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.
2

Outline Materi
• Konsep dasar analisis varians
• Klasifikasi satu arah ulangan sama
• Klasifikasi satu arah ulangan tidak sama
• Prosedur uji F
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
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.
H
0
:

1
=

2
=

3
= . . . =
 k
H a
: Not all population means are equal
• If H
0 is rejected, we cannot conclude that all population means are different.
• Rejecting H
0 means that at least two population means have different values.
5

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
H
0
: 
1
= 
2
= 
3
= . . . =  k
H a
: 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 H
0 if p-value < 
Critical Value Approach: Reject H
0 if F > F
 where the value of F and n
T
 is based on an
F distribution with k - 1 numerator d.f.
- k denominator d.f.

Sampling Distribution of MSTR/MSE

Rejection Region
Sampling Distribution of MSTR/MSE
Do Not Reject H
0
Reject H
0

F

Critical Value
MSTR/MSE

ANOVA Table
Source of
Variation
Treatment
Error
Total
Sum of
Squares
SSTR
SSE
SST
Degrees of
Freedom
k – 1 n
T
– k n
T
- 1
Mean
Squares
MSTR
MSE
F
MSTR/MSE
SST is partitioned into SSTR and SSE.
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 n
T
– 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:
SST  k n
 j j  1 i  1
( x ij
 x ) 2  

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
3
4
1
2
5
Sample Mean
Sample Variance
Plant 1
Buffalo
48
54
57
54
62
55
26.0
Plant 2
Pittsburgh
73
63
66
64
74
68
26.5
Plant 3
Detroit
51
63
61
54
56
57
24.5

Test for the Equality of k Population Means
 p -Value and Critical Value Approaches
1. Develop the hypotheses.
H
0
: 
1
= 
2
= 
3
H a
: Not all the means are equal where:



1
= mean number of hours worked per
2 week by the managers at Plant 1
= mean number of hours worked per
3 week by the managers at Plant 2
= 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
2
12
14
Mean
Squares
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 H
0
.
The p-value < .05, so we reject H
0
.
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 H
0 if F > 3.89
5. Determine whether to reject H
0
.
Because F = 9.55 > 3.89, we reject H
0
.
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
H
0 a
:
:
 
 i i
  j j
• Test Statistic t 
 j
MSE(
1 n i
 1 n j
)

Fisher’s LSD Procedure

Rejection Rule
p-value Approach:
Reject H
0 if p-value < 
Critical Value Approach:
Reject H
0 if t < -t
a/2 or t > t
a/2 where the value of t
a/2 is based on a
t distribution with n
T
- k degrees of freedom.


Fisher’s LSD Procedure
Based on the Test Statistic x i
Hypotheses
H
H
0 a
:
:
 
 i i
  j j
x j
• Test Statistic i j

Rejection Rule
Reject H
0 where
LSD  t
 /2
MSE(
1 n i
 1 n j
)

Fisher’s LSD Procedure
Based on the Test Statistic x i
- x j

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 x i
x j
For

= .05 and n
T
k
– 3 = 12 degrees of
= 15 freedom, t
.
025
= 2.179
LSD  t
 /2
MSE(
1 n i
 1 n j
)
LSD  .
.
.
.
( (  ) )  .
.
MSE value was computed earlier

Fisher’s LSD Procedure
Based on the Test Statistic x i
• LSD for Plants 1 and 2
•
Hypotheses (A) H
H
0 a
:  
: 
1
 
2
2
•
Rejection Rule
•
Reject H
0 if x
1
 x
2
> 6.98
Test Statistic x
1
 x = |55  68| = 13
x j
•
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 x i
- x j

•
LSD for Plants 1 and 3
•
Hypotheses (B) H
H
0 a
:
:
 

1
 
3
3
Rejection Rule
•
Reject H
0 if x
1
 x
3
> 6.98
Test Statistic x
1
 x = |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 x i
- x j

•
LSD for Plants 2 and 3
•
Hypotheses (C) H
H
0 a
:
:


2
2




3
3
Rejection Rule
•
Reject H
0 if x
2
 x
3
> 6.98
Test Statistic x
2
 x = |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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