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.