1. Lecture Slides Elementary Statistics Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola

2. Chapter 12 Analysis of Variance
12-1 Review and Preview
12-2 One-Way ANOVA
12-3 Two-Way ANOVA

3. Review
In chapter 9, we introduced methods for comparing the means from two independent samples.
Now we look to test the equality of three or more means by using the method of one-way analysis of variance (ANOVA).

4. ANOVA Methods Require the F Distribution
1. The F distribution is not symmetric; it is skewed to the right.
2. The values of F cannot be negative.
3. The exact shape of the F distribution depends on the two different degrees of freedom.

5. F Distribution

6. Chapter 12 Analysis of Variance
12-1 Review and Preview
12-2 One-Way ANOVA
12-3 Two-Way ANOVA

7. 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.
Because the calculations are very complicated, we emphasize the interpretation of results obtained by using technology.

8. Key Concept
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.
Develop an understanding of the underlying rationale by studying the examples in this section.

9. Part 1: Basics of One-Way Analysis of Variance

10. Definition
One-way analysis of variance (ANOVA) is a method of testing the equality of three or more population means by analyzing sample variances.
One-way analysis of variance is used with data categorized with one factor (or treatment), which is a characteristic that allows us to distinguish the different populations from one another.

11. One-Way ANOVA Requirements
1. The populations have approximately normal distributions.
2. The populations have the same variance σ2 (or standard deviation σ).
3. The samples are simple random samples of quantitative data.
4. The samples are independent of each other.
5. The different samples are from populations that are categorized in only one way.

12. Procedure
1. Use STATDISK, Minitab, Excel, StatCrunch, a TI-83/84 calculator, or any other technology to obtain results.
2. Identify the P-value from the display.
Form a conclusion based on these criteria:
If the P-value ≤ α, reject the null hypothesis of equal means. Conclude at least one mean is different from the others.
If the P-value > α, fail to reject the null hypothesis of equal means.

13. Caution
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 some of them are discussed in Part 2 of this section.

14. Example
Use the performance IQ scores listed in Table 12-1 and a significance level of α = to test the claim that the three samples come from populations with means that are all equal.

15. Example - Continued
Here are summary statistics from the collected data:

16. Example - Continued Requirement Check:
The three samples appear to come from populations that are approximately normal (normal quantile plots OK).
The three samples have standard deviations that are not dramatically different.
We can treat the samples as simple random samples.
The samples are independent of each other and the IQ scores are not matched in any way.
The three samples are categorized according to a single factor: low lead, medium lead, and high lead.

17. Example - Continued The hypotheses are:
The significance level is α = 0.05.
Technology results are presented on the next slides.

18. Example - Continued

19. Example - Continued

20. Example - Continued

21. Example - Continued
The displays all show that the P-value is when rounded.
Because the P-value is less than the significance level of α = 0.05, we can reject the null hypothesis.
There is sufficient evidence that the three samples come from populations with means that are different.
We cannot conclude formally that any particular mean is different from the others, but it appears that greater blood lead levels are associated with lower performance IQ scores.

22. P-Value and Test Statistic
Larger values of the test statistic result in smaller P-values, so the ANOVA test is right-tailed.
The figure on the next slide shows the relationship between the F test statistic and the P-value.
Assuming that the populations have the same variance σ2 (as required for the test), the F test statistic is the ratio of these two estimates of σ2:
variation between samples (based on variation among sample means)
variation within samples (based on the sample variances).

23. Relationship Between F Test Statistic and P-Value

24. Test Statistic for One-Way ANOVA

25. Caution
When testing for equality of three or more populations, use analysis of variance.
Do not use multiple hypothesis tests with two samples at a time.

26. Part 2: Calculations and Identifying Means That Are Different

27. Calculations with Equal or
Unequal Sample Sizes
The text beginning on page 605 provides a detailed discussion on the calculations and ramifications of sample size for the F statistic.
As the material doesn’t lend itself to clean PowerPoint slides, we leave it to the reader to reference the main text.

28. Designing the Experiment
When performing ANOVA, we use one factor as the basis for partitioning the data into several categories.
If we conclude that there is a significant difference among means, we can’t be absolutely certain the differences are explained by the factor being used.
One way to reduce the effect of extraneous factors is to run a completely randomized design, in which each sample value is given the same chance of belonging to different factor groups.
Another way to reduce the effect of extraneous factors is to use a rigorously controlled design, in which sample values are carefully chosen so that all other factors have no variability.

29. Identifying Which Means Are Different
After conducting ANOVA, there are several informal methods for determining which means are different:
Construct boxplots of the different samples to see if one or more of them is very different from the others.
Construct confidence interval estimates of the means for the different samples, then compare those confidence intervals to see if one or more of them does not overlap with the others.

30. Identifying Which Means Are Different
There are several formal procedures to determine which means are different.
Range tests allow us to identify subsets of means that are not significantly different from each other.
Multiple comparison tests use pairs of means, making adjustments to overcome the problem of having a significance level that increases as the number of tests increases.
There are many multiple comparisons tests, we introduce just one: The Bonferroni Multiple Comparison Test.

31. Bonferroni Multiple Comparison Test
Step 1: Do a separate t test for each pair of samples, but make the following adjustments:
Step 2: Use the value of MS(error), which uses all available sample data, as an estimate for the variance σ2. This value is obtained when conducting ANOVA.
Calculate the test statistic:

32. Bonferroni Multiple Comparison Test
Step 3: Make the following adjustments:
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 as per usual, but adjust the P-value by multiplying it by the number of different possible pairings of two samples (e.g. with three samples, there are three possible pairings, so multiply by 3).
Critical Value: When finding the critical value, adjust α by dividing it by the number of possible pairings of two samples.

33. Example
We previously concluded, for the IQ score test, that there is sufficient evidence to warrant rejection of the claim of equal means.
Use the Bonferroni test with a 0.05 level of significance to identify which mean is different from the others.

34. Example - Continued The null hypotheses to be tested are:
We begin with the first, and using the sample data presented earlier, arrive at (using technology):

35. Example - Continued
The test statistic:

36. Example - Continued
The test statistic of t = has N – k = 121 – 3 = 118 df.
The two-tailed P-value is , but it needs to be multiplied by 3 because there are three possible pairing of sample means.
Thus, the P-value = rounded.
Because this P-value is not small (less than 0.05), we fail to reject the null hypothesis.
It appears that Samples 1 and 2 do not have significantly different means.

37. Example - Continued
Technology can display Bonferroni test results:

38. Example - Continued Conclusion:
Although the ANOVA test tells us that at least one of the sample means is different from the others, the Bonferroni test results do not identify any one particular sample mean that is significantly different from the others.

39. Chapter 12 Analysis of Variance
12-1 Review and Preview
12-2 One-Way ANOVA
12-3 Two-Way ANOVA

40. Key Concepts
We introduce the method of two-way analysis of variance, which is used with data partitioned into categories according to two factors.
The methods of this section require that we begin by testing for an interaction between the two factors.
Then we test whether the row or column factors have effects.

41. Example The data in the table are categorized with two factors:
Sex: Male or Female
Blood Lead Level: Low, Medium, or High
The subcategories are called cells, and the response variable is IQ score.
The data are presented on the next slide:

42. Example

43. Definition
There is an interaction between two factors if the effect of one of the factors changes for different categories of the other factor.

44. Example
Let’s explore the IQ data in the table by calculating the mean for each cell and constructing an interaction graph.

45. Interpreting an Interaction Graph
An interaction effect is suggested if the line segments are far from being parallel.
No interaction effect is suggested if the line segments are approximately parallel.
For the IQ scores, it appears there is an interaction effect:
Females with high lead exposure appear to have lower IQ scores, while males with high lead exposure appear to have high IQ scores.

46. Requirements for Two-Way ANOVA
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 σ2.
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 (a balanced design – this section does not include methods for a design that is not balanced).

47. Technology and Two-Way ANOVA
Two-Way ANOVA calculations are quite involved, so we will assume that a software package is being used.
Minitab, Excel, StatCrunch, TI-83/4, STATDISK, or other software can be used.

48. Procedure for Two-Way ANOVA
Step 1: Interaction Effect - test the null hypothesis that there is no interaction
Step 2: Row/Column Effects - if we conclude there is no interaction effect, proceed with these two hypothesis tests
Row Factor: no effects from row
Column Factor: no effects from column
All tests use the F distribution and it is assumed technology will be used.

49. Procedure for Two-Way ANOVA

50. Procedure for Two-Way ANOVA

51. Example
Given the performance IQ scores in the table at the beginning of this section, use two-way ANOVA to test for an interaction effect, an effect from the row factor of gender, and an effect from the column factor of blood lead level.
Use a 0.05 level of significance.

52. Example - Continued Requirement Check:
For each cell, the sample values appear to be from a normally distributed population.
The variances of the cells are 95.3, 146.7, 130.8, 812.7, 142.3, and 143.8, which are considerably different from each other. We might have some reservations that the population variances are equal – but for the purposes of this example, we will assume the requirement is met.
The samples are simple random samples.
The samples are independent of each other.
The sample values are categorized in two ways.
All the cells have the same number of values.

53. Example - Continued
The technology output is displayed below:

54. Example - Continued
Step 1: Test that there is no interaction between the two factors.
The test statistic is F = 0.43 and the P-value is 0.655, so we fail to reject the null hypothesis.
It does not appear that the performance IQ scores are affected by an interaction between sex and blood lead level.
There does not appear to be an interaction effect, so we proceed to test for row and column effects.

55. Example - Continued Step 2: We now test:
For the row factor, F = 0.07 and the P-value is Fail to reject the null hypothesis, there is no evidence that IQ scores are affected by the gender of the subject.
For the column factor, F = 0.10 and the P-value is Fail to reject the null hypothesis, there is no evidence that IQ scores are effected by the level of lead exposure.

56. Example - Continued Interpretation:
Based on the sample data, we conclude that IQ scores do not appear to be affected by sex or blood lead level.

57. Caution
Two-way analysis of variance is not one-way analysis of variance done twice.
Be sure to test for an interaction between the two factors.

58. One Observation per Cell and
Special Case:
One Observation per Cell and
No Interaction
If our sample data consist of only one observation per cell, there is no variation within individual cells and sample variances cannot be calculated for individual cells.
If it seems reasonable to assume there is no interaction between the two factors, make that assumption and test separately:
(The mechanics of the tests are the same as presented earlier.)