Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 20
Analisis Varians Klasifikasi Dua Arah
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menjelaskan analisis
varians klasifikasi dua arah.
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Outline Materi
•
•
•
•
Hasil pengamatan
Sumber keragaman
Tabel ANOVA
Pengambilan keputusan
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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
4
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
F
Sum of
Squares
Degrees of
Freedom
Treatments
SSTR
k-1
Blocks
SSBL
b-1
Error
SSE
(k - 1)(b - 1)
Total
SST
nT - 1
Mean
Squares
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) Blocks
Blend X Blend Y Blend Z
Means
31
30
30
30.333
30
29
29
29.333
29
29
28
28.667
33
31
29
31.000
26
25
26
25.667
29.8
28.8
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
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the three gasoline blends.
• Selamat Belajar Semoga Sukses.
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