chapter 13 experimental design and analysis of variance
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1 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13 Experimental Design and Analysis of
Variance An Introduction to Experimental
Design and Analysis of Variance Analysis of Variance and the Completely Randomized Design Multiple Comparison
Procedures
2 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Statistical studies can be classified as being either experimental or observational.
In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. In an observational study, no attempt is made to control the factors.
Cause-and-effect relationships are easier to establish in experimental studies than in observational studies.
An Introduction to Experimental Designand Analysis of Variance
Analysis of variance (ANOVA) can be used to analyze the data obtained from experimental or observational studies.
3 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
An Introduction to Experimental Designand Analysis of Variance
In this chapter three types of experimental designs are introduced.• a completely randomized
design• a randomized block design• a factorial experiment
4 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
An Introduction to Experimental Designand Analysis of Variance
A factor is a variable that the experimenter has selected for investigation.
A treatment is a level of a factor. Experimental units are the objects of interest
in the experiment. A completely randomized design is an
experimental design in which the treatments are randomly assigned to the experimental units.
5 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Analysis of Variance: A Conceptual Overview
Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means.
Data obtained from observational or experimental studies can be used for the analysis.
We want to use the sample results to test the following hypotheses:
H0: 1=2=3=. . . = k
Ha: Not all population means are equal
6 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
H0: 1=2=3=. . . = k
Ha: Not all population means are equal
If H0 is rejected, we cannot conclude that all population means are different.
Rejecting H0 means that at least two population means have different values.
Analysis of Variance: A Conceptual Overview
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or duplicated, or posted to a publicly accessible website, in whole or in part.
For each population, the response (dependent) variable is normally distributed.
The variance of the response variable, denoted 2, is the same for all of the populations.
The observations must be independent.
Assumptions for Analysis of Variance
Analysis of Variance: A Conceptual Overview
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Sampling Distribution of Given H0 is Truex
1x 3x2x
Sample means are close together because there is only
one sampling distribution when H0 is true.
22x n
Analysis of Variance: A Conceptual Overview
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Sampling Distribution of Given H0 is Falsex
3 1x 2x3x 1 2
Sample means come fromdifferent sampling distributionsand are not as close together
when H0 is false.
Analysis of Variance: A Conceptual Overview
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Analysis of Variance andthe Completely Randomized Design
Between-Treatments Estimate of Population Variance Within-Treatments Estimate of Population Variance Comparing the Variance Estimates: The F Test ANOVA Table
11 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
2
1( )
MSTR 1
k
j jj
n x x
k
Between-Treatments Estimateof Population Variance 2
Denominator is thedegrees of freedom
associated with SSTR
Numerator is calledthe sum of squares
dueto treatments (SSTR)
The estimate of 2 based on the variation of the
sample means is called the mean square due to
treatments and is denoted by MSTR.
12 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
The estimate of 2 based on the variation of the
sample observations within each sample is called the mean square error and is denoted by MSE.
Within-Treatments Estimateof Population Variance 2
Denominator is thedegrees of freedom
associated with SSE
Numerator is called
the sum of squares
due to error (SSE)
MSE
( )n s
n k
j jj
k
T
1 21MSE
( )n s
n k
j jj
k
T
1 21
13 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Comparing the Variance Estimates: The F Test
If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with MSTR d.f. equal to k - 1 and MSE d.f. equal to nT - k. If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates 2. Hence, we will reject H0 if the resulting value of MSTR/MSE appears to be too large to have been selected at random from the appropriate F distribution.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Sampling Distribution of MSTR/MSE
Do Not Reject H0
Reject H0
MSTR/MSE
Critical ValueF
Sampling Distributionof MSTR/MSE
Comparing the Variance Estimates: The F Test
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or duplicated, or posted to a publicly accessible website, in whole or in part.
MSTR SSTR-
k 1MSTR SSTR-
k 1MSE SSE
-n kT
MSE SSE-
n kT
MSTRMSE
MSTRMSE
Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquare F
Treatments
Error
Total
k - 1
nT - 1
SSTR
SSE
SST
nT - k
SST is partitionedinto SSTR and
SSE.
SST’s degrees of freedom
(d.f.) are partitioned into
SSTR’s d.f. and SSE’s d.f.
ANOVA Tablefor a Completely Randomized Design
p-Value
16 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
SST divided by its degrees of freedom nT – 1 is the overall sample variance that would be obtained if we treated the entire set of observations as one data set.
With the entire data set as one sample, the formula for computing the total sum of squares, SST, is:
2
1 1SST ( ) SSTR SSE
jnk
ijj i
x x
ANOVA Tablefor a Completely Randomized Design
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or duplicated, or posted to a publicly accessible website, in whole or in part.
ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error.
Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal population means.
ANOVA Tablefor a Completely Randomized Design
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Test for the Equality of k Population Means
F = MSTR/MSE
H0: 1=2=3=. . . = k
Ha: Not all population means are equal
Hypotheses
Test Statistic
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Test for the Equality of k Population Means
Rejection Rule
where the value of F is based on anF distribution with k - 1 numerator d.f.and nT - k denominator d.f.
Reject H0 if p-value < p-value Approach:
Critical Value Approach: Reject H0 if F > F
20 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
AutoShine, Inc. is considering marketing a long-lasting car wax. Three different waxes (Type 1, Type 2,and Type 3) have been developed.
Example: AutoShine, Inc.
In order to test the durability of these waxes, 5 newcars were waxed with Type 1, 5 with Type 2, and 5with Type 3. Each car was then repeatedly runthrough an automatic carwash until the wax coatingshowed signs of deterioration.
Testing for the Equality of k Population Means:
A Completely Randomized Design
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or duplicated, or posted to a publicly accessible website, in whole or in part.
The number of times each car went through thecarwash before its wax deteriorated is shown on thenext slide. AutoShine, Inc. must decide which waxto market. Are the three waxes equally effective?
Example: AutoShine, Inc.
Testing for the Equality of k Population Means:
A Completely Randomized Design
Factor . . . Car waxTreatments . . . Type I, Type 2, Type 3Experimental units . . . Cars
Response variable . . . Number of washes
22 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
12345
2730292831
3328313030
2928303231
Sample MeanSample Variance
ObservationWax
Type 1Wax
Type 2Wax
Type 3
2.5 3.3 2.529.0 30.4 30.0
Testing for the Equality of k Population Means:
A Completely Randomized Design
23 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypotheses
where: 1 = mean number of washes using Type 1 wax2 = mean number of washes using Type 2 wax3 = mean number of washes using Type 3 wax
H0: 1=2=3
Ha: Not all the means are equal
Testing for the Equality of k Population Means:
A Completely Randomized Design
24 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Because the sample sizes are all equal:
MSE = 33.2/(15 - 3) = 2.77
MSTR = 5.2/(3 - 1) = 2.6
SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2
SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2
Mean Square Error
Mean Square Between Treatments
1 2 3( )/ 3x x x x = (29 + 30.4 + 30)/3 = 29.8
Testing for the Equality of k Population Means:
A Completely Randomized Design
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Rejection Rule
where F.05 = 3.89 is based on an F distributionwith 2 numerator degrees of freedom and 12denominator degrees of freedom
p-Value Approach: Reject H0 if p-value < .05Critical Value Approach: Reject H0 if F > 3.89
Testing for the Equality of k Population Means:
A Completely Randomized Design
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Test Statistic
There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same.
ConclusionF = MSTR/MSE = 2.60/2.77 = .939
The p-value is greater than .10, where F = 2.81. (Excel provides a p-value of .42.) Therefore, we cannot reject H0.
Testing for the Equality of k Population Means:
A Completely Randomized Design
27 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquares F
TreatmentsError
Total
2
14
5.233.2
38.4
122.602.77
.939
ANOVA Table
Testing for the Equality of k Population Means:
A Completely Randomized Design
p-Value
.42
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Example: Reed Manufacturing Janet Reed would like to know if there is anysignificant difference in the mean number of hoursworked per week for the department managers at herthree manufacturing plants (in Buffalo, Pittsburgh,and Detroit). An F test will be conducted using = .05.
Testing for the Equality of k Population Means:
An Observational Study
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Example: Reed Manufacturing A simple random sample of five managers fromeach of the three plants was taken and the number ofhours worked by each manager in the previous weekis shown on the next slide.
Testing for the Equality of k Population Means:
An Observational Study
Factor . . . Manufacturing plantTreatments . . . Buffalo, Pittsburgh, DetroitExperimental units . . . ManagersResponse variable . . . Number of hours worked
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or duplicated, or posted to a publicly accessible website, in whole or in part.
12345
4854575462
7363666474
5163615456
Plant 1Buffalo
Plant 2Pittsburgh
Plant 3DetroitObservation
Sample MeanSample Variance
55 68 5726.0 26.5 24.5
Testing for the Equality of k Population Means:
An Observational Study
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or duplicated, or posted to a publicly accessible website, in whole or in part.
H0: 1=2=3
Ha: Not all the means are equalwhere: 1 = mean number of hours worked per
week by the managers at Plant 1 2 = mean number of hours worked per week by the managers at Plant 23 = mean number of hours worked per week by the managers at Plant 3
1. Develop the hypotheses.
p -Value and Critical Value Approaches
Testing for the Equality of k Population Means:
An Observational Study
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or duplicated, or posted to a publicly accessible website, in whole or in part.
2. Specify the level of significance. = .05
p -Value and Critical Value Approaches
3. Compute the value of the test statistic.
MSTR = 490/(3 - 1) = 245SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490
= (55 + 68 + 57)/3 = 60x(Sample sizes are all equal.)
Mean Square Due to Treatments
Testing for the Equality of k Population Means:
An Observational Study
33 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
3. Compute the value of the test statistic.
MSE = 308/(15 - 3) = 25.667SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308Mean Square Due to Error
(con’t.)
F = MSTR/MSE = 245/25.667 = 9.55
p -Value and Critical Value Approaches
Testing for the Equality of k Population Means:
An Observational Study
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or duplicated, or posted to a publicly accessible website, in whole or in part.
TreatmentErrorTotal
490308798
21214
24525.667
Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquare
9.55F
ANOVA Table
Testing for the Equality of k Population Means:
An Observational Study
p-Value.0033
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or duplicated, or posted to a publicly accessible website, in whole or in part.
5. Determine whether to reject H0.
We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.
The p-value < .05, so we reject H0.
With 2 numerator d.f. and 12 denominator d.f.,the p-value is .01 for F = 6.93. Therefore, thep-value is less than .01 for F = 9.55.
p –Value Approach4. Compute the p –value.
Testing for the Equality of k Population Means:
An Observational Study
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or duplicated, or posted to a publicly accessible website, in whole or in part.
5. Determine whether to reject H0.Because F = 9.55 > 3.89, we reject H0.
Critical Value Approach4. Determine the critical value and rejection rule.
Reject H0 if F > 3.89
We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.
Based on an F distribution with 2 numeratord.f. and 12 denominator d.f., F.05 = 3.89.
Testing for the Equality of k Population Means:
An Observational Study
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Multiple Comparison Procedures
Suppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means.
Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Test Statistic
Fisher’s LSD ProcedureBased on the Test Statistic xi - xj
_ _
/ 21 1LSD MSE( )
i jt n n
where
i jx x
Reject H0 if > LSDi jx x
Hypotheses
Rejection Rule
0 : i jH : a i jH
39 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Fisher’s LSD ProcedureBased on the Test Statistic xi - xj
Example: Reed Manufacturing Recall that Janet Reed wants to know if there is anysignificant difference in the mean number of hoursworked per week for the department managers at herthree manufacturing plants. Analysis of variance has provided statisticalevidence to reject the null hypothesis of equalpopulation means. Fisher’s least significant difference(LSD) procedure can be used to determine where thedifferences occur.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
For = .05 and nT - k = 15 – 3 = 12
degrees of freedom, t.025 = 2.179
LSD 2 179 25 667 15 15 6 98. . ( ) .LSD 2 179 25 667 15 15 6 98. . ( ) .
/ 21 1LSD MSE( )
i jt n n
MSE value wascomputed earlier
Fisher’s LSD ProcedureBased on the Test Statistic xi - xj
41 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
LSD for Plants 1 and 2
Fisher’s LSD ProcedureBased on the Test Statistic xi - xj
• Conclusion
• Test Statistic1 2x x = |55 68| = 13
Reject H0 if > 6.981 2x x• Rejection Rule
0 1 2: H 1 2: aH
• Hypotheses (A)
The mean number of hours worked at Plant 1 isnot equal to the mean number worked at Plant 2.
42 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
LSD for Plants 1 and 3
Fisher’s LSD ProcedureBased on the Test Statistic xi - xj
• Conclusion
• Test Statistic1 3x x = |55 57| = 2
Reject H0 if > 6.981 3x x• Rejection Rule
0 1 3: H 1 3: aH
• Hypotheses (B)
There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3.
43 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
LSD for Plants 2 and 3
Fisher’s LSD ProcedureBased on the Test Statistic xi - xj
• Conclusion
• Test Statistic2 3x x = |68 57| = 11
Reject H0 if > 6.982 3x x• Rejection Rule
0 2 3: H 2 3: aH
• Hypotheses (C)
The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
The experiment-wise Type I error rate gets larger for problems with more populations (larger k).
Type I Error Rates
EW = 1 – (1 – )(k – 1)!
The comparison-wise Type I error rate indicates the level of significance associated with a single pairwise comparison.
The experiment-wise Type I error rate EW is the probability of making a Type I error on at least one of the (k – 1)! pairwise comparisons.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Experimental units are the objects of interest in the experiment.
A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.
Randomized Block Design
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or duplicated, or posted to a publicly accessible website, in whole or in part.
• For a randomized block design the sum of squares total (SST) is partitioned into three groups: sum of squares due to treatments, sum of squares due to blocks, and sum of squares due to error.
ANOVA Procedure
Randomized Block Design
SST = SSTR + SSBL + SSE• The total degrees of freedom, nT - 1, are
partitioned such that k - 1 degrees of freedom go to treatments, b - 1 go to blocks, and (k - 1)(b - 1) go to the error term.
47 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
MSTR SSTR-
k 1MSTR SSTR-
k 1MSTRMSE
MSTRMSE
Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquare F
Treatments
Error
Total
k - 1
nT - 1
SSTR
SSE
SST
Randomized Block Design ANOVA Table
Blocks SSBL b - 1
(k – 1)(b – 1)
SSBLMSBL -1b
MSE SSE
( )( )k b1 1MSE SSE ( )( )k b1 1
p-Value
48 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Randomized Block Design Example: Crescent Oil Co.
Crescent Oil has developed three new blends ofgasoline and must decide which blend or blends toproduce and distribute. A study of the miles pergallon ratings of the three blends is being conductedto determine if the mean ratings are the same for thethree blends.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Randomized Block Design Example: Crescent Oil Co.
Five automobiles have been tested using each ofthe three gasoline blends and the miles per gallonratings are shown on the next slide.Factor . . . Gasoline blend
Treatments . . . Blend X, Blend Y, Blend ZBlocks . . . Automobiles
Response variable . . . Miles per gallon
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Randomized Block Design
29.8 28.8 28.4TreatmentMeans
12345
3130293326
3029293125
3029282926
30.33329.33328.66731.00025.667
Type of Gasoline (Treatment) BlockMeansBlend X Blend Y Blend Z
Automobile(Block)
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Mean Square Due to Error
Randomized Block Design
MSE = 5.47/[(3 - 1)(5 - 1)] = .68SSE = 62 - 5.2 - 51.33 = 5.47
MSBL = 51.33/(5 - 1) = 12.8SSBL = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33
MSTR = 5.2/(3 - 1) = 2.6SSTR = 5[(29.8 - 29)2 + (28.8 - 29)2 + (28.4 - 29)2] = 5.2
The overall sample mean is 29. Thus, Mean Square Due to Treatments
Mean Square Due to Blocks
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquare F
Treatments
Error
Total
2
14
5.20
5.47
62.00
8
2.60
.68
3.82
ANOVA Table
Randomized Block Design
Blocks 51.33 12.804
p-Value
.07
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Rejection Rule
Randomized Block Design
For = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator)
p-Value Approach: Reject H0 if p-value < .05Critical Value Approach: Reject H0 if F > 4.46
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Conclusion
Randomized Block Design
There is insufficient evidence to conclude thatthe miles per gallon ratings differ for the threegasoline blends.
The p-value is greater than .05 (where F = 4.46) and less than .10 (where F = 3.11). (Excel provides a p-value of .07). Therefore, we cannot reject H0.
F = MSTR/MSE = 2.6/.68 = 3.82 Test Statistic
55 Slide© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
or duplicated, or posted to a publicly accessible website, in whole or in part.
Factorial Experiment In some experiments we want to draw
conclusions about more than one variable or factor. Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required.
For example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations.
The term factorial is used because the experimental conditions include all possible combinations of the factors.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
• The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment.
ANOVA Procedure
SST = SSA + SSB + SSAB + SSE
• The total degrees of freedom, nT - 1, are partitioned such that (a – 1) d.f go to Factor A, (b – 1) d.f go to Factor B, (a – 1)(b – 1) d.f. go to Interaction, and ab(r – 1) go to Error.
Two-Factor Factorial Experiment
• We again partition the sum of squares total (SST) into its sources.
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or duplicated, or posted to a publicly accessible website, in whole or in part.
SSAMSA -1a
MSAMSE
Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquare F
Factor A
Error
Total
a - 1
nT - 1
SSA
SSE
SST
Factor B SSB b - 1
ab (r – 1)
SSEMSE ( 1)ab r
Two-Factor Factorial Experiment
SSBMSB -1b
Interaction SSAB (a – 1)(b – 1) SSABMSAB ( 1)( 1)a b
MSBMSE
MSABMSE
p-Value
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Step 3 Compute the sum of squares for factor B
2
1 1 1SST = ( )
a b r
ijki j k
x x
2
1SSA = ( . )
a
ii
br x x
2
1SSB = ( . )
b
jj
ar x x
Two-Factor Factorial Experiment Step 1 Compute the total sum of squares
Step 2 Compute the sum of squares for factor A
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Step 4 Compute the sum of squares for interaction
2
1 1SSAB = ( . . )
a b
ij i ji j
r x x x x
Two-Factor Factorial Experiment
SSE = SST – SSA – SSB - SSAB
Step 5 Compute the sum of squares due to error
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A survey was conducted of hourly wages for asample of workers in two industries at three locationsin Ohio. Part of the purpose of the survey was todetermine if differences exist in both industry type and location. The sample data are shown on the next slide.
Example: State of Ohio Wage Survey
Two-Factor Factorial Experiment
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Example: State of Ohio Wage Survey
Two-Factor Factorial Experiment
Industry
Cincinnati Cleveland Columbu
sI $12.10 $11.80 $12.90I 11.80 11.20 12.70I 12.10 12.00 12.20II 12.40 12.60 13.00II 12.50 12.00 12.10II 12.00 12.50 12.70
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Factors
Two-Factor Factorial Experiment
• Each experimental condition is repeated 3 times
• Factor B: Location (3 levels)• Factor A: Industry Type (2 levels)
Replications
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Source ofVariation
Sum ofSquares
Degrees ofFreedom
MeanSquare F
Factor A
Error
Total
1
17
.50
1.43
3.42
12
.50
.12
4.19
ANOVA Table
Factor B 1.12 .562
Two-Factor Factorial Experiment
Interaction .37 .1924.691.55
p-Value
.06
.03
.25
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Conclusions Using the Critical Value Approach
Two-Factor Factorial Experiment
Interaction is not significant.• Interaction: F = 1.55 < F =
3.89
Mean wages differ by location.• Locations: F = 4.69 > F = 3.89
Mean wages do not differ by industry type.• Industries: F = 4.19 < F = 4.75
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Conclusions Using the p-Value Approach
Two-Factor Factorial Experiment
Interaction is not significant.• Interaction: p-value = .25 > = .05
Mean wages differ by location.• Locations: p-value = .03 < = .05
Mean wages do not differ by industry type.• Industries: p-value = .06 > = .05
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