Chapter 13 STATISTICS in PRACTICE
Transcript of Chapter 13 STATISTICS in PRACTICE
111SlideSlideSlide
Chapter 13Chapter 13
STATISTICSSTATISTICS in in PRACTICEPRACTICE� Burke Marketing Services,
Inc., is one of the most
experienced market research
firms in the industry.
� In one study, a firm retained
Burke to evaluate potential new versions of a
children’s dry cereal.
� Analysis of variance was the statistical method used to study the data obtained from the taste tests.
� The experimental design employed by Burke and the subsequent analysis of variance were helpful in making a product design recommendation.
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Chapter 13 Analysis of Variance and Chapter 13 Analysis of Variance and
Experimental DesignExperimental Design
� Introduction to Analysis of Variance
�Analysis of Variance: Testing for the Equality of k Population Means
�Multiple Comparison Procedures
� Factorial Experiments
�An Introduction to Experimental Design
�Completely Randomized Designs
�Randomized Block Design
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IntroductionIntroduction to Analysis of Varianceto Analysis of Variance
Analysis of Variance (ANOVA) can be used to test
for the equality of three or more population means.
Analysis of VarianceAnalysis of Variance (ANOVA) can be used to test(ANOVA) can be used to test
for the equality of three or more population means.for the equality of three or more population means.
Data obtained from observational or experimental
studies can be used for the analysis.
Data obtained from observational or experimentalData obtained from observational or experimental
studies can be used for the analysis.studies can be used for the analysis.
We want to use the sample results to test the
following hypotheses:
We want to use the sample results to test theWe want to use the sample results to test the
following hypotheses:following hypotheses:
HH00: : µµ11 == µµ22 == µµ33 == . . . . . . = = µµkk
HHaa: : Not all population means are equalNot all population means are equal
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Introduction to Analysis of VarianceIntroduction to Analysis of Variance
HH00: : µµ11 == µµ22 == µµ33 == . . . . . . = = µµkk
HHaa: : Not all population means are equalNot all population means are equal
IfH0 is rejected, we cannot conclude that all population
means are different.
IfIf HH00 is rejected, we cannot conclude that all populationis rejected, we cannot conclude that all population
means are different.means are different.
RejectingH0 means that at least two population means
have different values.
RejectingRejecting HH00 means that at least two population meansmeans that at least two population means
have different values.have different values.
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� Sampling Distribution of Given H0 is Truexx
Introduction to Analysis of VarianceIntroduction to Analysis of Variance
µµ 1x1x 3x3x2x2x
Sample means are close togetherSample means are close togetherbecause there is onlybecause there is only
one sampling distributionone sampling distributionwhen when HH00 is true.is true.
22x
n
σσ =
22x
n
σσ =
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Introduction to Analysis of VarianceIntroduction to Analysis of Variance
� Sampling Distribution of Given H0 is Falsexx
µ3µ3 1x1x 2x2x3x3x µ1µ1 µ2µ2
Sample means come fromSample means come fromdifferent sampling distributionsdifferent sampling distributions
and are not as close togetherand are not as close togetherwhen when HH00 is false.is false.
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For each population, the response variable is
normally distributed.
For each population, the response variable isFor each population, the response variable is
normally distributed.normally distributed.
Assumptions for Analysis of VarianceAssumptions for Analysis of Variance
The variance of the response variable, denoted σ 2,is the same for all of the populations.
The variance of the response variable, denotedThe variance of the response variable, denoted σσ 22,,is the same for all of the populationsis the same for all of the populations..
The observations must be independent.The observations must be independent.The observations must be independent.
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Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
� Between-Treatments Estimate of Population
Variance
� Within-Treatments Estimate of Population
Variance
� Comparing the Variance Estimates: The F Test
� ANOVA Table
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Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
� Analysis of variance can be used to test for the
equality of k population means.
� The hypotheses tested is
H0:
Ha: Not all population means are equal
where mean of the jth population.
kµµµ === L21
jµ
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Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
� Sample data
= value of observation i for treatment j
= number of observations for treatment j
= sample mean for treatment j
= sample variance for treatment j
= sample standard deviation for treatment j
ijx
jn
jx2
js
js
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� Statisitcs
� The sample mean for treatment j
� The sample variance for treatment j
Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
j
n
i
ij
jn
x
x
j
∑=
=1
1
)(1
2
2
−
−
=
∑=
j
n
i
jij
jn
xx
s
j
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Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
� The overall sample mean
where nT = n1 + n2 +. . . + nk
� If the size of each sample is n, nT = kn then
T
k
j
n
i
ij
n
x
x
j
∑∑= =
=1 1
kn
x
k
nx
kn
x
x
k
j
ij
k
j
n
i
ij
k
j
n
i
ij
jj
∑∑∑∑∑== == =
===11 11 1
/
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Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
� Between-Treatments Estimate of Population
Variance
� The sum of squares due to treatments (SSTR)
� The mean square due to treatments (MSTR)
1
)(1
2
−
−
=
∑=
k
xxn
MSTR
k
j
jj
∑=
−
k
j
jj xxn1
2)(
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Analysis of Variance:Analysis of Variance:
Testing for the Equality of Testing for the Equality of kk Population MeansPopulation Means
� Within-Treatments Estimate of Population Variance
� The sum of squares due to error (SSE)
� The mean square due to error (MSE)
∑=
−
k
j
jj sn1
2)1(
kn
sn
MSET
k
j
jj
−
−
=
∑=1
2)1(
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BetweenBetween--Treatments EstimateTreatments Estimate
of Population Varianceof Population Variance
� A between-treatment estimate of σσσσ 2 is called the
mean square treatment and is denoted MSTR.
2
1
( )
MSTR1
k
j jj
n x x
k
=
−
=−
∑2
1
( )
MSTR1
k
j jj
n x x
k
=
−
=−
∑
Denominator representsDenominator represents
the the degrees of freedomdegrees of freedom
associated with SSTRassociated with SSTR
Numerator is theNumerator is the
sum of squaressum of squares
due to treatmentsdue to treatments
and is denoted SSTRand is denoted SSTR
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� The estimate of σσσσ 2 based on the variation of the
sample observations within each sample is called
the mean square error and is denoted by MSE.
WithinWithin--Samples EstimateSamples Estimate
of Population Varianceof Population Variance
kn
sn
T
k
j
jj
−
−
=
∑=1
2)1(
MSEkn
sn
T
k
j
jj
−
−
=
∑=1
2)1(
MSE
Denominator representsDenominator represents
the the degrees of freedomdegrees of freedom
associated with SSEassociated with SSE
Numerator is theNumerator is the
sum of squaressum of squares
due to errordue to error
and is denoted SSEand is denoted SSE
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Comparing the Variance Estimates:Comparing the Variance Estimates: TheThe FF TestTest
� 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 σσσσ 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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Test for the Equality ofTest for the Equality of kk Population MeansPopulation Means
FF = MSTR/MSE= MSTR/MSE
HH00: : µµ11 == µµ22 == µµ33 == . . . . . . = = µµkk
HHaa: Not all population means are equal: Not all population means are equal
� Hypotheses
� Test Statistic
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Test for the Equality ofTest for the Equality of k k Population MeansPopulation Means
� Rejection Rule
where the value of Fα α α α
is based on an
F distribution with k - 1 numerator d.f.and nT - k denominator d.f.
Reject Reject HH00 if if pp--valuevalue << ααp-value Approach:
Critical Value Approach: RejectReject HH00 if if FF >> FFαα
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Sampling Distribution of MSTR/MSESampling Distribution of MSTR/MSE
� Rejection Region
Do Not Reject H0Do Not Reject H0
Reject H0Reject H0
MSTR/MSEMSTR/MSE
Critical ValueCritical Value
FαFα
Sampling DistributionSampling Distribution
of MSTR/MSEof MSTR/MSE
αα
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ANOVA TableANOVA Table
SST is partitionedSST is partitionedinto SSTR and SSE.into SSTR and SSE.
SSTSST’’s degrees of freedoms degrees of freedom
(d.f.) are partitioned into(d.f.) are partitioned into
SSTRSSTR’’ss d.f. and d.f. and SSESSE’’ss d.f.d.f.
TreatmentTreatment
ErrorError
TotalTotal
SSTRSSTR
SSESSE
SSTSST
kk –– 11
nnTT –– kk
nnTT -- 11
MSTRMSTR
MSEMSE
Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares
MSTR/MSEMSTR/MSE
FF
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ANOVA TableANOVA Table
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.
SST divided by its degrees of freedomSST divided by its degrees of freedom nnTT –– 1 1 is theis the
overall sample variance that would be obtained if weoverall sample variance that would be obtained if we
treated the entire set of observations as one data set.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:
With the entire data set as one sample, the formulaWith the entire data set as one sample, the formula
for computing the total sum of squares, SST, is:for computing the total sum of squares, SST, is:
2
1 1
SST ( ) SSTR SSEjnk
ijj i
x x= =
= − = +∑∑2
1 1
SST ( ) SSTR SSEjnk
ijj i
x x= =
= − = +∑∑
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ANOVA TableANOVA Table
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.
ANOVA can be viewed as the process of partitioningANOVA can be viewed as the process of partitioning
the total sum of squares and the degrees of freedomthe total sum of squares and the degrees of freedom
into their corresponding sources: treatments and error.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.
Dividing the sum of squares by the appropriate degrees Dividing the sum of squares by the appropriate degrees
of freedom provides the variance estimates and the of freedom provides the variance estimates and the FF
value used to test the hypothesis of equal population value used to test the hypothesis of equal population
means.means.
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Test for the Equality ofTest for the Equality of kk Population MeansPopulation Means
� Example:
� National Computer Products, Inc. (NCP),
manufactures printers and fax machines at plants
located in Atlanta, Dallas, and Seattle.
� Object: To measure how much employees at these
plants know about total quality management.
� A random sample of six employees was selected from
each plant and given a quality awareness examination.
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Test for the Equality ofTest for the Equality of kk Population MeansPopulation Means
� Data
� Let
= mean examination score for population 1
= mean examination score for population 2
= mean examination score for population 3
1µ
2µ
3µ
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TestTest for the Equality of for the Equality of kk Population MeansPopulation Means
� Hypotheses
H0: = =
Ha: Not all population means are equal
� In this example
1. dependent or response variable : examination score
2. independent variable or factor : plant location
3. levels of the factor or treatments : the values of a
factor selected for investigation, in the NCP
example the three treatments or three population
are Atlanta, Dallas, and Seattle.
1µ 2µ3µ
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Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
Three assumptions
1. For each population, the response variable is normally
distributed. The examination scores (response variable)
must be normally distributed at each plant.
2. The variance of the response variable, , is the same for
all of the populations. The variance of examination scores
must be the same for all three plants.
3. The observations must be independent. The examination
score for each employee must be independent of the
examination score for any other employee.
2σ
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Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
� ANOVA Table
� p-value = 0.003 < αααα = .05. We reject H0.
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� Example: Reed Manufacturing
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
Janet Reed would like to know if
there is any significant difference in
the mean number of hours worked per
week for the department managers
at her three manufacturing plants
(in Buffalo, Pittsburgh, and Detroit).
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� Example: Reed Manufacturing
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
A simple random sample of five
managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide.
Conduct an F test using αααα = .05.
313131SlideSlideSlide
11
22
33
44
55
4848
5454
5757
5454
6262
7373
6363
6666
6464
7474
5151
6363
6161
5454
5656
PlantPlant 11BuffaloBuffalo
Plant 2Plant 2PittsburghPittsburgh
PlantPlant 33DetroitDetroitObservationObservation
SampleSample MeanMean
Sample VarianceSample Variance
5555 6868 5757
26.026.0 26.526.5 24.524.5
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
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Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
H0: µµµµ 1 = µµµµ 2 = µµµµ 3
Ha: Not all the means are equal
where:
µµµµ 1 = mean number of hours worked per
week by the managers at Plant 1
µµµµ 2 = mean number of hours worked per
week by the managers at Plant 2
µµµµ 3 = mean number of hours worked per
week by the managers at Plant 3
1. Develop the hypotheses.1. Develop the hypotheses.
� p -Value and Critical Value Approaches
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2. Specify the level of significance.2. Specify the level of significance. α α α α = .05
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
� p -Value and Critical Value Approaches
3. Compute the value of the test statistic.3. Compute the value of the test statistic.
MSTR = 490/(3 - 1) = 245
SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490
(Sample sizes are all equal.)
Mean Square Due to Treatments
= (55 + 68 + 57)/3 = 60x
343434SlideSlideSlide
3. Compute the value of the test statistic.3. Compute the value of the test statistic.
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
MSE = 308/(15 - 3) = 25.667
SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308
Mean Square Due to Error
(continued)
F = MSTR/MSE = 245/25.667 = 9.55
� p -Value and Critical Value Approaches
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TreatmentTreatment
ErrorError
TotalTotal
490490
308308
798798
22
1212
1414
245245
25.66725.667
Source ofSource of
VariationVariation
Sum ofSum of
SquaresSquaresDegrees ofDegrees of
FreedomFreedom
MeanMean
SquaresSquares
9.559.55
FF
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
� ANOVA Table
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Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
5. Determine whether to reject5. Determine whether to reject HH00..
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, the
p-value is less than .01 for F = 9.55.
� p -Value Approach
4. Compute the 4. Compute the pp ––value.value.
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5. Determine whether to reject5. Determine whether to reject HH00..
Because F = 9.55 > 3.89, we reject H0.
� Critical Value Approach
4. Determine the critical value and rejection rule.4. Determine the critical value and rejection rule.
Reject H0 if F > 3.89
Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
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 numerator
d.f. and 12 denominator d.f., F.05 = 3.89.
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Test for the Equality of Test for the Equality of kk Population MeansPopulation Means
� Summary
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Multiple Comparison ProceduresMultiple 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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FisherFisher’’s LSD Procedures LSD Procedure
1 1MSE( )
i j
i j
x xt
n n
−=
+1 1MSE( )
i j
i j
x xt
n n
−=
+
� Test Statistic
� Hypotheses
µ µ−0 : i jH µ µ−0 : i jH
µ µ≠: a i jH µ µ≠: a i jH
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FisherFisher’’s LSD Procedures LSD Procedure
where the value of ta/2 is based on a
t distribution with nT - k degrees of freedom.
� Rejection Rule
Reject Reject HH00 if if pp--value value << aa
p-value Approach:
Critical Value Approach:
Reject Reject HH00 ifif tt < < --ttaa/2 /2 oror tt > > ttaa/2 /2
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� Test Statistic
FisherFisher’’s LSD Procedures LSD Procedure
Based on the Test StatisticBased on the Test Statistic xxii -- xxjj
/21 1LSD MSE( )
i jt n nα
= +/21 1LSD MSE( )
i jt n nα
= +where
−i jx x−i jx x
Reject Reject HH00 if > LSDif > LSD−i jx x−i jx x
�Hypotheses
�Rejection Rule
µ µ−0 : i jH µ µ−0 : i jH
µ µ≠: a i jH µ µ≠: a i jH
434343SlideSlideSlide
FisherFisher’’s LSD Procedures LSD Procedure
Based on the Test StatisticBased on the Test Statistic xxii -- xxjj
� Example: Reed Manufacturing
Recall that Janet Reed wants to know
if there is any significant difference in
the mean number of hours worked per
week for the department managers
at her three manufacturing plants.
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.
444444SlideSlideSlide
For αααα = .05 and nT - k = 15 – 3 = 12
degrees of freedom, t.025 = 2.179
98.6)5
15
1(667.25179.2LSD =+=
/21 1LSD MSE( )
i jt n nα
= +/21 1LSD MSE( )
i jt n nα
= +
MSE value wasMSE value was
computed earliercomputed earlier
FisherFisher’’s LSD Procedures LSD Procedure
Based on the Test StatisticBased on the Test Statistic xxii -- xxjj
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� LSD for Plants 1 and 2
FisherFisher’’s LSD Procedures LSD Procedure
Based on the Test StatisticBased on the Test Statistic xxii -- xxjj
• Conclusion
• Test Statistic
−1 2x x−1 2x x = |55 - 68| = 13
Reject H0 if > 6.98−1 2x x−1 2x x
• Rejection Rule
µ µ−0 1 2: H µ µ−0 1 2: H
µ µ≠1 2: aH µ µ≠1 2: aH
• Hypotheses (A)
The mean number of hours worked at Plant 1 is
not equal to the mean number worked at Plant 2.
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� LSD for Plants 1 and 3
FisherFisher’’s LSD Procedures LSD Procedure
Based on the Test StatisticBased on the Test Statistic xxii -- xxjj
• Conclusion
• Test Statistic
−1 3x x−1 3x x = |55 −−−− 57| = 2
Reject H0 if > 6.98−1 3x x−1 3x x
• Rejection Rule
µ µ−0 1 3: H µ µ−0 1 3: H
µ µ≠1 3: aH µ µ≠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.
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� LSD for Plants 2 and 3
FisherFisher’’s LSD Procedures LSD Procedure
Based on the Test StatisticBased on the Test Statistic xxii -- xxjj
• Conclusion
• Test Statistic
−2 3x x−2 3x x = |68 - 57| = 11
Reject H0 if > 6.98−2 3x x−2 3x x
• Rejection Rule
µ µ−0 2 3: H µ µ−0 2 3: H
µ µ≠2 3: aH µ µ≠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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� The experimentwise Type I error rate gets larger
for problems with more populations (larger k).
Type I Error RatesType I Error Rates
ααEWEW = 1 = 1 –– (1 (1 –– αα))((k k –– 1)!1)!
� The comparisonwise Type I error rate αααα indicates
the level of significance associated with a single
pairwise comparison.
� The experimentwise Type I error rate ααααEW is the
probability of making a Type I error on at least
one of the (k – 1)! pairwise comparisons.
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An Introduction to Experimental DesignAn Introduction to Experimental Design
� 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.
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An Introduction to Experimental DesignAn Introduction to Experimental Design
� 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.
� If the experimental units are heterogeneous,
blocking can be used to form homogeneous groups,
resulting in a randomized block design.
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The between-samples estimate of σ σ σ σ 2 is referred
to as the mean square due to treatments (MSTR).
2
1
( )
MSTR1
k
j jj
n x x
k
=
−
=−
∑2
1
( )
MSTR1
k
j jj
n x x
k
=
−
=−
∑
� Between-Treatments Estimate of Population Variance
Completely Randomized DesignCompletely Randomized Design
denominator is thedenominator is the
degrees of freedomdegrees of freedom
associated with SSTRassociated with SSTR
numerator is callednumerator is called
the the sum of squares duesum of squares due
to treatmentsto treatments (SSTR)(SSTR)
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The second estimate of σ σ σ σ 2, the within-samples estimate, is referred to as the mean square due to error (MSE).
�Within-Treatments Estimate of Population Variance
Completely Randomized DesignCompletely Randomized Design
denominator is thedenominator is thedegrees of freedomdegrees of freedomassociated with SSEassociated with SSE
numerator is callednumerator is calledthe the sum of squaressum of squaresdue to errordue to error (SSE)(SSE)
MSE =
−∑
−
=
( )n s
n k
j jj
k
T
1 2
1MSE =
−∑
−
=
( )n s
n k
j jj
k
T
1 2
1
535353SlideSlideSlide
MSTRSSTR
-=k 1
MSTRSSTR
-=k 1
MSESSE
-=n kT
MSESSE
-=n kT
MSTR
MSE
MSTR
MSE
�ANOVA Table
Completely Randomized DesignCompletely Randomized Design
Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
TreatmentsTreatments
ErrorError
TotalTotal
kk -- 11
nnTT -- 11
SSTRSSTR
SSESSE
SSTSST
nnTT -- kk
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AutoShine, Inc. is considering marketing a long-
lasting car wax. Three different waxes (Type 1, Type 2,
and Type 3) have been developed.
Completely Randomized DesignCompletely Randomized Design
� Example: AutoShine, Inc.
In order to test the durability
of these waxes, 5 new cars were
waxed with Type 1, 5 with Type
2, and 5 with Type 3. Each car was then
repeatedly run through an automatic carwash until the
wax coating showed signs of deterioration.
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Completely Randomized DesignCompletely Randomized Design
The number of times each car went through the
carwash is shown on the next slide. AutoShine,
Inc. must decide which wax to market. Are the
three waxes equally effective?
� Example: AutoShine, Inc.
565656SlideSlideSlide
11
22
33
44
55
2727
3030
2929
2828
3131
3333
2828
3131
3030
3030
2929
2828
3030
3232
3131
Sample MeanSample Mean
Sample VarianceSample Variance
ObservationObservationWaxWaxType 1Type 1
WaxWaxType 2Type 2
WaxWaxType 3Type 3
2.52.5 3.33.3 2.52.5
29.029.0 30.430.4 30.030.0
Completely Randomized DesignCompletely Randomized Design
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�Hypotheses
Completely Randomized DesignCompletely Randomized Design
where:
µµµµ1 = mean number of washes for Type 1 wax
µµµµ2 = mean number of washes for Type 2 wax
µµµµ3 = mean number of washes for Type 3 wax
H0: µµµµ1 = µµµµ2 = µµµµ3
Ha: Not all the means are equal
585858SlideSlideSlide
Because the sample sizes are all equal:
Completely Randomized DesignCompletely Randomized Design
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= + +1 2 3( )/3x x x x = (29 + 30.4 + 30)/3 = 29.8
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�Rejection Rule
Completely Randomized DesignCompletely Randomized Design
where F.05 = 3.89 is based on an F distribution
with 2 numerator degrees of freedom and 12
denominator degrees of freedom
p-Value Approach: Reject H0 if p-value < .05
Critical Value Approach: Reject H0 if F > 3.89
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� Test Statistic
Completely Randomized DesignCompletely Randomized Design
There is insufficient evidence to conclude that
the mean number of washes for the three wax
types are not all the same.
�Conclusion
F = MSTR/MSE = 2.6/2.77 = .939
The p-value is greater than .10, where F = 2.81.
(Excel provides a p-value of .42.)
Therefore, we cannot reject H0.
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Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
TreatmentsTreatments
ErrorError
TotalTotal
22
1414
5.25.2
33.233.2
38.438.4
1212
Completely Randomized DesignCompletely Randomized Design
2.602.60
2.772.77
.939.939
�ANOVA Table
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• 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 DesignRandomized Block Design
SST = SSTR + SSBL + SSESST = 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.
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MSTRSSTR
-=k 1
MSTRSSTR
-=k 1
MSTR
MSE
MSTR
MSE
Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
TreatmentsTreatments
ErrorError
TotalTotal
kk -- 11
nnTT -- 11
SSTRSSTR
SSESSE
SSTSST
Randomized Block DesignRandomized Block Design
�ANOVA Table
BlocksBlocks SSBLSSBL bb -- 11
((k k –– 1)(1)(bb –– 1)1)
=SSBL
MSBL-1b
=SSBL
MSBL-1b
MSESSE
=− −( )( )k b1 1
MSESSE
=− −( )( )k b1 1
646464SlideSlideSlide
Randomized Block DesignRandomized Block Design
� Example: Crescent Oil Co.
Crescent Oil has developed three
new blends of gasoline and must
decide which blend or blends to
produce and distribute. A study
of the miles per gallon ratings of the
three blends is being conducted to determine if the
mean ratings are the same for the three blends.
656565SlideSlideSlide
Randomized Block DesignRandomized Block Design
� Example: Crescent Oil Co.
Five automobiles have been
tested using each of the three
gasoline blends and the miles
per gallon ratings are shown on
the next slide.
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Randomized Block DesignRandomized Block Design
29.8 29.8 28.828.8 28.428.4TreatmentTreatmentMeansMeans
11
22
33
44
55
3131
3030
2929
3333
2626
3030
2929
2929
3131
2525
3030
2929
2828
2929
2626
30.33330.333
29.33329.333
28.66728.667
31.00031.000
25.66725.667
Type of Gasoline (Treatment)Type of Gasoline (Treatment)BlockBlockMeansMeansBlend XBlend X Blend YBlend Y Blend ZBlend Z
AutomobileAutomobile(Block)(Block)
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�Mean Square Due to Error
Randomized Block DesignRandomized Block Design
MSE = 5.47/[(3 - 1)(5 - 1)] = .68
SSE = 62 - 5.2 - 51.33 = 5.47
MSBL = 51.33/(5 - 1) = 12.8
SSBL = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33
MSTR = 5.2/(3 - 1) = 2.6
SSTR = 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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Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
TreatmentsTreatments
ErrorError
TotalTotal
22
1414
5.205.20
5.475.47
62.0062.00
88
2.602.60
.68.68
3.823.82
�ANOVA Table
Randomized Block DesignRandomized Block Design
BlocksBlocks 51.3351.33 12.8012.8044
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�Rejection Rule
Randomized Block DesignRandomized Block Design
For αααα = .05, F.05 = 4.46
(2 d.f. numerator and 8 d.f. denominator)
p-Value Approach: Reject H0 if p-value < .05
Critical Value Approach: Reject H0 if F > 4.46
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�Conclusion
Randomized Block DesignRandomized Block Design
There is insufficient evidence to conclude that
the miles per gallon ratings differ for the three
gasoline 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
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Factorial ExperimentsFactorial Experiments
� 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.
727272SlideSlideSlide
• 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 + SSESST = 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.
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
• We again partition the sum of squares total (SST)
into its sources.
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=SSA
MSA-1a
=SSA
MSA-1a
MSA
MSE
MSA
MSE
Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
Factor AFactor A
ErrorError
TotalTotal
aa -- 11
nnTT -- 11
SSASSA
SSESSE
SSTSST
Factor BFactor B SSBSSB bb -- 11
abab((rr –– 1)1) =−
SSEMSE
( 1)ab r=
−
SSEMSE
( 1)ab r
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
=SSB
MSB-1b
=SSB
MSB-1b
InteractionInteraction SSABSSAB ((a a –– 1)(1)(bb –– 1)1) =− −
SSABMSAB
( 1)( 1)a b=
− −
SSABMSAB
( 1)( 1)a b
MSB
MSE
MSB
MSE
MSAB
MSE
MSAB
MSE
747474SlideSlideSlide
� Step 3 Compute the sum of squares for factor B
2
1 1 1
SST = ( )a b r
ijki j k
x x= = =
−∑∑∑2
1 1 1
SST = ( )a b r
ijki j k
x x= = =
−∑∑∑
2
1
SSA = ( . )a
ii
br x x=
−∑2
1
SSA = ( . )a
ii
br x x=
−∑
2
1
SSB = ( . )b
jj
ar x x=
−∑2
1
SSB = ( . )b
jj
ar x x=
−∑
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
� Step 1 Compute the total sum of squares
� Step 2 Compute the sum of squares for factor A
757575SlideSlideSlide
� Step 4 Compute the sum of squares for interaction
2
1 1
SSAB = ( . . )a b
ij i ji j
r x x x x= =
− − +∑∑2
1 1
SSAB = ( . . )a b
ij i ji j
r x x x x= =
− − +∑∑
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
SSE = SST SSE = SST –– SSA SSA –– SSB SSB -- SSABSSAB
� Step 5 Compute the sum of squares due to error
767676SlideSlideSlide
A survey was conducted of hourly wages
for a sample of workers in two industries
at three locations in Ohio. Part of the
purpose of the survey was to
determine if differences exist
in both industry type and
location. The sample data are shown
on the next slide.
� Example: State of Ohio Wage Survey
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
777777SlideSlideSlide
� Example: State of Ohio Wage Survey
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
12.7012.5012.00II
12.1012.0012.50II
13.0012.6012.40II
12.2012.0012.10I
12.7011.2011.80I
12.9011.8012.10I
ColumbusClevelandCincinnatiIndustry
787878SlideSlideSlide
� Factors
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
• Each experimental condition is repeated 3 times
• Factor B: Location (3 levels)
• Factor A: Industry Type (2 levels)
�Replications
797979SlideSlideSlide
Source ofSource ofVariationVariation
Sum ofSum ofSquaresSquares
Degrees ofDegrees ofFreedomFreedom
MeanMeanSquaresSquares FF
Factor AFactor A
ErrorError
TotalTotal
11
1717
.50.50
1.431.43
3.423.42
1212
.50.50
.12.12
4.194.19
�ANOVA Table
Factor BFactor B 1.121.12 .56.5622
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
InteractionInteraction .37.37 .19.1922
4.694.69
1.551.55
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�Conclusions Using the p-Value Approach
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
(p-values were found using Excel)
Interaction is not significant.
•Interaction: p-value = .25 > αααα = .05
Mean wages differ by location.
•Locations: p-value = .03 < αααα = .05
Mean wages do not differ by industry type.
•Industries: p-value = .06 > αααα = .05
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�Conclusions Using the Critical Value
Approach
TwoTwo--Factor Factorial ExperimentFactor Factorial Experiment
Interaction is not significant.
•Interaction: F = 1.55 < Fαααα
= 3.89
Mean wages differ by location.
•Locations: F = 4.69 > Fαααα= 3.89
Mean wages do not differ by industry type.
•Industries: F = 4.19 < Fαααα
= 4.75