1

Pertemuan 11Analisis Ragam (Varians) - 2

Matakuliah : I0262 – Statistik Probabilitas

Tahun : 2007

Versi : Revisi

2

Learning Outcomes

Pada akhir pertemuan ini, diharapkan mahasiswa

akan mampu :

• Mahasiswa akan dapat membandingkan beberapa perlakuan.

3

Outline Materi

• Analisis varians klasifikasi dua arah

• Analisis varians bebas sebaran

4

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

5

• 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.

An Introduction to Analysis of Variance

6

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.

7

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

8

Randomized Block Design

• The ANOVA Procedure

• Computations and Conclusions

9

• The ANOVA procedure for the randomized block design requires us to partition the sum of squares total (SST) into three groups: sum of squares due to treatments, sum of squares due to blocks, and sum of squares due to error.

• The formula for this partitioning is•

SST = SSTR + SSBL + SSE

• The total degrees of freedom, nT - 1, are partitioned such that k - 1 degrees of freedom go to treatments,

b - 1 go to blocks, and (k - 1)(b - 1) go to the error term.

The ANOVA Procedure

10

ANOVA Table for aRandomized Block Design

Source of Sum of Degrees of Mean

Variation Squares Freedom Squares F

Treatments SSTR k - 1

Blocks SSBL b - 1

Error SSE (k - 1)(b - 1)

Total SST nT - 1

MSTRSSTR

-

k 1MSTR

SSTR-

k 1

MSESSE

( )( )k b1 1MSE

SSE ( )( )k b1 1

MSBLSSBL

-

b 1MSBL

SSBL-

b 1

MSTRMSE

MSTRMSE

11

Contoh Soal: Eastern Oil Co.

Eastern Oil has developed three new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the miles per gallon ratings of the three blends is being conducted to determine if the mean ratings are the same for the three blends.

Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide.

12

Contoh Soal: Eastern Oil Co.

Automobile Type of Gasoline (Treatment) Blocks

(Block) Blend X Blend Y Blend Z Means

1 31 30 30 30.333

2 30 29 29 29.333

3 29 29 28 28.667

4 33 31 29 31.000

5 26 25 26 25.667Treatment Means 29.8 28.8 28.4

13

Contoh Soal: Eastern Oil Co.

• Randomized Block Design– Mean Square Due to Treatments

The overall sample mean is 29. Thus,

SSTR = 5[(29.8 - 29)2 + (28.8 - 29)2 + (28.4 - 29)2] = 5.2

MSTR = 5.2/(3 - 1) = 2.6– Mean Square Due to Blocks

SSBL = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33

MSBL = 51.33/(5 - 1) = 12.8 – Mean Square Due to Error

SSE = 62 - 5.2 - 51.33 = 5.47

MSE = 5.47/[(3 - 1)(5 - 1)] = .68

14

Contoh Soal: Eastern Oil Co.

• Randomized Block Design– Rejection Rule

Assuming = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator). Reject H0 if F > 4.46.

– Test Statistic

F = MSTR/MSE = 2.6/.68 = 3.82– Conclusion

Since 3.82 < 4.46, we cannot reject H0. There is not sufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends.

15

• Selamat Belajar Semoga Sukses.