Matakuliah
Tahun
: I0184 – Teori Statistika II
: 2009
Rancangan Acak Lengkap (Analisis Varians
Klasifikasi Satu Arah)
Pertemuan 16
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa akan dapat menerapkan uji perbedaan ratarata lebih dari 2 populasi.
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Outline Materi
•
•
•
•
Konsep dasar analisis varians
Klasifikasi satu arah ulangan sama
Klasifikasi satu arah ulangan tidak sama
Prosedur uji F
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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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Test for the Equality of k Population Means

Hypotheses
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal

Test Statistic
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 < a
Critical Value Approach:
Reject H0 if F > Fa
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.
10
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 a = .05.
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Test for the Equality of k Population Means
Observation
1
2
3
4
5
Plant 1
Buffalo
48
54
57
54
62
Sample Mean
55
Sample Variance 26.0
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Plant 2
Pittsburgh
73
63
66
64
74
Plant 3
Detroit
51
63
61
54
56
68
26.5
57
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
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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. a = .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) =
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245
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Test for the Equality of k Population Means
 p -Value and Critical Value Approaches
3. Compute the value of the test statistic. (continued)
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 Degrees of
Squares Freedom
490
308
798
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.
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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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Multiple Comparison Procedures
• Suppose that analysis of variance has provided
statistical evidence to reject the null hypothesis of
equal population means.

Fisher’s least significant difference (LSD)
procedure can be used to determine where the
differences occur.
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Fisher’s LSD Procedure

Hypotheses
H 0 : i   j
H a : i   j
• Test Statistic
t
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xi  x j
MSE( 1 n  1 n )
i
j
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Fisher’s LSD Procedure

Rejection Rule
p-value Approach:
Reject H0 if p-value < 
Critical Value Approach:
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
H 0 : i   j
H a : i   j
• Test Statistic

xi  x j
Rejection Rule
Reject H0 if xi  x j > LSD
where
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LSD  t /2 MSE( 1 n  1 n )
i
j
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Fisher’s LSD Procedure
Based on the Test Statistic xi - xj

Example: Reed Manufacturing
Recall that Janet Reed wants to know
if there is any significant difference in
the mean number of hours worked per
week for the department managers
at her three manufacturing plants.
Analysis of variance has provided
statistical evidence to reject the null
hypothesis of equal population means.
Fisher’s least significant difference (LSD) procedure
can be used to determine where the differences occur.
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Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
For  = .05 and nT - k = 15 – 3 = 12
degrees of freedom, t.025 = 2.179
LSD  t /2 MSE( 1 n  1 n )
i
j
LSD  2. 179 25 . 667 ( 1 5  1 5 )  6. 98
MSE value was
computed earlier
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Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
• LSD for Plants 1 and 2
•
Hypotheses (A) H0 : 1  2
H a : 1   2
•
Rejection Rule
•
Reject H0 if x1  x2 > 6.98
Test Statistic
x1  x2 = |55  68| = 13
•
Conclusion
The mean number of hours worked at Plant 1 is
not equal to the mean number worked at Plant 2.
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Fisher’s LSD Procedure
Based on the Test Statistic xi - xj

LSD for Plants 1 and 3
• Hypotheses (B)
H 0 : 1   3
H a : 1   3
•
•
Rejection Rule
Reject H0 if x1  x3 > 6.98
Test Statistic
x1  x3 = |55  57| = 2
•
Conclusion
There is no significant difference between the mean
number of hours worked at Plant 1 and the mean
number of hours worked at Plant 3.
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Fisher’s LSD Procedure
Based on the Test Statistic xi - xj

LSD for Plants 2 and 3
•
Hypotheses (C)
H 0 : 2  3
H a : 2  3
•
Rejection Rule
Reject H0 if x2  x3 > 6.98
•
Test Statistic
•
x2  x3 = |68  57| = 11
Conclusion
The mean number of hours worked at Plant 2 is
not equal to the mean number worked at Plant 3.
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• Selamat Belajar Semoga Sukses.
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