Matakuliah
Tahun
Versi
: I0284 - Statistika
: 2008
: Revisi
Pertemuan 19
Analisis Varians Klasifikasi Satu Arah
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa akan dapat menerapkan uji
perbedaan rata-rata lebih dari 2 populasi.
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Outline Materi
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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
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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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
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.
Sampling Distribution of MSTR/MSE

Rejection Region
Sampling Distribution
of MSTR/MSE
Reject H0
Do Not Reject H0

F
Critical Value
MSTR/MSE
ANOVA Table
Source of
Variation
Treatment
Error
Total
Sum of
Squares
Degrees of
Freedom
Mean
Squares
SSTR
SSE
SST
k–1
nT – k
nT - 1
MSTR
MSE
SST is partitioned
into SSTR and SSE.
F
MSTR/MSE
SST’s degrees of freedom
(d.f.) are partitioned into
SSTR’s d.f. and SSE’s d.f.
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
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.
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.
Test for the Equality of k Population Means
Observation
1
2
3
4
5
Sample Mean
Sample Variance
Plant 1
Buffalo
48
54
57
54
62
Plant 2
Pittsburgh
73
63
66
64
74
Plant 3
Detroit
51
63
61
54
56
55
26.0
68
26.5
57
24.5
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
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
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
Test for the Equality of k Population Means
 ANOVA Table
Source of
Variation
Treatment
Error
Total
Sum of
Squares
490
308
798
Degrees of
Freedom
Mean
Squares
2
12
14
245
25.667
F
9.55
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.
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.
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.
Fisher’s LSD Procedure

Hypotheses
H 0 : i   j
H a : i   j
• Test Statistic
t
xi  x j
MSE( 1 n  1 n )
i
j
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.
Fisher’s LSD Procedure
_ _
Based on the Test Statistic xi - xj

Hypotheses
H 0 : i   j
H a : i   j
• Test Statistic

Rejection Rule
where
xi  x j
Reject H0 if xi  x j > LSD
LSD  t /2 MSE( 1 n  1 n )
i
j
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.
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
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
• LSD for Plants 1 and 2
•
Hypotheses (A)
•
Rejection Rule
H 0 : 1   2
H a : 1   2
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.
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj

LSD for Plants 1 and 3
•
Hypotheses (B)
•
Rejection Rule
H 0 : 1   3
H a : 1   3
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.
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj

LSD for Plants 2 and 3
•
Hypotheses (C)
•
Rejection Rule
H 0 : 2  3
H a : 2  3
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.
• Selamat Belajar Semoga Sukses.
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