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Analysis of Variance

Chapter 15

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15.1 Introduction

Analysis of variance compares two or more

populations of interval data.

Specifically, we are interested in determiningwhether differences exist between the population

means.

The procedure works by analyzing the samplevariance.

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The analysis of variance is a procedure thattests to determine whether differences exitsbetween two or more population means.

To do this, the technique analyzes the sample

variances

15.2 One Way Analysis of

Variance

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Example 15.1 An apple juice manufacturer is planning to develop a new

product -a liquid concentrate. The marketing manager has to decide how to market the

new product.

Three strategies are considered Emphasize convenience of using the product.

Emphasize the quality of the product.

Emphasize the products low price.

One Way Analysis of Variance

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Example 15.1 - continued

An experiment was conducted as follows:

In three cities an advertisement campaign was launched .

In each city only one of the three characteristics

(convenience, quality, and price) was emphasized.

The weekly sales were recorded for twenty weeks

following the beginning of the campaigns.


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Convnce Quality Price529 804 672658 630 531793 774 443514 717 596

663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691

663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532

See file

Xm15 -01

Weekly

sales

Weekly

sales

Weekly

sales
http://d/PP/PPP/Xm15-01.xlshttp://d/PP/PPP/Xm15-01.xlshttp://d/PP/PPP/Xm15-01.xlshttp://d/PP/PPP/Xm15-01.xls

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Solution

The data are interval The problem objective is to compare sales in three

cities.

We hypothesize that the three population means are

equal


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H0: m1 = m2= m3

H1: At least two means differ

To build the statistic needed to test thehypotheses use the following notation:

Solution

Defining the Hypotheses

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Independent samples are drawn from k populations (treatments).

1 2 k

X11

x21...

Xn1,1

1

1

x

n

X12x22...

Xn2,2

2

2

x

n

X1kx2k...

Xnk,k

k

k

x

n

Sample size

Sample mean

First observation,

first sample

Second observation,

second sample

X is the response variable.

The variables value are called responses.

Notation

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Terminology

In the context of this problem

Response variable weekly sales

Responses actual sale valuesExperimental unit weeks in the three cities when we

record sales figures.

Factor the criterion by which we classify the populations

(the treatments). In this problems the factor is the marketing

strategy.

Factor levels the population (treatment) names. In this

problem factor levels are the marketing trategies.

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Two types of variability are employed when

testing for the equality of the population

means

The rationale of the test statistic

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Graphical demonstration:

Employing two types of variability

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20

25

30

1

7

Treatment 1 Treatment 2 Treatment 3

10

12

19

9

Treatment 1 Treatment 2 Treatment 3

20

16

1514

11109

10x1

15x2

20x3

10x1

15x2

20x3

The sample means are the same as before,

but the larger within-sample variability

makes it harder to draw a conclusion

about the population means.

A small variability within

the samples makes it easier

to draw a conclusion about the

population means.

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The rationale behind the test statistic I

If the null hypothesis is true, we would expect all

the sample means to be close to one another

(and as a result, close to the grand mean). If the alternative hypothesis is true, at least some

of the sample means would differ.

Thus, we measure variability between samplemeans.

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The variability between the sample means is

measured as the sum of squared distances

between each mean and the grand mean.

This sum is called the

Sum ofSquares forTreatmentsSST

In our example treatments are

represented by the different

advertising strategies.

Variability between sample means

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2k

1j

jj)xx(nSST

There are k treatments

The size of sample j The mean of sample j

Sum of squares for treatments (SST)

Note: When the sample means are close to

one another, their distance from the grand

mean is small, leading to a small SST. Thus,

large SST indicates large variation between

sample means, which supports H1.

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Solution continued

Calculate SST

2k

1j

jj

321

)xx(nSST

65.608x00.653x577.55x

= 20(577.55 - 613.07)2 ++ 20(653.00 - 613.07)2 +

+ 20(608.65 - 613.07)2 =

= 57,512.23

The grand mean is calculated by

k21

kk2211

n...nn

xn...xnxnX


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Is SST = 57,512.23 large enough to

reject H0

in favor of H1?

See next.


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Large variability within the samples weakens the

ability of the sample means to represent their

corresponding population means. Therefore, even though sample means may

markedly differ from one another, SST must be

judged relative to the within samples variability.

The rationale behind test statistic II

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The variability within samples is measured by

adding all the squared distances between

observations and their sample means.

This sum is called the

Sum of Squares forError

SSEIn our example this is thesum of all squared differences

between sales in city j and the

sample mean of city j (over all

the three cities).

Within samples variability

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Solution continued

Calculate SSE

Sum of squares for errors (SSE)

k

j

jij

n

i

xxSSE

sss

j

1

2

1

23

22

21

)(

24.670,811,238,700.775,10

(n1 - 1)s12 + (n2 -1)s2

2 + (n3 -1)s32

= (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24

= 506,983.50

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Is SST = 57,512.23 large enough

relative to SSE = 506,983.50 to reject

the null hypothesis that specifies that

all the means are equal?

Sum of squares for errors (SSE)

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To perform the test we need to calculate

the mean squaresas follows:

The mean sum of squares

Calculation ofMST -Mean Square forTreatments

12.756,28

13

23.512,57

1

k

SSTMST

Calculation ofMSEMean Square forError

45.894,8

360

50.983,509

kn

SSEMSE

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Calculation of the test statistic

23.3

45.894,812.756,28

MSE

MSTF

with the following degrees of freedom:

v1=k -1 and v2=n-k

Required Conditions:

1. The populations tested

are normally distributed.2. The variances of all the

populations tested are

equal.

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And finally the hypothesis test:

H0: m1 = m2= =mk


Test statistic:

R.R: F>Fa,k-1,n-k

M SE

M STF

The F test rejection region

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The F test

Ho: m1 = m2= m3


Test statistic F= MST MSE= 3.23

15.3FFF:.R.R 360,13,05.0knk a 1

Since 3.23 > 3.15, there is sufficient evidence

to reject Ho in favor of H1,and argue that at least one

of the mean sales is different than the others.

23.3

17.894,8

12.756,28

MSE

MSTF

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-0.02

0

0.02

0.04

0.06

0.08

0.1

0 1 2 3 4

Use Excel to find the p-value

fx Statistical FDIST(3.23,2,57) = .0467

The F test p- value

p Value = P(F>3.23) = .0467

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Excel single factor ANOVA

SS(Total) = SST + SSE

Anova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Convenience 20 11551 577.55 10775.00

Quality 20 13060 653.00 7238.11Price 20 12173 608.65 8670.24

ANOVA

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

Between Groups 57512 2 28756 3.23 0.0468 3.16

Within Groups 506984 57 8894

Total 564496 59

Xm15-01.xls

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15.3 Analysis of Variance

Experimental Designs Several elements may distinguish between one

experimental design and others.

The number of factors. Each characteristic investigated is called a factor.

Each factor has several levels.

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Factor ALevel 1Level2

Level 1

Factor B

Level 3

Two - way ANOVA

Two factors

Level2

One - way ANOVA

Single factor

Treatment 3 (level 1)

Response

Response



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Groups of matched observations are formed into

blocks, in order to remove the effects of

unwanted variability.

By doing so we improve the chances of

detecting the variability of interest.

Independent samples or blocks

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Fixed effects

If all possible levels of a factor are included in our analysis

we have a fixed effect ANOVA.

The conclusion of a fixed effect ANOVA applies only to thelevels studied.

Random effects

If the levels included in our analysis represent a random

sample of all the possible levels, we have a random-effectANOVA.

The conclusion of the random-effect ANOVA applies to all the

levels (not only those studied).

Models ofFixed and Random Effects

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In some ANOVA models the test statistic of the fixed

effects case may differ from the test statistic of the

random effect case.

Fixed and random effects - examples

Fixed effects - The advertisement Example (15.1): All the

levels of the marketing strategies were included

Random effects - To determine if there is a difference in the

production rate of 50 machines, four machines are randomly

selected and there production recorded.

Models ofFixed and Random Effects.

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15.4 Randomized Blocks (Two-way)


The purpose of designing a randomized block

experiment is to reduce the within-treatments

variation thus increasing the relative amount of

between treatment variation.

This helps in detecting differences between the

treatment means more easily.

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Treatment 4

Treatment 3

Treatment 2

Treatment 1

Block 1Block3 Block2

Block all the observations with some

commonality across treatments

Randomized Blocks

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TreatmentBlock 1 2 k Block mean

1 X11 X12 . . . X1k

2 X21 X22 X2k

.

.

.

b Xb1 Xb2 Xbk

Treatment mean

1]B[x

2]B[x

b]B[x

1]T[x 2]T[x k]T[x

Block all the observations with some

commonality across treatments

Randomized Blocks

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The sum of square total is partitioned into three

sources of variation

Treatments

Blocks

Within samples (Error)

SS(Total) = SST + SSB + SSE

Sum of square for treatments Sum of square for blocks Sum of square for error

Recall.

For the independent

samples design we have:

SS(Total) = SST + SSE

Partitioning the total variability

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Calculating the sums of squares

Formulai for the calculation of the sums of squares


1 X11 X12 . . . X1k

2 X21 X22 X2k

.

.

.

b Xb1 Xb2 Xbk

Treatment mean

1]B[x

2]B[x

1]T[x 2]T[x k]T[x x2

1 X)]T[x(b

...X)]T[x(b2

2

2

k X)]T[x(b

SST =

2

1 X)]B[x(k

2

2 X)]B[x(k

2

k X)]B[x(k

SSB=

...)()(...

)()(...)()()(

22

21

222

212

221

211

XxXX

XxXxXxXxTotalSS

kk

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Calculating the sums of squares

Formulai for the calculation of the sums of squares


1 X11 X12 . . . X1k

2 X21 X22 X2k

.

.

.

b Xb1 Xb2 Xbk

Treatment mean

1]B[x

2]B[x

1]T[x 2]T[x k]T[x x2

1 X)]T[x(b

...X)]T[x(b2

2

2

k X)]T[x(b

SST =

2

1 X)]B[x(k

2

2 X)]B[x(k

2

k X)]B[x(k

SSB=

.. .)X]B[x]T[xx()X]B[x]T[xx(

...)X]B[x]T[xx()X]B[x]T[xx(

...)X]B[x]T[xx()X]B[x]T[xx(SSE

22kk2

21kk1

22222

21212

22121

21111

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To perform hypothesis tests for treatments and blocks we

need

Mean square for treatments

Mean square for blocks Mean square for error

Mean Squares

1k

SSTMST

1b

SSBMSB

1bkn

SSEMSE

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Test statistics for the randomized block

design ANOVA

MSEMSTF

MSE

MSBF

Test statistic for treatments

Test statistic for blocks

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Testing the mean responses for treatments

F > Fa,k-1,n-k-b+1

Testing the mean response for blocks

F> Fa,b-1,n-k-b+1

The F test rejection regions

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Example 15.2 Are there differences in the effectiveness of cholesterol

reduction drugs?

To answer this question the following experiment wasorganized:

25 groups of men with high cholesterol were matched by ageand weight. Each group consisted of 4 men.

Each person in a group received a different drug. The cholesterol level reduction in two months was recorded.

Can we infer from the data in Xm15-02 that there aredifferences in mean cholesterol reduction among the fourdrugs?

Randomized Blocks ANOVA - Example
http://d/PP/PPP/Xm15-02.xlshttp://d/PP/PPP/Xm15-02.xlshttp://d/PP/PPP/Xm15-02.xlshttp://d/PP/PPP/Xm15-02.xlshttp://d/PP/PPP/Xm15-02.xls

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Solution

Each drug can be considered a treatment.

Each 4 records (per group) can be blocked, becausethey are matched by age and weight.

This procedure eliminates the variability in

cholesterol reductionrelated to differentcombinations of age and weight.

This helps detect differences in the mean cholesterol

reduction attributed to the different drugs.


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BlocksTreatments b-1 MST / MSE MSB / MSE

Conclusion: At 5% significance level there is sufficient evidence

to infer that the mean cholesterol reduction gained by at least

two drugs are different.

K-1


ANOVA


Rows 3848.7 24 160.36 10.11 0.0000 1.67

Columns 196.0 3 65.32 4.12 0.0094 2.73

Error 1142.6 72 15.87

Total 5187.2 99

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Chapter 15 - continued

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15.5 Two-Factor Analysis of Variance -

Example 15.3

Suppose in Example 15.1, two factors are to be

examined: The effects of the marketing strategy on sales. Emphasis on convenience

Emphasis on quality

Emphasis on price

The effects of the selected media on sales. Advertise on TV

Advertise in newspapers

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Solution

We may attempt to analyze combinations of levels, one

from each factor using one-way ANOVA. The treatments will be: Treatment 1: Emphasize convenience and advertise in TV

Treatment 2: Emphasize convenience and advertise in

newspapers .

Treatment 6: Emphasize price and advertise in newspapers

Attempting one-way ANOVA

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Solution

The hypotheses tested are:

H0: m1= m2= m3= m4= m5= m6

H1: At least two means differ.


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City1 City2 City3 City4 City5 City6Convnce Convnce Quality Quality Price Price

TV Paper TV Paper TV Paper

In each one of six cities sales are recorded for ten

weeks.

In each city a different combination of marketingemphasis and media usage is employed.

Solution


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The p-value =.0452.

We conclude that there is evidence that differences

exist in the mean weekly sales among the six cities.

City1 City2 City3 City4 City5 City6Convnce Convnce Quality Quality Price Price

TV Paper TV Paper TV Paper

Solution

Xm15-03


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These result raises some questions: Are the differences in sales caused by the different

marketing strategies?

Are the differences in sales caused by the different

media used for advertising? Are there combinations of marketing strategy and

media that interact to affect the weekly sales?

Interesting questions no answers

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The current experimental design cannot provide

answers to these questions.

A new experimental design is needed.

Two-way ANOVA (two factors)

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City 1sales

City3sales

City 5sales

City 2

sales

City 4

sales

City 6

sales

TV

Newspapers

Convenience Quality Price

Are there differences in the mean sales

caused by different marketing strategies?

Factor A: Marketing strategy

FactorB:

Adv

ertisingmedia

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Test whether mean sales of Convenience, Quality,and Price significantly differ from one another.

H0: mConv.= mQuality = mPrice


Calculations are

based on the sum of

square for factor A

SS(A)


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City 1sales

City 3sales

City 5sales

City 2sales

City 4sales

City 6sales


FactorB:

Adv

ertisingmedia


caused by different advertising media?

TV

Newspapers


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Test whether mean sales of the TV, and Newspapers

significantly differ from one another.

H0: mTV = mNewspapersH

1

: The means differ

Calculations are based on

the sum of square for factor B

SS(B)


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City 1

sales

City 5

sales

City 2

sales

City 4

sales

City 6

sales

TV

Newspapers



Factor

B:

Advertising

media


caused by interaction between marketing

strategy and advertising medium?

City 3

sales

TV

Quality

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Test whether mean sales of certain cellsare different than the level expected.

Calculation are based on the sum of square forinteraction SS(AB)


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Graphical description of the possible

relationships between factors A and B.

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Levels of factor A

1 2 3

Level 1 of factor B

Level 2 of factor B

1 2 3

1 2 31 2 3

Level 1and 2 of factor B

Difference between the levels of factor A

No difference between the levels of factor B

Difference between the levels of factor A, and

difference between the levels of factor B; no

interaction

Levels of factor A

Levels of factor A Levels of factor A

No difference between the levels of factor A.

Difference between the levels of factor B

Interaction

M Re e

sa pn o

nse

M Re e

sa pn o

ns

e

M Re e

sa pn o

nse

M Re e

sa pn o

nse

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Sums of squares

a

1i

2i )x]A[x(rb)A(SS })()()){(2(10(

222. xxxxxx pricequalityconv

b

1j

2j )x]B[x(ra)B(SS })()){(3)(10(

22xxxx NewspaperTV

b

1j

2

jiij

a

1i )x]B[x]A[x]AB[x(r)AB(SS

r

k

ijijk

b

j

a

i

ABxxSSE

1

2

11

)][(

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F tests for the Two-way ANOVA

Test for the difference between the levels of the main

factors A and B

F= MS(A)MSE

F= MS(B)MSE

Rejection region: F > Fa,a-1 ,n-ab F > Fa, b-1, n-ab

Test for interaction between factors A and BF=

MS(AB)

MSE

Rejection region: F > Fa(a-1)(b-1),n-ab

SS(A)/(a-1) SS(B)/(b-1)

SS(AB)/(a-1)(b-1)

SSE/(n-ab)

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Required conditions:

1. The response distributions is normal

2. The treatment variances are equal.

3. The samples are independent.

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Example 15.3 continued( Xm15-03)


C on ven ien ce Qu al ity Pr ice

TV 491 677 575

TV 712 627 614

TV 558 590 706

TV 447 632 484

TV 479 683 478

TV 624 760 650

TV 546 690 583

TV 444 548 536

TV 582 579 579

TV 672 644 795

Newspaper 464 689 803

Newspaper 559 650 584Newspaper 759 704 525








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Example 15.3 continued

Test of the difference in mean sales between the three marketing

strategies

H0: mconv. = mquality = mpriceH1: At least two mean sales are different


ANOVA


Sample 13172.0 1 13172.0 1.42 0.2387 4.02

Columns 98838.6 2 49419.3 5.33 0.0077 3.17

Interaction 1609.6 2 804.8 0.09 0.9171 3.17

Within 501136.7 54 9280.3

Total 614757.0 59

Factor A Marketing strategies

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Example 15.3 continued

Test of the difference in mean sales between the three

marketing strategies

H0

: mconv.

= mquality

= mprice

H1: At least two mean sales are different

F = MS(Marketing strategy)/MSE = 5.33

Fcritical = Fa,a-1,n-ab = F.05,3-1,60-(3)(2) = 3.17; (p-value = .0077)

At 5% significance level there is evidence to infer that

differences in weekly sales exist among the marketing

strategies.


MS(A)MSE

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Example 15.3 - continued Test of the difference in mean sales between the two

advertising media

H0: mTV. = mNespaperH1: The two mean sales differ


Factor B = Advertising media

ANOVA


Sample 13172.0 1 13172.0 1.42 0.2387 4.02

Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17

Within 501136.7 54 9280.3

Total 614757.0 59

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Test of the difference in mean sales between the two

advertising media

H0: mTV. = mNespaper

H1: The two mean sales differ

F = MS(Media)/MSE = 1.42

Fcritical = Faa-1,n-ab = F.05,2-1,60-(3)(2) = 4.02 (p-value = .2387)

At 5% significance level there is insufficient evidence to infer

that differences in weekly sales exist between the two

advertising media.


MS(B)MSE

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Test for interaction between factors A and B

H0: mTV*conv. = mTV*quality==mnewsp.*price



Interaction AB = Marketing*Media

ANOVA


Sample 13172.0 1 13172.0 1.42 0.2387 4.02

Columns 98838.6 2 49419.3 5.33 0.0077 3.17

Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3

Total 614757.0 59

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Test for interaction between factor A and B

H0

: mTV*conv.

= mTV*quality

==mnewsp.*price


F = MS(Marketing*Media)/MSE = .09

Fcritical = Fa(a-1)(b-1),n-ab = F.05,(3-1)(2-1),60-(3)(2) = 3.17 (p-value= .9171)

At 5% significance level there is insufficient evidence to infer

that the two factors interact to affect the mean weekly sales.

MS(AB)MSE


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15.7 Multiple Comparisons

When the null hypothesis is rejected, it may bedesirable to find which mean(s) is (are) different,and at what ranking order.

Three statistical inference procedures, geared atdoing this, are presented: Fishers least significant difference (LSD) method

Bonferroni adjustment

Tukeys multiple comparison method

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Two means are considered different if the difference

between the corresponding sample means is larger

than a critical number. Then, the larger sample mean is

believed to be associated with a larger populationmean.

Conditions common to all the methods here:

The ANOVA model is the one way analysis of variance

The conditions required to perform the ANOVA are satisfied.

The experiment is fixed-effect

15.7 Multiple Comparisons

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Fisher Least Significant Different (LSD) Method

This method builds on the equal variances t-test of thedifference between two means.

The test statistic is improved by using MSE rather than sp2.

We can conclude that mi and mj differ (at a% significancelevel if |mi - mj| > LSD, where

kn.f.d

)n1

n1(MSEtLSD

ji2

a

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Experimentwise Type I error rate (aE)(the effective Type I error)

The Fishers method may result in an increased probability ofcommitting a type I error.

The experimentwise Type I error rate is the probability of

committing at least one Type I error at significance level ofa Itiscalculated byaE = 1-(1a)

C

where C is the number of pairwise comparisons (I.e.C = k(k-1)/2

The Bonferroni adjustment determines the required Type I errorprobability per pairwise comparison (a) ,to secure a pre-determined overall aE

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The procedure:

Compute the number of pairwise comparisons (C)

[C=k(k-1)/2], where k is the number of populations.

Set a = aE/C, where aE is the true probability of making atleast one Type I error (called experimentwise Type I error).

We can conclude that mi and mj differ (at a/C% significance

level if

kn.f.d

)n

1

n

1(M SEt

ji

)C2(ji

mm a

Bonferroni Adjustment

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35.4465.6080.653xx

10.3165.60855.577xx

45.750.65355.577xx

32

31

21


Rank the effectiveness of the marketing strategies

(based on mean weekly sales).

Use the Fishers method, and the Bonferroni adjustment method

Solution (the Fishers method)

The sample mean sales were 577.55, 653.0, 608.65.

Then,

71.59)20/1()20/1(8894t

)n

1

n

1(M SEt

2/05.

ji

2

a

Fisher and Bonferroni Methods

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Solution (the Bonferroni adjustment)

We calculate C=k(k-1)/2 to be 3(2)/2 = 3.

We set a = .05/3 = .0167, thus t.01672, 60-3 = 2.467 (Excel).

54.73)20/1()20/1(8894467.2

)n

1

n

1(M SEt

ji

2

a

Again, the significant difference is between m1 and m2.

35.4465.6080.653xx

10.3165.60855.577xx

45.750.65355.577xx

32

31

21

Fisher and Bonferroni Methods

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The test procedure:

Find a critical number w as follows:

gn

M SE),k(q w

a

k = the number of samples

=degrees of freedom = n - kng = number of observations per sample

(recall, all the sample sizes are the same)

a = significance level

qa(k,) = a critical value obtained from the studentized range table

Tukey Multiple Comparisons

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If the sample sizes are not extremely different, we can use the

above procedure with ng calculated as the harmonic mean of

the sample sizes. k21n1...n1n1

kgn

Repeat this procedure for each pair of samples.

Rank the means if possible.

Select a pair of means. Calculate the difference

between the larger and the smaller mean.

If there is sufficient evidence toconclude that mmax > mmin .

minmax xx

w minmax xx


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City 1 vs. City 2: 653 - 577.55 = 75.45

City 1 vs. City 3: 608.65 - 577.55 = 31.1

City 2 vs. City 3: 653 - 608.65 = 44.35

Example 15.1 - continued We had three populations(three marketing strategies).K = 3,

Sample sizes were equal. n1

= n2

= n3

= 20, = n-k = 60-3 = 57,MSE = 8894.

minmax xx

70.71

20

8894)57,3(.q

n

MSE),k(q 05

g

w a

Take q.05(3,60) from the table.

Population

Sales - City 1

Sales - City 2

Sales - City 3

Mean

577.55

653

698.65

w minmax xx


Excel Tukey and Fisher LSD method

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Excel Tukey and Fisher LSD method

Xm15-01

Fishers LDS

Bonferroni adjustments

a = .05

a = .05/3 = .0167

Multiple Comparisons

LSD Omega

Treatment Treatment Difference Alpha = 0.05 Alpha = 0.05

Convenience Quality -75.45 59.72 71.70

Price -31.1 59.72 71.70

Quality Price 44.35 59.72 71.70

Multiple Comparisons

LSD Omega

Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05

Convenience Quality 75 45 73 54 71 70

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