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

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

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

4. 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. 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.  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. An Introduction to Analysis of Variance

6. 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 F = MSTR/MSE H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal n Hypotheses n Test Statistic

8. Test for the Equality of k Population Means n Rejection Rule where the value of F  is based on an F distribution with k - 1 numerator d.f. and n T - k denominator d.f. Reject H 0 if p -value <  p -value Approach: Critical Value Approach: Reject H 0 if F > F 

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

10. ANOVA Table 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. Treatment Error Total SSTR SSE SST k– 1 n T n T – k nT nT nT nT - 1 MSTR MSE Source of Variation Sum of Squares Degrees of Freedom MeanSquares MSTR/MSE F

11. ANOVA Table SST divided by its degrees of freedom n T – 1 is the SST divided by its degrees of freedom n T – 1 is the overall sample variance that would be obtained if we overall sample variance that would be obtained if we treated the entire set of observations as one data set. treated the entire set of observations as one data set. SST divided by its degrees of freedom n T – 1 is the SST divided by its degrees of freedom n T – 1 is the overall sample variance that would be obtained if we overall sample variance that would be obtained if we treated the entire set of observations as one data set. treated the entire set of observations as one data set. With the entire data set as one sample, the formula With the entire data set as one sample, the formula for computing the total sum of squares, SST, is: for computing the total sum of squares, SST, is: With the entire data set as one sample, the formula With the entire data set as one sample, the formula for computing the total sum of squares, SST, is: for computing the total sum of squares, SST, is:

12. ANOVA Table ANOVA can be viewed as the process of partitioning ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error. into their corresponding sources: treatments and error. ANOVA can be viewed as the process of partitioning ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error. into their corresponding sources: treatments and error. Dividing the sum of squares by the appropriate Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal and the F value used to test the hypothesis of equal population means. population means. Dividing the sum of squares by the appropriate Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal and the F value used to test the hypothesis of equal population means. population means.

13. n Example: Reed Manufacturing Test for the Equality of k Population Means A simple random sample of five 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. Conduct an F test using  =.05.

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

15. H 0 :  1  =  2  =  3  H a : Not all the means are equal where:  1 = mean number of hours worked per week by the managers at Plant 1 week by the managers at Plant 1  2 = mean number of hours worked per  2 = mean number of hours worked per week by the managers at Plant 2 week by the managers at Plant 2  3 = mean number of hours worked per week by the managers at Plant 3 week by the managers at Plant 3 1. Develop the hypotheses. p -Value and Critical Value Approaches p -Value and Critical Value Approaches

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

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

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

19. Test for the Equality of k Population Means 5. Determine whether to 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. The p -value <.05, so we reject H 0. 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. p –Value Approach p –Value Approach 4. Compute the p –value.

20. 5. Determine whether to reject H 0. Because F = 9.55 > 3.89, we reject H 0. Critical Value Approach Critical Value Approach 4. Determine the critical value and rejection rule. Reject H 0 if F > 3.89 Test for the Equality of k Population Means 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. Based on an F distribution with 2 numerator d.f. and 12 denominator d.f., F.05 = 3.89.

21. Multiple Comparison Procedures Suppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. n Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.

22. Fisher’s LSD Procedure Test Statistic n Hypotheses

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

24. Test Statistic Fisher’s LSD Procedure Based on the Test Statistic x i - x j __ where Reject H 0 if > LSD n Hypotheses n Rejection Rule

25. Fisher’s LSD Procedure Based on the Test Statistic x i - x j n Example: Reed Manufacturing Recall that Janet Reed wants to know 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 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. For  =.05 and n T - k = 15 – 3 = 12 degrees of freedom, t.025 = 2.179 MSE value was computed earlier Fisher’s LSD Procedure Based on the Test Statistic x i - x j

27. LSD for Plants 1 and 2 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Conclusion Test Statistic Test Statistic = |55  68| = 13 Reject H 0 if > 6.98 Rejection Rule Rejection Rule Hypotheses (A) Hypotheses (A) The mean number of hours worked at Plant 1 is The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

28. n LSD for Plants 1 and 3 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Conclusion Test Statistic Test Statistic = |55  57| = 2 Reject H 0 if > 6.98 Rejection Rule Rejection Rule Hypotheses (B) Hypotheses (B) There is no significant difference between the mean There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3. number of hours worked at Plant 3.

29. n LSD for Plants 2 and 3 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Conclusion Test Statistic Test Statistic = |68  57| = 11 Reject H 0 if > 6.98 Rejection Rule Rejection Rule Hypotheses (C) Hypotheses (C) The mean number of hours worked at Plant 2 is The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3. not equal to the mean number worked at Plant 3.

30. 30 Selamat Belajar Semoga Sukses.