Presentation on theme: "Analisis Varians/Ragam Klasifikasi Dua Arah Pertemuan 18 Matakuliah: L0104 / Statistika Psikologi Tahun : 2008."— Presentation transcript:
Analisis Varians/Ragam Klasifikasi Dua Arah Pertemuan 18 Matakuliah: L0104 / Statistika Psikologi Tahun : 2008
Bina Nusantara Learning Outcomes 3 Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa akan dapat menyusun simpulan hasil analisis varians.
Bina Nusantara Outline Materi 4 Analisis varians rancangan kelompok Partisi jumlah kuadrat perlakuan Prosedur uji F Pembandingan ganda perlakuan
Bina Nusantara 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
Bina Nusantara 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 : μ 1 = μ 2 = μ 3 = ... = μ 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. An Introduction to Analysis of Variance
Bina Nusantara Assumptions for Analysis of Variance For each population, the response variable is normally distributed. The variance of the response variable, denoted σ 2, is the same for all of the populations. The observations must be independent.
Bina Nusantara 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
Bina Nusantara Randomized Block Design The ANOVA Procedure Computations and Conclusions
Bina Nusantara 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, n T - 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
Bina Nusantara ANOVA Table for a Randomized 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 n T - 1
Bina Nusantara 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.
Bina Nusantara Contoh Soal: Eastern Oil Co. Automobile Type of Gasoline (Treatment) Blocks (Block) Blend X Blend Y Blend Z Means Treatment Means
Bina Nusantara Contoh Soal: Eastern Oil Co. Randomized Block Design –Mean Square Due to Treatments The overall sample mean is 29. Thus, SSTR = 5[( ) 2 + ( ) 2 + ( ) 2 ] = 5.2 MSTR = 5.2/(3 - 1) = 2.6 –Mean Square Due to Blocks SSBL = 3[( ) ( ) 2 ] = MSBL = 51.33/(5 - 1) = 12.8 –Mean Square Due to Error SSE = = 5.47 MSE = 5.47/[(3 - 1)(5 - 1)] =.68
Bina Nusantara Contoh Soal: Eastern Oil Co. Randomized Block Design –Rejection Rule Assuming α =.05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator). Reject H 0 if F > –Test Statistic F = MSTR/MSE = 2.6/.68 = 3.82 –Conclusion Since 3.82 < 4.46, we cannot reject H 0. There is not sufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends.