13 Section Comparing Three or More Means OneWay Analysis of Variance 131 Objectives Verify the requirements to perform a oneway ANOVA Test a hypothesis regarding three or more means using oneway ANOVA ID: 657944
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Slide1
Chapter
Comparing Three or More Means
13Slide2
Section
Comparing Three or More Means (One-Way Analysis of Variance)
13.1Slide3
Objectives
Verify the requirements to perform a one-way ANOVA
Test a hypothesis regarding three or more means using one-way ANOVA
13-
3Slide4
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
. Slide5
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.Requirements of a One-Way ANOVA Test
13-
5Slide6
H
0: μ
1
=
μ
2
=
μ
3
H
1: At least one population mean is different from the othersTesting a Hypothesis Regarding k = 3
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6Slide7
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
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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
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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
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9Slide10
1990 1995 2000
2.50 2.52 2.50
2.50 2.54 2.48
2.49 2.50 2.49
2.53 2.48 2.50
2.46 2.52 2.48
2.50 2.50 2.52
2.47 2.49 2.51
2.53 2.53 2.49
2.51 2.48 2.51
2.49 2.55 2.502.48 2.49 2.5213-
10Slide11
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.
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11Slide12
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12Slide13
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13Slide14
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14Slide15
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
(2•0.0141 = 0.0282 > 0.02430)
so the requirement of equal population variances is considered satisfied.
Solution
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15Slide16
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
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16Slide17
Objective 2
Test a Hypothesis Regarding Three or More Means Using One-Way ANOVA
13-
17Slide18
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.
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18Slide19
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.
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The analysis of variance
F-test statistic is given by
ANOVA
F
-Test Statistic
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20Slide21
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
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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
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Compute the
F-test statistic for the penny data.
Parallel Example 2:
Computing the F-Test Statistic
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23Slide24
Solution
Step 1:
Compute the overall mean.
Step 2:
Find the sample variance for each treatment (year).
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Solution
Step 3:
Find the sample variance for each treatment (year).
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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
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Solution
Step 5:
Compute the mean square due to treatment, MST, and the mean square due to error, MSE.
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Solution
Step 6:
Compute the
F
-statistic.
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Solution
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
Total
0.0139
32
ANOVA Table:
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If the
P-
value is less than the level of significance,
α
, reject the null hypothesis.
Decision Rule in the One-Way ANOVA Test
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Section
Post Hoc Tests on One-Way Analysis of Variance
13.2Slide32
Objective
Perform the Tukey Test
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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.
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Objective 1
Perform the Tukey Test
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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 n2 is the sample size from population 2.
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35Slide36
The test statistic for Tukey’s test when testing
H0:
μ
1
=
μ
2
versus
H
1
: μ1≠ μ2 is given bywhere s2 is the mean square estimate of σ2 (MSE) from ANOVA n1 is the sample size from population 1
n
2
is the sample size from population 2
Test Statistic for Tukey’s Test
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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 comparedCritical Value for Tukey’s Test
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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
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Solution
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After rejecting the null hypothesis
H0:
μ
1
=
μ
2
= ··· =
μ
k
the following steps can be used to compare pairs of means for significant differences, provided thatThere 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.
Tukey’s Test
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Step 2:
Compute the pairwise differences
where .
Step 3:
Compute the test statistic,
for each pairwise difference.
Tukey’s Test
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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: μi = μj and conclude that the means are significantly different.Step 6:
Compare all pairwise differences to identify which means differ.
Tukey’s Test
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Suppose that there is sufficient evidence to reject
H0
:
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 n1= n2= n3=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
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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:
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Solution
Step 3:
Compute the test statistic
q
0
for each pairwise difference.
2-1:
2-3:
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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 q0.05,24,3 = 3.532.
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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
=
μ2 and H0: μ2=μ3 but not H0
:
μ
1
=
μ
3
.Step 6: The conclusions of Tukey’s test are
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Section
The Randomized Complete Block Design
13.3Slide49
Objectives
Conduct analysis of variance on the randomized complete block design
Perform the Tukey test
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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.
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The randomized block design is an experimental design that captures more information and therefore reduces experimental error.
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“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.
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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.
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Objective 1
Conduct Analysis of Variance on the Randomized Complete Block Design
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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
σ
2
.
Requirements for Analyzing Data from a
Randomized Complete Block Design
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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
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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
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Solution
We wish to test
H
0
:
μ
1
=
μ
2= 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.
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Solution
s
1
= 1.144
s
2
=1.775
s
3
=1.449 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.
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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
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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.
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Objective 2
Perform the Tukey Test
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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.
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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.
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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
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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
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Section
Two-Way Analysis of Variance
13.4Slide68
Objectives
Analyze a two-way ANOVA design
Draw interaction plots
Perform the Tukey test
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Objective 1
Analyze a Two-Way Analysis of Variance Design
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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.
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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.
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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?
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Fertilizer 1
Fertilizer 2
Fertilizer 3
Fertilizer 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
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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.
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The populations from which the samples are drawn must be normal.
The samples are independent.
The populations all have the same variance.
Requirements to Perform the Two-Way
Analysis of Variance
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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 variableH1: there is an effect of factor A on the response variableH0: there is no effect of factor B on the response variableH1: there is an effect of factor B on the response variable
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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.
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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 fertilizersthe 2 seed types
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Solution
The standard deviations for each treatment combination are given in the table below:
Fertilizer 1
Fertilizer 2
Fertilizer 3
Fertilizer 4
Seed Type A
0.283
0.424
0.354
0.495
Seed Type B
0.424
0.495
0.424
0.283
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.
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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.
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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.
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Objective 2
Draw Interaction Plots
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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
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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
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Parallel Example 4:
Drawing an Interaction Plot
Draw an interaction plot for the data from the rice farmer example.
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Solution
The cell means are given in the table below.
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
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86Slide87
Note that the lines have fairly similar slopes between points which indicates no interaction between fertilizer and seed type.
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Objective 3
Perform the Tukey Test
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88Slide89
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-
89Slide90
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.
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90Slide91
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.
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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 3.6Fertilizer = 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
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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.
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93
Solution Confidence Interval MethodSlide94
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.
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94
Solution Confidence Interval MethodSlide95
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 0.85943 1.0750 0.2867 3.7498 0.02344 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-
95Slide96
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