1. Slide Slide 1 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Lecture Slides Elementary Statistics Tenth Edition and the Triola Statistics Series by Mario F. Triola

2. Slide Slide 2 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Chapter 12 Analysis of Variance 12-1 Overview 12-2 One-Way ANOVA 12-3 Two-Way ANOVA

3. Slide Slide 3 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Section 12-1 & 12-2 Overview and One-Way ANOVA Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

4. Slide Slide 4 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept This section introduces the method of one-way analysis of variance, which is used for tests of hypotheses that three or more population means are all equal.

5. Slide Slide 5 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Overview  Analysis of variance (ANOVA) is a method for testing the hypothesis that three or more population means are equal.  For example: H 0 : µ 1 = µ 2 = µ 3 =... µ k H 1 : At least one mean is different

6. Slide Slide 6 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. ANOVA Methods Require the F-Distribution 1. The F- distribution is not symmetric; it is skewed to the right. 2. The values of F can be 0 or positive; they cannot be negative. 3. There is a different F-distribution for each pair of degrees of freedom for the numerator and denominator. Critical values of F are given in Table A-5

7. Slide Slide 7 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. F - distribution Figure 12-1

8. Slide Slide 8 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. One-Way ANOVA 1.Understand that a small P -value (such as 0.05 or less) leads to rejection of the null hypothesis of equal means. With a large P -value (such as greater than 0.05), fail to reject the null hypothesis of equal means. 2.Develop an understanding of the underlying rationale by studying the examples in this section. An Approach to Understanding ANOVA

9. Slide Slide 9 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. One-Way ANOVA 3.Become acquainted with the nature of the SS (sum of squares) and MS (mean square) values and their role in determining the F test statistic, but use statistical software packages or a calculator for finding those values. An Approach to Understanding ANOVA

10. Slide Slide 10 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definition Treatment (or factor) A treatment (or factor) is a property or characteristic that allows us to distinguish the different populations from one another. Use computer software or TI-83/84 for ANOVA calculations if possible.

11. Slide Slide 11 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. One-Way ANOVA Requirements 1. The populations have approximately normal distributions. 2. The populations have the same variance  2 (or standard deviation  ). 3. The samples are simple random samples. 4. The samples are independent of each other. 5. The different samples are from populations that are categorized in only one way.

12. Slide Slide 12 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Procedure for testing H o : µ1 = µ2 = µ3 =... 1. Use STATDISK, Minitab, Excel, or a TI-83/84 Calculator to obtain results. 2. Identify the P -value from the display. 3. Form a conclusion based on these criteria: If P -value  , reject the null hypothesis of equal means. If P -value > , fail to reject the null hypothesis of equal means.

13. Slide Slide 13 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Procedure for testing H o : µ1 = µ2 = µ3 =... Caution when interpreting ANOVA results: When we conclude that there is sufficient evidence to reject the claim of equal population means, we cannot conclude from ANOVA that any particular mean is different from the others. There are several other tests that can be used to identify the specific means that are different, and those procedures are called multiple comparison procedures, and they are discussed later in this section.

14. Slide Slide 14 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Do the samples come from populations with different means?

15. Slide Slide 15 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees For a significance level of  = 0.05, use STATDISK, Minitab, Excel, or a TI-83/84 calculator to test the claim that the four samples come from populations with means that are not all the same. H 0 :  1 =  2 =  3 =  4 H 1 : At least one of the means is different from the others. Do the samples come from populations with different means?

16. Slide Slide 16 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Do the samples come from populations with different means? STATDISK Minitab

17. Slide Slide 17 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Do the samples come from populations with different means? Excel TI-83/84 PLUS

18. Slide Slide 18 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees H 0 :  1 =  2 =  3 =  4 H 1 : At least one of the means is different from the others. Do the samples come from populations with different means? The displays all show a P-value of approximately 0.007. Because the P-value is less than the significance level of  = 0.05, we reject the null hypothesis of equal means. There is sufficient evidence to support the claim that the four population means are not all the same. We conclude that those weights come from populations having means that are not all the same.

19. Slide Slide 19 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Estimate the common value of  2 : 1. The variance between samples (also called variation due to treatment) is an estimate of the common population variance  2 that is based on the variability among the sample means. 2.The variance within samples (also called variation due to error) is an estimate of the common population variance  2 based on the sample variances. ANOVA Fundamental Concepts

20. Slide Slide 20 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. An excessively large F test statistic is evidence against equal population means. F = variance between samples variance within samples Test Statistic for One-Way ANOVA ANOVA Fundamental Concepts

21. Slide Slide 21 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Relationships Between the F Test Statistic and P-Value Figure 12-2

22. Slide Slide 22 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Calculations with Equal Sample Sizes where s p = pooled variance (or the mean of the sample variances) 2  Variance within samples = s p 2  Variance between samples = n where = variance of sample means sxsx 2 sxsx 2

23. Slide Slide 23 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Sample Calculations

24. Slide Slide 24 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 1. Find the variance between samples =. For the means 5.5, 6.0 & 6.0, the sample variance is = 0.0833 = 4 X 0.0833 = 0.3332 2. Estimate the variance within samples by calculating the mean of the sample variances.. Use Table 12-2 to calculate the variance between samples, variance within samples, and the F test statistic. Example: Sample Calculations 3.0 + 2.0 + 2.0 3 = = 2.3333

25. Slide Slide 25 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 3. Evaluate the F test statistic F = F = = 0.1428 variance between samples variance within samples Use Table 12-2 to calculate the variance between samples, variance within samples, and the F test statistic. Example: Sample Calculations 0.3332 2.3333

26. Slide Slide 26 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Critical Value of F  Right-tailed test  Degree of freedom with k samples of the same size n numerator df = k – 1 denominator df = k ( n – 1 )

27. Slide Slide 27 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. where x = mean of all sample scores combined k = number of population means being compared n i = number of values in the i th sample x i = mean of values in the i th sample s i = variance of values in the i th sample 2 Calculations with Unequal Sample Sizes F = = variance between samples variance within samples  (n i – 1 )s i  (n i – 1) 2  n i (x i – x ) 2 k – 1

28. Slide Slide 28 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Components of the ANOVA Method SS(total), or total sum of squares, is a measure of the total variation (around x ) in all the sample data combined. Formula 12-1 SS(total) =  (x – x) 2

29. Slide Slide 29 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Components of the ANOVA Method SS(treatment), also referred to as SS(factor) or SS(between groups) or SS(between samples), is a measure of the variation between the sample means. SS(treatment) = n 1 (x 1 – x) 2 + n 2 (x 2 – x) 2 +... n k (x k – x) 2 =  n i (x i - x) 2 Formula 12-2

30. Slide Slide 30 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. SS(error), (also referred to as SS(within groups) or SS(within samples), is a sum of squares representing the variability that is assumed to be common to all the populations being considered. SS(error) = (n 1 – 1 )s 1 + (n 2 – 1 )s 2 + (n 3 – 1 )s 3... n k (x k – 1 )s i =  (n i – 1 )s i Formula 12-3 2 2 2 2 2 Key Components of the ANOVA Method

31. Slide Slide 31 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Formula 12-4 SS(total) = SS(treatment) + SS(error) Key Components of the ANOVA Method Given the previous expressions for SS(total), SS(treatment), and SS(error), the following relationship will always hold.

32. Slide Slide 32 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Mean Squares (MS) MS(treatment) is a mean square for treatment, obtained as follows: Formula 12-5 MS(treatment) = SS (treatment) k – 1 MS(error) is a mean square for error, obtained as follows: Formula 12-6 MS(error) = SS (error) N – k

33. Slide Slide 33 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Mean Squares (MS) MS(total) is a mean square for the total variation, obtained as follows: Formula 12-7 MS(total) = SS(total) N – 1

34. Slide Slide 34 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Test Statistic for ANOVA with Unequal Sample Sizes  Numerator df = k – 1  Denominator df = N – k F = MS (treatment) MS (error) Formula 12-8

35. Slide Slide 35 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Table 12-3 has a format often used in computer displays.

36. Slide Slide 36 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Identifying Means That Are Different After conducting an analysis of variance test, we might conclude that there is sufficient evidence to reject a claim of equal population means, but we cannot conclude from ANOVA that any particular mean is different from the others.

37. Slide Slide 37 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Identifying Means That Are Different Informal procedures to identify the specific means that are different 1.Use the same scale for constructing boxplots of the data sets to see if one or more of the data sets are very different from the others. 2.Construct confidence interval estimates of the means from the data sets, then compare those confidence intervals to see if one or more of them do not overlap with the others.

38. Slide Slide 38 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Bonferroni Multiple Comparison Test Step 1. Do a separate t test for each pair of samples, but make the adjustments described in the following steps. Step 2. For an estimate of the variance σ 2 that is common to all of the involved populations, use the value of MS(error).

39. Slide Slide 39 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Bonferroni Multiple Comparison Test Step 2 (cont.) Using the value of MS(error), calculate the value of the test statistic, as shown below. (This example shows the comparison for Sample 1 and Sample 4.)

40. Slide Slide 40 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Bonferroni Multiple Comparison Test Step 2 (cont.) Change the subscripts and use another pair of samples until all of the different possible pairs of samples have been tested. Step 3. After calculating the value of the test statistic t for a particular pair of samples, find either the critical t value or the P-value, but make the following adjustment:

41. Slide Slide 41 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Bonferroni Multiple Comparison Test Step 3 (cont.) P-value: Use the test statistic t with df = N-k, where N is the total number of sample values and k is the number of samples. Find the P-value the usual way, but adjust the P-value by multiplying it by the number of different possible pairings of two samples. (For example, with four samples, there are six different possible pairings, so adjust the P-value by multiplying it by 6.)

42. Slide Slide 42 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Bonferroni Multiple Comparison Test Step 3 (cont.) Critical value: When finding the critical value, adjust the significance level α by dividing it by the number of different possible pairings of two samples. (For example, with four samples, there are six different possible pairings, so adjust the P-value by dividing it by 6.)

43. Slide Slide 43 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Using the Bonferroni Test Using the data in Table 12-1, we concluded that there is sufficient evidence to warrant rejection of the claim of equal means. The Bonferroni test requires a separate t test for each different possible pair of samples.

44. Slide Slide 44 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Using the Bonferroni Test We begin with testing H 0 : μ 1 = μ 4. From Table 12-1: x 1 = 0.184, n 1 = 5, x 4 = 1.334, n 4 = 5

45. Slide Slide 45 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees Using the Bonferroni Test Test statistic = – 3.484 df = N – k = 20 – 4 = 16 Two-tailed P-value is 0.003065, but adjust it by multiplying by 6 (the number of different possible pairs of samples) to get P-value = 0.01839. Because the adjusted P-value is less than α = 0.05, reject the null hypothesis. It appears that Samples 1 and 4 have significantly different means.

46. Slide Slide 46 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Weights of Poplar Trees SPSS Bonferroni Results

47. Slide Slide 47 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed:  One-way analysis of variance (ANOVA) Calculations with Equal Sample Sizes Calculations with Unequal Sample Sizes  Identifying Means That Are Different Bonferroni Multiple Comparison Test

48. Slide Slide 48 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Section 12-3 Two-Way ANOVA Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

49. Slide Slide 49 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concepts The analysis of variance procedure introduced in Section 12-2 is referred to as one-way analysis of variance because the data are categorized into groups according to a single factor (or treatment). In this section we introduce the method of two-way analysis of variance, which is used with data partitioned into categories according to two factors.

50. Slide Slide 50 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Two-Way Analysis of Variance Two-Way ANOVA involves two factors. The data are partitioned into subcategories called cells.

51. Slide Slide 51 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Poplar Tree Weights

52. Slide Slide 52 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. There is an interaction between two factors if the effect of one of the factors changes for different categories of the other factor. Definition

53. Slide Slide 53 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Poplar Tree Weights Exploring Data Calculate the mean for each cell.

54. Slide Slide 54 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Poplar Tree Weights Exploring Data Display the means on a graph. If a graph such as Figure 12-3 results in line segments that are approximately parallel, we have evidence that there is not an interaction between the row and column variables.

55. Slide Slide 55 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Poplar Tree Weights Minitab

56. Slide Slide 56 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Requirements 1. For each cell, the sample values come from a population with a distribution that is approximately normal. 2. The populations have the same variance  2. 3. The samples are simple random samples. 4. The samples are independent of each other. 5. The sample values are categorized two ways. 6. All of the cells have the same number of sample values.

57. Slide Slide 57 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Two-Way ANOVA calculations are quite involved, so we will assume that a software package is being used. Minitab, Excel, TI-83/4 or STATDISK can be used.

58. Slide Slide 58 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Figure 12- 4 Procedure for Two-Way ANOVAS

59. Slide Slide 59 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. The P- value is shown as 0.915, so we fail to reject the null hypothesis of no interaction between the two factors. F = = = 0.17 MS(interaction) 0.05721 MS(error) 0.33521 Using the Minitab display: Step 1: Test for interaction between the effects: H 0 : There are no interaction effects. H 1 : There are interaction effects. Example: Poplar Tree Weights

60. Slide Slide 60 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Step 2: Check for Row/Column Effects: H 0 : There are no effects from the row factors. H 1 : There are row effects. H 0 : There are no effects from the column factors. H 1 : There are column effects. The P- value is shown as 0.374, so we fail to reject the null hypothesis of no effects from site. Using the Minitab display: Example: Poplar Tree Weights F = = = 0.81 MS(site) 0.27225 MS(error) 0.33521

61. Slide Slide 61 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Special Case: One Observation per Cell and No Interaction If our sample data consist of only one observation per cell, we lose MS(interaction), SS(interaction), and df(interaction). If it seems reasonable to assume that there is no interaction between the two factors, make that assumption and then proceed as before to test the following two hypotheses separately: H 0 : There are no effects from the row factors. H 0 : There are no effects from the column factors.

62. Slide Slide 62 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Special Case: One Observation per Cell and No Interaction Minitab

63. Slide Slide 63 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Special Case: One Observation per Cell and No Interaction Row factor: F = = = 0.28 MS(site) 0.156800 MS(error) 0.562300 The P-value in the Minitab display is 0.634. Fail to reject the null hypothesis.

64. Slide Slide 64 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Special Case: One Observation per Cell and No Interaction Column factor: F = = = 0.73 MS(site) 0.412217 MS(error) 0.562300 The P-value in the Minitab display is 0.598. Fail to reject the null hypothesis.

65. Slide Slide 65 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed:  Two- way analysis of variance (ANOVA)  Special case: One observation per cell and no interaction