chapter 13 experimental design and analysis of variance

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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



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.




In this chapter three types of experimental designs are introduced.• a completely randomized

design• a randomized block design• a factorial experiment




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.



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



H0: 1=2=3=. . . = k


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.




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




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




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 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



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.



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



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.



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



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



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 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.




Test for the Equality of k Population Means

F = MSTR/MSE

H0: 1=2=3=. . . = k


Hypotheses

Test Statistic



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



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



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.



Factor . . . Car waxTreatments . . . Type I, Type 2, Type 3Experimental units . . . Cars

Response variable . . . Number of washes



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: 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





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





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 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.





Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

TreatmentsError

Total

2

14

5.233.2

38.4

122.602.77

.939

ANOVA Table



p-Value

.42



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.


An Observational Study



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.



Factor . . . Manufacturing plantTreatments . . . Buffalo, Pittsburgh, DetroitExperimental units . . . ManagersResponse variable . . . Number of hours worked



12345

4854575462

7363666474

5163615456

Plant 1Buffalo

Plant 2Pittsburgh

Plant 3DetroitObservation

Sample MeanSample Variance

55 68 5726.0 26.5 24.5





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





2. Specify the level of significance. = .05


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





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






TreatmentErrorTotal

490308798

21214

24525.667

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquare

9.55F

ANOVA Table



p-Value.0033



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.





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.





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.



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




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.



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




LSD for Plants 1 and 2


• 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.





• Conclusion



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.





• Conclusion



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.



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.



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



• 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


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 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



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.



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




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



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


Blocks 51.33 12.804

p-Value

.07



Rejection Rule


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



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 = 3.82 Test Statistic



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.



• 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

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


SSBMSB -1b

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

MSBMSE

MSABMSE

p-Value



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



Step 4 Compute the sum of squares for interaction

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 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




Example: State of Ohio Wage Survey


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



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

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


Interaction .37 .1924.691.55

p-Value

.06

.03

.25



Conclusions Using the Critical Value Approach


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



Conclusions Using the p-Value Approach


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

chapter 13 experimental design and analysis of variance

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observational studies

randomized block design