chapter 10: analysis of variance: comparing more than two means
TRANSCRIPT
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Chapter 10: Analysis of Variance: Comparing More Than Two Means
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Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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Where We’ve Been
Presented methods for estimating and testing hypotheses about a single population mean
Presented methods for comparing two population means
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Where We’re Going
Discuss the critical elements in the design of a sampling experiment
Investigate completely randomized, randomized block, and factorial designs
Show how to analyze data using a technique called analysis of variance
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Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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10.1: Elements of a Designed Experiment
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10.1: Elements of a Designed Experiment
Factor Levels are the values of the factors
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10.1: Elements of a Designed Experiment
Treatments are the factor-level combinations In the example above, a variety of
different GPA – Hours Studied combinations could occur within each subset (Yes or No) of the Study Group factor
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10.1: Elements of a Designed Experiment
An experimental unit is the object on which the response and factors are observed or measured In the example above, an individual
student would be the experimental unit
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Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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10.1: Elements of a Designed Experiment
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10.1: Elements of a Designed Experiment
The method by which the
experimental units are selected
determines the type of experiment
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Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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10.1: Elements of a Designed Experiment
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10.2: The Completely Randomized Design
The completely randomized design is a design in which treatments are randomly assigned to the experimental units or in which independent random samples of experimental units are selected for each treatment.
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Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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10.2: The Completely Randomized Design
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10.2: The Completely Randomized Design
Very often the object is to determine whether the varying treatments result in different means:
H0: µ1 = µ2 = µ3 = µ4 = ··· = µk
Ha: At least two of the k treatment means differ
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Testing the equity of the means involves comparing the variability among the different treatments as well as within the treatments, adjusted for degrees of freedom.
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10.2: The Completely Randomized Design
1 2
2
1
2 21 1 2 2
1 1
Variation between the treatment means:
Sum of Squares for Treatments (SST)
( )
Variation within the treatments:
Sum of Squares for Error (SSE)
( ) ( ) + (
k
i ii
n n
j jj j
SST n x x
SSE x x x x
2
1
2 2 21 1 2 2
)
( 1) ( 1) ( 1)
kn
kj kj
k k
x x
n s n s n s
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10.2: The Completely Randomized Design
Adjusting for degrees of freedom produces comparable measures of variability
Mean Square for Treatments (MST)
1Mean Square for Error (MSE)
SSTMST
k
SSEMSE
n k
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10.2: The Completely Randomized Design
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10.2: The Completely Randomized Design
The ratio of the variability among the treatment means to that within the treatment means is an F -statistic:
with k-1 numerator and n-k denominator degrees of freedom.
1,k n k
MSTF
MSE
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10.2: The CompletelyRandomized Design
If F* 1, the difference between the treatment means may be attributable to sampling error.
If F* > 1 (significantly), there is support for the alternative hypothesis that the treatments themselves produce different results.
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10.2: The CompletelyRandomized Design
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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0 1 2
ANOVA -Test to Compare Treatment Means:
Completely Randomized Design
:
: At least two treatment means differ.
Test Statistic: =
Rejection region: * , with 1 numerator
and
k
a
F k
H
H
MSTF
MSEF F k
- denominator degrees of freedom.n k
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10.2: The CompletelyRandomized Design
Conditions required for a Valid ANOVA F-Test: Completely Randomized Design1. The samples are randomly selected in an
independent manner from the k treatment populations.
2. All k sampled populations have distributions that are approximately normal.
3. The k population variances are equal.
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10.2: The Completely Randomized Design
The USGA compares the driving distance of four brands of golf balls. H0: µ1 = µ2 = µ3 = µ4
Ha: The mean distances differ for at least two of the brands
= .10 Test Statistic: F = MST/MSE Rejection region: F > 2.25 = F.10 with v1 = 3 and v2 = 36
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10.2: The Completely Randomized Design
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10.2: The Completely Randomized Design
The USGA compares the driving distance of four brands of golf balls. H0: µ1 = µ2 = µ3 = µ4
Ha: The mean distances differ for at least two of the brands = .10 Test Statistic: F = MST/MSE
Rejection region: F > 2.25 = F.10 with v1 = 3 and v2 = 36
Source Degrees of Freedom
Sum of Squares
Mean Square
F p-value
Brands 3 2,794.39 931.46 43.99 .000
Error 36 762.30 21.18
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10.2: The Completely Randomized Design
The USGA compares the driving distance of four brands of golf balls. H0: µ1 = µ2 = µ3 = µ4
Ha: The mean distances differ for at least two of the brands = .10 Test Statistic: F = MST/MSE
Rejection region: F > 2.25 = F.10 with v1 = 3 and v2 = 36
Source Degrees of Freedom
Sum of Squares
Mean Square
F p-value
Brands 3 2,794.39 931.46 43.99 .000
Error 36 762.30 21.18
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Since the calculated F > 2.25, we reject the null hypothesis.
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10.2: The CompletelyRandomized Design
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If the conditions for ANOVA are not met, a nonparametric procedure is recommended (see Chapter 14).
If the null hypothesis is not rejected, that is not conclusive proof that the treatment means are all equal.
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10.3: Multiple Comparisons of Means
Suppose the ANOVA F-test indicates differences in the means. To determine the differences, we would compare the differences of the means.
With k treatment means, there are
c = k(k – 1)/2
pairs of means to be compared, and each would have a significance level smaller than the overall .
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10.3: Multiple Comparisons of Means To retain the overall confidence level,
various techniques are available for pair wise comparisons: Tukey – treatment sample sizes are
equal Bonferroni - treatment sample sizes may
be unequal Scheffé – general procedure for all linear
combinations of treatment means
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10.3: Multiple Comparisons of Means
Let’s go back to the four brands of golf balls in the previous example: Rank the treatment means with an overall
95% level of confidence using Tukey’s procedure.
Estimate the highest ranked golf ball's mean driving distance.
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10.3: Multiple Comparisons of Means
Brand Comparison 95% Confidence Interval
µA - µB -15.82 < µA - µB < - 4.74
µA - µC -24.71 < µA - µC < -13.63
µA - µD -4.08 < µA - µD < 7.00
µB - µC -14.43 < µB - µC < -3.35
µB - µD 6.2 < µB - µD < 17.28
µC - µD 15.09 < µC - µD < 26.17
Pair wise Comparisons for Four Golf ball BrandsBased on a SAS ANOVA report (see pages 506-7)
Brand C outperforms each of the other brands.
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10.3: Multiple Comparisons of Means To construct a confidence interval on Brand
C, we can use the descriptive statistics from the ANOVA and a straightforward one-sample t-based confidence interval (see section 7.3):
/2, 36
/2
95% sure 1/ ,
where , 10 and 2,
270 (2)(4.60) 1/10 270 2.9
C dfx t s n
s MSE n t
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10.4: The Randomized Block Design
The randomized block design: Blocks (matched sets of experimental
units) are formed. Each of the b blocks has k experimental
units, one for each treatment. One experimental unit from each block is
randomly assigned to each treatment, for a total of n = bk responses.
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10.4: The Randomized Block Design To test the equity of the means, we use the
ratio MST/MSE ~ F
2
1
2 2 2
1 1 1
( )
1 1( )
1 1
( ) ( ) ( )
1
i
i i
k
Ti
n k b
i T Bi i i
b x xSST
MSTk kSSE SS total SST SSB
MSEn b k n b k
x x b x x k x x
n b k
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10.4: The Randomized Block Design
0 1 2
ANOVA -Test to Compare Treatment Means:
Randomized Block Design
:
: At least two treatment means differ
Test Statistic: =
Rejection Region: , with ( -1) numerator and
( -
p
a
F k
H
H
MSTF
MSEF F k
n b
- 1) [= ( -1)( -1)] denominator degrees of freedom.k b k
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10.4: The Randomized Block Design
Conditions required for a valid ANOVA F – Test The b blocks are randomly selected and all k
treatments are applied (in random order) to each block.
The distribution of observations corresponding to all bk block-treatment combinations are approximately normal.
The bk block-treatment distributions have equal variances.
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10.4: The Randomized Block Design
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Completely Randomized Design
Randomized Block Design
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10.4: The Randomized Block Design
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Suppose the golf balls analyzed above are analyzed again using ten real golfers instead of a machine. Each golfer is a block Each brand is a treatment assigned in random
order to each golfer The ten drives for each brand produce the
following means: Brand A Brand B Brand C Brand D
227 yards 233.2 yards 245.3 yards 220.7 yards
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10.4: The Randomized Block Design
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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Source df SS MS F p
Treatment (Brand) 3 2,298.7 1,099.6 54.31 .000
Block (Golfer) 9 12,073.9 1,341.5
Error 27 546.6 20.2
Total 39 15,919.2
ANOVA Table for the Golf Ball Tests
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10.4: The Randomized Block Design
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Equity of Means 95% Confidence Intervals for the Golf Balls’ Distance
µA - µB µA - µC µA - µD µB - µC µB - µD µC - µD
(-11.9,--.4) (-24.0, -2.6) (.6, 12.0) (-17.9, -6.4) (6.7, 18.2) (18.9, 30.3)
None of the confidence intervals contain zero, so we can be 95% certain all of the brand means differ.
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10.4: The Randomized Block Design
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
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Source df SS MS F p
Treatment (Brand) 3 2,298.7 1,099.6 54.31 .000
Block (Golfer) 9 12,073.9 1,341.5
Error 27 546.6 20.2
Total 39 15,919.2
ANOVA Table for the Golf Ball Tests
To test for block mean differences, use the ratio of MSB to MEE
MSB MS(Golfers) 1,341.566.26
MSE MS(Error) 20.2F
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10.5: Factorial Experiments
A complete factorial experiment is a factorial experiment in which every factor-level combination is utilized. That is, the number of treatments in the experiment equals the total number of factor-level combinations.
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10.5: Factorial Experiments
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10.5: Factorial Experiments
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Stage 1 Stage 2
10.5: Factorial Experiments
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10.5: Factorial Experiments Tests Conducted in Analyses of Factorial
Experiments: Completely Randomized Design, r Replicates per Treatment
0
Test for Treatment Means
: No difference among the treatment means
: At least two treatment means differ
MST : =
MSERejection Region : , with ( - 1) numerator
and ( - ) denom
a
H ab
H
Test Statistic F
F F ab
n ab
inator degrees of freedom
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10.5: Factorial Experiments Tests Conducted in Analyses of Factorial
Experiments: Completely Randomized Design, r Replicates per Treatment
0
Test for Factor Interaction
: Factors A and B do not interact to affect the response mean
: Factors A and B do interact to affect the response mean
MS( ) : =
MSERejection Region : ,
a
H
H
ABTest Statistic F
F F with ( - 1)( - 1) numerator and
( - ) denominator degrees of freedom
a b
n ab
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10.5: Factorial Experiments Tests Conducted in Analyses of Factorial
Experiments: Completely Randomized Design, r Replicates per Treatment
0
Test for Main Effect of Factor A
: No difference among the mean levels of factor
: At least two factor mean levels differ
MS( ) : =
MSERejection Region : , with ( - 1) numerat
a
H a A
H A
ATest Statistic F
F F a or and
( - ) denominator degrees of freedomn ab
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10.5: Factorial Experiments Tests Conducted in Analyses of Factorial
Experiments: Completely Randomized Design, r Replicates per Treatment
0
Test for Main Effect of Factor
: No difference among the mean levels of factor
: At least two factor mean levels differ
MS( ) : =
MSERejection Region : , with ( - 1) numerat
a
B
H b B
H B
BTest Statistic F
F F b or and
( - ) denominator degrees of freedomn ab
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10.5: Factorial Experiments Tests Conducted in Analyses of Factorial
Experiments: Completely Randomized Design, r Replicates per Treatment Conditions Required:
Response distribution for each factor-level combination is normal.
Response variance is constant for all treatments. Random and independent samples of
experimental units are associated with each treatment.
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10.5: Factorial Experiments
The four brands of golf balls are tested again, this time with a driver and a 5 iron. Each brand-club combination (eight in all) is assigned randomly to four experimental units in a sequence of swings by Iron Byron. Are the treatment means equal? Do the factors “brand“ and “club” interact?
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10.5: Factorial Experiments
Source df SS MS F
Model 7 33659.81 4808.54 140.35
Brand 1 32,092.11 32,093.11 936.75
Club 3 800.74 266.91 7.79
Interaction 3 765.96 255.32 7.45
Error 24 822.24 34.26
Total 31 34,482.05
TABLE 10.13: ANOVA Summary Table for Example 10.10
0 : The treatment means are equal.
: At least two of the eight means differ.
MST 33,659.81/ 7Test Statistic: 140.35; .0001
MSE 822.24 / 24
a
H
H
F p
![Page 51: Chapter 10: Analysis of Variance: Comparing More Than Two Means](https://reader035.vdocuments.site/reader035/viewer/2022062221/56649f3b5503460f94c59f9e/html5/thumbnails/51.jpg)
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
51
10.5: Factorial Experiments
Source df SS MS F
Model 7 33659.81 4808.54 140.35
Brand 1 32,092.11 32,093.11 936.75
Club 3 800.74 266.91 7.79
Interaction 3 765.96 255.32 7.45
Error 24 822.24 34.26
Total 31 34,482.05
TABLE 10.13: ANOVA Summary Table for Example 10.10
Reject the null hypothesis
0 : The treatment means are equal.
: At least two of the eight means differ.
MST 33,659.81/ 7Test Statistic: 140.35; .0001
MSE 822.24 / 24
a
H
H
F p
![Page 52: Chapter 10: Analysis of Variance: Comparing More Than Two Means](https://reader035.vdocuments.site/reader035/viewer/2022062221/56649f3b5503460f94c59f9e/html5/thumbnails/52.jpg)
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
52
10.5: Factorial Experiments
Source df SS MS F
Model 7 33659.81 4808.54 140.35
Brand 1 32,092.11 32,093.11 936.75
Club 3 800.74 266.91 7.79
Interaction 3 765.96 255.32 7.45
Error 24 822.24 34.26
Total 31 34,482.05
TABLE 10.13: ANOVA Summary Table for Example 10.10
0 : The factors Club and Brand do not interact to affect the mean response.
: The factors Club and Brand interact to affect the mean response
MS(AB) 255.32Test Statistic: 7.45; .0011
MSE 34.26
a
H
H
F p
![Page 53: Chapter 10: Analysis of Variance: Comparing More Than Two Means](https://reader035.vdocuments.site/reader035/viewer/2022062221/56649f3b5503460f94c59f9e/html5/thumbnails/53.jpg)
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
53
10.5: Factorial Experiments
Source df SS MS F
Model 7 33659.81 4808.54 140.35
Brand 1 32,092.11 32,093.11 936.75
Club 3 800.74 266.91 7.79
Interaction 3 765.96 255.32 7.45
Error 24 822.24 34.26
Total 31 34,482.05
TABLE 10.13: ANOVA Summary Table for Example 10.10
0 : The factors Club and Brand do not interact to affect the mean response.
: The factors Club and Brand interact to affect the mean response
MS(AB) 255.32Test Statistic: 7.45; .0011
MSE 34.26
a
H
H
F p
Reject the null hypothesis
![Page 54: Chapter 10: Analysis of Variance: Comparing More Than Two Means](https://reader035.vdocuments.site/reader035/viewer/2022062221/56649f3b5503460f94c59f9e/html5/thumbnails/54.jpg)
Statistics for Business and Economics, 11th ed. Chapter 10: Analysis of Variance
54
10.5: Factorial Experiments
Further analysis (see text) suggests that, although the factor “Club” clearly has an impact on distance, the results for “Brand “ are more ambiguous: Brand B hit with a 5 iron outdistances the others, but not when hit with the driver.