© 2005 thomson/south-western slide 1 · 1 © 2005 thomson/south-western slide 1 slides prepared by...

1

1Slide© 2005 Thomson/South-Western

Slides Prepared byJOHN S. LOUCKS

St. Edward’s University


Chapter 13, Part BAnalysis of Variance and Experimental Design

Factorial Experiments

An Introduction to Experimental DesignCompletely Randomized DesignsRandomized Block Design


An 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.

2


An 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.


The between-samples estimate of σ 2 is referredto 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 Design

denominator is thedegrees of freedom

associated with SSTR

numerator is calledthe sum of squares due

to treatments (SSTR)


The second estimate of σ 2, the within-samples estimate, is referred to as the mean square due to error (MSE).

Within-Treatments Estimate of Population Variance


denominator is thedegrees of freedomassociated with SSE

numerator is calledthe sum of squaresdue to error (SSE)

MSE =−∑

−=

( )n s

n k

j jj

k

T

1 2

1MSE =−∑

−=

( )n s

n k

j jj

k

T

1 2

1

3


MSTR SSTR-

=k 1

MSTR SSTR-

=k 1

MSE SSE-

=n kT

MSE SSE-

=n kT

MSTRMSE

MSTRMSE

ANOVA Table


Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Treatments

Error

Total

k - 1

nT - 1

SSTR

SSE

SST

nT - k


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 durabilityof these waxes, 5 new cars werewaxed with Type 1, 5 with Type2, and 5 with Type 3. Each car was thenrepeatedly run through an automatic carwash until thewax coating showed signs of deterioration.



The number of times each car went through thecarwash is shown on the next slide. AutoShine, Inc.must decide which wax to market. Are the threewaxes equally effective?

Example: AutoShine, Inc.

4


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



Hypotheses


where: µ1 = mean number of washes for Type 1 waxµ2 = mean number of washes for Type 2 waxµ3 = mean number of washes for Type 3 wax

H0: µ1 = µ2 = µ3 Ha: Not all the means are equal


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= + +1 2 3( )/3x x x x = (29 + 30.4 + 30)/3 = 29.8

5


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


Test Statistic


There is insufficient evidence to conclude thatthe mean number of washes for the three waxtypes 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.


Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Treatments

Error

Total

2

14

5.2

33.2

38.4

12


2.60

2.77

.939

ANOVA Table

6


• 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.


MSTR SSTR-

=k 1

MSTR SSTR-

=k 1

MSTRMSE

MSTRMSE

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Treatments

Error

Total

k - 1

nT - 1

SSTR

SSE

SST


ANOVA Table

Blocks SSBL b - 1

(k – 1)(b – 1)

=SSBLMSBL

-1b=

SSBLMSBL-1b

MSE SSE=

− −( )( )k b1 1MSE SSE

=− −( )( )k b1 1



Example: Crescent Oil Co.Crescent Oil has developed three

new blends of gasoline and mustdecide which blend or blends toproduce and distribute. A studyof the miles per gallon ratings of thethree blends is being conducted to determine if themean ratings are the same for the three blends.

7



Example: Crescent Oil Co.Five automobiles have been

tested using each of the threegasoline blends and the milesper gallon ratings are shown onthe next slide.



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)


Mean Square Due to Error


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

8


Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Treatments

Error

Total

2

14

5.20

5.47

62.00

8

2.60

.68

3.82

ANOVA Table


Blocks 51.33 12.804


Rejection Rule


For α = .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


Conclusion


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 = 3.82Test Statistic

9


Factorial 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.


• 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.


=SSAMSA

-1a=

SSAMSA-1a

MSAMSEMSAMSE

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Factor A

Error

Total

a - 1

nT - 1

SSA

SSE

SST

Factor B SSB b - 1

ab(r – 1) =−

SSEMSE( 1)ab r

=−

SSEMSE( 1)ab r


=SSBMSB

-1b=

SSBMSB-1b

Interaction SSAB (a – 1)(b – 1) =− −SSABMSAB

( 1)( 1)a b=

− −SSABMSAB

( 1)( 1)a b

MSBMSEMSBMSE

MSABMSE

MSABMSE

10


Step 3 Compute the sum of squares for factor B

2

1 1 1SST = ( )

a b r

ijki j k

x x= = =

−∑∑∑ 2

1 1 1SST = ( )

a b r

ijki j k

x x= = =

−∑∑∑

2

1SSA = ( . )

a

ii

br x x=

−∑ 2

1SSA = ( . )

a

ii

br x x=

−∑

2

1SSB = ( . )

b

jj

ar x x=

−∑ 2

1SSB = ( . )

b

jj

ar x x=

−∑


Step 1 Compute the total sum of squares

Step 2 Compute the sum of squares for factor A


Step 4 Compute the sum of squares for interaction

2

1 1SSAB = ( . . )

a b

ij i ji j

r x x x x= =

− − +∑∑ 2

1 1SSAB = ( . . )

a b

ij i ji j

r x x x x= =

− − +∑∑


SSE = SST – SSA – SSB - SSAB

Step 5 Compute the sum of squares due to error


A survey was conducted of hourly wagesfor a sample of workers in two industriesat three locations in Ohio. Part of thepurpose of the survey was todetermine if differences existin both industry type and

location. The sample data are shownon the next slide.

Example: State of Ohio Wage Survey


11


Example: State of Ohio Wage Survey


12.7012.5012.00II12.1012.0012.50II13.0012.6012.40II12.2012.0012.10I12.7011.2011.80I12.9011.8012.10I

ColumbusClevelandCincinnatiIndustry


Factors


•Each experimental condition is repeated 3 times

•Factor B: Location (3 levels)•Factor A: Industry Type (2 levels)

Replications


Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Factor A

Error

Total

1

17

.50

1.43

3.42

12

.50

.12

4.19

ANOVA Table

Factor B 1.12 .562


Interaction .37 .1924.691.55

12


Conclusions Using the p-Value Approach


(p-values were found using Excel)

Interaction is not significant.•Interaction: p-value = .25 > α = .05

Mean wages differ by location.•Locations: p-value = .03 < α = .05

Mean wages do not differ by industry type.•Industries: p-value = .06 > α = .05


Conclusions Using the Critical Value Approach


Interaction is not significant.•Interaction: F = 1.55 < Fα = 3.89

Mean wages differ by location.•Locations: F = 4.69 > Fα = 3.89

Mean wages do not differ by industry type.•Industries: F = 4.19 < Fα = 4.75


End of Chapter 13, Part B

© 2005 thomson/south-western slide 1 · 1 © 2005 thomson/south-western slide 1 slides prepared by...

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