chapter 3: analysis of variance
DESCRIPTION
5. Chapter 3: Analysis of Variance. 5. Chapter 3: Analysis of Variance. Objectives. Understand the use of sums of squares for comparing group means. Use JMP to perform a one-way ANOVA. Compare the one-way ANOVA with two groups to the two sample t -test. Sums of Squares. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 3: Analysis of Variance
3.1 One-Way ANOVA with Two Groups
3.2 One-Way ANOVA with More than Two Groups
3.3 Fitting an N-Way ANOVA Model
3.4 Contrasts in N-Way ANOVA (Self-Study)
3.5 Power and Sample Size
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Chapter 3: Analysis of Variance
3.1 One-Way ANOVA with Two Groups3.1 One-Way ANOVA with Two Groups
3.2 One-Way ANOVA with More than Two Groups
3.3 Fitting an N-Way ANOVA Model
3.4 Contrasts in N-Way ANOVA (Self-Study)
3.5 Power and Sample Size
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Objectives Understand the use of sums of squares for comparing
group means. Use JMP to perform a one-way ANOVA. Compare the one-way ANOVA with two groups to the
two sample t-test.
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Sums of Squares
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Total Sum of Squares
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SST = (3-6)2 + (4-6)2 + (5-6)2 + (7-6)2 + (8-6)2 + (9-6)2 = 28
(3-6)2
(7-6)2
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Error Sum of Squares
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SSE = (3-4)2 + (4-4)2 + (5-4)2 + (7-8)2 + (8-8)2 + (9-8)2 = 4
(5-4)2
8B
4A
(7-8)2
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Model Sum of Squares
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SSM = 3*(4-6)2 + 3*(8-6)2 = 24
8B
4A
3*(8-6)2
3*(4-6)2
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ANOVA Table
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Source Degrees
OfFreedom
Sum ofSquares
MeanSquare
F Ratio
Model
Error
Total
1k
kn kn
ESS
1n
1
SSM
k
2
1
( )k
j jj
n y y
k
j 1
2
1
)( jij
n
i
yyj
E
M
MS
MS
2
1
( )n
ii
y y
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The Reference Distribution
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E
M
MS
MSF
for your sample and compare the sample F to the reference distribution.
Compute
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3.01 Multiple Answer PollThe model sum of squares (SSM) describes which of the following?
a. The variability between the groupsb. The variability within the groupsc. The variability explained by the grouping variable
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3.01 Multiple Answer Poll – Correct AnswersThe model sum of squares (SSM) describes which of the following?
a. The variability between the groupsb. The variability within the groupsc. The variability explained by the grouping variable
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The One-Way ANOVA Model and Standard Assumptions
= + +
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Yik = µ + τi + εik
The errors are normally distributed with a mean of zero. The variance is the same for all groups. The errors occur independently of one another.
Strength ofConcrete
Base Level Type ofAdditive
UnexplainedVariation
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Concrete Example
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Reinforced Standard
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This demonstration illustrates the concepts discussed previously.
Comparing Two Means Using a One-Way ANOVA
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3.02 QuizMatch the significance test output on the right to the correct null hypothesis on the left.
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1. H0: σ21 = σ2
2
2. H0: μ1 = μ2
3. H0: data from normal distribution
A.
B.
C.
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3.02 Quiz – Correct AnswerMatch the significance test output on the right to the correct null hypothesis on the left.
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1-A, 2-C, 3-B
1. H0: σ21 = σ2
2
2. H0: μ1 = μ2
3. H0: data from normal distribution
A.
B.
C.
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3.03 QuizWhich of the following statements are true at α=0.05?
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1. There is not sufficient evidence to reject equal variances.
2. There is not sufficient evidence to reject equal means.
3. There is not sufficient evidence to reject normality.
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3.03 Quiz – Correct AnswerWhich of the following statements are true at α=0.05?
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1. There is not sufficient evidence to reject equal variances.
2. There is not sufficient evidence to reject equal means.
3. There is not sufficient evidence to reject normality.
(2 is false because the p-value is .0011, so there is sufficient evidence to reject H0: µ1 = µ2 at α = 0.05)
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Chapter 3: Analysis of Variance
3.1 One-Way ANOVA with Two Groups
3.2 One-Way ANOVA with More than Two Groups3.2 One-Way ANOVA with More than Two Groups
3.3 Fitting an N-Way ANOVA Model
3.4 Contrasts in N-Way ANOVA (Self-Study)
3.5 Power and Sample Size
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Objectives Fit a one-way ANOVA with more than two groups. Use multiple comparisons procedures to determine
which means are significantly different.
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Insulation ExampleDo additives differ in their effectiveness at enhancing the properties of an insulator?
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The One-Way ANOVA Hypothesis
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H0: All means equal
80
60
40
20
0
H1: At least onemean different
80
60
40
20
0
H0: μ1 = μ2 = μ3 H1: not (μ1 = μ2 = μ3)
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This demonstration illustrates the concepts discussed previously.
Fitting a One-Way ANOVA Model with More than Two Groups
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Multiple Comparisons Methods
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ControlComparisonwise
Error Rate
ControlExperimentwise
Error Rate
Compare Each PairPairwise t-Tests
Compare All PairsTukey-Kramer
Compare with ControlDunnett's Test
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Experimentwise Error Rates
Comparisonwise error rate = 0.05
EER 1-(1-)nc where nc = number of comparisons
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Properties of Multiple Comparisons Methods
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Procedure Comparisons Adjustment Conservative
t-test k(k-1)/2 None Least
Tukey-Kramer
k(k-1)/2 Largest Most
Dunnett’s (k-1) Smallest Less
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This demonstration illustrates the concepts discussed previously.
Generating Multiple Comparisons
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3.04 Multiple Choice PollA statistically conservative method tends to do which of the following?
a. Find more significant differences than might otherwise be found
b. Vary in its findings, depending on the observed datac. Find fewer significant differences than might otherwise
be found
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3.04 Multiple Choice Poll – Correct AnswerA statistically conservative method tends to do which of the following?
a. Find more significant differences than might otherwise be found
b. Vary in its findings, depending on the observed datac. Find fewer significant differences than might otherwise
be found
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This exercise reinforces the concepts discussed previously.
Exercise
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Chapter 3: Analysis of Variance
3.1 One-Way ANOVA with Two Groups
3.2 One-Way ANOVA with More than Two Groups
3.3 Fitting an N-Way ANOVA Model3.3 Fitting an N-Way ANOVA Model
3.4 Contrasts in N-Way ANOVA (Self-Study)
3.5 Power and Sample Size
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Objectives Fit an N-way ANOVA model. Understand interactions in ANOVA. Use JMP to fit a model with interactions. Compare response groups.
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N-Way ANOVA
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Neutralization ExampleDo different diluents and buffers affect the final pH of media?
Is there an interaction between these predictors?
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Interactions
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Interactions When the interactions are not significant, you can
usually delete the interactions and analyze just the main effects.
When the interactions are significant, you should not delete them or their corresponding main effects from the model. Try to understand the interactions by examining the interaction plots.
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The N-Way ANOVA Model and Standard Assumptions
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Yijk = µ + αi + τj + (ατ)ij + εijk
The errors are normally distributed with a mean of zero. The variance is the same for all groups. The errors occur independently of one another.
pHBaseLevel
Buffer ResidualDiluent Interaction= + + + +
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Residuals The residuals represent the random variation
in the statistical model. You expect them to do the following:
– display no particular pattern (unbiased)– be normally distributed– exhibit constant variance (independent
of response)– be independent (no correlation)
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N-Way ANOVA Hypothesis
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This demonstration illustrates the concepts discussed previously.
ANOVA with TwoCategorical Predictors
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This exercise reinforces the concepts discussed previously.
Exercise
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Chapter 3: Analysis of Variance
3.1 One-Way ANOVA with Two Groups
3.2 One-Way ANOVA with More than Two Groups
3.3 Fitting an N-Way ANOVA Model
3.4 Contrasts in N-Way ANOVA (Self-Study)3.4 Contrasts in N-Way ANOVA (Self-Study)
3.5 Power and Sample Size
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Objectives Understand the use of contrasts. Use JMP to construct contrasts in N-way ANOVA.
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Simple Contrasts
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Other Contrasts
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This demonstration illustrates the concepts discussed previously.
Examining a Contrast
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3.05 QuizWhat does the following contrast test?
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3.05 Quiz – Correct AnswerWhat does the following contrast test?
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The contrast tests if the average of diluent A across the first 3 levels of buffer is the same as the average of diluent B across the first 3 levels of buffer. H0: 1/3(μA/1 + μA/2 + μA/3) - 1/3(μB/1 + μB/2 + μB/3) = 0
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3.06 QuizRecall that the following contrast tests:
H0: ½ (μA/1 + μA/2) = ½ (μB/1 + μB/2)
H1: ½ (μA/1 + μA/2) ≠ ½ (μB/1 + μB/2)
What can you conclude from the p-value of the contrast?
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3.06 Quiz – Correct AnswerRecall that the following contrast tests:
H0: ½ (μA/1 + μA/2) = ½ (μB/1 + μB/2)
H1: ½ (μA/1 + μA/2) ≠ ½ (μB/1 + μB/2)
What can you conclude from the p-value of the contrast?
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There is sufficient evidence to reject the null hypothesis and say that a difference does exist between the average of diluent A across buffers 1 and 2 and the average of diluent B across buffers 1 and 2.
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Chapter 3: Analysis of Variance
3.1 One-Way ANOVA with Two Groups
3.2 One-Way ANOVA with More than Two Groups
3.3 Fitting an N-Way ANOVA Model
3.4 Contrasts in N-Way ANOVA (Self-Study)
3.5 Power and Sample Size3.5 Power and Sample Size
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Objectives Understand the relationship between power,
significance level, effect size, variability, and sample size.
Use JMP software’s Power and Sample Size platform to determine the required sample size for an experiment.
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Types of Errors and PowerYou perform a hypothesis test and make a decision, but was the decision correct?
The probability of a Type I error is denoted by .
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Types of Errors and PowerYou perform a hypothesis test and make a decision, but was the decision correct?
The probability of a Type II error is denoted by .
The power of the statistical test is 1 - .
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How Many Observations?
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Level of Significance
(Alpha)
Effect Size(Difference to
detect)
Power
Variability(Std Dev)Required
Sample Size
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3.07 Multiple Answer PollPower is defined to be which of the following?
a. The measure of the ability of the statistical hypothesis test to reject the null hypothesis when it is actually false
b. The probability of not committing a Type II errorc. The probability of failing to reject the null hypothesis
when it is actually false
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3.07 Multiple Answer Poll – Correct AnswersPower is defined to be which of the following?
a. The measure of the ability of the statistical hypothesis test to reject the null hypothesis when it is actually false
b. The probability of not committing a Type II errorc. The probability of failing to reject the null hypothesis
when it is actually false
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This demonstration illustrates the concepts discussed previously.
Power and Sample Size for Experiments
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3.08 QuizAn experimenter decides an effect size of 6 mm is no longer sufficient and instead he wants to be able to detect an effect size of 4 mm. If his sample size, alpha (significance level), and variability remain the same as before, will the power of the test increase or decrease? Why?
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3.08 Quiz – Correct AnswerAn experimenter decides an effect size of 6 mm is no longer sufficient and instead he wants to be able to detect an effect size of 4 mm. If his sample size, alpha (significance level), and variability remain the same as before, will the power of the test increase or decrease? Why?
The power of the test will decrease. The experimenter wants to be able to detect a smaller effect size. That is, he wants the significance test to detect a smaller difference, and this is harder to do. Smaller differences are harder to detect, so the power of the test decreases (holding other terms the same).
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