1. STATISTICS INFORMED DECISIONS USING DATA
Fifth Edition
Chapter 13
Comparing Three or More Means
Copyright © 2017, 2013, 2010 Pearson Education, Inc. All Rights Reserved

2. 13.1 Comparing Three or More Means (One-Way Analysis of Variance) Learning Objectives
1. Verify the requirements to perform a one-way ANOVA 2. Test a hypothesis regarding three or more means using one-way ANOVA

3. 13.1 Introduction to the Practice of Statistics Introduction (1 of 2)
Analysis of Variance (ANOVA) is an inferential method used to test the equality of three or more population means.

4. 13.1 Introduction to the Practice of Statistics Introduction (2 of 2)
CAUTION!
Do not test H0: μ1 = μ2 = μ3 by conducting three separate hypothesis tests, because the probability of making a Type I error will be much higher than α.

5. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (1 of 12)
Requirements of a One-Way ANOVA Test
There must be k simple random samples; one from each of k populations or a randomized experiment with k treatments.
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.
The populations are normally distributed.
The populations must have the same variance; that is, each treatment group has the population variance σ2.

6. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (2 of 12)
Testing a Hypothesis Regarding k = 3
H0: μ1 = μ2 = μ3
H1: At least one population mean is different from the others

7. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (3 of 12)
Testing Using ANOVA
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.

8. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (4 of 12)
Verifying the Requirement of Equal Population Variance
The one-way ANOVA procedures may be used if the largest sample standard deviation is no more than twice the smallest sample standard deviation.

9. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (5 of 12)
Parallel Example 1: Verifying the Requirements of ANOVA
The following data represent the weight (in grams) of pennies minted at the Denver mint in 1990,1995, and Verify that the requirements in order to perform a one-way ANOVA are satisfied.

10. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (6 of 12)
1990
1995
2000
2.50
2.52
2.54
2.48
2.49
2.53
2.46
2.47
2.51
2.55

11. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (7 of 12)
Solution
The 3 samples are simple random samples.
The samples were obtained independently.
Normal probability plots for the 3 years follow. All of the plots are roughly linear so the normality assumption is satisfied.

12. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (8 of 12)

13. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (9 of 12)

14. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (10 of 12)

15. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (11 of 12)
Solution
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 (2 • = > ) so the requirement of equal population variances is considered satisfied.

16. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Verify the requirements to perform a one-way ANOVA (12 of 12)
Descriptive Statistics
Variable N
Mean
Median
TrMean
StDev
SE Mean
1990 11
2.4964
2.5000
2.4967
0.0220
0.0066
1995
2.5091
2.5078
0.0243
0.0073
2000
0.0141
0.0043
Variable
Minimum
Maximum Q1 Q3
1990
2.4600
2.5300
2.4800
2.5100
1995
2.5500
2.4900
2000
2.5200

17. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (1 of 14)
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.

18. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (2 of 14)
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.

19. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (3 of 14)
ANOVA F-Test Statistic

20. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (4 of 14)
Computing the F-Test Statistic

21. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (5 of 14)
Computing the F-Test Statistic
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:

22. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (6 of 14)
Parallel Example 2: Computing the F-Test Statistic
Compute the F-test statistic for the penny data.

23. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (7 of 14)
Solution
Step 1: Compute the overall mean.

24. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (8 of 14)
Solution
Step 3: Find the sample variance for each treatment (year).

25. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (9 of 14)
Solution
Step 4: Compute the sum of squares due to treatment, SST, and the sum of squares due to error, SSE.
SST = 11( − )2 + 11( − ) (2.5 − )2 =
SSE = (11 − 1)(0.0005) + (11 − 1)(0.0006) + (11 − 1)(0.0002) = 0.013

26. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (10 of 14)
Solution
Step 5: Compute the mean square due to treatment, MST, and the mean square due to error, MSE.

27. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (11 of 14)
Solution
Step 6: Compute the F-statistic.

28. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (12 of 14)
Solution
The large value of the F-test statistic suggests that the sample means for each treatment differ.

29. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (13 of 14)
Solution
The results of the computations for the data that led to the F-test statistic are presented in the following table.
ANOVA Table:
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Squares
F-Test Statistic
Treatment
0.0009 2
0.0005
1.25
Error
0.013 30
0.0004
blank
Total
0.0139 32

30. 13. 1 Introduction to the Practice of Statistics 13. 1
13.1 Introduction to the Practice of Statistics Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA (14 of 14)
Decision Rule in the One-Way ANOVA Test
If the P-value is less than the level of significance, α, reject the null hypothesis.

31. 13.2 Post Hoc Tests on One-Way Analysis of Variance Learning Objective
1. Perform the Tukey Test

32. 13.2 Post Hoc Tests on One-Way Analysis of Variance Introduction
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.

33. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (1 of 13)
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

34. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (2 of 13)
The test statistic for Tukey’s test when testing H0: μ1 = μ2 versus H1: μ1 ≠ μ2 is given by

35. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (3 of 13)
The critical value for Tukey’s test using a familywise error rate α is given by
qα, v, k
where
ν is the degrees of freedom due to error (n − k)
k is the total number of means being compared

36. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (4 of 13)
Parallel Example 1: Finding the Critical Value from the Studentized Range Distribution
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.

37. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (5 of 13)
Solution

38. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (6 of 13)
Tukey’s Test
After rejecting the null hypothesis H0: μ1 = μ2 = ··· = μk the following steps can be used to compare pairs of means for significant differences, provided that
There are k simple random samples from k populations.
The k samples are independent of each other.
The populations are normally distributed.
The populations have the same variance.
Step 1: Arrange the sample means in ascending order.

39. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (7 of 13)
Tukey’s Test

40. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (8 of 13)
Tukey’s Test
Step 4: Determine the critical value, qα,v,k, where α is the level of significance (the familywise error rate).
Step 5: If q0 ≥ qα,v,k, reject the null hypothesis that H0: μi = μj and conclude that the means are significantly different.
Step 6: Compare all pairwise differences to identify which means differ.

41. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (9 of 13)
Parallel Example 2: Performing Tukey’s Test

42. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (10 of 13)
Solution
Step 1: The means, in ascending order, are

43. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (11 of 13)
Solution
Step 3: Compute the test statistic q0 for each pairwise difference.

44. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (12 of 13)
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 q0.05,24,3 =

45. 13. 2 Post Hoc Tests on One-Way Analysis of Variance 13. 2
13.2 Post Hoc Tests on One-Way Analysis of Variance Perform the Tukey Test (13 of 13)
Solution
Step 5: Since 6.22 and 4.42 are greater than 3.532, but 1.79 is less than 3.532, we reject H0: μ1 = μ2 and H0: μ2 = μ3 but not H0: μ1 = μ3.
Step 6: The conclusions of Tukey’s test are

46. 13.3 The Randomized Complete Block Design Learning Objectives
1. Conduct analysis of variance on the randomized complete block design 2. Perform the Tukey test

47. 13.3 The Randomized Complete Block Design Introduction (1 of 4)
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.

48. 13.3 The Randomized Complete Block Design Introduction (2 of 4)
The randomized block design is an experimental design that captures more information and therefore reduces experimental error.

49. 13.3 The Randomized Complete Block Design Introduction (3 of 4)
“In Other Words”
A block is a method for controlling experimental error. Blocks should form a homogenous group. For example, if gender is thought to explain some of the variability in the response variable, remove gender from the experimental error by forming blocks of experimental units with the same gender.

50. 13.3 The Randomized Complete Block Design Introduction (4 of 4)
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.

51. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (1 of 7)
Requirements for Analyzing Data from a Randomized Complete Block Design
The response variable for each of the k populations is normally distributed.
The response variable for each of the k populations has the same variance; that is, each treatment group has population variance σ2.

52. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (2 of 7)
Parallel Example 2: Analyzing the Randomized Complete Block Design
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?

53. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (3 of 7)
blank
Fertilizer 1
Fertilizer 2
Fertilizer 3
Fertilizer 4
Row 1
13.7
14.0
16.2
17.1
Row 2
13.6
14.4
15.3
16.9
Row 3
12.2
11.7
13.0
14.1
Row 4
15.0
16.0
15.9
17.3

54. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (4 of 7)
Solution
We wish to test
H0: μ1 = μ2 = μ3 = μ4 versus
H1: at least one of the means is different
We first verify the requirements for the test:
Normal probability plots for the data from each of the four fertilizers indicates that the normality requirement is satisfied.

55. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (5 of 7)
Solution
s1 = s2 = s3 = s4 = 1.509
Since the largest standard deviation is not more than twice the smallest standard deviation, the assumption of equal variances is not violated.

56. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (6 of 7)
Solution
ANOVA output from Minitab:
Two-way ANOVA: Fruiting period versus Fertilizer, Row
Source DF SS MS F P
Fertilizer 3
17.885
22.40
0.000
Row
24.110
30.20
Error 9
2.395
blank
Total 15
44.390

57. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Conduct Analysis of Variance on the Randomized Complete Block Design (7 of 7)
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.

58. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Perform the Tukey Test (1 of 6)
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 qα,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.

59. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Perform the Tukey Test (2 of 6)
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.

60. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Perform the Tukey Test (3 of 6)

61. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Perform the Tukey Test (4 of 6)

62. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Perform the Tukey Test (5 of 6)
Parallel Example 3: Multiple Comparisons Using Tukey’s Test
The interval for the mean of fertilizer 3 − the mean of fertilizer 1 does not include 0, therefore, reject the null hypothesis that the fruiting period for the two fertilizers are equal.
The interval for the mean of fertilizer 4 − the mean of fertilizer 1 does not include 0, therefore, reject the null hypothesis that the fruiting period for the two fertilizers are equal.

63. 13. 3 The Randomized Complete Block Design 13. 3
13.3 The Randomized Complete Block Design Perform the Tukey Test (6 of 6)
Parallel Example 3: Multiple Comparisons Using Tukey’s Test
The interval for the mean of fertilizer 4 − the mean of fertilizer 2 does not include 0, therefore, reject the null hypothesis that the fruiting period for the two fertilizers are equal.
The interval for the mean of fertilizer 4 − the mean of fertilizer 3 does not include 0, therefore, reject the null hypothesis that the fruiting period for the two fertilizers are equal.

64. 13.4 Two-Way Analysis of Variance Learning Objectives
1. Analyze a two-way ANOVA design 2. Draw interaction plots 3. Perform the Tukey test

65. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (1 of 12)
Recall, there are two ways to deal with factors:
control the factors by fixing them at a single level or by fixing them at different levels, and
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.

66. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (2 of 12)
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 n × m factorial design.

67. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (3 of 12)
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?

68. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (4 of 12)
blank
Fertilizer 1
Fertilizer 2
Fertilizer 3
Fertilizer 4
Seed Type A
13.5
13.9
14.1
15.2
14.7
17.1
16.4
Seed Type B
14.4
15.0
15.4
15.3
15.9
16.9
17.3

69. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (5 of 12)
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.

70. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (6 of 12)
Requirements to Perform the Two-Way Analysis of Variance
The populations from which the samples are drawn must be normal.
The samples are independent.
The populations all have the same variance.

71. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (7 of 12)
In a two-way ANOVA, we test three separate hypotheses.
Hypotheses Regarding Interaction Effect H0: there is no interaction effect between the factors H1: there is interaction between the factors
Hypotheses Regarding Main Effects H0: there is no effect of factor A on the response variable H1: there is an effect of factor A on the response variable
H0: there is no effect of factor B on the response variable H1: there is an effect of factor B on the response variable

72. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (8 of 12)
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.

73. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (9 of 12)
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.
Verify the requirement of equal population variances.
Use MINITAB to test whether there is an interaction effect between fertilizer type and seed type.
If there is no significant interaction, determine if there is a significant difference in the means for
the 4 fertilizers
the 2 seed types

74. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (10 of 12)
Solution
The standard deviations for each treatment combination are given in the table below:
blank
Fertilizer 1
Fertilizer 2
Fertilizer 3
Fertilizer 4
Seed Type A
0.283
0.424
0.354
0.495
Seed Type B

75. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (11 of 12)
Solution
MINITAB output: Analysis of Variance for Fruiting period
Source DF SS MS F P
Fertilizer 3
6.1090
37.16
0.000
Seed 1
2.6406
16.06
0.004
Fert*Seed
0.4669
0.1556
0.95
0.463
Error 8
1.3150
0.1644
blank
Total 15

76. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Analyze a Two-Way Analysis of Variance Design (12 of 12)
Solution c)
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.
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.

77. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Draw Interaction Plots (1 of 5)
Constructing Interaction Plots
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.

78. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Draw Interaction Plots (2 of 5)
Constructing Interaction Plots
Step 2: In a Cartesian plane, label the horizontal axis for each level of factor A. Let the vertical axis 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.

79. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Draw Interaction Plots (3 of 5)
Parallel Example 4: Drawing an Interaction Plot
Draw an interaction plot for the data from the rice farmer example.

80. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Draw Interaction Plots (4 of 5)
Solution
The cell means are given in the table below.
blank
Fertilizer 1
Fertilizer 2
Fertilizer 3
Fertilizer 4
Seed Type A
13.7
13.8
14.95
16.75
Seed Type B
14.7
15.05
15.6
17.1

81. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Draw Interaction Plots (5 of 5)
Note that the lines have fairly similar slopes between points which indicates no interaction between fertilizer and seed type.

82. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (1 of 8)
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 qα,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

83. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (2 of 8)
where m is the product of the number of levels for the factor and the number of observations within each cell.

84. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (3 of 8)
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.

85. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (4 of 8)
Solution Confidence Interval Method

86. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (5 of 8)
Solution Confidence Interval Method
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.

87. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (6 of 8)
Solution Confidence Interval Method
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.

88. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (7 of 8)
Solution P-value Method

89. 13. 4 Two-Way Analysis of Variance 13. 4
13.4 Two-Way Analysis of Variance Perform the Tukey Test (8 of 8)
Solution P-value Method
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