1. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Chapter Comparing Three or More Means 13

2. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section Comparing Three or More Means (One-Way Analysis of Variance) 13.1

3. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objectives 1.Verify the requirements to perform a one- way ANOVA 2.Test a hypothesis regarding three or more means using one-way ANOVA 13-3

4. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Analysis of Variance (ANOVA) is an inferential method used to test the equality of three or more population means. 13-4 CAUTION! Do not test H 0 : μ 1 = μ 2 = μ 3 by conducting three separate hypothesis tests, because the probability of making a Type I error will be much higher than α.

5. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 1.There must be k simple random samples; one from each of k populations or a randomized experiment with k treatments. 2.The k samples are independent of each other; that is, the subjects in one group cannot be related in any way to subjects in a second group. 3.The populations are normally distributed. 4.The populations must have the same variance; that is, each treatment group has the population variance σ 2. Requirements of a One-Way ANOVA Test 13-5

6. Copyright © 2013, 20 and 2007 Pearson Education, Inc. H 0 : μ 1 = μ 2 = μ 3 H 1 : At least one population mean is different from the others Testing a Hypothesis Regarding k = 3 13-6

7. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The methods of one-way ANOVA are robust, so small departures from the normality requirement will not significantly affect the results of the procedure. In addition, the requirement of equal population variances does not need to be strictly adhered to, especially if the sample size for each treatment group is the same. Therefore, it is worthwhile to design an experiment in which the samples from the populations are roughly equal in size. Testing Using ANOVE 13-7

8. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The one-way ANOVA procedures may be used if the largest sample standard deviation is no more than twice the smallest sample standard deviation. Verifying the Requirement of Equal Population Variance 13-8

9. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The following data represent the weight (in grams) of pennies minted at the Denver mint in 1990,1995, and 2000. Verify that the requirements in order to perform a one-way ANOVA are satisfied. Parallel Example 1: Verifying the Requirements of ANOVA 13-9

10. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 1990 19952000 2.50 2.522.50 2.50 2.542.48 2.49 2.502.49 2.53 2.482.50 2.46 2.522.48 2.50 2.502.52 2.47 2.492.51 2.53 2.532.49 2.51 2.482.51 2.49 2.552.50 2.48 2.492.52 13-10

11. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution 1.The 3 samples are simple random samples. 2.The samples were obtained independently. 3.Normal probability plots for the 3 years follow. All of the plots are roughly linear so the normality assumption is satisfied. 13-11

12. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 13-12

13. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 13-13

14. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 13-14

15. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 4.The sample standard deviations are computed for each sample using Minitab and shown on the following slide. The largest standard deviation is not more than twice the smallest standard deviation (20.0141 = 0.0282 > 0.02430) so the requirement of equal population variances is considered satisfied. Solution 13-15

16. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Descriptive Statistics Variable N Mean Median TrMean StDev SE Mean 1990 11 2.4964 2.5000 2.4967 0.0220 0.0066 1995 11 2.5091 2.5000 2.5078 0.0243 0.0073 2000 11 2.5000 2.5000 2.5000 0.0141 0.0043 Variable Minimum Maximum Q1 Q3 1990 2.4600 2.5300 2.4800 2.5130 1995 2.4800 2.5500 2.4900 2.5300 2000 2.4800 2.5200 2.4900 2.5100 13-16

17. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 2 Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA 13-17

18. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The basic idea in one-way ANOVA is to determine if the sample data could come from populations with the same mean, μ, or suggests that at least one sample comes from a population whose mean is different from the others. To make this decision, we compare the variability among the sample means to the variability within each sample. 13-18

19. Copyright © 2013, 20 and 2007 Pearson Education, Inc. We call the variability among the sample means the between-sample variability, and the variability of each sample the within-sample variability. If the between-sample variability is large relative to the within-sample variability, we have evidence to suggest that the samples come from populations with different means. 13-19

20. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The analysis of variance F-test statistic is given by ANOVA F-Test Statistic 13-20

21. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Step 1: Compute the sample mean of the combined data set by adding up all the observations and dividing by the number of observations. Call this value. Step 2: Find the sample mean for each sample (or treatment). Let represent the sample mean of sample 1, represent the sample mean of sample 2, and so on. Step 3: Find the sample variance for each sample (or treatment). Let represent the sample variance for sample 1, represent the sample variance for sample 2, and so on. Computing the F-Test Statistic 13-21

22. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Step 4: Compute the sum of squares due to treatments, SST, and the sum of squares due to error, SSE. Step 5: Divide each sum of squares by its corresponding degrees of freedom (k – 1 and n – k, respectively) to obtain the mean squares MST and MSE. Step 6: Compute the F-test statistic: Computing the F-Test Statistic 13-22

23. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Compute the F-test statistic for the penny data. Parallel Example 2: Computing the F-Test Statistic 13-23

24. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 1: Compute the overall mean. Step 2: Find the sample variance for each treatment (year). 13-24

25. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 3: Find the sample variance for each treatment (year). 13-25

26. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 4: Compute the sum of squares due to treatment, SST, and the sum of squares due to error, SSE. SST =11(2.4964–2.5018) 2 + 11(2.5091– 2.5018) 2 + 11(2.5–2.5018) 2 = 0.0009 SSE =(11–1)(0.0005) + (11–1)(0.0006) + (11–1)(0.0002) = 0.013 13-26

27. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 5: Compute the mean square due to treatment, MST, and the mean square due to error, MSE. 13-27

28. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 6: Compute the F-statistic. 13-28

29. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Source of Variation Sum of Squares Degrees of Freedom Mean Squares F-Test Statistic Treatment0.000920.00051.25 Error0.013300.0004 Total0.013932 ANOVA Table: 13-29

30. Copyright © 2013, 20 and 2007 Pearson Education, Inc. If the P-value is less than the level of significance, α, reject the null hypothesis. Decision Rule in the One-Way ANOVA Test 13-30

31. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section Post Hoc Tests on One-Way Analysis of Variance 13.2

32. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 1.Perform the Tukey Test 13-32

33. Copyright © 2013, 20 and 2007 Pearson Education, Inc. When the results from a one-way ANOVA lead us to conclude that at least one population mean is different from the others, we can make additional comparisons between the means to determine which means differ significantly. The procedures for making these comparisons are called multiple comparison methods. 13-33

34. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 1 Perform the Tukey Test 13-34

35. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The computation of the test statistic for Tukey’s test for comparing pairs of means follows the same logic as the test for comparing two means from independent sampling. However, the standard error that is used is where s 2 is the mean square error estimate (MSE) of σ 2 from the one-way ANOVA, n 1 is the sample size from population 1, and n 2 is the sample size from population 2. 13-35

36. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The test statistic for Tukey’s test when testing H 0 : μ 1 = μ 2 versus H 1 : μ 1 ≠ μ 2 is given by where s 2 is the mean square estimate of σ 2 (MSE) from ANOVA n 1 is the sample size from population 1 n 2 is the sample size from population 2 Test Statistic for Tukey’s Test 13-36

37. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The critical value for Tukey’s test using a familywise error rate α is given by where ν is the degrees of freedom due to error (n – k) k is the total number of means being compared Critical Value for Tukey’s Test 13-37

38. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Find the critical value from the Studentized range distribution with v = 13 degrees of freedom and k = 4 degrees of freedom with a familywise error rate α = 0.01. Find the critical value from the Studentized range distribution with v = 64 degrees of freedom and k = 6 degrees of freedom with a familywise error rate α = 0.05. Parallel Example 1: Finding the Critical Value from the Studentized Range Distribution 13-38

39. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution a) b) 13-39

40. Copyright © 2013, 20 and 2007 Pearson Education, Inc. After rejecting the null hypothesis H 0 : μ 1 = μ 2 = ··· = μ k the following steps can be used to compare pairs of means for significant differences, provided that 1.There are k simple random samples from k populations. 2.The k samples are independent of each other. 3.The populations are normally distributed. 4.The populations have the same variance. Step 1: Arrange the sample means in ascending order. Tukey’s Test 13-40

41. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Step 2: Compute the pairwise differences where. Step 3: Compute the test statistic, for each pairwise difference. Tukey’s Test 13-41

42. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Step 4: Determine the critical value,, where α is the level of significance (the familywise error rate). Step 5: If, reject the null hypothesis that H 0 : μ i = μ j and conclude that the means are significantly different. Step 6: Compare all pairwise differences to identify which means differ. Tukey’s Test 13-42

43. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Suppose that there is sufficient evidence to reject H 0 : μ 1 = μ 2 = μ 3 using a one-way ANOVA. The mean square error from ANOVA is found to be 28.7. The sample means are, and, with n 1 = n 2 = n 3 =9. Use Tukey’s test to determine which pairwise means are significantly different using a familywise error rate of α = 0.05. Parallel Example 2: Performing Tukey’s Test 13-43

44. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 1: The means, in ascending order, are Step 2: We next compute the pairwise differences for each pair, subtracting the smaller sample mean from the larger sample mean: 13-44

45. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 3: Compute the test statistic q 0 for each pairwise difference. 2-1: 2-3: 13-45

46. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution 3-1: Step 4: Find the critical value using an α = 0.05 familywise error rate with ν = n – k =27 – 3 = 24 and k = 3. Then q 0.05,24,3 = 3.532. 13-46

47. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Step 5: Since 6.22 and 4.42 are greater than 3.532, but 1.79 is less than 3.532, we reject H 0 : μ 1 =μ 2 and H 0 : μ 2 =μ 3 but not H 0 : μ 1 =μ 3. Step 6: The conclusions of Tukey’s test are 13-47

48. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section The Randomized Complete Block Design 13.3

49. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objectives 1.Conduct analysis of variance on the randomized complete block design 2.Perform the Tukey test 13-49

50. Copyright © 2013, 20 and 2007 Pearson Education, Inc. In the completely randomized design, the researcher manipulates a single factor and fixes it at two or more levels and then randomly assigns experimental units to a treatment. This design is not always sufficient because the researcher may be aware of additional factors that cannot be fixed at a single level throughout the experiment. 13-50

51. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The randomized block design is an experimental design that captures more information and therefore reduces experimental error. 13-51

52. Copyright © 2013, 20 and 2007 Pearson Education, Inc. “In Other Words” A block is a method for controlling experimental error. Blocks should form a homogenous group. For example, if age is thought to explain some of the variability in the response variable, we can remove age from the experimental error by forming blocks of experimental units with the same age. Gender is another common block. 13-52

53. Copyright © 2013, 20 and 2007 Pearson Education, Inc. CAUTION! When we block, we are not interested in determining whether the block is significant. We only want to remove experimental error to reduce the mean square error. 13-53

54. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 1 Conduct Analysis of Variance on the Randomized Complete Block Design 13-54

55. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 1.The response variable for each of the k populations is normally distributed. 2.The response variable for each of the k populations has the same variance; that is, each treatment group has population variance σ 2. Requirements for Analyzing Data from a Randomized Complete Block Design 13-55

56. Copyright © 2013, 20 and 2007 Pearson Education, Inc. A rice farmer is interested in the effect of four fertilizers on fruiting period. He randomly selects four rows from his field that have been planted with the same seed and divides each row into four segments. The fertilizers are then randomly assigned to the four segments. Assume that the environmental conditions are the same for each of the four rows. The data given in the next slide represent the fruiting period, in days, for each row/fertilizer combination. Is there sufficient evidence to conclude that the fruiting period for the four fertilizers differs at the α = 0.05 level of significance? Parallel Example 2: Analyzing the Randomized Complete Block Design 13-56

57. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Fertilizer 1Fertilizer 2Fertilizer 3Fertilizer 4 Row 113.714.016.217.1 Row 213.614.415.316.9 Row 312.211.713.014.1 Row 415.016.015.917.3 13-57

58. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution We wish to test H 0 : μ 1 = μ 2 = μ 3 = μ 4 versus H 1 : at least one of the means is different We first verify the requirements for the test: 1.Normal probability plots for the data from each of the four fertilizers indicates that the normality requirement is satisfied. 13-58

59. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution 2. s 1 = 1.144 s 2 =1.775 s 3 =1.449 s 4 =1.509 Since the largest standard deviation is not more than twice the smallest standard deviation, the assumption of equal variances is not violated. 13-59

60. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution ANOVA output from Minitab: Two-way ANOVA: Fruiting period versus Fertilizer, Row Source DF SS MS F P Fertilizer 3 17.885 5.96167 22.40 0.000 Row 3 24.110 8.03667 30.20 0.000 Error 9 2.395 0.26611 Total 15 44.390 13-60

61. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Since the P-value is < 0.001, we reject the null hypothesis and conclude that there is a difference in fruiting period for the four fertilizers. Note: We are not interested in testing whether the fruiting periods among the blocks are equal. 13-61

62. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 2 Perform the Tukey Test 13-62

63. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Once the null hypothesis of equal population means is rejected, we can proceed to determine which means differ significantly using Tukey’s test. The steps are identical to those for comparing means in the one-way ANOVA. The critical value is q α,ν,k using a familywise error rate of α with ν = (r – 1)(c – 1) = the error degrees of freedom (r is the number of blocks and c is the number of treatments) and k is the number of means being tested. 13-63

64. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Parallel Example 3: Multiple Comparisons Using Tukey’s Test Use Tukey’s test to determine which pairwise means differ for the data presented in Example 2 with a familywise error rate of α = 0.05, using MINITAB. 13-64

65. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Tukey Simultaneous Tests Response Variable Fruiting period All Pairwise Comparisons among Levels of Fertilizer Fertilizer = 1 subtracted from: Fertilizer Lower Center Upper -----+---------+---------+---------+- 2 -0.7400 0.4000 1.540 (-------*------) 3 0.3350 1.4750 2.615 (-------*------) 4 1.5850 2.7250 3.865 (------*-------) -----+---------+---------+---------+- 0.0 1.5 3.0 4.5 Fertilizer = 2 subtracted from: Fertilizer Lower Center Upper -----+---------+---------+---------+- 3 -0.06505 1.075 2.215 (------*-------) 4 1.18495 2.325 3.465 (-------*------) -----+---------+---------+---------+- 0.0 1.5 3.0 4.5 Fertilizer = 3 subtracted from: Fertilizer Lower Center Upper -----+---------+---------+---------+- 4 0.1100 1.250 2.390 (------*-------) -----+---------+---------+---------+- 0.0 1.5 3.0 4.5 13-65

66. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Tukey Simultaneous Tests Response Variable Fruiting period All Pairwise Comparisons among Levels of Fertilizer Fertilizer = 1 subtracted from: Difference SE of Adjusted Fertilizer of Means Difference T-Value P-Value 2 0.4000 0.3648 1.097 0.7003 3 1.4750 0.3648 4.044 0.0127 4 2.7250 0.3648 7.471 0.0002 Fertilizer = 2 subtracted from: Difference SE of Adjusted Fertilizer of Means Difference T-Value P-Value 3 1.075 0.3648 2.947 0.0650 4 2.325 0.3648 6.374 0.0006 Fertilizer = 3 subtracted from: Difference SE of Adjusted Fertilizer of Means Difference T-Value P-Value 4 1.250 0.3648 3.427 0.0316 13-66

67. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section Two-Way Analysis of Variance 13.4

68. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objectives 1.Analyze a two-way ANOVA design 2.Draw interaction plots 3.Perform the Tukey test 13-68

69. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 1 Analyze a Two-Way Analysis of Variance Design 13-69

70. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Recall, there are two ways to deal with factors: (1) control the factors by fixing them at a single level or by fixing them at different levels, and (2) randomize so that their effect on the response variable is minimized. In both the completely randomized design and the randomized complete block design, we manipulated one factor to see how varying it affected the response variable. 13-70

71. Copyright © 2013, 20 and 2007 Pearson Education, Inc. In a Two-Way Analysis of Variance design, two factors are used to explain the variability in the response variable. We deal with the two factors by fixing them at different levels. We refer to the two factors as factor A and factor B. If factor A has n levels and factor B has m levels, we refer to the design as an factorial design. 13-71

72. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Parallel Example 1: A 2 x 4 Factorial Design Suppose the rice farmer is interested in comparing the fruiting period for not only the four fertilizer types, but for two different seed types as well. The farmer divides his plot into 16 identical subplots. He randomly assigns each seed/fertilizer combination to two of the subplots and obtains the fruiting periods shown on the following slide. Identify the main effects. What does it mean to say there is an interaction effect between the two factors? 13-72

73. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Fertilizer 1Fertilizer 2Fertilizer 3Fertilizer 4 Seed Type A 13.5 13.9 13.5 14.1 15.2 14.7 17.1 16.4 Seed Type B 14.4 15.0 14.7 15.4 15.3 15.9 16.9 17.3 13-73

74. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution The two factors are A: fertilizer type and B: seed type. Since all levels of factor A are combined with all levels of factor B, we say that the factors are crossed. The main effect of factor A is the change in fruiting period that results from changing the fertilizer type. The main effect of factor B is the change in fruiting period that results from changing the seed type. We say that there is an interaction effect if the effect of fertilizer on fruiting period varies with seed type. 13-74

75. Copyright © 2013, 20 and 2007 Pearson Education, Inc. 1. The populations from which the samples are drawn must be normal. 2. The samples are independent. 3. The populations all have the same variance. Requirements to Perform the Two-Way Analysis of Variance 13-75

76. Copyright © 2013, 20 and 2007 Pearson Education, Inc. In a two-way ANOVA, we test three separate hypotheses. Hypotheses Regarding Interaction Effect H 0 : there is no interaction effect between the factors H 1 : there is interaction between the factors Hypotheses Regarding Main Effects H 0 : there is no effect of factor A on the response variable H 1 : there is an effect of factor A on the response variable H 0 : there is no effect of factor B on the response variable H 1 : there is an effect of factor B on the response variable 13-76

77. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Whenever conducting a two-way ANOVA, we always first test the hypothesis regarding interaction effect. If the null hypothesis of no interaction is rejected, we do not interpret the result of the hypotheses involving the main effects. This is because the interaction clouds the interpretation of the main effects. 13-77

78. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Parallel Example 3: Examining a Two-Way ANOVA Recall the rice farmer who is interested in determining the effect of fertilizer and seed type on the fruiting period of rice. Assume that probability plots indicate that it is reasonable to assume that the data come from populations that are normally distributed. a)Verify the requirement of equal population variances. b)Use MINITAB to test whether there is an interaction effect between fertilizer type and seed type. c)If there is no significant interaction, determine if there is a significant difference in the means for i)the 4 fertilizers ii)the 2 seed types 13-78

79. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution a)The standard deviations for each treatment combination are given in the table below: Fertilizer 1Fertilizer 2Fertilizer 3Fertilizer 4 Seed Type A 0.2830.4240.3540.495 Seed Type B 0.4240.4950.4240.283 Since the largest standard deviation, 0.495, is not more than twice the smallest standard deviation, 0.283, the assumption of equal variances is met. 13-79

80. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution MINITAB output: Analysis of Variance for Fruiting period Source DF SS MS F P Fertilizer 3 18.3269 6.1090 37.16 0.000 Seed 1 2.6406 2.6406 16.06 0.004 Fert*Seed 3 0.4669 0.1556 0.95 0.463 Error 8 1.3150 1.3150 0.1644 Total 15 22.7494 b)The P-value for the interaction term is 0.463 > 0.05, so we fail to reject the null hypothesis and conclude that there is no interaction effect. 13-80

81. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution c) i) Since the P-value for fertilizer is given as 0.000, we reject the null hypothesis and conclude that the mean fruiting period is different for at least one of the 4 types of fertilizer. ii) Since the P-value for seed type is found to be 0.004, we reject the null hypothesis and conclude that the mean fruiting period is different for the two seed types. 13-81

82. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 2 Draw Interaction Plots 13-82

83. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Step 1: Compute the mean value of the response variable within each cell. In addition, compute the row mean value of the response variable and the column mean value of the response variable with each level of each factor. Constructing Interaction Plots 13-83

84. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Step 2: In a Cartesian plane, label the horizontal axis for each level of factor A. The vertical axis will represent the mean value of the response variable. For each level of factor A, plot the mean value of the response variable for each level of factor B. Draw straight lines connecting the points for the common level of factor B. You should have as many lines as there are levels of factor B. The more difference there is in the slopes of the two lines, the stronger the evidence of interaction. Constructing Interaction Plots 13-84

85. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Parallel Example 4: Drawing an Interaction Plot Draw an interaction plot for the data from the rice farmer example. 13-85

86. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution The cell means are given in the table below. Fertilizer 1Fertilizer 2Fertilizer 3Fertilizer 4 Seed Type A 13.713.814.9516.75 Seed Type B 14.715.0515.617.1 13-86

87. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Note that the lines have fairly similar slopes between points which indicates no interaction between fertilizer and seed type. 13-87

88. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Objective 3 Perform the Tukey Test 13-88

89. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Once we reject the hypothesis of equal population means for either factor, we proceed to determine which means differ significantly using Tukey’s test. The steps are identical to those for one-way ANOVA. However, the critical value is q α,ν,k where α is the familywise error rate ν = N – ab where N is the total sample size, a is the number of levels for factor A and b is the number of levels for factor B k is the number of means being tested for the factor 13-89

90. Copyright © 2013, 20 and 2007 Pearson Education, Inc. The standard error is where m is the product of the number of levels for the factor and the number of observations within each cell. 13-90

91. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Parallel Example 7: Multiple Comparisons Using Tukey’s Test For the rice farming example, use MINITAB to perform Tukey’s test to determine which means differ for the four types of fertilizer using a familywise error rate of α = 0.05. 13-91

92. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution Confidence Interval Method Tukey 95.0% Simultaneous Confidence Intervals Response Variable Fruiting period All Pairwise Comparisons among Levels of Fertilizer Fertilizer = 1 subtracted from: Fertilizer Lower Center Upper ------+---------+---------+---------+ 2 -0.6933 0.2250 1.143 (-------*-------) 3 0.1567 1.0750 1.993 (-------*-------) 4 1.8067 2.7250 3.643 (-------*------) ------+---------+---------+---------+ 0.0 1.2 2.4 3.6 Fertilizer = 2 subtracted from: Fertilizer Lower Center Upper ------+---------+---------+---------+ 3 -0.06830 0.8500 1.768 (-------*-------) 4 1.58170 2.5000 3.418 (-------*------) ------+---------+---------+---------+ 0.0 1.2 2.4 3.6 Fertilizer = 3 subtracted from: Fertilizer Lower Center Upper ------+---------+---------+---------+ 4 0.7317 1.650 2.568 (-------*------) ------+---------+---------+---------+ 0.0 1.2 2.4 3.6 13-92

93. Copyright © 2013, 20 and 2007 Pearson Education, Inc. For Fertilizer 2 – Fertilizer 1 the confidence interval contains 0, so we do not reject the null hypothesis that means for Fertilizer 1 and 2 are equal. For Fertilizer 3 – Fertilizer 1 the confidence interval does not contain 0, so we reject the null hypothesis that means for Fertilizer 1 and 3 are equal. For Fertilizer 4 – Fertilizer 3 the confidence interval does not contain 0, so we reject the null hypothesis that means for Fertilizer 3 and 4 are equal. 13-93 Solution Confidence Interval Method

94. Copyright © 2013, 20 and 2007 Pearson Education, Inc. For Fertilizer 3 – Fertilizer 2 the confidence interval contains 0, so we do not reject the null hypothesis that means for Fertilizer 2 and 3 are equal. For Fertilizer 4 – Fertilizer 2 the confidence interval does not contain 0, so we reject the null hypothesis that means for Fertilizer 2 and 4 are equal. For Fertilizer 4 – Fertilizer 1 the confidence interval does not contain 0, so we reject the null hypothesis that means for Fertilizer 1 and 4 are equal. 13-94 Solution Confidence Interval Method

95. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Solution P-Value Method Tukey Simultaneous Tests Response Variable Fruiting period All Pairwise Comparisons among Levels of Fertilizer Fertilizer = 1 subtracted from: Difference SE of Adjusted Fertilizer of Means Difference T-Value P-Value 2 0.2250 0.2867 0.7848 0.8594 3 1.0750 0.2867 3.7498 0.0234 4 2.7250 0.2867 9.5053 0.0001 Fertilizer = 2 subtracted from: Difference SE of Adjusted Fertilizer of Means Difference T-Value P-Value 3 0.8500 0.2867 2.965 0.0699 4 2.5000 0.2867 8.720 0.0001 Fertilizer = 3 subtracted from: Difference SE of Adjusted Fertilizer of Means Difference T-Value P-Value 4 1.650 0.2867 5.755 0.0019 13-95

96. Copyright © 2013, 20 and 2007 Pearson Education, Inc. Using the P-value method yields the same results: we do not reject the null hypothesis that means for Fertilizer 1 and 2 are equal we reject the null hypothesis that means for Fertilizer 1 and 3 are equal we reject the null hypothesis that means for Fertilizer 3 and 4 are equal we do not reject the null hypothesis that means for Fertilizer 2 and 3 are equal we reject the null hypothesis that means for Fertilizer 2 and 4 are equal we reject the null hypothesis that means for Fertilizer 1 and 4 are equal 13-96 Solution P-Value Method