © 2011 Pearson Education, Inc

ANOVA F-Test Critical Value
If means are equal, F = MST / MSE  1. Only reject large F! Reject H  Do Not Reject H F F(α; k – 1, n – k) Always One-Tail! © T/Maker Co. © 2011 Pearson Education, Inc

ANOVA F-Test Example As production manager, you want to see if three filling machines have different mean filling times. You assign 15 similarly trained and experienced workers, 5 per machine, to the machines. At the .05 level of significance, is there a difference in mean filling times? Mach1 Mach2 Mach © 2011 Pearson Education, Inc

Summary Table Solution
From Computer Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment (Machines) 3 – 1 = 2 25.60 Error 15 – 3 = 12 .9211 Total 15 – 1 = 14 © 2011 Pearson Education, Inc

ANOVA F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 Not all equal Test Statistic: Decision: Conclusion: .05 F MST MSE  23 5820 9211 25.6 . F 3.89  = .05 Reject at  = .05 There is evidence population means are different © 2011 Pearson Education, Inc

ANOVA F-Test Thinking Challenge
You’re a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods ( =.05)? M1 M2 M3 M Use the following table. You assign randomly 3 people to each method, making sure that they are similar in intelligence etc. © T/Maker Co. © 2011 Pearson Education, Inc

Summary Table (Partially Completed)
Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment 348 (Methods) Error 80 Total © 2011 Pearson Education, Inc

ANOVA F-Test Solution H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 = 4 Not all equal Test Statistic: Decision: Conclusion: F MST MSE  116 10 11 6 . .05 F 4.07  = .05 Reject at  = .05 There is evidence population means are different © 2011 Pearson Education, Inc

Multiple Comparisons of Means
8.3 Multiple Comparisons of Means :1, 1, 3 © 2011 Pearson Education, Inc

Determining the Number of Pairwise Comparisons of Treatment Means
In general, if there are k treatment means, there are pairs of means that can be compared. © 2011 Pearson Education, Inc

Tukey Methods Tukey methods. For each of these procedures, the risk of making a Type I error applies to the comparisons of the treatment means in the experiment; thus, the value of  selected is called an experimentwise error rate (in contrast to a comparisonwise error rate). © 2011 Pearson Education, Inc

Bonferroni Method The Bonferroni method (see Miller, 1981), like the Tukey procedure, can be applied when pairwise comparisons are of interest; however, Bonferroni’s method does not require equal sample sizes. © 2011 Pearson Education, Inc

Scheffé Method Scheffé (1953) developed a more general procedure for comparing all possible linear combinations of treatment means (called contrasts). Consequently, when making pairwise comparisons, the confidence intervals produced by Scheffé’s method will generally be wider than the Tukey or Bonferroni confidence intervals. © 2011 Pearson Education, Inc

Tukey Procedure Tells which population means are significantly different Example: μ1 = μ 2 ≠ μ 3 Post hoc procedure Done after rejection of equal means in ANOVA Output from many statistical computer programs x f(x) m 1 = 2 3 2 Groupings © 2011 Pearson Education, Inc 105

The Randomized Block Design
8.4 The Randomized Block Design :1, 1, 3 © 2011 Pearson Education, Inc

Randomized Block Design
The randomized block design consists of a two- step procedure: 1. Matched sets of experimental units, called blocks, are formed, each block consisting of k experimental units (where k is the number of treatments). The b blocks should consist of experimental units that are as similar as possible. 2. One experimental unit from each block is randomly assigned to each treatment, resulting in a total of n = bk responses. © 2011 Pearson Education, Inc

Randomized Block Design
Reduces sampling variability (MSE) Matched sets of experimental units (blocks) One experimental unit from each block is randomly assigned to each treatment © 2011 Pearson Education, Inc

Randomized Block Design Total Variation Partitioning
© 2011 Pearson Education, Inc

ANOVA F-Test to Compare k Treatment Means: Randomized Block Design
H0: µ1 = µ2 = … = µk Ha: At least two treatment means differ Rejection region: F > F, where F is based on (k – 1) numerator degrees of freedom and (n – b – k +1) denominator degrees of freedom. Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

Conditions Required for a Valid ANOVA F-test: Randomized Block Design
The b blocks are randomly selected, and all k treatments are applied (in random order) to each block. The distributions of observations corresponding to all bk block-treatment combinations are approximately normal. All bk block-treatment distributions have equal variances. © 2011 Pearson Education, Inc

Randomized Block Design F-Test Test Statistic
F = MST / MSE MST is Mean Square for Treatment MSE is Mean Square for Error Degrees of Freedom 1 = k – 1 (numerator) 2 = n – k – b + 1 (denominator) k = Number of groups n = Total sample size b = Number of blocks © 2011 Pearson Education, Inc

Randomized Block Design Example
A production manager wants to see if three assembly methods have different mean assembly times (in minutes). Five employees were selected at random and assigned to use each assembly method. At the .05 level of significance, is there a difference in mean assembly times? Employee Method 1 Method 2 Method 3 © 2011 Pearson Education, Inc

Random Block Design F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 Not all equal .05 F 4.46  = .05 © 2011 Pearson Education, Inc

Degrees of Freedom Mean Square (Variance) Source of Variation Sum of Squares F Treatment (Methods) 3 – 1 = 2 5.43 2.71 12.9 Block (Employee) 5 – 1 = 4 10.69 2.67 12.7 Error 15 – 3 – = 8 1.68 .21 Total 15 – 1 = 14 17.8 © 2011 Pearson Education, Inc

Random Block Design F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): 1 = 2 = 3 Not all equal Test Statistic: Decision: Conclusion: F MST MSE  2.71 .21 12.9 .05 F 4.46  = .05 Reject at  = .05 There is evidence population means are different © 2011 Pearson Education, Inc

Steps for Conducting an ANOVA for a Randomized Block Design
1. Be sure the design consists of blocks (preferably, blocks of homogeneous experimental units) and that each treatment is randomly assigned to one experimental unit in each block. 2. If possible, check the assumptions of normality and equal variances for all block-treatment combinations. [Note: This may be difficult to do because the design will likely have only one observation for each block-treatment combination.] © 2011 Pearson Education, Inc

3. Create an ANOVA summary table that specifies the variability attributable to Treatments, Blocks, and Error, which leads to the calculation of the F-statistic to test the null hypothesis that the treatment means are equal in the population. Use a statistical software package or the calculation formulas in Appendix C to obtain the necessary numerical ingredients. © 2011 Pearson Education, Inc

4. If the F-test leads to the conclusion that the means differ, use the Bonferroni, Tukey, or similar procedure to conduct multiple comparisons of as many of the pairs of means as you wish. Use the results to summarize the statistically significant differences among the treatment means. Remember that, in general, the randomized block design cannot be used to form confidence intervals for individual treatment means. © 2011 Pearson Education, Inc

5. If the F-test leads to the nonrejection of the null hypothesis that the treatment means are equal, consider the following possibilities: a. The treatment means are equal–that is, the null hypothesis is true. © 2011 Pearson Education, Inc

b. The treatment means really differ, but other important factors affecting the response are not accounted for by the randomized block design. These factors inflate the sampling variability, as measured by MSE, resulting in smaller values of the F-statistic. Either increase the sample size for each treatment or conduct an experiment that accounts for the other factors affecting the response. Do not automatically reach the former conclusion because the possibility of a Type II error must be considered if you accept H0. © 2011 Pearson Education, Inc

6. If desired, conduct the F-test of the null hypothesis that the block means are equal. Rejection of this hypothesis lends statistical support to using the randomized block design. © 2011 Pearson Education, Inc

Note: It is often difficult to check whether the assumptions for a randomized block design are satisfied. There is usually only one observation for each block-treatment combination. When you feel these assumptions are likely to be violated, a nonparametric procedure is advisable. © 2011 Pearson Education, Inc

What Do You Do When the Assumptions Are Not Satisfied for an ANOVA for a Completely Randomized Design? Answer: Use a nonparametric statistical method such as the Friedman Fr test. © 2011 Pearson Education, Inc

Factorial Experiments: Two Factors
8.5 Factorial Experiments: Two Factors :1, 1, 3 © 2011 Pearson Education, Inc

Factorial Design A complete factorial experiment is one in which every factor-level combination is employed–that is, the number of treatments in the experiment equals the total number of factor-level combinations. Also referred to as a two-way classification. © 2011 Pearson Education, Inc

Factorial Design To determine the nature of the treatment effect, if any, on the response in a factorial experiment, we need to break the treatment variability into three components: Interaction between Factors A and B, Main Effect of Factor A, and Main Effect of Factor B. The Factor Interaction component is used to test whether the factors combine to affect the response, while the Factor Main Effect components are used to determine whether the factors separately affect the response. © 2011 Pearson Education, Inc

Factorial Design Experimental units (subjects) are assigned randomly to treatments Subjects are assumed homogeneous Two or more factors or independent variables Each has two or more treatments (levels) Analyzed by two-way ANOVA © 2011 Pearson Education, Inc

Procedure for Analysis of Two-Factor Factorial Experiment
1. Partition the Total Sum of Squares into the Treatments and Error components. Use either a statistical software package or the calculation formulas in Appendix C to accomplish the partitioning. © 2011 Pearson Education, Inc

Use the F-ratio of Mean Square for Treatments to Mean Square for Error to test the null hypothesis that the treatment means are equal. a. If the test results in nonrejection of the null hypothesis, consider refining the experiment by increasing the number of replications or introducing other factors. Also consider the possibility that the response is unrelated to the two factors. b. If the test results in rejection of the null hypothesis, then proceed to step 3. © 2011 Pearson Education, Inc

3. Partition the Treatments Sum of Squares into the Main Effect and Interaction Sum of Squares. Use either a statistical software package or the calculation formulas in Appendix C to accomplish the partitioning. © 2011 Pearson Education, Inc

4. Test the null hypothesis that factors A and B do not interact to affect the response by computing the F-ratio of the Mean Square for Interaction to the Mean Square for Error. a. If the test results in nonrejection of the null hypothesis, proceed to step 5. b. If the test results in rejection of the null hypothesis, conclude that the two factors interact to affect the mean response. Then proceed to step 6a. © 2011 Pearson Education, Inc

5. Conduct tests of two null hypotheses that the mean response is the same at each level of factor A and factor B. Compute two F-ratios by comparing the Mean Square for each Factor Main Effect to the Mean Square for Error. a. If one or both tests result in rejection of the null hypothesis, conclude that the factor affects the mean response. Proceed to step 6b. © 2011 Pearson Education, Inc

b. If both tests result in nonrejection, an apparent contradiction has occurred. Although the treatment means apparently differ (step 2 test), the interaction (step 4) and main effect (step 5) tests have not supported that result. Further experimentation is advised. © 2011 Pearson Education, Inc

6. Compare the means: a. If the test for interaction (step 4) is significant, use a multiple comparisons procedure to compare any or all pairs of the treatment means. b. If the test for one or both main effects (step 5) is significant, use a multiple comparisons procedure to compare the pairs of means corresponding to the levels of the significant factor(s). © 2011 Pearson Education, Inc

ANOVA Tests Conducted for Factorial Experiments: Completely Randomized Design, r Replicates per Treatment Test for Treatment Means H0: No difference among the ab treatment means Ha: At least two treatment means differ Rejection region: F > F, based on (ab – 1) numerator and (n – ab) denominator degrees of freedom [Note: n = abr.] Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

ANOVA Tests Conducted for Factorial Experiments: Completely Randomized Design, r Replicates per Treatment Test for Factor Interaction H0: Factors A and B do not interact to affect the response mean Ha: Factors A and B do interact to affect the response mean Rejection region: F > F, based on (a – 1)(b – 1) numerator and (n – ab) denominator degrees of freedom Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

ANOVA Tests Conducted for Factorial Experiments: Completely Randomized Design, r Replicates per Treatment Test for Main Effect of Factor A H0: No difference among the a mean levels of factor A Ha: At least two factor A mean levels differ Rejection region: F > F, based on (a – 1) numerator and (n – ab) denominator degrees of freedom Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

ANOVA Tests Conducted for Factorial Experiments: Completely Randomized Design, r Replicates per Treatment Test for Main Effect of Factor B H0: No difference among the b mean levels of factor B Ha: At least two factor B mean levels differ Rejection region: F > F, based on (b – 1) numerator and (n – ab) denominator degrees of freedom Note: There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable. Ask, what are nominal & interval scales? © 2011 Pearson Education, Inc

Conditions Required for Valid F-tests in Factorial Experiments
1. The response distribution for each factor-level combination (treatment) is normal. 2. The response variance is constant for all treatments. 3. Random and independent samples of experimental units are associated with each treatment. © 2011 Pearson Education, Inc

ANOVA Data Table Factor Factor B A 1 2 ... b Observation k 1 x111 x121 ... x1b1 xijk x112 x122 ... x1b2 2 x211 x221 ... x2b1 x212 x222 ... x2b2 Level i Factor A Level j Factor B : : : : : Treatment a xa11 xa21 ... xab1 xa12 xa22 ... xab2 © 2011 Pearson Education, Inc

Factorial Design Example
Factor 2 (Training Method) Factor Levels Level 1 Level 2 Level 3    Level 1 15 hr. 10 hr. 22 hr. (High)    Factor 1 (Motivation) 11 hr. 12 hr. 17 hr.   Treatment  Level 2 27 hr. 15 hr. 31 hr. (Low)    29 hr. 17 hr. 49 hr. © 2011 Pearson Education, Inc

Advantages of Factorial Designs
Saves time and effort e.g., Could use separate completely randomized designs for each variable Controls confounding effects by putting other variables into model Can explore interaction between variables © 2011 Pearson Education, Inc

Two-Way ANOVA Tests the equality of two or more population means when several independent variables are used Same results as separate one-way ANOVA on each variable No interaction can be tested Used to analyze factorial designs © 2011 Pearson Education, Inc

Interaction Occurs when effects of one factor vary according to levels of other factor When significant, interpretation of main effects (A and B) is complicated Can be detected In data table, pattern of cell means in one row differs from another row In graph of cell means, lines cross © 2011 Pearson Education, Inc

Graphs of Interaction Effects of motivation (high or low) and training method (A, B, C) on mean learning time Interaction No Interaction Average Average Response High Response High Low Low A B C A B C © 2011 Pearson Education, Inc

Two-Way ANOVA Summary Table
Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) a – 1 SS(A) MS(A) MS(A) MSE B (Column) b – 1 SS(B) MS(B) MS(B) MSE AB (Interaction) (a – 1)(b – 1) SS(AB) MS(AB) MS(AB) MSE Error n – ab SSE MSE Same as other designs Total n – 1 SS(Total) © 2011 Pearson Education, Inc

Factorial Design Example
Human Resources wants to determine if training time is different based on motivation level and training method. Conduct the appropriate ANOVA tests. Use α = .05 for each test. Training Method Factor Levels Self– paced Classroom Computer 15 hr. 10 hr. 22 hr. Motivation High 11 hr. 12 hr. 17 hr. 27 hr. 31 hr. Low 29 hr. 49 hr. © 2011 Pearson Education, Inc

Treatment Means F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): The 6 treatment means are equal At least 2 differ .05 F 4.39  = .05 © 2011 Pearson Education, Inc

Treatment Means F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): The 6 treatment means are equal At least 2 differ Test Statistic: Decision: Conclusion: .05 Reject at  = .05 F 4.39  = .05 There is evidence population means are different © 2011 Pearson Education, Inc

Factor Interaction F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): The factors do not interact The factors interact .05 F 5.14  = .05 © 2011 Pearson Education, Inc

Source of Variation Degrees of Freedom Sum of Squares Mean Square F A (Row) 1 546.75 546.75 17.40 B (Column) 2 531.5 265.75 8.46 AB (Interaction) 2 123.5 61.76 1.97 Error 6 188.5 31.42 Same as other designs Total 11 SS(Total) © 2011 Pearson Education, Inc

Factor Interaction F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): The factors do not interact The factors interact Test Statistic: Decision: Conclusion: .05 Do not reject at  = .05 F 5.14  = .05 There is no evidence the factors interact © 2011 Pearson Education, Inc

Main Factor A F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): No difference between motivation levels Motivation levels differ .05 F 5.99  = .05 © 2011 Pearson Education, Inc

Main Factor A F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): No difference between motivation levels Motivation levels differ Test Statistic: Decision: Conclusion: .05 Reject at  = .05 F 5.99  = .05 There is evidence that motivations levels differ © 2011 Pearson Education, Inc

Main Factor B F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): No difference between training methods Training methods differ .05 F 5.14  = .05 © 2011 Pearson Education, Inc

Main Factor B F-Test Solution
H0: Ha:  = 1 = 2 = Critical Value(s): No difference between training methods Training methods differ Test Statistic: Decision: Conclusion: .05 Reject at  = .05 F 5.14  = .05 There is evidence training methods differ © 2011 Pearson Education, Inc

Key Ideas Key Elements of a Designed Experiment 1. Response (dependent) variable – quantitative 2. Factors (independent variables) – quantitative or qualitative 3. Factor levels (values of factors) – selected by the experimenter 4. Treatments – combinations of factor levels 5. Experimental units – assign treatments to experimental units and measure response for each As a result of this class, you will be able to... © 2011 Pearson Education, Inc

Key Ideas Key Elements of a Designed Experiment Balanced Design Sample sizes for each treatment are equal. Test for main effects in a factorial design Only appropriate if the test for factor interaction is nonsignificant. Robust method Slight to moderate departures from normality do not have impact on validity of the ANOVA results. As a result of this class, you will be able to... © 2011 Pearson Education, Inc

Key Ideas Conditions Required for Valid F-Test in a Completely Randomized Design 1. All k treatment populations are approximately normal. 2. As a result of this class, you will be able to... © 2011 Pearson Education, Inc

Key Ideas Conditions Required for Valid F-Test in a Randomized Block Design 1. All treatment-block populations are approximately normal. 2. All treatment-block populations have the same variance. As a result of this class, you will be able to... © 2011 Pearson Education, Inc

Key Ideas Conditions Required for Valid F-Test in a Complete Factorial Design 1. All treatment populations are approximately normal. 2. All treatment populations have the same variance. Experimentwise error rate Risk of making at least one Type I error when making multiple comparisons of means in ANOVA As a result of this class, you will be able to... © 2011 Pearson Education, Inc

Key Ideas Multiple Comparison of Means Methods Bonferroni method: 1. Either balanced or unbalanced design 2. Pairwise comparison of means As a result of this class, you will be able to... © 2011 Pearson Education, Inc

Key Ideas Multiple Comparison of Means Methods Scheffé method: 1. Either balanced or unbalanced design 2. General contrast of means As a result of this class, you will be able to... © 2011 Pearson Education, Inc

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