Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 19
Analisis Varians Klasifikasi Satu Arah
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menjelaskan analisis
varians klasifikasi satu arah.
2
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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Between-Samples Estimate
of Population Variance
• A between-samples estimate of 2 is
called the mean square_ between
(MSB).
=
k
MSB 
2
 n j ( x j  x) 2
j1
k1
• The numerator of MSB is called the sum of
squares between (SSB).
• The denominator of MSB represents the
degrees of freedom associated with SSB. 8
Within-Samples Estimate
of Population Variance
• The estimate of 2 based on the variation of the
sample observations within each sample is
called the mean square within (MSW).
k
MSW 
2
 (n j  1) s 2j
j1
nT  k
• The numerator of MSW is called the sum of
squares within (SSW).
• The denominator of MSW represents the
degrees of freedom associated with SSW.
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Comparing the Variance Estimates: The
F Test
• If the null hypothesis is true and the ANOVA
assumptions are valid, the sampling
distribution of MSB/MSW is an F distribution
with MSB d.f. equal to k - 1 and MSW d.f.
equal to nT - k.
• If the means of the k populations are not
equal, the value of MSB/MSW will be inflated
because MSB overestimates 2.
• Hence, we will reject H0 if the resulting value
of MSB/MSW appears to be too large to have
been selected at random from the appropriate
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F distribution.
Test for the Equality of k
Population Means
• Hypotheses
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal
• Test Statistic
F = MSB/MSW
• Rejection Rule
Reject H0 if F > F
where the value of F is based on an F distribution
with k - 1 numerator degrees of freedom and nT - 1
denominator degrees of freedom.
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Sampling Distribution of
MSTR/MSE
• The figure below shows the
rejection region associated
with a level of significance
equal to  where F denotes
the critical value.
Do Not Reject H0
Reject H0
F
Critical Value
MSTR/MSE
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The ANOVA Table
Source of
Sum of
Variation Squares
Treatment SSTR
Error
SSE
Total
SST
Degrees of
Freedom
k-1
nT - k
nT - 1
Mean
Squares
F
MSTR MSTR/MSE
MSE
SST divided by its degrees of freedom nT - 1 is simply the
overall sample variance that would be obtained if we
treated the entire nT observations as one data set.
k
nj
SST   ( xij  x) 2  SSTR  SSE
j 1 i 1
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Fisher’s LSD Procedure
• Hypotheses
H0: i = j
Ha: i  j
• Test Statistic
xi  x j
t
MSW( 1 n  1 n )
i
j
• Rejection Rule
Reject H0 if t < -ta/2 or t > ta/2
where the value of ta/2 is based on a t distribution
with nT - k degrees of freedom.
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Fisher’s LSD Procedure_ _
Based on the Test Statistic xi - xj
• Hypotheses
• Test Statistic
• Rejection Rule
H0: i = j
Ha: i  j
_
_
xi - xj
_
_
Reject H0 if |xi - xj| > LSD
where
LSD  t  /2 MSW ( 1 n  1 n )
i
j
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ANOVA Table for a
Completely Randomized Design
Source of
Variation
F
Sum of
Squares
Degrees of
Freedom
Treatments
SSTR
k-1
Error
SSE
nT - k
Total
SST
nT - 1
Mean
Squares
SSTR
MSTR 
k-1
MSE 
MSTR
MSE
SSE
nT - k
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• Selamat Belajar Semoga Sukses.
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