Feb 11, 2016

Pertemuan 1 9 Analisis Varians Klasifikasi Satu Arah. Matakuliah: I0284 - Statistika Tahun: 200 8 Versi: Revisi. Learning Outcomes. Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa akan dapat men erapkan uji perbedaan rata-rata lebih dari 2 populasi . - PowerPoint PPT Presentation

Welcome message from author

This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript

1

Pertemuan 19Analisis Varians Klasifikasi Satu Arah

Matakuliah : I0284 - StatistikaTahun : 2008Versi : Revisi

2

Learning Outcomes

Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu :• Mahasiswa akan dapat menerapkan uji

perbedaan rata-rata lebih dari 2 populasi.

3

Outline Materi

• Konsep dasar analisis varians• Klasifikasi satu arah ulangan sama• Klasifikasi satu arah ulangan tidak sama• Prosedur uji F

4

Analysis of Variance and Experimental Design

• An Introduction to Analysis of Variance • Analysis of Variance: Testing for the Equality of k Population Means• Multiple Comparison Procedures• An Introduction to Experimental Design• Completely Randomized Designs• Randomized Block Design

5

• Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means using data obtained from observational or experimental studies.

• We want to use the sample results to test the following hypotheses.

H0: 1=2=3=. . . = k

Ha: Not all population means are equal

• If H0 is rejected, we cannot conclude that all population means are different.

• Rejecting H0 means that at least two population means have different values.

An Introduction to Analysis of Variance

6

Assumptions for Analysis of Variance

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

• The variance of the response variable, denoted 2, is the same for all of the populations.

• The observations must be independent.

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

FF = MSTR/MSE = MSTR/MSE

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

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

HypothesesHypotheses

Test StatisticTest Statistic

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

Sampling Distribution of MSTR/MSESampling Distribution of MSTR/MSE Rejection RegionRejection 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 partitionedpartitionedinto SSTR and into SSTR and SSE.SSE.

SST’s degrees of SST’s degrees of freedomfreedom(d.f.) are partitioned (d.f.) are partitioned intointoSSTR’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

ANOVA TableANOVA Table

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:

2

1 1SST ( ) SSTR SSE

jnk

ijj i

x x

ANOVA TableANOVA Table

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.

Example: Reed ManufacturingExample: Reed Manufacturing

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

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.

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

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

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

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

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

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

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

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 = 60x(Sample sizes are all equal.)(Sample sizes are all equal.)

Mean Square Due to TreatmentsMean Square Due to Treatments

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

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

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

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

TreatmentTreatmentErrorErrorTotalTotal

490490308308798798

2212121414

24524525.66725.667

Source ofSource ofVariationVariation

Sum ofSum ofSquaresSquares

Degrees ofDegrees ofFreedomFreedom

MeanMeanSquaresSquares

9.559.55FF

Test for the Equality of Test for the Equality of kk Population Population MeansMeans ANOVA TableANOVA Table

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

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 Approach4. Compute the 4. Compute the pp –value. –value.

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 Approach4. 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

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

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.

Multiple Comparison Procedures• Suppose that analysis of variance has provided

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

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

Fisher’s LSD Procedure

1 1MSE( )i j

i j

x xt

n n

• Test Statistic

HypothesesHypotheses 0 : i jH : a i jH

Fisher’s LSD Procedure

where the value of where the value of ttaa/2 /2 is based on ais based on att distribution with distribution with nnTT - - kk degrees of freedom. degrees of freedom.

Rejection RuleRejection Rule

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

Critical Value Approach:Critical Value Approach:Reject Reject HH00 if if tt < - < -ttaa/2 /2 or or tt > > ttaa/2 /2

• Test Statistic

Fisher’s LSD ProcedureBased on the Test Statistic xi - xj

__ __

/ 21 1LSD MSE( )

i jt n n

wherewhere

i jx x

Reject Reject HH00 if > LSD if > LSDi jx x

HypothesesHypotheses

Rejection RuleRejection Rule

0 : i jH : a i jH

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

Example: Reed ManufacturingExample: Reed Manufacturing Recall that Janet Reed wants to knowRecall that Janet Reed wants to knowif there is any significant difference inif there 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 plants. at her three manufacturing plants.

Analysis of variance has providedAnalysis of variance has providedstatistical evidence to reject the nullstatistical evidence to reject the nullhypothesis of equal population means.hypothesis of equal population means.Fisher’s least significant difference (LSD) Fisher’s least significant difference (LSD) procedureprocedurecan be used to determine where the differences can be used to determine where the differences occur.occur.

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

LSD 2 179 25 667 15 15 6 98. . ( ) .LSD 2 179 25 667 15 15 6 98. . ( ) .

/ 21 1LSD MSE( )

i jt n n

MSE value wasMSE value wascomputed earliercomputed earlier

Fisher’s LSD ProcedureBased on the Test Statistic xi - xj

• LSD for Plants 1 and 2

Fisher’s LSD ProcedureBased on the Test Statistic xi - xj

• ConclusionConclusion

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

Reject Reject HH00 if if > 6.98 > 6.981 2x x• Rejection RuleRejection Rule

0 1 2: H 1 2: aH

• Hypotheses (A)Hypotheses (A)

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

LSD for Plants 1 and 3LSD for Plants 1 and 3

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

• ConclusionConclusion

• Test StatisticTest Statistic1 3x x = |55 = |55 57| = 2 57| = 2

Reject Reject HH00 if if > 6.98 > 6.981 3x x• Rejection RuleRejection Rule

0 1 3: H 1 3: aH

• Hypotheses (B)Hypotheses (B)

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

LSD for Plants 2 and 3LSD for Plants 2 and 3

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

• ConclusionConclusion

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

Reject Reject HH00 if if > 6.98 > 6.982 3x x• Rejection RuleRejection Rule

0 2 3: H 2 3: aH

• Hypotheses (C)Hypotheses (C)

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

30

• Selamat Belajar Semoga Sukses.

Related Documents