1 1 slide © 2006 thomson/south-western slides prepared by john s. loucks st. edward’s university...

1 1 Slide

Slide

© 2006 Thomson/South-Western© 2006 Thomson/South-Western

Slides Prepared bySlides Prepared by

JOHN S. LOUCKSJOHN S. LOUCKSSt. Edward’s UniversitySt. Edward’s University

Slides Prepared bySlides Prepared by

JOHN S. LOUCKSJOHN S. LOUCKSSt. Edward’s UniversitySt. Edward’s University

2 2 Slide

Slide


Chapter 10Chapter 10 Comparisons Involving Means Comparisons Involving Means

Part BPart B Introduction to Analysis of Variance Introduction to Analysis of Variance Analysis of Variance: Testing for the Equality of Analysis of Variance: Testing for the Equality of k k Population Means Population Means

3 3 Slide

Slide


Introduction to Analysis of VarianceIntroduction to 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. 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. 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: 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 equal: Not all population means are equal

4 4 Slide

Slide



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


If If HH00 is rejected, we cannot conclude that is rejected, we cannot conclude that allall population means are different.population means are different.

If If HH00 is rejected, we cannot conclude that is rejected, we cannot conclude that allall population means are different.population means are different.

Rejecting Rejecting HH00 means that at least two population means that at least two population means have different values.means have different values.

Rejecting Rejecting HH00 means that at least two population means that at least two population means have different values.means have different values.

5 5 Slide

Slide


Sampling Distribution of Given Sampling Distribution of Given HH00 is True is Truexx


1x1x 3x3x2x2x

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

2

2x n

6 6 Slide

Slide



Sampling Distribution of Given Sampling Distribution of Given HH00 is False is Falsexx

33 1x1x 2x2x3x3x 11 22

Sample means come fromSample means come fromdifferent sampling distributionsdifferent sampling distributionsand are not as close togetherand are not as close together

when when HH00 is false. is false.

7 7 Slide

Slide


For each population, the response variable isFor each population, the response variable is normally distributed.normally distributed. 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, denoted The variance of the response variable, denoted 22,, is the same for all of the populations.is the same for all of the populations. The variance of the response variable, denoted The variance of the response variable, denoted 22,, is the same for all of the populations.is the same for all of the populations.

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

8 8 Slide

Slide


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

MeansMeans Between-Treatments Estimate of Population VarianceBetween-Treatments Estimate of Population Variance

Within-Treatments Estimate of Population VarianceWithin-Treatments Estimate of Population Variance Comparing the Variance Estimates: The Comparing the Variance Estimates: The F F Test Test ANOVA TableANOVA Table

9 9 Slide

Slide


Between-Treatments EstimateBetween-Treatments Estimateof Population Varianceof Population Variance

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

2

1

( )

MSTR1

k

j jj

n x x

k

2

1

( )

MSTR1

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

10 10 Slide

Slide


The estimate of The estimate of 22 based on the variation of the based on the variation of the sample observations within each sample is called sample observations within each sample is called the the mean square errormean square error and is denoted by MSE. and is denoted by MSE.

Within-Samples EstimateWithin-Samples Estimateof Population Varianceof Population Variance

kn

sn

T

k

jjj

1

2)1(

MSEkn

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

11 11 Slide

Slide


Comparing the Variance Estimates: The Comparing the Variance Estimates: The FF TestTest

If the null hypothesis is true and the ANOVAIf the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution ofassumptions are valid, the sampling distribution of MSTR/MSE is an MSTR/MSE is an FF distribution with MSTR d.f. distribution with MSTR d.f. equal to equal to kk - 1 and MSE d.f. equal to - 1 and MSE d.f. equal to nnTT - - kk..

If the means of the If the means of the kk populations are not equal, the populations are not equal, the value of MSTR/MSE will be inflated because MSTRvalue of MSTR/MSE will be inflated because MSTR overestimates overestimates 22.. Hence, we will reject Hence, we will reject HH00 if the resulting value of if the resulting value of MSTR/MSE appears to be too large to have beenMSTR/MSE appears to be too large to have been selected at random from the appropriate selected at random from the appropriate FF distribution.distribution.

12 12 Slide

Slide


Test for the Equality of Test for the Equality of kk Population Population MeansMeans

FF = MSTR/MSE = MSTR/MSE

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


HypothesesHypotheses

Test StatisticTest Statistic

13 13 Slide

Slide



Rejection RuleRejection Rule

where the value of where the value of FF is based on an is based on anFF distribution with distribution with kk - 1 numerator d.f. - 1 numerator d.f.and and nnTT - - kk denominator d.f. denominator d.f.

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

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

14 14 Slide

Slide


Sampling Distribution of MSTR/MSESampling Distribution of MSTR/MSE

Rejection RegionRejection Region

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSE

Critical ValueCritical ValueFF

Sampling DistributionSampling Distributionof MSTR/MSEof MSTR/MSE

15 15 Slide

Slide


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.

TreatmentTreatment

ErrorError

TotalTotal

SSTRSSTR

SSESSE

SSTSST

kk – 1 – 1

nnT T – – kk

nnTT - 1 - 1

MSTRMSTR

MSEMSE

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquaresSquares

MSTR/MSEMSTR/MSE

FF

16 16 Slide

Slide



SST divided by its degrees of freedom SST divided by its degrees of freedom nnTT – 1 is the – 1 is 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.

SST divided by its degrees of freedom SST divided by its degrees of freedom nnTT – 1 is the – 1 is 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: 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 1

SST ( ) SSTR SSEjnk

ijj i

x x

2

1 1

SST ( ) SSTR SSEjnk

ijj i

x x

17 17 Slide

Slide



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.

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

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

18 18 Slide

Slide


Example: Reed ManufacturingExample: Reed Manufacturing


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

19 19 Slide

Slide


Example: Reed ManufacturingExample: Reed Manufacturing


A simple random sample of fiveA simple random sample of fivemanagers from each of the three plantsmanagers from each of the three plantswas taken and the number of hourswas taken and the number of hoursworked by each manager for theworked by each manager for theprevious week is shown on the nextprevious week is shown on the nextslide.slide. Conduct an Conduct an FF test using test using = .05. = .05.

20 20 Slide

Slide


1122334455

48485454575754546262

73736363666664647474

51516363616154545656

Plant 1Plant 1BuffaloBuffalo

Plant 2Plant 2PittsburghPittsburgh

Plant 3Plant 3DetroitDetroitObservationObservation

Sample MeanSample MeanSample VarianceSample Variance

5555 68 68 57 5726.026.0 26.5 26.5 24.5 24.5


21 21 Slide

Slide



HH00: : 11==22==33

HHaa: Not all the means are equal: Not all the means are equalwhere: where: 1 1 = mean number of hours worked per= mean number of hours worked per

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

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

pp -Value and Critical Value Approaches -Value and Critical Value Approaches

22 22 Slide

Slide


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



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

MSTR = 490/(3 - 1) = 245MSTR = 490/(3 - 1) = 245SSTR = 5(55 - 60)SSTR = 5(55 - 60)22 + 5(68 - 60) + 5(68 - 60)22 + 5(57 - 60) + 5(57 - 60)22 = 490 = 490

= (55 + 68 + 57)/3 = 60= (55 + 68 + 57)/3 = 60xx(Sample sizes are all equal.)(Sample sizes are all equal.)

Mean Square Due to TreatmentsMean Square Due to Treatments

23 23 Slide

Slide


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


MSE = 308/(15 - 3) = 25.667MSE = 308/(15 - 3) = 25.667

SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308Mean Square Due to ErrorMean Square Due to Error

(continued)(continued)

FF = MSTR/MSE = 245/25.667 = 9.55 = MSTR/MSE = 245/25.667 = 9.55


24 24 Slide

Slide


TreatmentTreatment

ErrorError

TotalTotal

490490

308308

798798

22

1212

1414

245245

25.66725.667

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquaresSquares

9.559.55

FF



25 25 Slide

Slide



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

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

The The pp-value -value << .05, .05, so we reject so we reject HH00..

With 2 numerator d.f. and 12 With 2 numerator d.f. and 12 denominator d.f.,denominator d.f.,the the pp-value is .01 for -value is .01 for FF = 6.93. = 6.93. Therefore, theTherefore, thepp-value is less than .01 for -value is less than .01 for FF = 9.55. = 9.55.

pp –Value Approach –Value Approach

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

26 26 Slide

Slide


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

Because Because FF = 9.55 = 9.55 >> 3.89, we reject 3.89, we reject HH00..

Critical Value ApproachCritical Value Approach

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

Reject Reject HH00 if if FF >> 3.89 3.89


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

Based on an Based on an FF distribution with 2 numerator distribution with 2 numeratord.f. and 12 denominator d.f., d.f. and 12 denominator d.f., FF.05.05 = 3.89. = 3.89.

27 27 Slide

Slide


End of Chapter 10End of Chapter 10Part BPart B

1 1 slide © 2006 thomson/south-western slides prepared by john s. loucks st. edward’s university...

Documents