© 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 1 Slide Slide Slides Prepared by Juei-Chao Chen Fu Jen Catholic University Slides Prepared

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Slides Prepared bySlides Prepared byJuei-Chao ChenJuei-Chao ChenFu Jen Catholic UniversityFu Jen Catholic University


Chapter 13Chapter 13STATISTICSSTATISTICS in in PRACTICEPRACTICE

Burke Marketing Services, Inc., is one of the most experienced market research firms in the industry. In one study, a firm retained Burke to evaluate potential new versions of a children’s dry cereal. Analysis of variance was the statistical method used to

study the data obtained from the taste tests. The experimental design employed by Burke and the

subsequent analysis of variance were helpful in making a product design recommendation.


Chapter 13, Part AChapter 13, Part A Analysis of Variance and Experimental Design Analysis of Variance and Experimental Design Introduction to Analysis of Variance Analysis of Variance: Testing for the Equality

of k Population Means Multiple Comparison Procedures


IntroductionIntroduction to Analysis of Varianceto Analysis of Variance

Analysis of VarianceAnalysis of Variance (ANOVA) can be used to test (ANOVA) can be used to test for the equality of three or more population means.for the equality of three or more population means.

Data obtained from observational or experimentalData obtained from observational or experimental studies can be used for the analysis.studies can be used for the analysis.

We want to use the sample results to test theWe want to use the sample results to test the following hypotheses:following hypotheses:

HH00: : 11==22==33==. . . . . . = = kk

HHaa: : Not all population means are equalNot all population means are equal


Introduction to Analysis of VarianceIntroduction to Analysis of Variance

HH00: : 11==22==33==. . . . . . = = kk

HHaa: : Not all population means are equalNot all population means are equal

IfIf HH00 is rejected, we cannot conclude that all populationis rejected, we cannot conclude that all population means are different.means are different.

RejectingRejecting HH00 means that at least two population meansmeans that at least two population means have different values.have different values.


Sampling Distribution of Given H0 is Truex


1x 3x2x

Sample means are close togetherSample means are close together because there is onlybecause there is only

one sampling distributionone sampling distribution when when HH00 is true. is true.

22x n



Sampling Distribution of Given H0 is Falsex

3 1x 2x3x 1 2

Sample means come fromSample means come fromdifferent sampling distributionsdifferent sampling distributions

and are not as close togetherand are not as close together when when HH00 is false. is false.


For each population, the response variable isFor each population, the response variable is normally distributed.normally distributed.

Assumptions for Analysis of VarianceAssumptions for Analysis of Variance

The variance of the response variable, denotedThe variance of the response variable, denoted 22,, is the same for all of the populationsis the same for all of the populations..

The observations must be independent.The observations must be independent.


Analysis of Variance:Analysis of Variance:Testing for the Equality of Testing for the Equality of kk Population Means Population Means

Between-Treatments Estimate of Population Variance

Within-Treatments Estimate of Population Variance

Comparing the Variance Estimates: The F Test

ANOVA Table



Analysis of variance can be used to test for the equality of k population means.

The hypotheses tested is H0: Ha: Not all population means are equal where mean of the jth population.

k 21

j



Sample data

= value of observation i for treatment j = number of observations for treatment j = sample mean for treatment j = sample variance for treatment j = sample standard deviation for treatment j

ijx

jn

jx2js

js


Statisitcs The sample mean for treatment j

The sample variance for treatment j


j

n

iij

j n

xx

j

1

1

)(1

2

2

j

n

ijij

j n

xxs

j



The overall sample mean

where nT = n1 + n2 +. . . + nk

If the size of each sample is n, nT = kn then

T

k

j

n

iij

n

xx

j

1 1

kn

x

k

nx

kn

xx

k

jij

k

j

n

iij

k

j

n

iij

jj

11 11 1

/



Between-Treatments Estimate of Population Variance

The sum of squares due to treatments (SSTR)

The mean square due to treatments (MSTR)

1

)(1

2

k

xxnMSTR

k

jjj

k

jjj xxn

1

2)(



Within-Treatments Estimate of Population Variance The sum of squares due to error (SSE)

The mean square due to error (MSE)

k

jjj sn

1

2)1(

kn

snMSE

T

k

jjj

1

2)1(


Between-Treatments EstimateBetween-Treatments Estimateof Population Varianceof Population Variance

A between-treatment estimate of 2 is called the mean square treatment and is denoted MSTR.

2

1( )

MSTR 1

k

j jj

n x x

k

Denominator representsDenominator represents the the degrees of freedomdegrees of freedom associated with SSTRassociated with SSTR

Numerator is theNumerator is the sum of squaressum of squares

due to treatmentsdue to treatmentsand is denoted SSTRand is denoted SSTR


The estimate of 2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE.

Within-Samples EstimateWithin-Samples Estimateof Population Varianceof Population Variance

kn

sn

T

k

jjj

1

2)1(MSE

Denominator representsDenominator represents the the degrees of freedomdegrees of freedom

associated with SSEassociated with SSE

Numerator is theNumerator is the sum of squaressum of squares

due to errordue to errorand is denoted SSEand is denoted SSE


Comparing the Variance Estimates: The Comparing the Variance Estimates: The FF Test Test

If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with MSTR d.f. equal to k - 1 and MSE d.f. equal to nT - k. If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates 2. Hence, we will reject H0 if the resulting value of MSTR/MSE appears to be too large to have been selected at random from the appropriate F distribution.


Test for the Equality ofTest for the Equality of kk Population MeansPopulation Means

FF = MSTR/MSE = MSTR/MSE

HH00: : 11==22==33==. . . . . . = = kk

HHaa: Not all population means are equal: Not all population means are equal

Hypotheses

Test Statistic


Test for the Equality ofTest for the Equality of k k Population MeansPopulation Means

Rejection Rule

where the value of F is based on anF distribution with k - 1 numerator d.f.and nT - k denominator d.f.

Reject Reject HH00 if if pp-value-value << p-value Approach:

Critical Value Approach: RejectReject HH00 if if FF >> FF


Sampling Distribution of MSTR/MSESampling Distribution of MSTR/MSE Rejection Region

Do Not Reject H0

Reject H0

MSTR/MSE

Critical ValueF

Sampling DistributionSampling Distributionof MSTR/MSEof MSTR/MSE


ANOVA TableANOVA Table

SST is SST is partitionedpartitioned

into SSTR and into SSTR and SSE.SSE.

SST’s degrees of SST’s degrees of freedomfreedom

(d.f.) are partitioned (d.f.) are partitioned intointo

SSTR’s d.f. and SSE’s SSTR’s d.f. and SSE’s d.f.d.f.

TreatmentTreatmentErrorErrorTotalTotal

SSTRSSTRSSESSESSTSST

kk – 1 – 1nnT T – – kknnTT - 1 - 1

MSTRMSTRMSEMSE

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquaresSquares

MSTR/MSEMSTR/MSEFF



SST divided by its degrees of freedomSST divided by its degrees of freedom nnTT – 1 – 1 is theis the overall sample variance that would be obtained if weoverall sample variance that would be obtained if we treated the entire set of observations as one data set.treated the entire set of observations as one data set.

With the entire data set as one sample, the formulaWith the entire data set as one sample, the formula for computing the total sum of squares, SST, is:for computing the total sum of squares, SST, is:

2

1 1SST ( ) SSTR SSE

jnk

ijj i

x x



ANOVA can be viewed as the process of partitioningANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedomthe total sum of squares and the degrees of freedom into their corresponding sources: treatments and error.into their corresponding sources: treatments and error.

Dividing the sum of squares by the appropriate degrees Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the of freedom provides the variance estimates and the FF value used to test the hypothesis of equal population value used to test the hypothesis of equal population means.means.


Test for the Equality of Test for the Equality of kk Population Means Population Means

Example: National Computer Products, Inc. (NCP),

manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle.

Object: To measure how much employees at these plants know about total quality management. A random sample of six employees was selected from

each plant and given a quality awareness examination.



Data

Let = mean examination score for population 1 = mean examination score for population 2 = mean examination score for population 3

123


Test for the Equality of Test for the Equality of kk Population Means Population Means Hypotheses H0: = = Ha: Not all population means are equal

In this example 1. dependent or response variable : examination score 2. independent variable or factor : plant location 3. levels of the factor or treatments : the values of a factor selected for investigation, in the NCP example the three treatments or three population are Atlanta, Dallas, and Seattle.

1 2 3



Three assumptions1. For each population, the response variable is normally

distributed. The examination scores (response variable) must be normally distributed at each plant.

2. The variance of the response variable, , is the same for all of the populations. The variance of examination scores must be the same for all three plants.

3. The observations must be independent. The examination score for each employee must be independent of the examination score for any other employee.

2



ANOVA Table

p-value = 0.003 < α = .05. We reject H0.


Example: Reed Manufacturing


Janet Reed would like to know ifthere is any significant difference inthe mean number of hours worked per week for the department managersat her three manufacturing plants(in Buffalo, Pittsburgh, and Detroit).


Example: Reed Manufacturing


A simple random sample of fivemanagers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Conduct an F test using α = .05.


1122334455

48485454575754546262

73736363666664647474

51516363616154545656

PlantPlant 1 1BuffaloBuffalo

Plant 2Plant 2PittsburghPittsburgh

PlantPlant 3 3DetroitDetroitObservationObservation

SampleSample Mean MeanSample VarianceSample Variance

5555 68 68 575726.026.0 26.5 26.5 24.524.5




H0: 1=2=3

Ha: Not all the means are equal

where: 1 = mean number of hours worked per week by the managers at Plant 1 2 = mean number of hours worked per week by the managers at Plant 23 = mean number of hours worked per week by the managers at Plant 3

1. Develop the hypotheses.1. Develop the hypotheses.

p -Value and Critical Value Approaches


2. Specify the level of significance.2. Specify the level of significance. = .05



3. Compute the value of the test statistic.3. Compute the value of the test statistic.

MSTR = 490/(3 - 1) = 245SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490

(Sample sizes are all equal.)Mean Square Due to Treatments

= (55 + 68 + 57)/3 = 60x


3. Compute the value of the test statistic.3. Compute the value of the test statistic.


MSE = 308/(15 - 3) = 25.667SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308

Mean Square Due to Error

(continued)

F = MSTR/MSE = 245/25.667 = 9.55



TreatmentTreatmentErrorErrorTotalTotal

490490308308798798

2212121414

24524525.66725.667

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquaresSquares

9.559.55FF


ANOVA Table



5. Determine whether to reject5. Determine whether to reject HH00..

We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.

The p-value < .05, so we reject H0.

With 2 numerator d.f. and 12 denominator d.f.,the p-value is .01 for F = 6.93. Therefore, the p-value is less than .01 for F = 9.55.

p - Value Approach

4. Compute the 4. Compute the pp –value. –value.


5. Determine whether to reject5. Determine whether to reject HH00..Because F = 9.55 > 3.89, we reject H0.

Critical Value Approach

4. Determine the critical value and rejection rule.4. Determine the critical value and rejection rule.

Reject H0 if F > 3.89


We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.

Based on an F distribution with 2 numeratord.f. and 12 denominator d.f., F.05 = 3.89.



Summary


Multiple Comparison ProceduresMultiple Comparison Procedures Suppose that analysis of variance has provided

statistical evidence to reject the null hypothesis of equal population means.

Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.


Fisher’s LSD ProcedureFisher’s LSD Procedure

1 1MSE( )i j

i j

x xt

n n

Test Statistic

Hypotheses 0 : i jH : a i jH


Fisher’s LSD ProcedureFisher’s LSD Procedure

where the value of ta/2 is based on at distribution with nT - k degrees of freedom.

Rejection Rule

Reject Reject HH00 if if pp-value -value << aa

p-value Approach:

Critical Value Approach:

Reject Reject HH00 ifif tt < - < -ttaa/2 /2 oror tt > > ttaa/2 /2


Test Statistic

Fisher’s LSD ProcedureFisher’s LSD ProcedureBased on the Test Statistic Based on the Test Statistic xxii - - xxjj

/ 21 1LSD MSE( )

i jt n n where

i jx x

Reject Reject HH00 if > LSD if > LSDi jx x

Hypotheses

Rejection Rule

0 : i jH : a i jH



Example: Reed Manufacturing Recall that Janet Reed wants to knowif there is any significant difference inthe mean number of hours worked per week for the department managersat her three manufacturing plants. Analysis of variance has providedstatistical evidence to reject the nullhypothesis of equal population means.Fisher’s least significant difference (LSD) procedurecan be used to determine where the differences occur.


For = .05 and nT - k = 15 – 3 = 12 degrees of freedom, t.025 = 2.179

/ 21 1LSD MSE( )

i jt n n

MSE value wasMSE value wascomputed earliercomputed earlier



LSD for Plants 1 and 2


• Conclusion

• Test Statistic1 2x x = |55 - 68| = 13

Reject H0 if > 6.981 2x x• Rejection Rule

0 1 2: H 1 2: aH

• Hypotheses (A)

The mean number of hours worked at Plant 1 isnot equal to the mean number worked at Plant 2.




• Conclusion

• Test Statistic1 3x x = |55 57| = 2


0 1 3: H 1 3: aH

• Hypotheses (B)

There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3.




• Conclusion

• Test Statistic2 3x x = |68 - 57| = 11


0 2 3: H 2 3: aH

• Hypotheses (C)

The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3.


The experimentwise Type I error rate gets larger for problems with more populations (larger k).

Type I Error RatesType I Error Rates

EWEW = 1 – (1 – = 1 – (1 – ))((k k – 1)!– 1)!

The comparisonwise Type I error rate indicates the level of significance associated with a single pairwise comparison.

The experimentwise Type I error rate EW is the probability of making a Type I error on at least one of the (k – 1)! pairwise comparisons.


End of Chapter 13, Part AEnd of Chapter 13, Part A

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© 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 1 Slide Slide Slides Prepared by Juei-Chao Chen Fu Jen Catholic University Slides Prepared