1. 1 1 Slide © 2003 South-Western/Thomson Learning™ Slides Prepared by JOHN S. LOUCKS St. Edward’s University

2. 2 2 Slide © 2003 South-Western/Thomson Learning™ Chapter 13 Analysis of Variance and Experimental Design n An Introduction to Analysis of Variance n Analysis of Variance: Testing for the Equality of k Population Means k Population Means n Multiple Comparison Procedures n An Introduction to Experimental Design n Completely Randomized Designs n Randomized Block Design n Factorial Experiments

3. 3 3 Slide © 2003 South-Western/Thomson Learning™ n 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. n 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 n If H 0 is rejected, we cannot conclude that all population means are different. n Rejecting H 0 means that at least two population means have different values. An Introduction to Analysis of Variance

4. 4 4 Slide © 2003 South-Western/Thomson Learning™ Assumptions for Analysis of Variance n 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 variance of the response variable, denoted  2, is the same for all of the populations. n The observations must be independent.

5. 5 5 Slide © 2003 South-Western/Thomson Learning™ Analysis of Variance: Testing for the Equality of K Population Means n Between-Treatments Estimate of Population Variance n Within-Treatments Estimate of Population Variance n Comparing the Variance Estimates: The F Test n ANOVA Table

6. 6 6 Slide © 2003 South-Western/Thomson Learning™ A between-treatments estimate of  2 is called the mean square due to treatments (MSTR). A between-treatments estimate of  2 is called the mean square due to treatments (MSTR). n The numerator of MSTR is called the sum of squares due to treatments (SSTR). n The denominator of MSTR represents the degrees of freedom associated with SSTR. Between-Treatments Estimate of Population Variance

7. 7 7 Slide © 2003 South-Western/Thomson Learning™ The estimate of  2 based on the variation of the sample observations within each treatment is called the mean square due to error (MSE). The estimate of  2 based on the variation of the sample observations within each treatment is called the mean square due to error (MSE). n The numerator of MSE is called the sum of squares due to error (SSE). n The denominator of MSE represents the degrees of freedom associated with SSE. Within-Treatments Estimate of Population Variance

8. 8 8 Slide © 2003 South-Western/Thomson Learning™ Comparing the Variance Estimates: The F Test n 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. If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates  2. n 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.

9. 9 9 Slide © 2003 South-Western/Thomson Learning™ Test for the Equality of k Population Means n Hypotheses H 0 :  1  =  2  =  3  = ... =  k  H 0 :  1  =  2  =  3  = ... =  k   H a : Not all population means are equal n Test Statistic F = MSTR/MSE

10. 10 Slide © 2003 South-Western/Thomson Learning™ Test for the Equality of k Population Means n Rejection Rule Using test statistic: Reject H 0 if F > F  Using test statistic: Reject H 0 if F > F  Using p -value: Reject H 0 if p -value <  where the value of F  is based on an F distribution with k - 1 numerator degrees of freedom and n T - k denominator degrees of freedom

11. 11 Slide © 2003 South-Western/Thomson Learning™ The figure below shows the rejection region associated with a level of significance equal to  where F  denotes the critical value. The figure below shows the rejection region associated with a level of significance equal to  where F  denotes the critical value. Sampling Distribution of MSTR/MSE Do Not Reject H 0 Reject H 0 MSTR/MSE Critical Value FF FF

12. 12 Slide © 2003 South-Western/Thomson Learning™ ANOVA Table Source of Sum of Degrees of Mean Variation Squares Freedom Squares F TreatmentSSTR k - 1 MSTR MSTR/MSE Error SSE n T - k MSE Total SST n T - 1 SST divided by its degrees of freedom n T - 1 is simply the overall sample variance that would be obtained if we treated the entire n T observations as one data set.

13. 13 Slide © 2003 South-Western/Thomson Learning™ Example: Reed Manufacturing n Analysis of Variance J. R. Reed would like to know if the mean number of hours worked per week is the same for the department managers at her three manufacturing plants (Buffalo, Pittsburgh, and Detroit). A simple random sample of 5 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.

14. 14 Slide © 2003 South-Western/Thomson Learning™ n Sample Data Plant 1Plant 2Plant 3 ObservationBuffalo Pittsburgh Detroit ObservationBuffalo Pittsburgh Detroit 1 48 73 51 1 48 73 51 2 54 63 63 2 54 63 63 3 57 66 61 3 57 66 61 4 54 64 54 4 54 64 54 5 62 74 56 5 62 74 56 Sample Mean 55 68 57 Sample Mean 55 68 57 Sample Variance 26.0 26.5 24.5 Sample Variance 26.0 26.5 24.5 Example: Reed Manufacturing

15. 15 Slide © 2003 South-Western/Thomson Learning™ n Hypotheses H 0 :  1  =  2  =  3  H a : Not all the means are equal where: 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  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 Example: Reed Manufacturing

16. 16 Slide © 2003 South-Western/Thomson Learning™ n Mean Square Due to Treatments Since the sample sizes are all equal Since the sample sizes are all equal x = (55 + 68 + 57)/3 = 60 x = (55 + 68 + 57)/3 = 60 SSTR = 5(55 - 60) 2 + 5(68 - 60) 2 + 5(57 - 60) 2 = 490 SSTR = 5(55 - 60) 2 + 5(68 - 60) 2 + 5(57 - 60) 2 = 490 MSTR = 490/(3 - 1) = 245 MSTR = 490/(3 - 1) = 245 n Mean Square Due to Error SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 MSE = 308/(15 - 3) = 25.667 MSE = 308/(15 - 3) = 25.667 = = Example: Reed Manufacturing

17. 17 Slide © 2003 South-Western/Thomson Learning™ n F - Test If H 0 is true, the ratio MSTR/MSE should be If H 0 is true, the ratio MSTR/MSE should be near 1 because both MSTR and MSE are estimating  2. If H a is true, the ratio should be significantly larger than 1 because MSTR tends to overestimate  2. Example: Reed Manufacturing

18. 18 Slide © 2003 South-Western/Thomson Learning™ Example: Reed Manufacturing n Rejection Rule Using test statistic: Reject H 0 if F > 3.89 Using test statistic: Reject H 0 if F > 3.89 Using p -value: Reject H 0 if p -value <.05 where F.05 = 3.89 is based on an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom

19. 19 Slide © 2003 South-Western/Thomson Learning™ Example: Reed Manufacturing n Test Statistic F = MSTR/MSE = 245/25.667 = 9.55 F = MSTR/MSE = 245/25.667 = 9.55 n Conclusion F = 9.55 > F.05 = 3.89, so we reject H 0. The mean F = 9.55 > F.05 = 3.89, so we reject H 0. The mean number of hours worked per week by department number of hours worked per week by department managers is not the same at each plant. managers is not the same at each plant.

20. 20 Slide © 2003 South-Western/Thomson Learning™ n ANOVA Table Source of Sum of Degrees of Mean Source of Sum of Degrees of Mean Variation Squares Freedom Square F Variation Squares Freedom Square F Treatments 490 2 245 9.55 Treatments 490 2 245 9.55 Error 308 12 25.667 Error 308 12 25.667 Total 798 14 Total 798 14 Example: Reed Manufacturing

21. 21 Slide © 2003 South-Western/Thomson Learning™ n Step 1 Select the Tools pull-down menu n Step 2 Choose the Data Analysis option n Step 3 Choose Anova: Single Factor from the list of Analysis Tools … continued Using Excel’s Anova: Single Factor Tool

22. 22 Slide © 2003 South-Western/Thomson Learning™ n Step 4 When the Anova: Single Factor dialog box appears: Enter B1:D6 in the Input Range box Enter B1:D6 in the Input Range box Select Grouped By Columns Select Grouped By Columns Select Labels in First Row Select Labels in First Row Enter.05 in the Alpha box Enter.05 in the Alpha box Select Output Range Select Output Range Enter A8 (your choice) in the Output Range box Enter A8 (your choice) in the Output Range box Click OK Click OK Using Excel’s Anova: Single Factor Tool

23. 23 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (top portion) Using Excel’s Anova: Single Factor Tool

24. 24 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (bottom portion) Using Excel’s Anova: Single Factor Tool

25. 25 Slide © 2003 South-Western/Thomson Learning™ n Using the p -Value The value worksheet shows that the p -value is.00331 The value worksheet shows that the p -value is.00331 The rejection rule is “Reject H 0 if p -value <.05” The rejection rule is “Reject H 0 if p -value <.05” Thus, we reject H 0 because the p -value =.00331 <  =.05 Thus, we reject H 0 because the p -value =.00331 <  =.05 We conclude that the mean number of hours worked per week by the managers differ among the three plants We conclude that the mean number of hours worked per week by the managers differ among the three plants Using Excel’s Anova: Single Factor Tool

26. 26 Slide © 2003 South-Western/Thomson Learning™ Multiple Comparison Procedures Suppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. Fisher’s least significance difference (LSD) procedure can be used to determine where the differences occur.

27. 27 Slide © 2003 South-Western/Thomson Learning™ Fisher’s LSD Procedure n Hypotheses H 0 :  i =  j H 0 :  i =  j H a :  i  j H a :  i  j n Test Statistic

28. 28 Slide © 2003 South-Western/Thomson Learning™ Fisher’s LSD Procedure n Rejection Rule Using test statistic: Reject H 0 if t t a /2 Using p -value: Reject H 0 if p -value <  where the value of t a /2 is based on a t distribution with n T - k degrees of freedom.

29. 29 Slide © 2003 South-Western/Thomson Learning™ Fisher’s LSD Procedure based on the Test Statistic x i - x j __ _ _ _ _ n Hypotheses H 0 :  i =  j H 0 :  i =  j H a :  i  j H a :  i  j n Test Statistic x i - x j x i - x j n Rejection Rule Reject H 0 if | x i - x j | > LSD Reject H 0 if | x i - x j | > LSDwhere

30. 30 Slide © 2003 South-Western/Thomson Learning™ n Fisher’s LSD Assuming  =.05, Hypotheses (A) H 0 :  1 =  2 Hypotheses (A) H 0 :  1 =  2 H a :  1  2 H a :  1  2 Test Statistic Test Statistic | x 1 - x 2 | = |55 - 68| = 13 | x 1 - x 2 | = |55 - 68| = 13 Conclusion Conclusion The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2. __ Example: Reed Manufacturing

31. 31 Slide © 2003 South-Western/Thomson Learning™ n Fisher’s LSD Hypotheses (B) Hypotheses (B) H 0 :  1 =  3 H 0 :  1 =  3 H a :  1  3 H a :  1  3 Test Statistic Test Statistic | x 1 - x 3 | = |55 - 57| = 2 | x 1 - x 3 | = |55 - 57| = 2 Conclusion 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. __ Example: Reed Manufacturing

32. 32 Slide © 2003 South-Western/Thomson Learning™ n Fisher’s LSD Hypotheses (C) Hypotheses (C) H 0 :  2 =  3 H 0 :  2 =  3 H a :  2  3 H a :  2  3 Test Statistic Test Statistic | x 2 - x 3 | = |68 - 57| = 11 | x 2 - x 3 | = |68 - 57| = 11 Conclusion Conclusion The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3. __ Example: Reed Manufacturing

33. 33 Slide © 2003 South-Western/Thomson Learning™ The comparisonwise Type I error rate  indicates the level of significance associated with a single pairwise comparison 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 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  EW = 1 – (1 –  ) ( k – 1)! n The experimentwise Type I error rate gets larger for problems with more populations (larger k ) Type I Error Rates

34. 34 Slide © 2003 South-Western/Thomson Learning™ An Introduction to Experimental Design n Statistical studies can be classified as being either experimental or observational. n In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. n In an observational study, no attempt is made to control the factors. n Cause-and-effect relationships are easier to establish in experimental studies than in observational studies.

35. 35 Slide © 2003 South-Western/Thomson Learning™ An Introduction to Experimental Design n A factor is a variable that the experimenter has selected for investigation. n A treatment is a level of a factor. n Experimental units are the objects of interest in the experiment. n A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. n If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.

36. 36 Slide © 2003 South-Western/Thomson Learning™ Completely Randomized Designs n Between-Treatments Estimate of Population Variance n Within-Treatments Estimate of Population Variance n Comparing the Variance Estimates: The F Test n ANOVA Table n Pairwise Comparisons

37. 37 Slide © 2003 South-Western/Thomson Learning™ In the context of experimental design, the between- samples estimate of  2 is referred to as the mean square due to treatments (MSTR). In the context of experimental design, the between- samples estimate of  2 is referred to as the mean square due to treatments (MSTR). n The formula for MSTR is n The numerator is called the sum of squares due to treatments (SSTR). n The denominator k - 1 represents the degrees of freedom associated with SSTR. Between-Treatments Estimate of Population Variance

38. 38 Slide © 2003 South-Western/Thomson Learning™ The second estimate of  2, the within-samples estimate, is referred to as the mean square due to error (MSE). The second estimate of  2, the within-samples estimate, is referred to as the mean square due to error (MSE). n The formula for MSE is n The numerator is called the sum of squares due to error (SSE). n The denominator n T - k represents the degrees of freedom associated with SSE. Within-Treatments Estimate of Population Variance

39. 39 Slide © 2003 South-Western/Thomson Learning™ ANOVA Table for a Completely Randomized Design Source of Sum of Degrees of Mean Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Variation Squares Freedom Squares F Treatments SSTR k - 1 Treatments SSTR k - 1 Error SSE n T - k Error SSE n T - k Total SST n T - 1 Total SST n T - 1

40. 40 Slide © 2003 South-Western/Thomson Learning™ Example: Home Products, Inc. n Completely Randomized Design Home Products, Inc. is considering marketing a long-lasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration. The number of times each car went through the carwash is shown on the next slide. Home Products, Inc. must decide which wax to market. Are the three waxes equally effective?

41. 41 Slide © 2003 South-Western/Thomson Learning™ Example: Home Products, Inc. Wax Wax Wax Wax Wax Wax ObservationType 1 Type 2 Type 3 ObservationType 1 Type 2 Type 3 1 27 33 29 1 27 33 29 2 30 28 28 2 30 28 28 3 29 31 30 3 29 31 30 4 28 30 32 4 28 30 32 5 31 30 31 5 31 30 31 Sample Mean 29.0 30.4 30.0 Sample Mean 29.0 30.4 30.0 Sample Variance 2.5 3.3 2.5 Sample Variance 2.5 3.3 2.5

42. 42 Slide © 2003 South-Western/Thomson Learning™ n Hypotheses H 0 :  1  =  2  =  3  H a : Not all the means are equal where: where:  1 = mean number of washes for Type 1 wax  2 = mean number of washes for Type 2 wax  2 = mean number of washes for Type 2 wax  3 = mean number of washes for Type 3 wax Example: Home Products, Inc.

43. 43 Slide © 2003 South-Western/Thomson Learning™ n Mean Square Between Treatments Since the sample sizes are all equal: Since the sample sizes are all equal: x = (x 1 + x 2 + x 3 )/3 = (29 + 30.4 + 30)/3 = 29.8 x = (x 1 + x 2 + x 3 )/3 = (29 + 30.4 + 30)/3 = 29.8 SSTR = 5(29–29.8) 2 + 5(30.4–29.8) 2 + 5(30–29.8) 2 = 5.2 SSTR = 5(29–29.8) 2 + 5(30.4–29.8) 2 + 5(30–29.8) 2 = 5.2 MSTR = 5.2/(3 - 1) = 2.6 MSTR = 5.2/(3 - 1) = 2.6 n Mean Square Error SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2 SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2 MSE = 33.2/(15 - 3) = 2.77 MSE = 33.2/(15 - 3) = 2.77 = ___ Example: Home Products, Inc.

44. 44 Slide © 2003 South-Western/Thomson Learning™ n Rejection Rule Using test statistic: Reject H 0 if F > 3.89 Using test statistic: Reject H 0 if F > 3.89 Using p -value: Reject H 0 if p -value <.05 where F.05 = 3.89 is based on an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom Example: Home Products, Inc.

45. 45 Slide © 2003 South-Western/Thomson Learning™ Example: Home Products, Inc. n Test Statistic F = MSTR/MSE = 2.6/2.77 =.939 F = MSTR/MSE = 2.6/2.77 =.939 n Conclusion Since F =.939 < F.05 = 3.89, we cannot reject H 0. There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same.

46. 46 Slide © 2003 South-Western/Thomson Learning™ n ANOVA Table Source of Sum of Degrees of Mean Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Variation Squares Freedom Squares F Treatments 5.2 2 2.60.9398 Treatments 5.2 2 2.60.9398 Error 33.2 12 2.77 Error 33.2 12 2.77 Total 38.4 14 Total 38.4 14 Example: Home Products, Inc.

47. 47 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (top portion) Using Excel’s ANOVA: Single Factor Tool

48. 48 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (bottom portion) Using Excel’s ANOVA: Single Factor Tool

49. 49 Slide © 2003 South-Western/Thomson Learning™ n Conclusion Using the p -Value The value worksheet shows a p -value of.418 The value worksheet shows a p -value of.418 The rejection rule is “Reject H 0 if p -value <.05” The rejection rule is “Reject H 0 if p -value <.05” Because.418 >.05, we cannot reject H 0. There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same. Because.418 >.05, we cannot reject H 0. There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same. Using Excel’s ANOVA: Single Factor Tool

50. 50 Slide © 2003 South-Western/Thomson Learning™ Randomized Block Design n ANOVA Procedure n Computations and Conclusions

51. 51 Slide © 2003 South-Western/Thomson Learning™ ANOVA Procedure n 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. n The formula for this partitioning is SST = SSTR + SSBL + SSE SST = SSTR + SSBL + SSE n 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.

52. 52 Slide © 2003 South-Western/Thomson Learning™ 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

53. 53 Slide © 2003 South-Western/Thomson Learning™ Example: Eastern Oil Co. n Randomized Block Design 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.

54. 54 Slide © 2003 South-Western/Thomson Learning™ Example: Eastern Oil Co. Automobile Type of Gasoline (Treatment) Blocks Automobile Type of Gasoline (Treatment) Blocks (Block) Blend X Blend Y Blend Z Means (Block) Blend X Blend Y Blend Z Means 131 3030 30.333 131 3030 30.333 230 2929 29.333 230 2929 29.333 329 2928 28.667 329 2928 28.667 433 3129 31.000 433 3129 31.000 526 2526 25.667 526 2526 25.667Treatment Means 29.8 28.8 28.4 Means 29.8 28.8 28.4

55. 55 Slide © 2003 South-Western/Thomson Learning™ Example: Eastern Oil Co. n Mean Square Due to Treatments The overall sample mean is 29. Thus, The overall sample mean is 29. Thus, SSTR = 5[(29.8 - 29) 2 + (28.8 - 29) 2 + (28.4 - 29) 2 ] = 5.2 MSTR = 5.2/(3 - 1) = 2.6 MSTR = 5.2/(3 - 1) = 2.6 n Mean Square Due to Blocks SSBL = 3[(30.333 - 29) 2 +... + (25.667 - 29) 2 ] = 51.33 MSBL = 51.33/(5 - 1) = 12.8 n Mean Square Due to Error SSE = 62 - 5.2 - 51.33 = 5.47 MSE = 5.47/[(3 - 1)(5 - 1)] =.68 MSE = 5.47/[(3 - 1)(5 - 1)] =.68

56. 56 Slide © 2003 South-Western/Thomson Learning™ n Rejection Rule Using test statistic: Reject H 0 if F > 4.46 Using p -value: Reject H 0 if p -value <.05 Assuming  =.05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator) Example: Eastern Oil Co.

57. 57 Slide © 2003 South-Western/Thomson Learning™ Example: Eastern Oil Co. n Test Statistic F = MSTR/MSE = 2.6/.68 = 3.82 F = MSTR/MSE = 2.6/.68 = 3.82 n 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.

58. 58 Slide © 2003 South-Western/Thomson Learning™ Using Excel’s Anova: Two-Factor Without Replication Tool n Step 1 Select the Tools pull-down menu n Step 2 Choose the Data Analysis option n Step 3 Choose Anova: Two Factor Without Replication from the list of Analysis Tools … continued

59. 59 Slide © 2003 South-Western/Thomson Learning™ n Step 4 When the Anova: Two Factor Without Replication dialog box appears: Enter A1:D6 in the Input Range box Enter A1:D6 in the Input Range box Select Labels Select Labels Enter.05 in the Alpha box Enter.05 in the Alpha box Select Output Range Select Output Range Enter A8 (your choice) in the Output Range box Enter A8 (your choice) in the Output Range box Click OK Click OK Using Excel’s Anova: Two-Factor Without Replication Tool

60. 60 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (top portion) Using Excel’s Anova: Two-Factor Without Replication Tool

61. 61 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (middle portion) Using Excel’s Anova: Two-Factor Without Replication Tool

62. 62 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (bottom portion) Using Excel’s Anova: Two-Factor Without Replication Tool

63. 63 Slide © 2003 South-Western/Thomson Learning™ n Conclusion Using the p-Value The value worksheet shows that the p -value is.06899 The value worksheet shows that the p -value is.06899 The rejection rule is “Reject H 0 if p -value <.05” The rejection rule is “Reject H 0 if p -value <.05” Thus, we cannot reject H 0 because the p -value =.06899 >  =.05 Thus, we cannot reject H 0 because the p -value =.06899 >  =.05 There is not sufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends There is not sufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends Using Excel’s Anova: Two-Factor Without Replication Tool

64. 64 Slide © 2003 South-Western/Thomson Learning™ Factorial Experiments n ANOVA Procedure n Computations and Conclusions

65. 65 Slide © 2003 South-Western/Thomson Learning™ Factorial Experiments n In some experiments we want to draw conclusions about more than one variable or factor. n Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required. n The term factorial is used because the experimental conditions include all possible combinations of the factors. n For example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations.

66. 66 Slide © 2003 South-Western/Thomson Learning™ ANOVA Procedure n The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment n We again partition the sum of squares total (SST) into its sources: SST = SSA + SSB + SSAB + SSE SST = SSA + SSB + SSAB + SSE n The total degrees of freedom, n T - 1, are partitioned such that ( a – 1) d.f go to Factor A, ( b – 1) d.f go to Factor B, ( a – 1)( b – 1) d.f. go to Interaction, and ab ( r – 1) go to Error

67. 67 Slide © 2003 South-Western/Thomson Learning™ Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Factor A SSA a - 1 Factor B SSB b - 1 InteractionSSAB ( a – 1)( b – 1) Error SSE ab ( r - 1) Total SST n T - 1 ANOVA Table for the Two-Factor Factorial Experiment with r Replications

68. 68 Slide © 2003 South-Western/Thomson Learning™ Computations n Step 1 Compute the total sum of squares n Step 2 Compute the sum of squares for factor A n Step 3 Compute the sum of squares for factor B

69. 69 Slide © 2003 South-Western/Thomson Learning™ Computations n Step 4 Compute the sum of squares for interaction n Step 5 Compute the sum of squares due to error SSE = SST – SSA – SSB - SSAB SSE = SST – SSA – SSB - SSAB

70. 70 Slide © 2003 South-Western/Thomson Learning™ Example: State of Ohio Wage Survey The following data show the hourly wages for a sample of workers in two industries at three locations in Ohio. Part of the purpose of the wage survey was to determine if differences exist in both industry type and location. IndustryCincinnatiClevelandColumbusI5.505.105.90 I5.805.006.20 I6.105.506.10 II6.405.806.50 II6.506.006.00 II6.005.606.10

71. 71 Slide © 2003 South-Western/Thomson Learning™ Example: State of Ohio Wage Survey n Factors Factor A: Industry Type (2 levels) Factor A: Industry Type (2 levels) Factor B: Location (3 levels) Factor B: Location (3 levels) n Replications Each experimental condition is repeated 3 times Each experimental condition is repeated 3 times

72. 72 Slide © 2003 South-Western/Thomson Learning™ n ANOVA Table Example: State of Ohio Wage Survey Source of Variation Sum of Squares Degrees of FreedomMeanSquareF Factor A.761.7612.6 Factor B 1.422.7111.7 Interaction.182.091.5 Error.7312.06 Total3.0917

73. 73 Slide © 2003 South-Western/Thomson Learning™ Example: State of Ohio Wage Survey n Conclusions Industries: F = 12.7 > F  = 4.75 Industries: F = 12.7 > F  = 4.75 Mean wages differ by industry type Mean wages differ by industry type Locations: F = 11.8 > F  = 3.89 Locations: F = 11.8 > F  = 3.89 Mean wages differ by location Mean wages differ by location Interaction: F = 1.5 < F  = 3.89 Interaction: F = 1.5 < F  = 3.89 Interaction is not significant Interaction is not significant

74. 74 Slide © 2003 South-Western/Thomson Learning™ n Step 1 Select the Tools pull-down menu n Step 2 Choose the Data Analysis option n Step 3 Choose Anova: Two Factor With Replication from the list of Analysis Tools … continued Using Excel’s Anova: Two-Factor with Replication Tool

75. 75 Slide © 2003 South-Western/Thomson Learning™ n Step 4 When the Anova: Two Factor With Replication dialog box appears: Enter A1:D7 in the Input Range box Enter A1:D7 in the Input Range box Enter 3 in the Rows per sample box Enter 3 in the Rows per sample box Enter.05 in the Alpha box Enter.05 in the Alpha box Enter A9 in the Output Range box Enter A9 in the Output Range box Click OK Click OK Using Excel’s Anova: Two-Factor with Replication Tool

76. 76 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (top portion) Using Excel’s Anova: Two-Factor with Replication Tool Note: Rows 9-38 not shown.

77. 77 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (middle portion) Using Excel’s Anova: Two-Factor with Replication Tool Note: Rows 1-8 and 23-38 not shown.

78. 78 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (middle portion) Using Excel’s Anova: Two-Factor with Replication Tool Note: Rows 1-23 and 31-38 not shown.

79. 79 Slide © 2003 South-Western/Thomson Learning™ n Value Worksheet (bottom portion) Using Excel’s Anova: Two-Factor With Replication Tool Note: Rows 1-30 not shown.

80. 80 Slide © 2003 South-Western/Thomson Learning™ n Conclusions Using the p -Value Industries: p -value =.004 <  =.05 Industries: p -value =.004 <  =.05 Mean wages differ by industry type Mean wages differ by industry type Locations: p -value =.0015 <  =.05 Locations: p -value =.0015 <  =.05 Mean wages differ by location Mean wages differ by location Interaction: p -value =.263 >  =.05 Interaction: p -value =.263 >  =.05 Interaction is not significant Interaction is not significant Using Excel’s Anova: Two-Factor With Replication Tool

81. 81 Slide © 2003 South-Western/Thomson Learning™ End of Chapter 13