© 2003 prentice-hall, inc.chap 11-1 analysis of variance & post-anova analysis ie 340/440...

© 2003 Prentice-Hall, Inc. Chap 11-1

Analysis of Variance&

Post-ANOVA ANALYSIS

IE 340/440

PROCESS IMPROVEMENT THROUGH PLANNED EXPERIMENTATION

Dr. Xueping LiUniversity of Tennessee


What If There Are More Than Two 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


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


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


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

In general, there will be a levels of the factor, or a treatments, and n replicates of the experiment, run in random order…a completely randomized design (CRD)

N = an total runs We consider the fixed effects case…the random effects

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

treatment means


Models for the Data

There are several ways to write a model for the 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


The Analysis of Variance

The name “analysis of variance” stems from a partitioning of the total variability in the response variable into components that are consistent with a model 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 of squares:

The basic ANOVA partitioning is:

2..

1 1

( )a n

T iji j

SS y y

2 2.. . .. .

1 1 1 1

2 2. .. .

1 1 1

( ) [( ) ( )]

( ) ( )

a n a n

ij i ij ii j i j

a a n

i ij ii i j

T Treatments E

y y y y y y

n y y y y

SS SS SS



A large value of SSTreatments reflects large differences in treatment means

A small value of SSTreatments likely indicates no differences in treatment means

Formal statistical hypotheses are:

T Treatments ESS SS SS

0 1 2

1

:

: At least one mean is different aH

H

Chap 11-10


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

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

If the treatment means are equal, the treatment and error mean 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 ETreatments E

df df df

an a a n

SS SSMS MS

a a n


The Analysis of Variance is Summarized in a Table

Computing…see 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 ANOVA

F Statistic The F Statistic is the Ratio of the Among

Estimate of Variance and the Within Estimate of 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 is True


Features of One-Way ANOVA

F 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 Å/min) from the Plasma Etching Experiment)


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


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

Coding the observations

More about manual calculation p.70-71

Chap 11-18

Graphical View of the ResultsDESIGN-EXPERT Plot

Strength X = A: Cotton Weight %

Design Points

A: Cotton Weight %

Str

en

gth

One Factor Plot

15 20 25 30 35

7

11.5

16

20.5

25

22

22

22 22

22 22

22


Model Adequacy Checking in the ANOVA

Text 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

some of these assumptions are violated


Model Adequacy Checking in the ANOVA

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

Design-Expert generates the residuals

Residual plots are very useful

Normal probability plot of residuals

.

ˆij ij ij

ij i

e y y

y y


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


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


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


Figure 3.6 (p. 79)Plot of residuals versus fitted values.


Other Important Residual PlotsDESIGN-EXPERT PlotStrength

22

22

22

22

22

22

22

Predicted

Re

sid

ua

ls

Residuals vs. Predicted

-3.8

-1.55

0.7

2.95

5.2

9.80 12.75 15.70 18.65 21.60

DESIGN-EXPERT PlotStrength

Run Number

Re

sid

ua

ls

Residuals vs. Run

-3.8

-1.55

0.7

2.95

5.2

1 4 7 10 13 16 19 22 25


Post-ANOVA Comparison of Means

The analysis of variance tests the hypothesis of equal treatment means

Assume that residual analysis is satisfactory If that hypothesis is rejected, we don’t know which

specific means are different Determining which specific means differ following an

ANOVA is called the multiple comparisons problem There are lots of ways to do this…see text, Section 3-5,

pg. 86 We will use pairwise t-tests on means…sometimes called

Fisher’s Least Significant Difference (or Fisher’s LSD) Method


Tukey’s 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),(

Tyy ji ..


The Tukey-Kramer Procedure

Tells which Population Means are Significantly Different E.g., 1 = 2 3

2 groups whose means may be significantly different

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


The Tukey-Kramer Procedure: Example

1. Compute absolute mean differences:Machine1 Machine2 Machine3

25.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 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

2. Compute critical range:

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

( , )'

1 1Critical Range 1.618

2U c n cj j

MSWQ

n n


Fisher’s LSD

H0: Mu_i = Mu_j Least Significant Difference

Whether Where

Eji

aN MSnn

tLSD )11

(,2/

n

MStLSD E

aN

2,2/

LSDyy ji ..


Design-Expert Output

Treatment Means (Adjusted, If Necessary)Estimated StandardMean 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 H0Treatment 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 0.2347 2 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


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


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


Graphical Comparison of MeansText, pg. 89


For the Case of Quantitative Factors, a Regression Model is often

Useful

Response:Strength ANOVA for Response Surface Cubic ModelAnalysis of variance table [Partial sum of squares]

Sum of Mean FSource Squares DF Square Value Prob > FModel 441.81 3 147.27 15.85 < 0.0001A 90.84 1 90.84 9.78 0.0051A2 343.21 1 343.21 36.93 < 0.0001A3 64.98 1 64.98 6.99 0.0152Residual 195.15 21 9.29Lack of Fit 33.95 1 33.95 4.21 0.0535Pure Error 161.20 20 8.06Cor Total 636.96 24

Coefficient Standard 95% CI 95% CIFactor 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

Chap 11-37

The Regression Model

Final Equation in Terms of Actual Factors:

Strength = +62.61143-9.01143* Cotton Weight % +0.48143 * Cotton Weight %^2 -7.60000E-003 * Cotton Weight %^3

This is an empirical model of the experimental results

DESIGN-EXPERT Plot

Strength

X = A: Cotton Weight %

Design Points

15.00 20.00 25.00 30.00 35.00

7

11.5

16

20.5

25

A: Cotton Weight %

Str

en

gth

One Factor Plot

22

22

22 22

22 22

22


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


Table 3.9 (p. 83)Variance-Stabilizing Transformations


Figure 3.8 (p. 84)Plot of log Si versus logfor the peak discharge data from Example 3.5.

.iy


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


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


Display on page 103 y


Example 3-1 p70 EX3-1 p112


One-Way ANOVA F Test Example

As production manager, you want to see if 3 filling machines have different mean filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the .05 significance level, is there a difference in mean filling times?

Machine1 Machine2 Machine325.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 22.75 20.6025.10 21.60 20.40


One-Way ANOVA Example: Scatter Diagram

27

26

25

24

23

22

21

20

19

••

•••

•••••

••••

•

Time in SecondsMachine1 Machine2 Machine325.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 22.75 20.6025.10 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 Computations

Machine1 Machine2 Machine3

25.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 22.75 20.6025.10 21.60 20.40

2 2 25 24.93 22.71 22.61 22.71 20.59 22.71

47.164

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


Summary Table

Source of

Variation

Degrees of

Freedom

Sum ofSquares

Mean Squares

(Variance)

FStatistic

Among(Factor)

3-1=2 47.1640 23.5820MSA/MSW

=25.60

Within(Error)

15-3=12

11.0532 .9211

Total15-

1=1458.2172


One-Way ANOVA Example Solution

F0 3.89

H0: 1 = 2 = 3

H1: Not All Equal = .05df1= 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

921125 6

.

..


Solution in Excel

Use Tools | Data Analysis | ANOVA: Single Factor

Excel Worksheet that Performs the One-Factor ANOVA of the Example

Microsoft Excel Worksheet



Tells which Population Means are Significantly Different E.g., 1 = 2 3

2 groups whose means may be significantly different

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




25.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 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 than the critical range. There is a significant difference between each pair of means at the 5% level of significance.

( , )'

1 1Critical Range 1.618

2U c n cj j

MSWQ

n n


Solution in PHStat

Use PHStat | c-Sample Tests | Tukey-Kramer Procedure …

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



Levene’s Test for Homogeneity of Variance

The Null Hypothesis The c population variances are all equal

The Alternative Hypothesis Not all the c population variances are equal

2 2 20 1 2: cH

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


Levene’s Test for Homogeneity of Variance:

Procedure

1. For each observation in each group, obtain the absolute value of the difference between each observation and the median of the group.

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


Levene’s Test for Homogeneity of Variances:

Example

As production manager, you want to see if 3 filling machines have different variance in filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the .05 significance level, is there a difference in the variance in filling times?



Levene’s Test: Absolute Difference from the

Median

Machine1 Machine2 Machine3 Machine1 Machine2 Machine325.4 23.4 20 0.3 0.65 0.4

26.31 21.8 22.2 1.21 0.95 1.824.1 23.5 19.75 1 0.75 0.65

23.74 22.75 20.6 1.36 0 0.225.1 21.6 20.4 0 1.15 0

median 25.1 22.75 20.4

abs(Time - median(Time))Time


Summary Table

SUMMARYGroups Count Sum Average Variance

Machine1 5 3.87 0.774 0.35208Machine2 5 3.5 0.7 0.19Machine3 5 3.05 0.61 0.5005

ANOVASource 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


Levene’s Test Example:Solution

F0 3.89

H0:


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 21 2 3

0.03370.0970

0.3475

MSAF

MSW

2j


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 the effect of block before testing)

Response of Each Treatment Group is Obtained

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

blocks


Randomized Blocked Design(Example)

Factor (Training Method)

Factor Levels

(Groups)

BlockedExperiment

Units

Dependent Variable

(Response)

21 hrs 17 hrs 31 hrs




Randomized Block Design(Partition of Total Variation)

Variation Due to Group

SSA

Variation Due to Group

SSA

Variation Among Blocks SSBL

Variation Among Blocks SSBL

Variation Among All

Observations

SST

Variation Among All

Observations

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

Unexplained

Commonly referred to as:

Sum of Squares Among

Among Groups Variation

=

+

+Variation Due

to Random Sampling

SSW

Variation Due to Random Sampling

SSW

Commonly referred to as:

Sum of Squares Among Block


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

ijj 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


Among-Group Variation

2

1

1 (treatment group means)

1

1

c

jj

r

iji

j

SSA r X X

XX

rdf c

SSAMSA

c


Among-Block Variation

2

1

1 (block means)

1

1

r

ii

c

ijj

i

SSBL c X X

X

Xc

df r

SSBLMSBL

r


Random Error

2

1 1

1 1

1 1

c r

ij i jj i

SSE X X X X

df r c

SSEMSE

r c


Randomized Block F Test for Differences 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

UF F

Reject


Summary Table

Source of

Variation

Degrees of

Freedom

Sum ofSquare

s

Mean Squares

FStatisti

c

AmongGroup

c – 1 SSAMSA =

SSA/(c – 1)MSA/MSE

AmongBlock

r – 1 SSBLMSBL =

SSBL/(r – 1)MSBL/MSE

Error (r – 1) c – 1)

SSEMSE =

SSE/[(r – 1)(c– 1)]

Total rc – 1 SST


Randomized Block Design: Example

As production manager, you want to see if 3 filling machines have different mean filling times. You assign 15 workers with varied experience into 5 groups of 3 based on similarity of their experience, and assigned each group of 3 workers with similar experience to the machines. At the .05 significance level, is there a difference in mean filling times?



Randomized Block Design Example Computation

Machine1 Machine2 Machine3

25.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 22.75 20.6025.10 21.60 20.40

2 2 25 24.93 22.71 22.61 22.71 20.59 22.71

47.164

8.4025

/( -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


Randomized Block Design Example: Summary Table

Source of

Variation

Degrees of

Freedom

Sum ofSquare

s

Mean Square

s

FStatistic

AmongGroup

2SSA=

47.164MSA = 23.582

23.582/1.0503=22.452

AmongBlock

4SSBL=2.6507

MSBL =.6627

.6627/1.0503=.6039

Error SSE=8.4025

MSE = 1.0503

Total 14SST=

58.2172


Randomized Block Design Example: Solution

F0 4.46

H0: 1 = 2 = 3


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

1.050322.45

.


Randomized Block Designin Excel

Tools | Data Analysis | ANOVA: Two Factor Without Replication

Example Solution in Excel Spreadsheet




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

Critical Range

( , 1 1 )Critical Range U c r c

MSEQ

r




25.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 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. There is a significance difference between each pair of means at 5% level of significance.

( , 1 1 )

1.0503Critical Range 4.04 1.8516

5U c r c

MSEQ

r


The Tukey-Kramer Procedurein PHStat

PHStat | c-Sample Tests | Tukey-Kramer Procedure …

Example in Excel Spreadsheet



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 is set?


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)


SSE

Two-Way ANOVA Total Variation Partitioning

Variation Due to Factor A

Variation Due to Factor A

Variation Due to Random SamplingVariation Due to

Random Sampling

Variation Due to Interaction

Variation Due to Interaction

SSA

SSABSST

Variation Due to Factor B

Variation Due to Factor B

SSB

Total VariationTotal 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 Total 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


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

ijki j k

r c n r c n

ijk ijki j k i j k

SST X X

X X

Xrcn n


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


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


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


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

ijijki j k

SSE X X


Two-Way ANOVA:The F Test Statistic

F Test for Factor B Main Effect

F Test 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 = ••• = c

H1: Not all . j are equal

Reject if F > FU

Reject if F > FU

Reject if F > FU

1

MSA SSAF MSA

MSE r

F Test for Factor A Main Effect

1

MSB SSBF MSB

MSE c

1 1

MSAB SSABF MSAB

MSE r c


Two-Way ANOVASummary Table

Source ofVariation

Degrees of Freedom

Sum ofSquare

s

Mean Squares

FStatisti

c

Factor A(Row)

r – 1 SSAMSA =

SSA/(r – 1)MSA/MSE

Factor B(Column)

c – 1 SSBMSB =

SSB/(c – 1)MSB/MSE

AB(Interaction)

(r – 1)(c – 1) SSABMSAB =

SSAB/ [(r – 1)(c – 1)]MSAB/MSE

Error rc n’ – 1) SSEMSE =

SSE/[rc n’ – 1)]

Total rc n’ – 1 SST


Features of Two-Way ANOVA

F 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 the Same but the Numerator is Different

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


Kruskal-Wallis Rank Test 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 Randomized

Experimental Designs Use 2 Distribution to Approximate if Each

Sample Group Size nj > 5 df = c – 1


Kruskal-Wallis Rank Test

Assumptions Independent random samples are drawn Continuous dependent variable Data may be ranked both within and among

samples 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


Kruskal-Wallis Rank Test Procedure

Obtain Ranks In event of tie, each of the tied values gets

their average rank Add the Ranks for Data from Each of the

c Groups Square to obtain Tj

2

2

1

123( 1)

( 1)

cj

j j

TH n

n n n

1 2 cn n n n



Compute Test Statistic

# of observation in j –th sample H may be approximated by chi-square

distribution with df = c –1 when each nj >5

(continued)

2

1

123( 1)

( 1)

cj

j j

TH n

n n n

1 2 cn n n n

jn



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

2U



Kruskal-Wallis Rank Test: Example

As production manager, you want to see if 3 filling machines have different median filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the .05 significance level, is there a difference in median filling times?


Machine1 Machine2 Machine3 14 9 2 15 6 7 12 10 1 11 8 4 13 5 3

Example Solution: Step 1 Obtaining a Ranking

Raw Data Ranks

65 38 17



Example Solution: Step 2 Test Statistic Computation

212

3( 1)( 1) 1

2 2 212 65 38 173(15 1)

15(15 1) 5 5 5

11.58

Tc jH n

n n nj j


Kruskal-Wallis Test Example Solution

H0: M1 = M2 = M3

H1: Not all equal = .05df = 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

0 5.991


Kruskal-Wallis Test in PHStat

PHStat | c-Sample Tests | Kruskal-Wallis Rank Sum Test …

Example Solution in Excel Spreadsheet



Friedman Rank Test for 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 Use 2 Distribution to Approximate if the

Number of Blocks r > 5 df = c – 1


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:Procedure

Replace the c observations by their ranks in each of the r blocks; assign average rank for ties

Test statistic:

R.j2 is the square of the rank total for group j

FR can be approximated by a chi-square distribution with (c –1) degrees of freedom

The rejection region is in the right tail

2

1

123 1

1

c

R jj

F R r crc c


Friedman Rank Test: Example

As production manager, you want to see if 3 filling machines have different median filling times. You assign 15 workers with varied experience into 5 groups of 3 based on similarity of their experience, and assigned each group of 3 workers with similar experience to the machines. At the .05 significance level, is there a difference in median filling times?



Timing RankMachine 1 Machine 2 Machine 3 Machine 1 Machine 2 Machine 3

25.4 23.4 20 3 2 126.31 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 2 115 9 6

225 81 36

Friedman Rank Test: Computation Table

2. jR. jR

2

1

123 1

1

12 342 3 5 4 8.4

5 3 4

c

R jj

F R r crc c


Friedman Rank Test Example Solution

H0: M1 = M2 = M3

H1: Not all equal = .05df = c - 1 = 3 - 1 = 2Critical Value: Reject at

Test Statistic:

Decision:

Conclusion:There is evidence that population medians are not all equal.

= .05

= .05

FR = 8.4

0 5.991


Chapter Summary

Described the Completely Randomized Design: One-Way Analysis of Variance ANOVA Assumptions F Test for Difference in c Means The Tukey-Kramer Procedure Levene’s Test for Homogeneity of Variance

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


Chapter Summary

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

Discussed Kruskal-Wallis Rank Test for Differences in c Medians

Illustrated Friedman Rank Test for Differences in c Medians

(continued)

© 2003 prentice-hall, inc.chap 11-1 analysis of variance & post-anova analysis ie 340/440...

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