Chap 11-1 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 11-1 Chapter 11...

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 11-1Chap 11-1

Chapter 11

Analysis of Variance

Basic Business Statistics12th Edition

Chap 11-2Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 11-2

Learning Objectives

In this chapter, you learn: The basic concepts of experimental design How to use one-way analysis of variance to test for differences

among the means of several populations (also referred to as “groups” in this chapter)

To learn the basic structure and use of a randomized block design How to use two-way analysis of variance and interpret the

interaction effect How to perform multiple comparisons in a one-way analysis of

variance and a two-way analysis of variance


Chapter Overview

Analysis of Variance (ANOVA)

F-test

Tukey-Kramer Multiple

Comparisons

One-Way ANOVA

Two-Way ANOVA

InteractionEffects

Randomized Block Design

Tukey Multiple Comparisons

Levene TestFor

Homogeneityof Variance

Tukey Multiple Comparisons

DCOVA


General ANOVA Setting

Investigator controls one or more factors of interest Each factor contains two or more levels Levels can be numerical or categorical Different levels produce different groups Think of each group as a sample from a different

population Observe effects on the dependent variable

Are the groups the same? Experimental design: the plan used to collect the data

DCOVA


Completely Randomized Design

Experimental units (subjects) are assigned randomly to groups Subjects are assumed homogeneous

Only one factor or independent variable With two or more levels

Analyzed by one-factor analysis of variance (ANOVA)

DCOVA


One-Way Analysis of Variance

Evaluate the difference among the means of three or more groups

Examples: Accident rates for 1st, 2nd, and 3rd shift

Expected mileage for five brands of tires

Assumptions Populations are normally distributed Populations have equal variances Samples are randomly and independently drawn

DCOVA


Hypotheses of One-Way ANOVA

All population means are equal i.e., no factor effect (no variation in means among

groups)

At least one population mean is different i.e., there is a factor effect Does not mean that all population means are

different (some pairs may be the same)

c3210 μμμμ:H

same the are means population the of all Not:H1

DCOVA


One-Way ANOVA

The Null Hypothesis is TrueAll Means are the same:

(No Factor Effect)

c3210 μμμμ:H same the are μ all Not:H j1

321 μμμ

DCOVA


One-Way ANOVA

The Null Hypothesis is NOT true At least one of the means is different

(Factor Effect is present)

c3210 μμμμ:H same the are μ all Not:H j1

321 μμμ 321 μμμ

or

(continued)

DCOVA


Partitioning the Variation

Total variation can be split into two parts:

SST = Total Sum of Squares (Total variation)

SSA = Sum of Squares Among Groups (Among-group variation)

SSW = Sum of Squares Within Groups (Within-group variation)

SST = SSA + SSW

DCOVA



Total Variation = the aggregate variation of the individual data values across the various factor levels (SST)

Within-Group Variation = variation that exists among the data values within a particular factor level (SSW)

Among-Group Variation = variation among the factor sample means (SSA)

SST = SSA + SSW

(continued)

DCOVA


Partition of Total Variation

Variation Due to Factor (SSA)

Variation Due to Random Error (SSW)

Total Variation (SST)

= +

DCOVA


Total Sum of Squares

c

j

n

iij

j

XXSST1 1

2)(Where:

SST = Total sum of squares

c = number of groups or levels

nj = number of observations in group j

Xij = ith observation from group j

X = grand mean (mean of all data values)

SST = SSA + SSW

DCOVA


Total Variation

Group 1 Group 2 Group 3

Response, X

X

2212

211 )()()( XXXXXXSST

ccn

(continued)

DCOVA


Among-Group Variation

Where:

SSA = Sum of squares among groups

c = number of groups

nj = sample size from group j

Xj = sample mean from group j


2

1

)( XXnSSA j

c

jj

SST = SSA + SSW

DCOVA



Variation Due to Differences Among Groups

i j

2

1

)( XXnSSA j

c

jj

1

c

SSAMSA

Mean Square Among =

SSA/degrees of freedom

(continued)

DCOVA




Response, X

X1X

2X

2222

211 )()()( XXnXXnXXnSSA cc

(continued)

3X

DCOVA


Within-Group Variation

Where:

SSW = Sum of squares within groups


nj = sample size from group j

Xj = sample mean from group j

Xij = ith observation in group j

2

11

)( jij

n

i

c

j

XXSSWj

SST = SSA + SSW

DCOVA



Summing the variation within each group and then adding over all groups cn

SSWMSW

Mean Square Within =

SSW/degrees of freedom

2

11

)( jij

n

i

c

j

XXSSWj

(continued)

jμ

DCOVA




Response, X

1X2X

3X

22212

2111 )()()( ccn XXXXXXSSW

c

(continued)

DCOVA


Obtaining the Mean Squares

cn

SSWMSW

1

c

SSAMSA

1

n

SSTMST

The Mean Squares are obtained by dividing the various sum of squares by their associated degrees of freedom

Mean Square Among(d.f. = c-1)

Mean Square Within(d.f. = n-c)

Mean Square Total(d.f. = n-1)

DCOVA


One-Way ANOVA Table

Source of Variation

Sum OfSquares

Degrees ofFreedom

Mean Square(Variance)

Among Groups

c - 1 MSA =

Within Groups

SSWn - c MSW =

Total SSTn – 1

SSA

MSA

MSW

F

c = number of groupsn = sum of the sample sizes from all groupsdf = degrees of freedom

SSA

c - 1

SSW

n - c

FSTAT =

DCOVA


One-Way ANOVAF Test Statistic

Test statistic

MSA is mean squares among groups

MSW is mean squares within groups

Degrees of freedom df1 = c – 1 (c = number of groups) df2 = n – c (n = sum of sample sizes from all populations)

MSW

MSAFSTAT

H0: μ1= μ2 = … = μc

H1: At least two population means are different

DCOVA


Interpreting 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 = c -1 will typically be small df2 = n - c will typically be large

Decision Rule: Reject H0 if FSTAT > Fα,

otherwise do not reject H0

0

Reject H0Do not reject H0

Fα

DCOVA


One-Way ANOVA F Test Example

You want to see if when three different golf clubs are used, they hit the ball different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the 0.05 significance level, is there a difference in mean distance?

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204

DCOVA


••••

•

One-Way ANOVA Example: Scatter Plot

270

260

250

240

230

220

210

200

190

••

•••

•••••

Distance

1X

2X

3X

X

227.0 X

205.8 X 226.0X 249.2X 321

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204

Club1 2 3

DCOVA


One-Way ANOVA Example Computations

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204

X1 = 249.2

X2 = 226.0

X3 = 205.8

X = 227.0

n1 = 5

n2 = 5

n3 = 5

n = 15

c = 3

SSA = 5 (249.2 – 227)2 + 5 (226 – 227)2 + 5 (205.8 – 227)2 = 4,716.4

SSW = (254 – 249.2)2 + (263 – 249.2)2 +…+ (204 – 205.8)2 = 1,119.6

MSA = 4,716.4 / (3-1) = 2,358.2

MSW = 1,119.6 / (15-3) = 93.325.275

93.3

2,358.2FSTAT

DCOVA


FSTAT = 25.275

One-Way ANOVA Example Solution

H0: μ1 = μ2 = μ3

H1: μj not all equal

= 0.05

df1= 2 df2 = 12

Test Statistic:

Decision:

Conclusion:Reject H0 at = 0.05

There is evidence that at least one μj differs from the rest

0

= .05

Fα = 3.89Reject H0Do not

reject H0

25.27593.3

2358.2FSTAT

MSW

MSA

Critical Value:

Fα = 3.89

DCOVA


SUMMARY

Groups Count Sum Average Variance

Club 1 5 1246 249.2 108.2

Club 2 5 1130 226 77.5

Club 3 5 1029 205.8 94.2

ANOVA

Source of Variation

SS df MS F P-value F crit

Between Groups

4716.4 2 2358.2 25.275 4.99E-05 3.89

Within Groups

1119.6 12 93.3

Total 5836.0 14

One-Way ANOVA Excel Output DCOVA

Chap 11-30Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

One-Way ANOVA Minitab Output

Chap 11-30

One-way ANOVA: Distance versus Club

Source DF SS MS F PClub 2 4716.4 2358.2 25.28 0.000Error 12 1119.6 93.3Total 14 5836.0

S = 9.659 R-Sq = 80.82% R-Sq(adj) = 77.62%

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev -------+---------+---------+---------+--1 5 249.20 10.40 (-----*-----)2 5 226.00 8.80 (-----*-----)3 5 205.80 9.71 (-----*-----) -------+---------+---------+---------+-- 208 224 240 256Pooled StDev = 9.66

DCOVA


The Tukey-Kramer Procedure

Tells which population means are significantly different e.g.: μ1 = μ2 μ3

Done after rejection of equal means in ANOVA Allows paired comparisons

Compare absolute mean differences with critical range

xμ1 = μ 2

μ3

DCOVA


Tukey-Kramer Critical Range

where:Qα = Upper Tail Critical Value from Studentized

Range Distribution with c and n - c degrees of freedom (see appendix E.7 table)

MSW = Mean Square Within nj and nj’ = Sample sizes from groups j and j’

j'jα nn

MSWQangeCritical R

11

2

DCOVA


The Tukey-Kramer Procedure: Example

1. Compute absolute mean differences:Club 1 Club 2 Club 3

254 234 200263 218 222241 235 197237 227 206251 216 204 20.2205.8226.0xx

43.4205.8249.2xx

23.2226.0249.2xx

32

31

21

2. Find the Qα value from the table in appendix E.7 with c = 3 and (n – c) = (15 – 3) = 12 degrees of freedom:

3.77Q α

DCOVA


The Tukey-Kramer Procedure: Example

5. All of the absolute mean differences are greater than critical range. Therefore there is a significant difference between each pair of means at 5% level of significance. Thus, with 95% confidence we can conclude that the mean distance for club 1 is greater than club 2 and 3, and club 2 is greater than club 3.

285165

1

5

1

2

393773

11

2.

..

nn

MSWQangeCritical R

j'jα

3. Compute Critical Range:

20.2xx

43.4xx

23.2xx

32

31

21

4. Compare:

(continued)

DCOVA


ANOVA Assumptions

Randomness and Independence Select random samples from the c groups (or

randomly assign the levels) Normality

The sample values for each group are from a normal population

Homogeneity of Variance All populations sampled from have the same

variance Can be tested with Levene’s Test

DCOVA


ANOVA AssumptionsLevene’s Test

Tests the assumption that the variances of each population are equal.

First, define the null and alternative hypotheses: H0: σ2

1 = σ22 = …=σ2

c

H1: Not all σ2j are equal

Second, compute the absolute value of the difference between each value and the median of each group.

Third, perform a one-way ANOVA on these absolute differences.

DCOVA


Levene Homogeneity Of Variance Test Example

Calculate Medians

Club 1 Club 2 Club 3

237 216 197

241 218 200

251 227 204 Median

254 234 206

263 235 222

Calculate Absolute Differences

Club 1 Club 2 Club 3

14 11 7

10 9 4

0 0 0

3 7 2

12 8 18

H0: σ21 = σ2

2 = σ23

H1: Not all σ2j are equal

DCOVA


Levene Homogeneity Of Variance Test Example (Excel) (continued)

Anova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Club 1 5 39 7.8 36.2

Club 2 5 35 7 17.5

Club 3 5 31 6.2 50.2

Source of Variation SS df MS FP-

value F crit

Between Groups 6.4 2 3.2 0.092 0.912 3.885

Within Groups 415.6 12 34.6

Total 422 14

Since the p-value is greater than 0.05 there is insufficient evidence of a difference in the variances

DCOVA


Levene Homogeneity Of Variance Test Example (Minitab)

Chap 11-39

(continued)DCOVA

One-way ANOVA: Abs. Diff versus Club

Source DF SS MS F PClub 2 6.4 3.2 0.09 0.912Error 12 415.6 34.6Total 14 422.0

S = 5.885 R-Sq = 1.52% R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ---------+---------+---------+---------+1 5 7.800 6.017 (---------------*----------------)2 5 7.000 4.183 (---------------*---------------)3 5 6.200 7.085 (----------------*---------------) ---------+---------+---------+---------+ 3.5 7.0 10.5 14.0

Pooled StDev = 5.885

Since the p-value is greater than 0.05 there is insufficient evidence of a difference in the variances


The Randomized Block Design

Like One-Way ANOVA, we test for equal population means (for different factor levels, for example)...

...but we want to control for possible variation from a second factor (with two or more levels)

Levels of the secondary factor are called blocks

DCOVA



Total variation can now be split into three parts:

SST = Total variationSSA = Among-Group variationSSBL = Among-Block variationSSE = Random variation

SST = SSA + SSBL + SSE

DCOVA


Sum of Squares for Blocks

Where:


r = number of blocks

Xi. = mean of all values in block i


r

i

i. )XX(cSSBL1

2


DCOVA



Total variation can now be split into three parts:

SST and SSA are computed as they were in One-Way ANOVA


SSE = SST – (SSA + SSBL)

DCOVA


Mean Squares

1cgroups among square Mean

SSAMSA

1rblocking square Mean

SSBLMSBL

)1)(1(error square Mean

cr

SSEMSE

DCOVA


Randomized Block ANOVA Table

Source of Variation

dfSS MS

Among Blocks

SSBL MSBL

Error (r–1)(c-1)SSE MSE

Total rc - 1SST

r - 1 MSBL

MSE

F

c = number of populations rc = total number of observationsr = number of blocks df = degrees of freedom

Among Groups SSA c - 1 MSA

MSA

MSE

DCOVA


Main Factor test: df1 = c – 1

df2 = (r – 1)(c – 1)

MSA

MSE

c..3.2.10 μμμμ:H

equal are means population allNot :H1

FSTAT =

Reject H0 if FSTAT > Fα

Testing For Factor EffectDCOVA


Test For Block Effect

Blocking test: df1 = r – 1

df2 = (r – 1)(c – 1)

MSBL

MSE

r.3.2.1.0 ...:H μμμμ

equal are means block all Not:H1

FSTAT =

Reject H0 if FSTAT > Fα

DCOVA


Randomized Block Design Example

Chap 11-48

RESTAURANTS

RATERS A B C D Totals Means

1 70 61 82 74 287 71.75

2 77 75 88 76 316 79.00

3 76 67 90 80 313 78.25

4 80 63 96 76 315 78.75

5 84 66 92 84 326 81.50

6 78 68 98 86 330 82.50

Totals 465 400 546 476 1,887

Means 77.50 66.67 91.00 79.33 78.625

Ratings at Four Restaurants of a Fast-Food Chain

Raters are the blocksso r = 6.

Restaurants are thegroups of interest soc = 4.

n = rc = 24

1 1 1,88778.625

24

c r

ijj i

X

Xrc

DCOVA


Hypothesis Tests For This Example

Chap 11-49

DCOVA

To decide whether there is a difference in average ratingamong the restaurants:

H0: μA= μB= μC= μD vs H1: At least one of the μ’s is different

To decide whether there is a difference in average ratingamong the raters and the blocking has reduced error:

H0: μ1= μ2= μ3= μ4 = μ5= μ6 vsH1: At least one of the μ’s is different


ANOVA Output From Excel

Chap 11-50

DCOVA

Do the restaurants differ inaverage rating?

Since the p-value (0.0000) <0.05 conclude there is adifference in avg. rating.

Do the raters differ in averagerating?

Since the p-value (0.0205) <0.05 conclude there is adifference in the avg. rating ofraters. This indicates theblocking has reduced error.


ANOVA Output From Minitab

Chap 11-51

DCOVA

Do the restaurants differ inaverage rating?

Since the p-value (0.0000) <0.05 conclude there is adifference in avg. rating.

Do the raters differ in averagerating?

Since the p-value (0.0205) <0.05 conclude there is adifference in the avg. rating ofraters. This indicates theblocking has reduced error.


Factorial Design:Two-Way ANOVA

Examines the effect of Two factors of interest 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 carbonation

level depend on at which level the line speed is set?

DCOVA


Two-Way ANOVA

Assumptions

Populations are normally distributed Populations have equal variances Independent random samples are

drawn

(continued)

DCOVA


Two-Way ANOVA Sources of Variation

Two Factors of interest: A and B

r = number of levels of factor A

c = number of levels of factor B

n’ = number of replications for each cell

n = total number of observations in all cellsn = (r)(c)(n’)

Xijk = value of the kth observation of level i of factor A and level j of factor B

DCOVA


Two-Way ANOVA Sources of Variation

SSTTotal Variation

SSAFactor A Variation

SSBFactor B Variation

SSABVariation due to interaction

between A and B

SSERandom variation (Error)

Degrees of Freedom:

r – 1

c – 1

(r – 1)(c – 1)

rc(n’ – 1)

n - 1

SST = SSA + SSB + SSAB + SSE

(continued)

DCOVA


Two-Way ANOVA Equations

r

i

c

j

n

kijk XXSST

1 1 1

2)(

2

1

.. )( XXncSSAr

i

i

2

1

.. )( XXnrSSBc

j

j

Total Variation:

Factor A Variation:

Factor B Variation:

DCOVA



2r

1i

c

1j

.j.i..ij. )XXXX(n

SSAB

r

i

c

j

n

k

ijijk XXSSE1 1 1

2. )(

Interaction Variation:

Sum of Squares Error:

(continued)

DCOVA



where:Mean Grand

nrc

X

X

r

1i

c

1j

n

1kijk

r) ..., 2, 1, (i A factor of level i of Meannc

X

X th

c

1j

n

1kijk

..i

c) ..., 2, 1, (j B factor of level j of Meannr

XX th

r

1i

n

1kijk

.j.

ij cell of Meann

XX

n

1k

ijk.ij

r = number of levels of factor A

c = number of levels of factor B

n’ = number of replications in each cell

(continued)

DCOVA


Mean Square Calculations

1factor A square Mean

r

SSAMSA

1Bfactor square Mean

c

SSBMSB

)1)(1(ninteractio square Mean

cr

SSABMSAB

)1'(error square Mean

nrc

SSEMSE

DCOVA


Two-Way ANOVA:The F Test Statistics

F Test for Factor B Effect

F Test for Interaction Effect

H0: μ1..= μ2.. = μ3..= • • = µr..

H1: Not all μi.. are equal

H0: the interaction of A and B is equal to zero

H1: interaction of A and B is not zero

F Test for Factor A Effect

H0: μ.1. = μ.2. = μ.3.= • • = µ.c.

H1: Not all μ.j. are equal

Reject H0 if

FSTAT > FαMSE

MSAFSTAT

MSE

MSBFSTAT

MSE

MSABFSTAT

Reject H0 if

FSTAT > Fα

Reject H0 if

FSTAT > Fα

DCOVA


Two-Way ANOVASummary Table

Source ofVariation

Degrees Of

Freedom

Sum of Squares

Mean Squares F

Factor A r – 1 SSAMSA

= SSA /(r – 1)MSAMSE

Factor B c - 1 SSBMSB

= SSB /(c – 1)MSBMSE

AB(Interaction) (r–1)(c-1) SSAB

MSAB= SSAB / (r – 1)(c – 1)

MSABMSE

Error rc(n’ – 1) SSEMSE =

SSE/rc(n’ – 1)

Total n - 1 SST

DCOVA


Features of Two-Way ANOVA F Test

Degrees of freedom always add up n-1 = rc(n’-1) + (r-1) + (c-1) + (r-1)(c-1)

Total = error + factor A + factor B + interaction

The denominators of the F Test are always the same but the numerators are different

The sums of squares always add up SST = SSA + SSB + SSAB + SSE

Total = factor A + factor B + interaction + error

DCOVA


Examples:Interaction vs. No Interaction

No interaction: line segments are parallel

Factor B Level 1

Factor B Level 3

Factor B Level 2

Factor A Levels

Factor B Level 1

Factor B Level 3

Factor B Level 2

Factor A Levels

Mea

n R

espo

nse

Mea

n R

espo

nse

Interaction is present: some line segments not parallel

DCOVA


Multiple Comparisons: The Tukey Procedure

Unless there is a significant interaction, you can determine the levels that are significantly different using the Tukey procedure

Consider all absolute mean differences and compare to the calculated critical range

Example: Absolute differences

for factor A, assuming three levels:

3..2..

3..1..

2..1..

XX

XX

XX

DCOVA


Multiple Comparisons: The Tukey Procedure

Critical Range for Factor A:

(where Qα is from Table E.7 with r and rc(n’–1) d.f.)

Critical Range for Factor B:

(where Qα is from Table E.7 with c and rc(n’–1) d.f.)

n'cRange Critical

MSEQ

n'rRange Critical

MSEQ

DCOVA


Do ACT Prep Course Type & Length Impact Average ACT Scores

Chap 11-66

DCOVA

LENGTH OF COURSE

TYPE OF COURSE Condensed Regular

Traditional 26 18 34 28





Online 27 21 24 21

Online 29 32 16 19

Online 30 20 22 19

Online 24 28 20 24

Online 30 29 23 25

ACT Scores for Different Types and Lengths of Courses


Plotting Cell Means Shows A Strong Interaction

Chap 11-67

DCOVA

Nonparallel lines indicatethe effect of condensingthe course depends onwhether the course istaught in the traditionalclassroom or by onlinedistance learning

The online course yields higher scores when condensed while the traditional course yields higher scores when not condensed (regular).


Excel Analysis Of ACT Prep Course Data

Chap 11-68

DCOVA

The interaction between courselength & type is significantbecause its p-value is 0.0000.

While the p-values associatedwith both course length & course type are not significant,because the interaction issignificant you cannot directlyconclude they have no effect.


Minitab Analysis Of ACT Prep Course Data

Chap 11-69

DCOVAThe interaction between courselength & type is significantbecause its p-value is 0.0000.

While the p-values associatedwith both course length & course type are not significant,because the interaction issignificant you cannot directlyconclude they have no effect.


With The Significant Interaction Collapse The Data Into Four Groups

After collapsing into four groups do a one way ANOVA

The four groups are1. Traditional course condensed

2. Traditional course regular length

3. Online course condensed

4. Online course regular length

Chap 11-70

DCOVA


Excel Analysis Of Collapsed Data

Chap 11-71

DCOVA

Group is a significant effect.p-value of 0.0003 < 0.05

1. Traditional regular > Traditional condensed2. Online condensed > Traditional condensed3. Traditional regular > Online regular4. Online condensed > Online regular

If the course is take online should use thecondensed version and if the course is takenby traditional method should use the regular.


Minitab Analysis Of Collapsed Data Shows Same Conclusions

Chap 11-72

DCOVA


Chapter Summary

Described one-way analysis of variance The logic of ANOVA ANOVA assumptions F test for difference in c means The Tukey-Kramer procedure for multiple comparisons The Levene test for homogeneity of variance

Examined the basic structure and use of a randomized block design

Described two-way analysis of variance Examined effects of multiple factors Examined interaction between factors

Chap 11-1 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 11-1 Chapter 11...

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