Matakuliah
Tahun
: I0134 – Metode Statistika
: 2007
Analisis Varians Klasifikasi Dua arah
Pertemuan 24
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Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa akan dapat menyusun simpulan hasil
analisis varians.
2
Outline Materi
•
•
•
•
Analisis varians rancangan kelompok
Partisi jumlah kuadrat perlakuan
Prosedur uji F
Pembandingan ganda perlakuan
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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
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An Introduction to Analysis of Variance
• 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.
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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.
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Analysis of Variance:
Testing for the Equality of K Population Means
•
•
•
•
Between-Samples Estimate of Population Variance
Within-Samples Estimate of Population Variance
Comparing the Variance Estimates: The F Test
The ANOVA Table
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Randomized Block Design
• The ANOVA Procedure
• Computations and Conclusions
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The ANOVA Procedure
•
•
•
The ANOVA procedure for the randomized block design requires us to
partition the sum of squares total (SST) into three groups: sum of
squares due to treatments, sum of squares due to blocks, and sum of
squares due to error.
The formula for this partitioning is
SST = SSTR + SSBL + SSE
•
The total degrees of freedom, nT - 1, are partitioned such that k - 1
degrees of freedom go to treatments,
b - 1 go to blocks, and (k - 1)(b - 1) go to the error term.
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ANOVA Table for a
Randomized Block Design
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Squares
Treatments
SSTR
k-1
Blocks
SSBL
b-1
Error
SSE
Total
SST
(k - 1)(b - 1)
nT - 1
F
SSTR MSTR
k - 1 MSE
SSBL
MSBL 
b-1
MSTR 
SSE
MSE 
( k  1)(b  1)
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Contoh Soal: Eastern Oil Co.
Eastern Oil has developed three new blends of
gasoline and must decide which blend or blends to
produce and distribute. A study of the miles per gallon
ratings of the three blends is being conducted to
determine if the mean ratings are the same for the three
blends.
Five automobiles have been tested using each of
the three gasoline blends and the miles per gallon
ratings are shown on the next slide.
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Contoh Soal: Eastern Oil Co.
Automobile
(Block)
1
2
3
4
5
Treatment
Means
Type of Gasoline (Treatment)
Blend X Blend Y
Blend Z
31
30
30
30
29
29
29
29
28
33
31
29
26
25
26
29.8
28.8
Blocks
Means
30.333
29.333
28.667
31.000
25.667
28.4
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Contoh Soal: Eastern Oil Co.
•
Randomized Block Design
– Mean Square Due to Treatments
The overall sample mean is 29. Thus,
SSTR = 5[(29.8 - 29)2 + (28.8 - 29)2 + (28.4 - 29)2] = 5.2
MSTR = 5.2/(3 - 1) = 2.6
– Mean Square Due to Blocks
SSBL = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33
MSBL = 51.33/(5 - 1) = 12.8
– Mean Square Due to Error
SSE = 62 - 5.2 - 51.33 = 5.47
MSE = 5.47/[(3 - 1)(5 - 1)] = .68
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Contoh Soal: Eastern Oil Co.
• Randomized Block Design
–
–
–
Rejection Rule
Assuming  = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f.
denominator). Reject H0 if F > 4.46.
Test Statistic
F = MSTR/MSE = 2.6/.68 = 3.82
Conclusion
Since 3.82 < 4.46, we cannot reject H0. There is not
sufficient evidence to conclude that the miles per gallon
ratings differ for the three gasoline blends.
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• Selamat Belajar Semoga Sukses.
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