1. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 1 Slide Slide Slides Prepared by Juei-Chao Chen Fu Jen Catholic University Slides Prepared by Juei-Chao Chen Fu Jen Catholic University

2. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 2 Slide Slide Chapter 13 STATISTICS in PRACTICE nBurke Marketing Services, Inc., is one of the most experienced market research firms in the industry. n In one study, a firm retained Burke to evaluate potential new versions of a children’s dry cereal. nAnalysis of variance was the statistical method used to study the data obtained from the taste tests. nThe experimental design employed by Burke and the subsequent analysis of variance were helpful in making a product design recommendation.

3. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 3 Slide Slide Chapter 13, Part A Analysis of Variance and Experimental Design n nIntroduction to Analysis of Variance n nAnalysis of Variance: Testing for the Equality of k Population Means n nMultiple Comparison Procedures

4. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 4 Slide Slide Introduction to Analysis of Variance Analysis of Variance (ANOVA) can be used to test Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. for the equality of three or more population means. Analysis of Variance (ANOVA) can be used to test Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. for the equality of three or more population means. Data obtained from observational or experimental Data obtained from observational or experimental studies can be used for the analysis. studies can be used for the analysis. Data obtained from observational or experimental Data obtained from observational or experimental studies can be used for the analysis. studies can be used for the analysis. We want to use the sample results to test the We want to use the sample results to test the following hypotheses: following hypotheses: We want to use the sample results to test the We want to use the sample results to test the following hypotheses: following hypotheses: H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal

5. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 5 Slide Slide Introduction to Analysis of Variance 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 If H 0 is rejected, we cannot conclude that all population means are different. means are different. If H 0 is rejected, we cannot conclude that all population If H 0 is rejected, we cannot conclude that all population means are different. means are different. Rejecting H 0 means that at least two population means Rejecting H 0 means that at least two population means have different values. have different values. Rejecting H 0 means that at least two population means Rejecting H 0 means that at least two population means have different values. have different values.

6. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 6 Slide Slide n n Sampling Distribution of Given H 0 is True Introduction to Analysis of Variance   Sample means are close together because there is only because there is only one sampling distribution one sampling distribution when H 0 is true. when H 0 is true.

7. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 7 Slide Slide Introduction to Analysis of Variance n n Sampling Distribution of Given H 0 is False 33 33 11 11 22 22 Sample means come from different sampling distributions and are not as close together when H 0 is false. when H 0 is false.

8. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 8 Slide Slide For each population, the response variable is For each population, the response variable is normally distributed. normally distributed. For each population, the response variable is For each population, the response variable is normally distributed. normally distributed. Assumptions for Analysis of Variance The variance of the response variable, denoted  2, The variance of the response variable, denoted  2, is the same for all of the populations. is the same for all of the populations. The variance of the response variable, denoted  2, The variance of the response variable, denoted  2, is the same for all of the populations. is the same for all of the populations. The observations must be independent. The observations must be independent.

9. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 9 Slide Slide Analysis of Variance: Testing for the Equality of k Population Means n nBetween-Treatments Estimate of Population Variance n nWithin-Treatments Estimate of Population Variance n nComparing the Variance Estimates: The F Test n nANOVA Table

10. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 10 Slide Slide Analysis of Variance: Testing for the Equality of k Population Means nAnalysis of variance can be used to test for the equality of k population means. nThe hypotheses tested is H 0 : H a : Not all population means are equal where mean of the jth population.

11. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 11 Slide Slide Analysis of Variance: Testing for the Equality of k Population Means nSample data = value of observation i for treatment j = number of observations for treatment j = sample mean for treatment j = sample variance for treatment j = sample standard deviation for treatment j

12. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 12 Slide Slide n Statisitcs n The sample mean for treatment j n The sample variance for treatment j Analysis of Variance: Testing for the Equality of k Population Means

13. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 13 Slide Slide Analysis of Variance: Testing for the Equality of k Population Means nThe overall sample mean where n T = n 1 + n 2 +... + n k nIf the size of each sample is n, n T = kn then

14. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 14 Slide Slide Analysis of Variance: Testing for the Equality of k Population Means nBetween-Treatments Estimate of Population Variance nThe sum of squares due to treatments (SSTR) nThe mean square due to treatments (MSTR)

15. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 15 Slide Slide Analysis of Variance: Testing for the Equality of k Population Means nWithin-Treatments Estimate of Population Variance nThe sum of squares due to error (SSE) nThe mean square due to error (MSE)

16. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 16 Slide Slide Between-Treatments Estimate of Population Variance A between-treatment estimate of  2 is called the mean square treatment and is denoted MSTR. Denominator represents the degrees of freedom the degrees of freedom associated with SSTR associated with SSTR Numerator is the sum of squares sum of squares due to treatments due to treatments and is denoted SSTR

17. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 17 Slide Slide The estimate of  2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE. Within-Samples Estimate of Population Variance Denominator represents the degrees of freedom the degrees of freedom associated with SSE associated with SSE Numerator is the sum of squares sum of squares due to error and is denoted SSE

18. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 18 Slide Slide Comparing the Variance Estimates: The F Test If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with MSTR d.f. equal to k - 1 and MSE d.f. equal to n T - k. If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates  2. Hence, we will reject H 0 if the resulting value of MSTR/MSE appears to be too large to have been selected at random from the appropriate F distribution.

19. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 19 Slide Slide 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 n Hypotheses n n Test Statistic

20. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 20 Slide Slide Test for the Equality of k Population Means n 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 

21. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 21 Slide Slide Sampling Distribution of MSTR/MSE n nRejection Region Do Not Reject H 0 Reject H 0 MSTR/MSE Critical Value FF FF Sampling Distribution of MSTR/MSE 

22. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 22 Slide Slide 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

23. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 23 Slide Slide 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:

24. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 24 Slide Slide 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 degrees Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the F of freedom provides the variance estimates and the F value used to test the hypothesis of equal population value used to test the hypothesis of equal population means. means. Dividing the sum of squares by the appropriate degrees Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the F of freedom provides the variance estimates and the F value used to test the hypothesis of equal population value used to test the hypothesis of equal population means. means.

25. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 25 Slide Slide Test for the Equality of k Population Means nExample: nNational Computer Products, Inc. (NCP), manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. nObject: To measure how much employees at these plants know about total quality management. nA random sample of six employees was selected from each plant and given a quality awareness examination.

26. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 26 Slide Slide Test for the Equality of k Population Means nData nLet = mean examination score for population 1 = mean examination score for population 2 = mean examination score for population 3

27. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 27 Slide Slide Test for the Equality of k Population Means nHypotheses H 0 : = = H a : Not all population means are equal nIn this example 1. dependent or response variable : examination score 2. independent variable or factor : plant location 3. levels of the factor or treatments : the values of a factor selected for investigation, in the NCP example the three treatments or three population are Atlanta, Dallas, and Seattle.

28. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 28 Slide Slide Test for the Equality of k Population Means Three assumptions 1. For each population, the response variable is normally distributed. The examination scores (response variable) must be normally distributed at each plant. 2. The variance of the response variable,, is the same for all of the populations. The variance of examination scores must be the same for all three plants. 3. The observations must be independent. The examination score for each employee must be independent of the examination score for any other employee.

29. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 29 Slide Slide Test for the Equality of k Population Means nANOVA Table np-value = 0.003 < α =.05. We reject H 0.

30. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 30 Slide Slide n nExample: Reed Manufacturing Test for the Equality of k Population Means Janet Reed would like 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 (in Buffalo, Pittsburgh, and Detroit).

31. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 31 Slide Slide n nExample: Reed Manufacturing Test for the Equality of k Population Means 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 α =.05.

32. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 32 Slide Slide 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

33. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 33 Slide Slide Test for the Equality of k Population Means 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  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 1. Develop the hypotheses. p -Value and Critical Value Approaches

34. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 34 Slide Slide 2. Specify the level of significance.  =.05 Test for the Equality of k Population Means 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 (Sample sizes are all equal.) Mean Square Due to Treatments = (55 + 68 + 57)/3 = 60

35. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 35 Slide Slide 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

36. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 36 Slide Slide 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

37. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 37 Slide Slide 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 4. Compute the p –value.

38. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 38 Slide Slide 5. Determine whether to reject H 0. Because F = 9.55 > 3.89, we reject H 0. 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.

39. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 39 Slide Slide Test for the Equality of k Population Means nSummary

40. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 40 Slide Slide Multiple Comparison Procedures nSuppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. n nFisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.

41. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 41 Slide Slide Fisher’s LSD Procedure nTest Statistic n nHypotheses

42. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 42 Slide Slide Fisher’s LSD Procedure where the value of t a /2 is based on a t distribution with n T - k degrees of freedom. n n Rejection Rule Reject H 0 if p-value < a p-value Approach: Critical Value Approach: Reject H 0 if t t a /2

43. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 43 Slide Slide nTest Statistic Fisher’s LSD Procedure Based on the Test Statistic x i - x j where Reject H 0 if > LSD n nHypotheses n nRejection Rule

44. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 44 Slide Slide Fisher’s LSD Procedure Based on the Test Statistic x i - x j n nExample: 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.

45. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 45 Slide Slide 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

46. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 46 Slide Slide nLSD for Plants 1 and 2 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Test Statistic = |55 - 68| = 13 Reject H 0 if > 6.98 Rejection Rule Hypotheses (A) The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

47. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 47 Slide Slide n nLSD for Plants 1 and 3 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Test Statistic = |55  57| = 2 Reject H 0 if > 6.98 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.

48. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 48 Slide Slide n nLSD for Plants 2 and 3 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Test Statistic = |68 - 57| = 11 Reject H 0 if > 6.98 Rejection Rule Hypotheses (C) The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3.

49. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 49 Slide Slide nThe experimentwise Type I error rate gets larger for problems with more populations (larger k). Type I Error Rates  EW = 1 – (1 –  ) ( k – 1)! The comparisonwise Type I error rate  indicates the level of significance associated with a single pairwise comparison. The experimentwise Type I error rate  EW is the probability of making a Type I error on at least one of the (k – 1)! pairwise comparisons.

50. © 2006 by Thomson Learning, a division of Thomson Asia Pte Ltd.. 50 Slide Slide End of Chapter 13, Part A