2Randomized Blocks (Two-way) Analysis of Variance The purpose of designing a randomized block experiment is to reduce the within-treatments variation thus increasing the relative amount of between treatment variation.This helps in detecting differences between the treatment means more easily.
3Block all the observations with some commonality across treatments Randomized BlocksBlock all the observations with some commonality across treatments
4Partitioning the total variability The sum of square total is partitioned into three sources of variationTreatmentsBlocksWithin samples (Error)Recall. For the independent samples design we have:SS(Total) = SST + SSESS(Total) = SST + SSB + SSESum of square for treatmentsSum of square for blocksSum of square for error
5Calculating the sums of squares Formula for the calculation of the sums of squaresSSB=SST =
6Calculating the sums of squares Formula for the calculation of the sums of squaresSSB=SST =
7Mean Squares and Test statistics To perform hypothesis tests for treatments and blocks we needMean square for treatmentsMean square for blocksMean square for errorTest statistics for treatmentsTest statistics for blocks
8The F test Rejection Regions Testing the mean responses for treatmentsF > Fa,k-1,n-k-b+1Testing the mean response for blocksF> Fa,b-1,n-k-b+1
11Example 1A randomized block experiment produced the following statistics:k=5 b=12 SST= SSB=1000SS(Total)=3500a. Test to determine whether the treatment means differ. (Use =0.01)b. Test to determine whether the blocks means differ. (Use =0.01)
12Solution 1ANOVA tablea. Rejection region: F>F α,k-1,n-k-b+1=F0.01,4,44 ≒3.77F=16.50 >3.77There is enough evidence to conclude that the treatment means differb. Rejection region: F>F α,b-1,n-k-b+1=F0.01,11,44 ≒2.67Conclusion: F=4.00>2.67There is enough evidence to conclude that the block means differ
13Example 2As an experiment to understand measurement error, a statistics professor asks four students to measure the height of the professor, a male student, and a female student. The differences (in centimeters) between the correct dimension and the ones produced by the students are listed here. Can we infer at the 5% significance level that there are differences in the errors between the subjects being measured?studentProfessorMale StudentFemale Student126.96.36.199188.8.131.52184.108.40.206.62.9
14Solution 2 H0: μ1= μ2= μ3 H1: At least two means differ Rejection region: F>F α,k-1,n-k-b+1=F0.05,2,6 =5.14K=3, b=4, grand mean=2.38=7.3
16Two-Factor Analysis of Variance Example 2Suppose in Example 1, two factors are to be examined:The effects of the marketing strategy on sales.Emphasis on convenienceEmphasis on qualityEmphasis on priceThe effects of the selected media on sales.Advertise on TVAdvertise in newspapers
18Interaction Difference between the levels of factor A, and difference between the levels of factor B; nointeractionDifference between the levels of factor ANo difference between the levels of factor BM Re esa pn oneM Re esa pn oneLevel 1 of factor BLevel 1and 2 of factor BLevel 2 of factor BLevels of factor ALevels of factor A123123M Re esa pn oneM Re esa pn oneNo difference between the levels of factor A.Difference between the levels of factor BInteractionLevels of factor ALevels of factor A123123
19Interaction Without interaction With interaction B+ B+ B+ B- B- B- B-
20TerminologyA complete factorial experiment is an experiment in which the data for all possible combinations of the levels of the factors are gathered. This is also known as a two-way classification.The two factors are usually labeled A & B, with the number of levels of each factor denoted by a & b respectively.The number of observations for each combination is called a replicate, and is denoted by r. For our purposes, the number of replicates will be the same for each treatment, that is they are balanced.
21HypothesisH0: Factor A and Factor B do not interact to affect the mean responsesH1: Factor A and Factor B do interact to affect the mean responsesH0: The means of the a levels of factor A are equalH1: At least two means differH0: The means of the b levels of factor B are equal
25F tests for the Two-way ANOVA Test for the difference between the levels of the main factors A and BSS(A)/(a-1)SS(B)/(b-1)F=MS(A)MSEF=MS(B)MSESSE/(n-ab)Rejection region: F > Fa,a-1 ,n-ab F > Fa, b-1, n-abTest for interaction between factors A and BF=MS(AB)MSESS(AB)/(a-1)(b-1)Rejection region: F > Fa,(a-1)(b-1),n-ab
27Example 3The following data were generated from a 2 X 2 factorial experiment with 3 replicates.
28Example 3 - continueda. Test at the 5% significance level to determine whether factors A and B interact.b. Test at the 5% significance level to determine whether differences exists between the levels of factor A.c. Test at the 5% significance level to determine whether differences exist between the levels of factor B.
30Solution 3 - continueda F = .31, p-value = There is not enough evidence to conclude that factors A and B interact.b F = 1.23, p-value = There is not enough evidence to conclude that differences exist between the levels of factor A.c F = 13.00, p-value = There is enough evidence to conclude that differences exist between the levels of factor B.
32Example 4The required conditions for a two-factor ANOVA are that the distribution of the response is __________________ distributed; the variance for each treatment is ________; and the samples are _______ .(a) normally; equal; independent(b) normally; the same; independent(c) normally; identical; independent
33Multiple ComparisonsTwo means are considered different if the difference between the corresponding sample means is larger than a critical number. Then, the larger sample mean is believed to be associated with a larger population mean.Conditions common to all the methods here:The ANOVA model is the one way analysis of varianceThe conditions required to perform the ANOVA are satisfied.
34Inference about m1 – m2: Equal variances RecallConstruct the t-statistic as follows:Build a confidence interval
35Fisher Least Significant Different (LSD) Method This method builds on the equal variances t-test of the difference between two means.The test statistic is improved by using MSE rather than sp2.We can conclude that mi and mj differ (at a% significance level if > LSD, where
36Experimentwise Type I error rate (aE) (the effective Type I error) The Fisher’s method may result in an increased probability of committing a type I error.The experimentwise Type I error rate is the probability of committing at least one Type I error at significance level of a. It is calculated byaE = 1-(1 – a)Cwhere C is the number of pairwise comparisons(i.e. C = k(k-1)/2)The Bonferroni adjustment determines the required Type I error probability per pairwise comparison (a) , to secure a pre-determined overall aE.
37Bonferroni Adjustment The procedure:Compute the number of pairwise comparisons (C) [C=k(k-1)/2], where k is the number of populations.Set a = aE/C, where aE is the true probability of making at least one Type I error (called experimentwise Type I error).We can conclude that mi and mj differ (at a/C% significance level if
38Fisher and Bonferroni Methods Example 1 - continuedRank the effectiveness of the marketing strategies (based on mean weekly sales).Use the Fisher’s method, and the Bonferroni adjustment methodSolution (the Fisher’s method)The sample mean sales were , 653.0,Then,
39Fisher and Bonferroni Methods Solution (the Bonferroni adjustment)We calculate C=k(k-1)/2 to be 3(2)/2 = 3.We set a = .05/3 = .0167, thus t.0167/2, 60-3 = (Excel).Again, the significant difference is between m1 and m2.
40Tukey Multiple Comparisons The test procedure:Find a critical number w as follows:k = the number of treatmentsn =degrees of freedom = n - kng = number of observations per sample(recall, all the sample sizes are the same)a = significance levelqa(k,n) = a critical value obtained from the studentized range table
41Tukey Multiple Comparisons Select a pair of means. Calculate the difference between the larger and the smaller mean.If there is sufficient evidence to conclude that mmax > mmin .Repeat this procedure for each pair of samples. Rank the means if possible.If the sample sizes are not extremely different, we can use the above procedure with ng calculated as the harmonic mean ofthe sample sizes.
42Which Multiple Comparison Method to Use If you have identified two or three pairwise comparison, use the Bonferroni method.If you plan to compare all possible combinations, use Turkey.If the purpose of the analysis is to point to areas that should be investigated further, Fisher’s LSD method is indicated.
43Example 5a. Use Fisher’s LSD procedure with =0.05 to determine which population means differ given the following statistics.b. Repeat part a using the Bonferroni adjustment.c. Repeat part a using Tukey’s multiple comparison method
47Example 6Which of the following statements about multiple comparison methods is false?a. They are to be use once the F-test in ANOVA has been rejected.b. They are used to determine which particular population means differ.c. There are many different multiple comparison methods but all yield the same conclusions.d. All of these choices are true.