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Pertemuan 1 9 Analisis Varians Klasifikasi Satu Arah

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Pertemuan 1 9 Analisis Varians Klasifikasi Satu Arah. Matakuliah: I0284 - Statistika Tahun: 200 8 Versi: Revisi. Learning Outcomes. Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa akan dapat men erapkan uji perbedaan rata-rata lebih dari 2 populasi . - PowerPoint PPT Presentation
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1 Pertemuan 19 Analisis Varians Klasifikasi Satu Arah Matakuliah : I0284 - Statistika Tahun : 2008 Versi : Revisi
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  • *Pertemuan 19Analisis Varians Klasifikasi Satu ArahMatakuliah: I0284 - StatistikaTahun: 2008Versi: Revisi

  • *Learning OutcomesPada akhir pertemuan ini, diharapkan mahasiswa akan mampu :Mahasiswa akan dapat menerapkan uji perbedaan rata-rata lebih dari 2 populasi.

  • *Outline MateriKonsep dasar analisis variansKlasifikasi satu arah ulangan samaKlasifikasi satu arah ulangan tidak samaProsedur uji F

  • *

    Analysis of Variance and Experimental DesignAn Introduction to Analysis of Variance Analysis of Variance: Testing for the Equality of k Population MeansMultiple Comparison ProceduresAn Introduction to Experimental DesignCompletely Randomized DesignsRandomized Block Design

  • *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

  • *Assumptions for Analysis of VarianceFor 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.

  • Test for the Equality of k Population MeansF = MSTR/MSEH0: 1=2=3=. . . = kHa: Not all population means are equalHypothesesTest Statistic

  • Test for the Equality of k Population MeansRejection Rulewhere the value of F is based on anF distribution with k - 1 numerator d.f.and nT - k denominator d.f.Reject H0 if p-value < ap-value Approach:Critical Value Approach:Reject H0 if F > Fa

  • Sampling Distribution of MSTR/MSERejection RegionDo Not Reject H0Reject H0MSTR/MSECritical ValueFSampling Distributionof MSTR/MSEa

  • ANOVA TableSST is partitionedinto SSTR and SSE.SSTs degrees of freedom(d.f.) are partitioned intoSSTRs d.f. and SSEs d.f.TreatmentErrorTotalSSTRSSESSTk 1nT knT - 1MSTRMSESource ofVariationSum ofSquaresDegrees ofFreedomMeanSquaresMSTR/MSEF

  • 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:

  • 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.

  • Example: Reed ManufacturingTest for the Equality of k Population Means A simple random sample of fivemanagers from each of the three plantswas taken and the number of hoursworked by each manager for theprevious week is shown on the nextslide. Conduct an F test using a = .05.

  • 12345485457546273636664745163615456Plant 1BuffaloPlant 2PittsburghPlant 3DetroitObservationSample MeanSample Variance55 68 5726.0 26.5 24.5Test for the Equality of k Population Means

  • Test for the Equality of k Population MeansH0: 1= 2= 3Ha: Not all the means are equalwhere: 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 31. Develop the hypotheses. p -Value and Critical Value Approaches

  • 2. Specify the level of significance.a = .05Test for the Equality of k Population Means p -Value and Critical Value Approaches3. Compute the value of the test statistic.MSTR = 490/(3 - 1) = 245SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490(Sample sizes are all equal.)Mean Square Due to Treatments

  • 3. Compute the value of the test statistic.Test for the Equality of k Population MeansMSE = 308/(15 - 3) = 25.667SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308Mean Square Due to Error(continued)F = MSTR/MSE = 245/25.667 = 9.55 p -Value and Critical Value Approaches

  • TreatmentErrorTotal4903087982121424525.667Source ofVariationSum ofSquaresDegrees ofFreedomMeanSquares9.55FTest for the Equality of k Population Means ANOVA Table

  • Test for the Equality of k Population Means5. Determine whether to 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.The p-value < .05, so we reject H0.With 2 numerator d.f. and 12 denominator d.f.,the p-value is .01 for F = 6.93. Therefore, thep-value is less than .01 for F = 9.55. p Value Approach4. Compute the p value.

  • 5. Determine whether to reject H0.Because F = 9.55 > 3.89, we reject H0. Critical Value Approach4. Determine the critical value and rejection rule.Reject H0 if F > 3.89Test for the Equality of k Population MeansWe 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.Based on an F distribution with 2 numeratord.f. and 12 denominator d.f., F.05 = 3.89.

  • Multiple Comparison ProceduresSuppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means.Fishers least significant difference (LSD) procedure can be used to determine where the differences occur.

  • Fishers LSD ProcedureTest StatisticHypotheses

  • Fishers LSD Procedure where the value of ta/2 is based on at distribution with nT - k degrees of freedom.Rejection RuleReject H0 if p-value < ap-value Approach:Critical Value Approach:Reject H0 if t < -ta/2 or t > ta/2

  • Test StatisticFishers LSD ProcedureBased on the Test Statistic xi - xj__whereHypothesesRejection Rule

  • Fishers LSD Procedure Based on the Test Statistic xi - xjExample: Reed Manufacturing Recall that Janet Reed wants to knowif there is any significant difference inthe mean number of hours worked per week for the department managersat her three manufacturing plants. Analysis of variance has providedstatistical evidence to reject the nullhypothesis of equal population means.Fishers least significant difference (LSD) procedurecan be used to determine where the differences occur.

  • For = .05 and nT - k = 15 3 = 12 degrees of freedom, t.025 = 2.179

    MSE value wascomputed earlierFishers LSD ProcedureBased on the Test Statistic xi - xj

  • LSD for Plants 1 and 2Fishers LSD ProcedureBased on the Test Statistic xi - xj Conclusion Test Statistic Rejection Rule Hypotheses (A) The mean number of hours worked at Plant 1 isnot equal to the mean number worked at Plant 2.

  • LSD for Plants 1 and 3Fishers LSD Procedure Based on the Test Statistic xi - xj Conclusion Test Statistic Rejection Rule Hypotheses (B) 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.

  • LSD for Plants 2 and 3Fishers LSD Procedure Based on the Test Statistic xi - xj Conclusion Test Statistic Rejection Rule Hypotheses (C) 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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