Matakuliah
Tahun
: I0272 - STATISTIK PROBABILITAS
: 2009
ANALISIS RAGAM (ANALISIS VARIAN)
Pertemuan 13
Materi
• Analisis varians klasifikasi satu arah
• Ulangan sama dan tidak sama
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3
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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Test for the Equality of k Population Means

Hypotheses
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal

Test Statistic
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F = MSTR/MSE
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Test for the Equality of k Population Means

Rejection Rule
p-value Approach:
Reject H0 if p-value < 
Critical Value Approach:
Reject H0 if F > F
where the value of F is based on an
F distribution with k - 1 numerator d.f.
and nT - k denominator d.f.
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Sampling Distribution of MSTR/MSE

Rejection Region
Sampling Distribution
of MSTR/MSE
Reject H0

Do Not Reject H0
F
Critical Value
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MSTR/MSE
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ANOVA Table
Source of
Variation
Treatment
Error
Total
Sum of Degrees of
Squares Freedom
SSTR
SSE
SST
SST is partitioned
into SSTR and SSE.
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k–1
nT – k
nT - 1
Mean
Squares
MSTR
MSE
F
MSTR/MSE
SST’s degrees of freedom
(d.f.) are partitioned into
SSTR’s d.f. and SSE’s d.f.
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ANOVA Table
SST divided by its degrees of freedom nT – 1 is the
overall sample variance that would be obtained if we
treated the entire set of observations as one data set.
With the entire data set as one sample, the formula
for computing the total sum of squares, SST, is:
k
nj
SST   ( xij  x )2  SSTR  SSE
j 1 i 1
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ANOVA Table
ANOVA can be viewed as the process of partitioning
the total sum of squares and the degrees of freedom
into their corresponding sources: treatments and error.
Dividing the sum of squares by the appropriate
degrees of freedom provides the variance estimates
and the F value used to test the hypothesis of equal
population means.
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Test for the Equality of k Population Means

Example: Reed Manufacturing
A simple random sample of five
managers from each of the three
plants
was taken and the number of hours
worked by each manager for the
previous week is shown on the next
slide.
Conduct an F test using  = .05.
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Test for the Equality of k Population Means
Observation
1
2
3
4
5
Plant 1
Buffalo
Plant 2
Pittsburgh
48
54
57
54
62
73
63
66
64
74
Plant 3
Detroit
51
63
61
54
56
Sample Mean
55
68
57
Sample Variance
26.0
26.5
24.5
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Test for the Equality of k Population Means
 p -Value and Critical Value Approaches
1. Develop the hypotheses.
H0:  1 =  2 =  3
Ha: Not all the means are equal
where:
 1 = mean number of hours worked per
week by the managers at Plant 1
 2 = mean number of hours worked per
week by the managers at Plant 2
 3 = mean number of hours worked per
week by the managers at Plant 3
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Test for the Equality of k Population Means
 p -Value and Critical Value Approaches
2. Specify the level of significance.
 = .05
3. Compute the value of the test statistic.
Mean Square Due to Treatments
(Sample sizes are all equal.)
x = (55 + 68 + 57)/3 = 60
SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490
MSTR = 490/(3 - 1) = 245
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Test for the Equality of k Population Means (continued)
 p -Value and Critical Value Approaches
3. Compute the value of the test statistic.
Mean Square Due to Error
SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308
MSE = 308/(15 - 3) = 25.667
F = MSTR/MSE = 245/25.667 = 9.55
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Test for the Equality of k Population Means
 ANOVA Table
Source of
Variation
Treatment
Error
Total
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Sum of
Squares
490
308
798
Degrees of
Freedom
2
12
14
Mean
Squares
245
25.667
F
9.55
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Test for the Equality of k Population Means
 p –Value Approach
4. Compute the p –value.
With 2 numerator d.f. and 12 denominator d.f.,
the p-value is .01 for F = 6.93. Therefore, the
p-value is less than .01 for F = 9.55.
5. Determine whether to reject H0.
The p-value < .05, so we reject H0.
We have sufficient evidence to conclude that the
mean number of hours worked per week by
department managers is not the same at all 3 plant.
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Test for the Equality of k Population Means
 Critical Value Approach
4. Determine the critical value and rejection rule.
Based on an F distribution with 2 numerator
d.f. and 12 denominator d.f., F.05 = 3.89.
Reject H0 if F > 3.89
5. Determine whether to reject H0.
Because F = 9.55 > 3.89, we reject H0.
We have sufficient evidence to conclude that the
mean number of hours worked per week by
department managers is not the same at all 3 plant.
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