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Ch 14 實習 (2). 2 Randomized Blocks (Two-way) Analysis of Variance The purpose of designing a randomized block experiment is to reduce the within- treatments.

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Presentation on theme: "Ch 14 實習 (2). 2 Randomized Blocks (Two-way) Analysis of Variance The purpose of designing a randomized block experiment is to reduce the within- treatments."— Presentation transcript:

1 Ch 14 實習 (2)

2 2 Randomized 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.

3 3 Block all the observations with some commonality across treatments Randomized Blocks

4 4 The sum of square total is partitioned into three sources of variation Treatments Blocks Within samples (Error) SS(Total) = SST + SSB + SSE Sum of square for treatmentsSum of square for blocksSum of square for error Recall. For the independent samples design we have: SS(Total) = SST + SSE Partitioning the total variability

5 5 Calculating the sums of squares Formula for the calculation of the sums of squares SST = SSB=

6 6 Calculating the sums of squares Formula for the calculation of the sums of squares SST = SSB=

7 7 To perform hypothesis tests for treatments and blocks we need Mean square for treatments Mean square for blocks Mean square for error Test statistics for treatments Test statistics for blocks Mean Squares and Test statistics

8 8 Testing the mean responses for treatments F > F ,k-1,n-k-b+1 Testing the mean response for blocks F> F ,b-1,n-k-b+1 The F test Rejection Regions

9 9 [ 補充 ] Randomized Blocks ANOVA

10 10 ANOVA Table

11 11 Example 1 A randomized block experiment produced the following statistics: k=5 b=12 SST=1500 SSB=1000 SS(Total)=3500 a. Test to determine whether the treatment means differ. (Use =0.01) b. Test to determine whether the blocks means differ. (Use =0.01)

12 12 Solution 1 ANOVA table a. Rejection region: F>F α,k-1,n-k-b+1 =F 0.01,4,44 ≒ 3.77 F=16.50 >3.77 There is enough evidence to conclude that the treatment means differ b. Rejection region: F>F α,b-1,n-k-b+1 =F 0.01,11,44 ≒ 2.67 Conclusion: F=4.00>2.67 There is enough evidence to conclude that the block means differ

13 13 Example 2 As 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 Student Female Student

14 14 Solution 2 H 0 : μ 1 = μ 2 = μ 3 H 1 : At least two means differ Rejection region: F>F α,k-1,n-k-b+1 =F 0.05,2,6 =5.14 K=3, b=4, grand mean=2.38 =7.3

15 Solution 2 15

16 16 Two-Factor Analysis of Variance Example 2 Suppose in Example 1, two factors are to be examined: The effects of the marketing strategy on sales.  Emphasis on convenience  Emphasis on quality  Emphasis on price The effects of the selected media on sales.  Advertise on TV  Advertise in newspapers

17 17 Two-way ANOVA (two factors) City 1 sales City3 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers ConvenienceQualityPrice Factor A: Marketing strategy Factor B: Advertising media

18 18 Levels of factor A 123 Level 1 of factor B Level 2 of factor B Level 1and 2 of factor B Difference between the levels of factor A No difference between the levels of factor B Difference between the levels of factor A, and difference between the levels of factor B; no interaction Levels of factor A No difference between the levels of factor A. Difference between the levels of factor B Interaction M R e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e Interaction

19 19 Interaction Without interaction With interaction B-B- B-B- B+B+ B+B+ B+B+ B+B+ B-B- B-B-

20 20 Terminology A 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.

21 21 Hypothesis H 0 : Factor A and Factor B do not interact to affect the mean responses H 1 : Factor A and Factor B do interact to affect the mean responses H 0 : The means of the a levels of factor A are equal H 1 : At least two means differ H 0 : The means of the b levels of factor B are equal H 1 : At least two means differ

22 22 Sums of squares

23 23 Sums of squares

24 補充 24

25 25 F tests for the Two-way ANOVA Test for the difference between the levels of the main factors A and B F= MS(A) MSE F= MS(B) MSE Rejection region: F > F ,a-1,n-ab F > F , b-1, n-ab Test for interaction between factors A and B F= MS(AB) MSE Rejection region: F > F  a-1)(b-1),n-ab SS(A)/(a-1) SS(B)/(b-1) SS(AB)/(a-1)(b-1) SSE/(n-ab)

26 26 ANOVA Table n = abr

27 27 Example 3 The following data were generated from a 2 X 2 factorial experiment with 3 replicates.

28 28 Example 3 - continued a. 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.

29 29 Solution 3

30 30 Solution 3 - continued a 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.

31 Solution 3 - continued 31

32 Example 4 The 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 32

33 33 Two 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 variance The conditions required to perform the ANOVA are satisfied. Multiple Comparisons

34 34 Inference about    –   : Equal variances Recall Construct the t-statistic as follows: Build a confidence interval

35 35 Fisher 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 s p 2. We can conclude that  i and  j differ (at  % significance level if > LSD, where

36 36 Experimentwise Type I error rate (  E ) (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  It is  calculated by  E = 1-(1 –  ) C where 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 (  ), to secure a pre-determined overall  E 

37 37 The procedure: Compute the number of pairwise comparisons (C) [C=k(k-1)/2], where k is the number of populations. Set  =  E /C, where  E is the true probability of making at least one Type I error (called experimentwise Type I error). We can conclude that  i and  j differ (at  /C% significance level if Bonferroni Adjustment

38 38 Example 1 - continued Rank the effectiveness of the marketing strategies (based on mean weekly sales). Use the Fisher ’ s method, and the Bonferroni adjustment method Solution (the Fisher ’ s method) The sample mean sales were , 653.0, Then, Fisher and Bonferroni Methods

39 39 Solution (the Bonferroni adjustment) We calculate C=k(k-1)/2 to be 3(2)/2 = 3. We set  =.05/3 =.0167, thus t.0167  2, 60-3 = (Excel). Again, the significant difference is between  1 and  2. Fisher and Bonferroni Methods

40 40 The test procedure: Find a critical number  as follows: k = the number of treatments =degrees of freedom = n - k n g = number of observations per sample (recall, all the sample sizes are the same)  = significance level q  (k, ) = a critical value obtained from the studentized range table Tukey Multiple Comparisons

41 41 If the sample sizes are not extremely different, we can use the above procedure with n g calculated as the harmonic mean of the sample sizes. Repeat this procedure for each pair of samples. Rank the means if possible. Select a pair of means. Calculate the difference between the larger and the smaller mean. If there is sufficient evidence to conclude that  max >  min. Tukey Multiple Comparisons

42 42 Which 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.

43 43 Example 5 a. 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

44 44 Solution 5

45 45 Solution 5 - continued

46 46 Solution 5 - continued

47 Example 6 Which 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. 47


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