post-anova comparison of means.ppt

7/27/2019 Post-ANOVA Comparison of Means.ppt

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2003 Prentice-Hall, Inc. Chap 11-1

Analysis of Variance&

Post-ANOVA ANALYSIS

IE 340/440

PROCESS IMPROVEMENTTHROUGH PLANNED EXPERIMENTATION

Dr. Xueping LiUniversity of Tennessee


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What If There Are More ThanTwo Factor Levels?

The t-test does not directly apply There are lots of practical situations where there are

either more than two levels of interest, or there are

several factors of simultaneous interest The analysis of variance (ANOVA) is the appropriate

analysis engine for these types of experiments Chapter 3, textbook

The ANOVA was developed by Fisher in the early

1920s, and initially applied to agricultural experiments Used extensively today for industrial experiments


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Figure 3.1 (p. 61)A single-wafer plasma etching tool.


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Table 3.1 (p. 62)Etch Rate Data (in /min) from the Plasma Etching Experiment)


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The Analysis of Variance (Sec. 3-3, pg. 65)

In general, there will be alevels of the factor, or atreatments,andnreplicates of the experiment, run in randomorderacompletely randomized design (CRD)

N = antotal runs We consider the fixed effectscasethe random effects case will

be discussed later Objective is to test hypotheses about the equality of the a

treatment means


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Models for the Data

There are several ways to write a model forthe data:

is called the effects model

Let , then

is called the means model

Regression models can also be employed

ij i ij

i i

ij i ij

y

y


7/107 2003 Prentice-Hall, Inc. Chap 11-7

The Analysis of Variance

The name analysis of variance stems from apartitioning of the total variability in the responsevariable into components that are consistent with amodel for the experiment

The basic single-factor ANOVA model is

2

1, 2,...,,

1,2,...,

an overall mean, treatment effect,

experimental error, (0, )

ij i ij

i

ij

i ay

j n

ith

NID




Total variability is measured by the total sum ofsquares:

The basic ANOVA partitioning is:

2

..

1 1

( )a n

T ij

i j

SS y y

2 2

.. . .. .

1 1 1 1

2 2

. .. .

1 1 1

( ) [( ) ( )]

( ) ( )

a n a n

ij i ij i

i j i j

a a n

i ij i

i i j

T Treatments E

y y y y y y

n y y y y

SS SS SS




A large value ofSSTreatmentsreflects large differences in

treatment means A small value ofSSTreatments likely indicates no differences in

treatment means Formal statistical hypotheses are:

T Treatments E SS SS SS

0 1 2

1

:

: At least one mean is different

aH

H


10/107Chap 11-10


While sums of squares cannot be directly compared to testthe hypothesis of equal means, mean squares can becompared.

A mean square is a sum of squares divided by its degrees offreedom:

If the treatment means are equal, the treatment and errormean squares will be (theoretically) equal.

If treatment means differ, the treatment mean square will be

larger than the error mean square.

1 1 ( 1)

,

1 ( 1)

Total Treatments Error

Treatments E Treatments E

df df df

an a a n

SS SS MS MS

a a n



The Analysis of Variance isSummarized in a Table

Computingsee text, pp 70 73 The reference distribution for F0 is the Fa-1, a(n-1) distribution Reject the null hypothesis (equal treatment means) if

0 , 1, ( 1)a a nF F



Features of One-Way ANOVAF Statistic

The F Statistic is the Ratio of the AmongEstimate of Variance and the Within Estimateof Variance The ratio must always be positive df1 = a-1 will typically be small df2 = N- c will typically be large

The Ratio Should Be Close to 1 if the Null isTrue



Features of One-Way ANOVAF Statistic

If the Null Hypothesis is False The numerator should be greater than the

denominator The ratio should be larger than 1

(continued)



The Reference Distribution:



Table 3.1 (p. 62)Etch Rate Data (in /min) from the Plasma Etching Experiment)



Table 3.4 (p. 71)ANOVA for the Plasma Etching Experiment


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Table 3.5 (p. 72)Coded Etch Rate Data for Example 3.2

Coding the observations

More about manualcalculation p.70-71


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Chap 11-18

Graphical View of the ResultsDESIGN-EXPERT Plot

Strength

X = A: Cotton Weight %

Design Points

A: Cotton Weight %

Strength

One Factor Plot

15 20 25 30 35

7

11.5

16

20.5

25

22

22

22 22

22 22

22


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Model Adequacy Checking in the ANOVAText reference, Section 3-4, pg. 76

Checking assumptions is important Normality

Constant variance Independence Have we fit the right model?

Later we will talk about what to do if someof these assumptions are violated


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Model Adequacy Checking in the ANOVA

Examination ofresiduals(see text, Sec. 3-4, pg.76)

Design-Expert generatesthe residuals

Residual plots are veryuseful

Normal probability plotof residuals

.

ij ij ij

ij i

e y y

y y

Residual

Normal%p

roba

bility

-3.8 -1.55 0.7 2.95 5.2

1

5

10

20

30

50

7080

90

95

99


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Table 3.6 (p. 76)Etch Rate Data and Residuals from Example 3.1a.


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Figure 3.4 (p. 77)Normal probability plot ofresiduals for Example 3-1.


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Figure 3.5 (p. 78)Plot of residuals versus runorder or time.


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Figure 3.6 (p. 79)Plot of residuals versus fittedvalues.


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Other Important Residual Plots

22

22

22

22

22

22

22

Predicted

Residuals

.

-3.8

-1.55

0.7

2.95

5.2

9.80 12.75 15.70 18.65 21.60

Run Number

Residu

als

-3.8

-1.55

0.7

2.95

5.2

1 4 7 10 13 16 19 22 25


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Post-ANOVA Comparison of Means

The analysis of variance tests the hypothesis of equal treatmentmeans

Assume that residual analysis is satisfactory If that hypothesis is rejected, we dont know whichspecific

means are different Determining which specific means differ following an ANOVA is

called the multiple comparisons problem There are lotsof ways to do thissee text, Section 3-5, pg. 86 We will use pairwise t-tests on meanssometimes called Fishers

Least Significant Difference (or Fishers LSD) Method


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Tukeys Test

H0: Mu_i = Mu_j ; H1: Mu_i Mu_j T statistic

Whether Where

f is the DF of MSE a is the number of groups

n

MSfaqT E),(

Tyyji

..


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The Tukey-Kramer Procedure

Tells which Population Means are SignificantlyDifferent E.g., 1=23

2 groups whose meansmay be significantlydifferent

Post Hoc (A Posteriori) Procedure Done after rejection of equal means in ANOVA

Pairwise Comparisons Compare absolute mean differences with critical

range

X

f(X)

1 = 2 3


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The Tukey-Kramer Procedure:Example

1. Compute absolute meandifferences:Machine1Machine2 Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

1 2

1 3

2 3

24.93 22.61 2.32

24.93 20.59 4.34

22.61 20.59 2.02

X X

X X

X X

2. Compute critical range:

3. All of the absolute mean differences are greater than thecritical range. There is a significant difference between

each pair of means at the 5% level of significance.

( , )

'

1 1Critical Range 1.6182

U c n c

j j

MSWQ n n


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Fishers LSD

H0: Mu_i = Mu_j Least Significant Difference

Whether Where

E

ji

aN MSnn

tLSD )11

(,2/

n

MStLSD EaN

2,2/

LSDyyji

..


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Design-Expert Output

Treatment Means (Adjusted, If Necessary)Estimated Standard

Mean Error

1-15 9.80 1.27

2-20 15.40 1.27

3-25 17.60 1.27

4-30 21.60 1.27

5-35 10.80 1.27

Mean Standard t for H0

Treatment Difference DF Error Coeff=0 Prob > |t|

1 vs 2 -5.60 1 1.80 -3.12 0.0054

1 vs 3 -7.80 1 1.80 -4.34 0.0003

1 vs 4 -11.80 1 1.80 -6.57 < 0.0001

1 vs 5 -1.00 1 1.80 -0.56 0.5838

2 vs 3 -2.20 1 1.80 -1.23 0.23472 vs 4 -6.20 1 1.80 -3.45 0.0025

2 vs 5 4.60 1 1.80 2.56 0.0186

3 vs 4 -4.00 1 1.80 -2.23 0.0375

3 vs 5 6.80 1 1.80 3.79 0.0012

4 vs 5 10.80 1 1.80 6.01 < 0.0001

Fi 3 12 ( 99)


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Figure 3.12 (p. 99)Design-Expert computer output forExample 3-1.

Figure 3 13 (p 100)


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Figure 3.13 (p. 100)Minitab computer output forExample 3-1.


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Graphical Comparison of MeansText, pg. 89


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For the Case ofQuantitative Factors, aRegression Model is often Useful

Response:StrengthANOVA for Response Surface Cubic Model

Analysis of variance table [Partial sum of squares]

Sum of Mean F

Source Squares DF Square Value Prob > F

Model 441.81 3 147.27 15.85 < 0.0001

A 90.84 1 90.84 9.78 0.0051

A2 343.21 1 343.21 36.93 < 0.0001A3 64.98 1 64.98 6.99 0.0152

Residual 195.15 21 9.29

Lack of Fit 33.95 1 33.95 4.21 0.0535

Pure Error161.20 20 8.06

Cor Total 636.96 24

Coefficient Standard 95% CI 95% CI

Factor Estimate DF Error Low High VIF

Intercept 19.47 1 0.95 17.49 21.44

A-Cotton % 8.10 1 2.59 2.71 13.49 9.03

A2 -8.86 1 1.46 -11.89 -5.83 1.00

A3 -7.60 1 2.87 -13.58 -1.62 9.03


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Chap 11-37

The Regression Model

Final Equation in Termsof Actual Factors:

Strength = +62.61143

-9.01143* Cotton Weight% +0.48143 * CottonWeight %^2 -7.60000E-003 * Cotton Weight %^3

This is an empirical modelof the experimental results

15.00 20.00 25.00 30.00 35.00

7

11.5

16

20.5

25

A: Cotton Weight %

Strength

22

22

22 22

22 22

22


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Figure 3.7 (p. 83)Plot of residuals versus ij for Example 3-5.


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Table 3.9 (p. 83)Variance-Stabilizing Transformations


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Figure 3.8 (p. 84)Plot of log Siversus logfor the peak discharge datafrom Example 3.5.

.iy

Figure 3 12 (p 99)


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Figure 3.12 (p. 99)Design-Expert computer output forExample 3-1.

Figure 3.13 (p. 100)


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g (p )Minitab computer output forExample 3-1.


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Display on page 103 y


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Example 3-1 p70 EX3-1 p112


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One Way ANOVA F Test


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One-Way ANOVA FTestExample

As production manager, youwant to see if 3 fillingmachines have different mean

filling times. You assign 15similarly trained &experienced workers, 5 permachine, to the machines. At

the .05 significance level, isthere a difference in meanfilling times?

Machine1Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.2024.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

One Wa ANOVA E ample


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One-Way ANOVA Example:Scatter Diagram

27

26

25

24

23

22

21

20

19

Machine1Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

1 2

3

24.93 22.61

20.59 22.71

X X

X X

1X

2X

3X

X

One Way ANOVA Example


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One-Way ANOVA ExampleComputations

Machine1Machine2 Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.6025.10 21.60 20.40

2 2 2

5 24.93 22.71 22.61 22.71 20.59 22.71

47.164

4.2592 3.112 3.682 11.0532

/( -1) 47.16 / 2 23.5820

/( - ) 11.0532 /12 .9211

SSA

SSW

MSA SSA c

MSW SSW n c

1

2

3

24.93

22.61

20.59

22.71

X

X

X

X

5

3

15

jn

c

n


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Summary Table

Source ofVariation

Degreesof

Freedom

Sum ofSquares

MeanSquares

(Variance)

FStatistic

Among(Factor)

3-1=2 47.1640 23.5820MSA/MSW

=25.60

Within

(Error) 15-3=12 11.0532 .9211Total 15-1=14 58.2172

One Way ANOVA Example


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One-Way ANOVA ExampleSolution

F0 3.89

H0: 1=2=3H1: Not All Equal= .05

df1= 2 df2 = 12

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Reject at = 0.05.

There is evidence that atleast one i differs fromthe rest.

= 0.05

FMSA

MSW

23 5820

9211

25 6.

.

.


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Solution in Excel

Use Tools | Data Analysis | ANOVA: SingleFactor

Excel Worksheet that Performs the One-FactorANOVA of the Example


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Tells which Population Means are SignificantlyDifferent E.g., 1=23

2 groups whose meansmay be significantlydifferent

Post Hoc (A Posteriori) Procedure

Done after rejection of equal means in ANOVA Pairwise Comparisons

Compare absolute mean differences with criticalrange

X

f(X)

1 = 2 3

The T ke K ame P oced e


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The Tukey-Kramer Procedure:Example


25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

1 2

1 3

2 3

24.93 22.61 2.32

24.93 20.59 4.3422.61 20.59 2.02

X X

X XX X


3. All of the absolute mean differences are greater than thecritical range. There is a significant difference betweeneach pair of means at the 5% level of significance.

( , )

'

1 1Critical Range 1.6182

U c n c

j j

MSWQ n n


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Solution in PHStat

Use PHStat | c-Sample Tests | Tukey-KramerProcedure

Excel Worksheet that Performs the Tukey-Kramer Procedure for the Previous Example

Levenes Test for


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Levene s Test forHomogeneity of Variance

The Null Hypothesis The cpopulation variances are all equal

The Alternative Hypothesis Not all the cpopulation variances are equal

2 2 2

0 1 2: cH

2

1 : Not all are equal ( 1,2, , )jH j c

Levenes Test for Homogeneity


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Levene s Test for Homogeneityof Variance: Procedure

1. For each observation in each group, obtainthe absolute value of the difference betweeneach observation and the median of thegroup.

2. Perform a one-way analysis of variance onthese absolute differences.

Levenes Test for Homogeneity


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Levene s Test for Homogeneityof Variances: Example

As production manager, youwant to see if 3 fillingmachines have different

variance in filling times. Youassign 15 similarly trained &experienced workers, 5 permachine, to the machines. At

the .05 significance level, isthere a difference in thevariance in filling times?

Machine1Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.2024.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

Levenes Test:


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Levene s Test:Absolute Difference from the Median

Machine1 Machine2 Machine3 Machine1 Machine2 Machine3

25.4 23.4 20 0.3 0.65 0.4

26.31 21.8 22.2 1.21 0.95 1.8

24.1 23.5 19.75 1 0.75 0.65

23.74 22.75 20.6 1.36 0 0.2

25.1 21.6 20.4 0 1.15 0

median 25.1 22.75 20.4

abs(Time - median(Time))Time


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Summary Table

SUMMARY

Groups Count Sum Average Variance

Machine1 5 3.87 0.774 0.35208

Machine2 5 3.5 0.7 0.19

Machine3 5 3.05 0.61 0.5005

ANOVA

Source of Variation SS df MS F P-value F crit

Between Groups 0.067453 2 0.033727 0.097048 0.908218 3.88529Within Groups 4.17032 12 0.347527

Total 4.237773 14

Levenes Test Example:


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Levene s Test Example:Solution

F0 3.89

H0:H1: Not All Equal= .05

df1= 2 df2 = 12

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Do not reject at = 0.05.

There is no evidence that

at least one differs

from the rest.

= 0.05

2 2 2

1 2 3

0.03370.0970

0.3475

MSAF

MSW

2

j


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Randomized Blocked Design Items are Divided into Blocks

Individual items in different samples are matched,or repeated measurements are taken

Reduced within group variation (i.e., remove theeffect of block before testing)

Response of Each Treatment Group isObtained

Assumptions Same as completely randomized design No interaction effect between treatments and

blocks

Randomized Blocked Design


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Randomized Blocked Design(Example)

Factor (Training Method)

Factor Levels(Groups)

BlockedExperiment

Units

DependentVariable

(Response)

21 hrs 17 hrs 31 hrs27 hrs 25 hrs 28 hrs

29 hrs 20 hrs 22 hrs

Randomized Block Design


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Randomized Block Design(Partition of Total Variation)

Variation Dueto Group

SSA

VariationAmong

Blocks

SSBL

Variation

Among All

Observations

SST

Commonly referred to as: Sum of Squares Error Sum of Squares

Unexplained

Commonly referred to as: Sum of Squares AmongAmong Groups Variation

=

+

+Variation Dueto Random

Sampling

SSW

Commonly referred to as: Sum of Squares Among

Block


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Total Variation

2

1

the number of blocks

the number of groups or levels

the total number of observations

the value in the -th block for the -th treatment level

the mean of all val

c r

ij

j i

ij

i

SST X X

r

c

n n rc

X i j

X

ues in block

the mean of all values for treatment level

1

j

i

X j

df n


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Among-Group Variation

2

1

1 (treatment group means)

1

1

c

j

j

r

ij

ij

SSA r X X

XX

r

df cSSA

MSAc


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Among-Block Variation

2

1

1(block means)

1

1

r

i

i

c

ij

j

i

SSBL c X X

X

Xc

df rSSBL

MSBLr


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Random Error

2

1 1

1 1

1 1

c r

ij i j

j i

SSE X X X X

df r c

SSEMSE

r c

Randomized Block F Test for


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Randomized BlockF Test forDifferences in c Means

No treatment effect

Test Statistic

Degrees of Freedom

0 1 2: cH

1 : Not all are equaljH

MSAF

MSE

1 1df c

2 1 1df r c 0

UFF

Reject


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Summary Table

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquares

FStatistic

AmongGroup c1 SSA MSA =SSA/(c1) MSA/MSE

AmongBlock

r1 SSBLMSBL =

SSBL/(r1)

MSBL/MSE

Error (r1)c1) SSEMSE =

SSE/[(r1)(c1)]

Total rc1 SST

Randomized Block Design:


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Randomized Block Design:Example

As production manager, youwant to see if 3 filling machineshave different mean fillingtimes. You assign 15 workers

with varied experience into 5groups of 3 based on similarityof their experience, andassigned each group of 3

workers with similar experienceto the machines. At the .05significance level, is there adifference in mean filling

times?

Machine1Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.2024.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40



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Randomized Block DesignExample Computation

Machine1Machine2 Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.6025.10 21.60 20.40

2 2 25 24.93 22.71 22.61 22.71 20.59 22.71

47.1648.4025

/( -1) 47.16 / 2 23.5820

/ ( -1) 1 8.4025 / 8 1.0503

SSA

SSE

MSA SSA c

MSE SSE r c

1

2

3

24.93

22.61

20.59

22.71

X

X

X

X

5

3

15

r

c

n



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Randomized Block DesignExample: Summary Table

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquares

FStatistic

AmongGroup2 SSA=

47.164MSA =23.582

23.582/1.0503=22.452

AmongBlock

4SSBL=2.6507

MSBL =

.6627

.6627/1.0503

=.6039

Error 8 SSE=8.4025

MSE =

1.0503

Total 14 SST=58.2172



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Randomized Block DesignExample: Solution

F0 4.46

H0: 1=2=3H1: Not All Equal= .05

df1= 2 df2 = 8

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Reject at = 0.05.

There is evidence that atleast one i differs fromthe rest.

= 0.05

FMSA

MSE

23 582

1.0503

22.45.



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Randomized Block Designin Excel

Tools | Data Analysis | ANOVA: Two FactorWithout Replication

Example Solution in Excel Spreadsheet


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Similar to the Tukey-Kramer Procedure for theCompletely Randomized Design Case

Critical Range

( , 1 1 )Critical Range U c r cMSE

Qr

The Tukey-Kramer Procedure:


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The Tukey Kramer Procedure:Example


25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.6025.10 21.60 20.40

1 2

1 3

2 3

24.93 22.61 2.32

24.93 20.59 4.34

22.61 20.59 2.02

X X

X X

X X


3. All of the absolute mean differences are greater. Thereis a significance difference between each pair of means at5% level of significance.

( , 1 1 )

1.0503

Critical Range 4.04 1.85165U c r c

MSE

Q r



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The Tukey Kramer Procedurein PHStat

PHStat | c-Sample Tests | Tukey-KramerProcedure

Example in Excel Spreadsheet


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Two-Way ANOVA

Examines the Effect of: Two factors on the dependent variable

E.g., Percent carbonation and line speed on soft

drink bottling process Interaction between the different levels of these

two factors E.g., Does the effect of one particular percentage of

carbonation depend on which level the line speed isset?


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Two-Way ANOVA

Assumptions Normality

Populations are normally distributed Homogeneity of Variance

Populations have equal variances

Independence of Errors Independent random samples are drawn

(continued)

Two-Way ANOVA


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SSE

o ay OTotal Variation Partitioning

Variation Due to

Factor A

Variation Due to

Random Sampling

Variation Due to

Interaction

SSA

SSABSST

Variation Due toFactor B

SSBTotal Variation

d.f.= n-1

d.f.= r-1

=

+

+d.f.= c-1

+

d.f.= (r-1)(c-1)

d.f.= rc(n-1)

Two-Way ANOVA


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yTotal Variation Partitioning

'

the number of levels of factor A

the number of levels of factor B

the number of values (replications) for each cell

the total number of observations in the experiment

the value of the -th oijk

r

c

n

n

X k

bservation for level of

factor A and level of factor B

i

j


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Total Variation

' '

' 2

1 1 1

1 1 1 1 1 1

'

Sum of Squares Total

= total variation among all

observations around the grand mean

the overall or grand mean

r c n

ijk

i j k

r c n r c n

ijk ijk

i j k i j k

SST X X

X X

Xrcn n


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Factor A Variation

2

'

1

r

i

i

SSA cn X X

Sum of Squares Due to Factor A

= the difference among the various

levels of factor A and the grand

mean


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Factor B Variation

2

'

1

c

j

j

SSB rn X X

Sum of Squares Due to Factor B

= the difference among the various

levels of factor B and the grand mean


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Interaction Variation

2

'

1 1

r c

ij i j

i j

SSAB n X X X X

Sum of Squares Due to Interaction between A and B

= the effect of the combinations of factor A and

factor B


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Random Error

Sum of Squares Error

= the differences among the observations within

each cell and the corresponding cell means

' 2

1 1 1

r c n

ijijk

i j k

SSE X X

Two-Way ANOVA:


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yThe F Test Statistic

FTest for Factor B Main Effect

FTest for Interaction Effect

H0: 1 .= 2 . = = r .H1: Not all i . are equal

H0: ij = 0 (for all i and j)H1: ij 0

H0: 1 = . 2 = = cH1: Not all . j are equal

Reject if

F > FU

Reject if

F > FU

Reject if

F > FU

1

MSA SSAF MSA

MSE r

FTest for Factor A Main Effect

1

MSB SSBF MSB

MSE c

1 1

MSAB SSABF MSAB

MSE r c

Two-Way ANOVA


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ySummary Table

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquares

FStatistic

Factor A

(Row)

r1 SSAMSA =

SSA/(r1)MSA/MSE

Factor B(Column)

c1 SSBMSB =

SSB/(c1)

MSB/MSE

AB

(Interaction)

(r1)(c1) SSABMSAB =

SSAB/ [(r1)(c1)]MSAB/MSE

Error rcn1) SSEMSE =

SSE/[rcn1)]

Total rcn1 SST

Features of Two-Way ANOVA


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Features of Two Way ANOVAF Test

Degrees of Freedom Always Add Up rcn-1=rc(n-1)+(c-1)+(r-1)+(c-1)(r-1) Total=Error+Column+Row+Interaction

The Denominator of the F Test is Always theSame but the Numerator is Different

The Sums of Squares Always Add Up Total=Error+Column+Row+Interaction

Kruskal-Wallis Rank Test


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for c Medians

Extension of Wilcoxon Rank Sum Test Tests the equality of more than 2 (c)

population medians

Distribution-Free Test Procedure

Used to Analyze Completely RandomizedExperimental Designs

Use 2 Distribution to Approximate if EachSample Group Size nj> 5

df= c 1


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Assumptions Independent random samples are drawn Continuous dependent variable Data may be ranked both within and amongsamples Populations have same variability Populations have same shape

Robust with Regard to Last 2 Conditions Use F test in completely randomized designs and

when the more stringent assumptions hold



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Procedure Obtain Ranks

In event of tie, each of the tied values gets theiraverage rank

Add the Ranks for Data from Each of the cGroups Square to obtain Tj2

2

1

123( 1)( 1)

cj

j j

TH nn n n

1 2 cn n n n



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Procedure

Compute Test Statistic

# of observation injth sample Hmay be approximated by chi-square distribution

with df = c1when each nj>5

(continued)

2

1

123( 1)

( 1)

cj

j j

TH n

n n n

1 2 cn n n n

jn



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Procedure

Critical Value for a Given Upper tail

Decision Rule Reject H0: M1= M2= = Mc if test statistic

H> Otherwise, do not reject H0

(continued)

2U

2

U

Kruskal-Wallis Rank Test:


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Machine1Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.7523.74 22.75 20.60

25.10 21.60 20.40

Example

As production manager, youwant to see if 3 filling machineshave different median filling

times. You assign 15 similarlytrained & experienced workers,5 per machine, to themachines. At the .05

significance level, is there adifference in median fillingtimes?

Example Solution: Step 1


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Machine1Machine2

Machine314 9 2

15 6 7

12 10 1

11 8 413 5 3

Obtaining a Ranking

Raw Data Ranks

65 38 17

Machine1Machine2

Machine325.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.6025.10 21.60 20.40

Example Solution: Step 2


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Test Statistic Computation

212

3( 1)

( 1) 1

2 2 212 65 38 17

3(15 1)15(15 1) 5 5 5

11.58

Tc jH n

n n nj j

Kruskal-Wallis Test Example


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Solution

H0: M1 = M2 = M3H1: Not all equal= .05

df= c- 1 = 3 - 1 = 2Critical Value(s): Reject at

Test Statistic:

Decision:

Conclusion:

There is evidence that

population medians are

not all equal.

= .05

= .05.

H= 11.58

05.991

k l ll


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Kruskal-Wallis Test in PHStat

PHStat | c-Sample Tests | Kruskal-Wallis RankSum Test

Example Solution in Excel Spreadsheet

Friedman Rank Test foriff i di


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Differences in c Medians

Tests the equality of more than 2 (c)population medians

Distribution-Free Test Procedure Used to Analyze Randomized Block

Experimental Designs

Use2

Distribution to Approximate if theNumber of Blocks r> 5

df= c 1

F i d R k T


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Friedman Rank Test

Assumptions The r blocks are independent The random variable is continuous The data constitute at least an ordinal scale of

measurement No interaction between the r blocks and the c

treatment levels The c populations have the same variability The c populations have the same shape

Friedman Rank Test:P d


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Procedure

Replace the c observations by their ranks ineach of the r blocks; assign average rankfor ties

Test statistic:

R.j2 is the square of the rank total for groupj

FR can be approximated by a chi-squaredistribution with (c1) degrees of freedom The rejection region is in the right tail

2

1

12 3 11

c

R j

j

F R r crc c

F i d R k T t E l


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Friedman Rank Test: Example

As production manager, youwant to see if 3 fillingmachines have differentmedian filling times. Youassign 15 workers with variedexperience into 5 groups of 3based on similarity of theirexperience, and assigned eachgroup of 3 workers with similar

experience to the machines. Atthe .05 significance level, isthere a difference in medianfilling times?

Machine1Machine2

Machine3

25.40 23.40 20.00

26.31 21.80 22.2024.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

Friedman Rank Test:C t ti T bl


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Timing Rank

Machine 1 Machine 2 Machine 3 Machine 1 Machine 2 Machine 3

25.4 23.4 20 3 2 1

26.31 21.8 22.2 3 1 2

24.1 23.5 19.75 3 2 123.74 22.75 20.6 3 2 1

25.1 21.6 20.4 3 2 1

15 9 6

225 81 36

Computation Table

2

. jR

. jR

2

1

123 1

1

12342 3 5 4 8.4

5 3 4

c

R j

j

F R r crc c

Friedman Rank Test Example


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Solution

H0: M1 = M2 = M3H1: Not all equal= .05

df= c- 1 = 3 - 1 = 2Critical Value: Reject at

Test Statistic:

Decision:

Conclusion:

There is evidence thatpopulation medians are

not all equal.

= .05

= .05

FR= 8.4

0 5.991

Ch t S


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Chapter Summary

Described the Completely RandomizedDesign: One-Way Analysis of Variance ANOVA Assumptions F Test for Difference in c Means The Tukey-Kramer Procedure Levenes Test for Homogeneity of Variance

Discussed the Randomized Block Design F Test for the Difference in c Means The Tukey Procedure

Ch t S


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Chapter Summary

Described the Factorial Design: Two-WayAnalysis of Variance Examine effects of factors and interaction

Discussed Kruskal-Wallis Rank Test forDifferences in c Medians

Illustrated Friedman Rank Test for Differencesin c Medians

(continued)

post-anova comparison of means.ppt

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