Pertemuan 1 9 Analisis Varians Klasifikasi Satu Arah

# Pertemuan 1 9 Analisis Varians Klasifikasi Satu Arah

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## Pertemuan 1 9 Analisis Varians Klasifikasi Satu Arah

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1. Pertemuan 19Analisis Varians Klasifikasi Satu Arah Matakuliah : I0284 - Statistika Tahun : 2008 Versi : Revisi

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

3. Outline Materi • Konsep dasar analisis varians • Klasifikasi satu arah ulangan sama • Klasifikasi satu arah ulangan tidak sama • Prosedur uji F

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. 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.

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. 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

8. Test for the Equality of k Population Means • Rejection Rule Reject H0 if p-value <a p-value Approach: Critical Value Approach: Reject H0 if F>Fa where the value of F is based on an F distribution with k - 1 numerator d.f. and nT - k denominator d.f.

9. Sampling Distribution of MSTR/MSE • Rejection Region Sampling Distribution of MSTR/MSE Reject H0 a Do Not Reject H0 MSTR/MSE F Critical Value

10. ANOVA Table Source of Variation Sum of Squares Mean Squares Degrees of Freedom F SSTR SSE SST k – 1 nT – k nT - 1 Treatment Error Total MSTR MSE MSTR/MSE SST’s degrees of freedom (d.f.) are partitioned into SSTR’s d.f. and SSE’s d.f. SST is partitioned into SSTR and SSE.

11. 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:

12. 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.

13. 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 a = .05.

14. Test for the Equality of k Population Means Plant 3 Detroit Plant 2 Pittsburgh Plant 1 Buffalo Observation 1 2 3 4 5 48 54 57 54 62 51 63 61 54 56 73 63 66 64 74 Sample Mean 55 68 57 Sample Variance 26.0 26.5 24.5

15. 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

16. = (55 + 68 + 57)/3 = 60 Test for the Equality of k Population Means • p -Value and Critical Value Approaches a = .05 2. Specify the level of significance. 3. Compute the value of the test statistic. Mean Square Due to Treatments (Sample sizes are all equal.) SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490 MSTR = 490/(3 - 1) = 245

17. 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

18. Test for the Equality of k Population Means • ANOVA Table Source of Variation Sum of Squares Mean Squares Degrees of Freedom F 490 308 798 2 12 14 Treatment Error Total 245 25.667 9.55

19. 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.

20. 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.

21. 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.

22. Fisher’s LSD Procedure • Hypotheses • Test Statistic

23. Fisher’s LSD Procedure • Rejection Rule p-value Approach: Reject H0 if p-value <a 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.

24. Reject H0 if > LSD Fisher’s LSD ProcedureBased on the Test Statistic xi - xj _ _ • Hypotheses • Test Statistic • Rejection Rule where

25. Fisher’s LSD ProcedureBased 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.

26. Fisher’s LSD ProcedureBased on the Test Statistic xi - xj For  = .05 and nT - k = 15 – 3 = 12 degrees of freedom, t.025 = 2.179 MSE value was computed earlier

27. Reject H0 if > 6.98 = |55 - 68| = 13 Fisher’s LSD ProcedureBased on the Test Statistic xi - xj • LSD for Plants 1 and 2 • Hypotheses (A) • Rejection Rule • Test Statistic • Conclusion • The mean number of hours worked at Plant 1 is • not equal to the mean number worked at Plant 2.

28. Reject H0 if > 6.98 = |55 - 57| = 2 Fisher’s LSD ProcedureBased on the Test Statistic xi - xj • LSD for Plants 1 and 3 • Hypotheses (B) • Rejection Rule • Test Statistic • 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.

29. Reject H0 if > 6.98 = |68 - 57| = 11 Fisher’s LSD ProcedureBased on the Test Statistic xi - xj • LSD for Plants 2 and 3 • Hypotheses (C) • Rejection Rule • Test Statistic • Conclusion The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3.

30. Selamat Belajar Semoga Sukses.