1. STAT 312 10.1 - Single-Factor ANOVA
Chapter 10 - Analysis of Variance (ANOVA)
Introduction
Single-Factor ANOVA
Multiple Comparisons in ANOVA
More on Single-Factor ANOVA

2. 1 2 k =
H0:
…etc…
Idea: Test all possible pairwise comparisons, each via a two-sample t-test.
Example : Suppose there are k = 5 treatment groups.
There are such comparisons.
PROBLEM???

3. SPURIOUS SIGNIFICANCE!!!
Suppose we wish to make n independent comparisons, each at significance level .
As the number of comparisons increases, so does the probability of rejection, just by chance!
SPURIOUS SIGNIFICANCE!!!
When n = 14, this probability is > 50%.

5. SPURIOUS SIGNIFICANCE!!!
Suppose we wish to make n independent comparisons, each at significance level .
As the number of comparisons increases, so does the probability of rejection, just by chance!
SPURIOUS SIGNIFICANCE!!!
When n = 14, this probability is > 50%.
One remedy: Make each t-test comparison more conservative, i.e., harder to reject!

6. SPURIOUS SIGNIFICANCE!!!
p-value
Suppose we wish to make n independent comparisons, each at significance level .
As the number of comparisons increases, so does the probability of rejection, just by chance!
SPURIOUS SIGNIFICANCE!!!
When n = 14, this probability is > 50%.
One remedy: Make each t-test comparison more conservative, i.e., harder to reject!
* =  /n

7. p-value  * =  /n H0: = PROBLEM??? …etc…1 2 k =
H0:
…etc…
Idea: Test all possible pairwise comparisons, each via a two-sample t-test.
Example : Suppose there are k = 5 treatment groups.
There are such comparisons.
PROBLEM???
* =  /n

8. * = .05 /10 = .005 H0: = PROBLEM??? …etc…1 2 k =
H0:
…etc…
Idea: Test all possible pairwise comparisons, each via a two-sample t-test.
Example : Suppose there are k = 5 treatment groups.
There are such comparisons.
PROBLEM???
* = .05 /10 = .005

9. BONFERRONI CORRECTION1 2 k =
H0:
…etc…
Idea: Test all possible pairwise comparisons, each via a two-sample t-test.
Example : Suppose there are k = 5 treatment groups.
There are such comparisons.
BONFERRONI CORRECTION
Holm-Bonferroni
Tukey’s Honest Significant Difference,…
* = .05 /10 = .005

10. Analysis of Variance (ANOVA)
Alternate method ~
Analysis of Variance (ANOVA)
Main Idea: Among several (k  2) independent, equivariant, normally-distributed “treatment groups”…
MODEL ASSUMPTIONS? 1 2 k =
H0:

11. Analysis of Variance (ANOVA)
Alternate method ~
Analysis of Variance (ANOVA)
Main Idea: Among several (k  2) independent, equivariant, normally-distributed “treatment groups”…
Equivariance can be tested via very similar “two variances” F-test in (but this is very sensitive to normality assumption), or others. If violated, can extend Welch Test for two means. 1 2 k =
H0:

12. Analysis of Variance (ANOVA)
Alternate method ~
Analysis of Variance (ANOVA)
Main Idea: Among several (k  2) independent, equivariant, normally-distributed “treatment groups”…
Normality can be tested via usual methods.
If violated, use nonparametric Kruskal-Wallis Test. 1 2 k =
H0:

13. Analysis of Variance (ANOVA)
Alternate method ~
Analysis of Variance (ANOVA)
Main Idea: Among several (k  2) independent, equivariant, normally-distributed “treatment groups”…
Extensions of ANOVA for data in matched “blocks” designs, repeated measures, multiple factor levels within groups, etc. 1 2 k =
H0:

14. Analysis of Variance (ANOVA)
Alternate method ~
Analysis of Variance (ANOVA)
Main Idea: Among several (k  2) independent, equivariant, normally-distributed “treatment groups”…
How to identify significant group(s)? Pairwise testing, with correction (e.g., Bonferroni) for spurious significance.
Example: k = 5 groups result in 10 such tests, so let each α* = α / 10. 1 2 k =
H0:

16. “spurious significance”

17. Example
Different methods of curing concrete are being compared. A total of 27 slabs are selected; 9 are randomly assigned to one of three curing methods. The number of days it takes to be fully cured is recorded:
Cure A:
Cure B:
Cure C:
> days = c(4,5,4,3,2,4,3,4,4,6,8,4,5,4,6,5,8,6,6,7,6,6,7,5,6,5,5)
> cure = c(rep("A",9), rep("B",9), rep("C",9))
> concrete = data.frame(days,cure)
> concrete
days cure A A A A
... C C

18. Example
Different methods of curing concrete are being compared. A total of 27 slabs are selected; 9 are randomly assigned to one of three curing methods. The number of days it takes to be fully cured is recorded:
Cure A:
Cure B:
Cure C:
> days = c(4,5,4,3,2,4,3,4,4,6,8,4,5,4,6,5,8,6,6,7,6,6,7,5,6,5,5)
> cure = c(rep("A",9), rep("B",9), rep("C",9))
> concrete = data.frame(days,cure)
> boxplot(days ~ cure, data = concrete)

19. Example
Different methods of curing concrete are being compared. A total of 27 slabs are selected; 9 are randomly assigned to one of three curing methods. The number of days it takes to be fully cured is recorded:
Cure A:
Cure B:
Cure C:
> days = c(4,5,4,3,2,4,3,4,4,6,8,4,5,4,6,5,8,6,6,7,6,6,7,5,6,5,5)
> cure = c(rep("A",9), rep("B",9), rep("C",9))
> concrete = data.frame(days,cure)
> boxplot(days ~ cure, data = concrete)
Method A seems the fastest; B and C seem comparable, both taking a longer time.

20. Example
Different methods of curing concrete are being compared. A total of 27 slabs are selected; 9 are randomly assigned to one of three curing methods. The number of days it takes to be fully cured is recorded:
Cure A:
Cure B:
Cure C:
STRONGLY REJECTED
> days = c(4,5,4,3,2,4,3,4,4,6,8,4,5,4,6,5,8,6,6,7,6,6,7,5,6,5,5)
> cure = c(rep("A",9), rep("B",9), rep("C",9))
> concrete = data.frame(days,cure)
> results = aov(days ~ cure, data = concrete)
> summary(results)
Df Sum Sq Mean Sq F value Pr(>F)
cure ***
Residuals
---
Signif. codes: 0 ‘***’ ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

21. Pairwise comparisons using t tests with pooled SD
Example
Different methods of curing concrete are being compared. A total of 27 slabs are selected; 9 are randomly assigned to one of three curing methods. The number of days it takes to be fully cured is recorded:
Cure A:
Cure B:
Cure C:
A B C
STRONGLY REJECTED
> days = c(4,5,4,3,2,4,3,4,4,6,8,4,5,4,6,5,8,6,6,7,6,6,7,5,6,5,5)
> cure = c(rep("A",9), rep("B",9), rep("C",9))
> concrete = data.frame(days,cure)
> results = aov(days ~ cure, data = concrete)
Multiple Comparisons?
> pairwise.t.test(days, cure, p.adjust="bonferroni")
Pairwise comparisons using t tests with pooled SD
data: days and cure
A B B C
P value adjustment method: bonferroni

22. Example
Different methods of curing concrete are being compared. A total of 27 slabs are selected; 9 are randomly assigned to one of three curing methods. The number of days it takes to be fully cured is recorded:
Cure A:
Cure B:
Cure C:
A B C
STRONGLY REJECTED
> days = c(4,5,4,3,2,4,3,4,4,6,8,4,5,4,6,5,8,6,6,7,6,6,7,5,6,5,5)
> cure = c(rep("A",9), rep("B",9), rep("C",9))
> concrete = data.frame(days,cure)
> results = aov(days ~ cure, data = concrete)
Multiple Comparisons?
Alt Method: Tukey’s Honest Significant Difference (HSD)
> TukeyHSD(results, conf.level = 0.95)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = days ~ cure, data = concrete)
$cure
diff lwr upr p adj
B-A
C-A
C-B