1. Statistics for Managers Using Microsoft® Excel 5th Edition
Chapter 11
Analysis of Variance
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

2. Learning Objectives
After completing this chapter, you should be able to:
Recognize situations in which to use analysis of variance (ANOVA)
Understand different analysis of variance designs
Evaluate assumptions of the model
Perform a single-factor ANOVA and interpret the results
Conduct and interpret a Tukey-Kramer post-analysis to determine which means are different
Analyze two-factor analysis of variance tests
Conduct and interpret a Tukey-Kramer post-analysis procedure to determine which factors are different
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

3. General ANOVA Analysis
Investigator controls one or more independent variables
Called factors or treatment variables
One factor contains three or more levels or groups or categories/classifications
Other factors contains two or more levels or groups or categories/classifications
Experimental design: the plan used to test the hypothesis
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

4. One-Factor ANOVA
Also known as Completely Randomized Design and One-way ANOVA
Experimental units (subjects) are assigned randomly to treatments
Subjects are assumed homogeneous
Only one factor or independent variable
With three or more treatment levels
Analyzed by one-factor analysis of variance
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

5. One-Factor Analysis of Variance
Evaluates the difference among the means of three or more groups
Examples: Accident rates for 1st, 2nd, and 3rd shift
Expected mileage for five brands of tires
Assumptions
Populations are normally distributed (test with Box plot or Normal Probability Plot)
Populations have equal variances (use Levene’s Test for Homogeneity of Variance)
Samples are randomly and independently drawn
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

6. Why Analysis of Variance?
We could compare the means in pairs using a t test for difference of means
Each t test contains Type 1 error
The total Type 1 error with k pairs of means is 1- (1 - a) k
If there are 5 means and you use a = .05
Must perform 10 comparisons
Type I error is 1 – (.95) 10 = .40
40% of the time you will reject the null hypothesis of equal means in favor of the alternative even when the null is true!
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

7. Hypotheses: One-Factor ANOVA
All population means are equal
i.e., no treatment effect (no variation in means among groups)
At least one population mean is different
i.e., there is a treatment (groups) effect
Does not mean that all population means are different (at least one of the means is different from the others)
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

8. Hypotheses: One-Factor ANOVA
All Means are the same:
The Null Hypothesis is True
(No Group Effect)
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

9. Hypotheses: One-Factor ANOVA
At least one mean is different:
The Null Hypothesis is NOT true
(Treatment Effect is present) or
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

10. One-Factor ANOVA Table
Source of Variation df SS MS
(Variance)
P-value
F-Ratio
Between Groups
c-1
SSA
MSA
P(X=F)
Within Groups
n-c
SSW
MSW
Total
n-1
SST = SSA+SSW
c = number of groups
n = sum of the sample sizes from all groups
df = degrees of freedom
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

11. One-Factor ANOVA Test Statistic
H0: μ1= μ2 = … = μc
H1: At least two population means are different
Test statistic
MSA is mean squares among variances
MSW is mean squares within variances
Degrees of freedom
df1 = c – (c = number of groups)
df2 = n – c (n = sum of all sample sizes)
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

12. One-Factor ANOVA Test Statistic
The F statistic is the ratio of the among variance to the within variance
The ratio must always be positive
df1 = c -1 will typically be small
df2 = n - c will typically be large
Decision Rule:
Reject H0 if F > FU, otherwise do not reject H0
 = .05
Do not
reject H0
Reject H0 FU
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

13. One-Factor ANOVA F Test Example
Club Club 2 Club
You want to see if three different golf clubs yield different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the .05 significance level, is there a difference in mean driving distance?
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

14. One-Way ANOVA Example
Distance
270
260
250
240
230
220
210
200
190
Club Club 2 Club • • • • • • • • • • • • • • •
Club
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

15. ANOVA -- Single Factor: Excel Output
EXCEL: Tools | Data Analysis | ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Club 1 5
1246
249.2
108.2
Club 2
1130
226
77.5
Club 3
1029
205.8
94.2
ANOVA
Source of Variation SS df MS F
P-value
F crit
Between Groups
4716.4 2
2358.2
25.275
4.99E-05
3.89
Within
1119.6 12
93.3
Total
5836.0 14
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

16. One-Factor ANOVA Example Solution
H0: μ1 = μ2 = μ3
H1: μi not all equal
 = .05
p-value: 4.99E-05
Decision:
Conclusion:
Reject H0 at  = 0.05
There is evidence that at least one μi differs from the rest
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

17. The Tukey-Kramer Procedure
Tells which population means are significantly different
e.g.: μ1 = μ2 ≠ μ3
Done after rejection of equal means in ANOVA
Allows pair-wise comparisons
Compare absolute mean differences with critical range μ μ μ x = 1 2 3
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

18. Tukey-Kramer Critical Range
where:
QU = Value from Studentized Range Distribution with c and n - c degrees of freedom for the desired level of  (see appendix E.9 table)
MSW = Mean Square Within
nj and nj’ = Sample sizes from groups j and j’
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

19. The Tukey-Kramer Procedure: Example
1. PhStat computes the absolute mean differences:
Club Club 2 Club
2. You find a QU value from the table in appendix E.9 with
c = 3 (across the table) and n – c = 15 – 3 = 12 degrees of freedom (down the table) for the desired level of  ( = .05 used here):
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

20. The Tukey-Kramer Procedure: Example
(continued)
3. PhStat computes the Critical Range:
4. Compare:
5. All of the absolute mean differences are greater than critical range. Therefore there is a significant difference between each pair of means at 5% level of significance.
PhStat does all the calculations for you but you must input the Q value
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

21. Tukey-Kramer in PHStat
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

22. ANOVA Assumptions Levene’s Test
Tests the assumption that the variances of each group are equal.
First, define the null and alternative hypotheses:
H0: σ21 = σ22 = …=σ2c
H1: Not all σ2j are equal
Second, compute the absolute value of the difference between each value and the median of each group.
Third, perform a one-way ANOVA on these absolute differences.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

23. Two-Factor ANOVA Examines the effect of
Two factors of interest on the dependent variable
e.g., Percent carbonation and line speed on soft drink bottling process
Interaction between the different levels of these two factors
e.g., Does the effect of one particular carbonation level depend on which level the line speed is set?
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

24. Two-Factor ANOVA Assumptions Populations are normally distributed
Populations have equal variances
Independent random samples are selected
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

25. Two-Factor ANOVA Sources of Variation
Two Factors of interest: A and B
r = number of levels of factor A
c = number of levels of factor B
n/ = number of replications for each cell
n = total number of observations in all cells (n = rcn/)
Xijk = value of the kth observation of level i of factor A and level j of factor B
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

26. Two-Factor ANOVA Summary Table With Replication
Source of Variation
Degrees of Freedom
Sum of Squares
Mean Squares
F Statistic
p-value
Sample Factor A (Row)
r – 1
SSA
MSA = SSA/(r – 1)
MSA/ MSE
f (FA)
Columns
Factor B
c – 1
SSB
MSB =
SSB/(c – 1)
MSB/ MSE
f (FB)
Interaction (AB)
(r – 1)(c – 1)
SSAB
MSAB = SSAB/ [(r – 1)(c – 1)]
MSAB/ MSE
f (FA&B)
Within Error
rc (n’ – 1)
SSE
MSE =
SSE/[rc (n’ – 1)]
Total
rc n’ – 1
SST
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

27. Two-Factor ANOVA: The F Test Statistic
F Test for Factor A Effect
H0: μ1A = μ2A = μ3A = • • •
H1: Not all means are equal
F Test for Factor B Effect
H0: μ1B = μ2B = μ3B = • • •
H1: Not all means are equal
F Test for Interaction Effect
H0: Interaction of A and B is equal to zero
H1: Interaction of A and B is not zero
Only for Two-Factor with replication
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

28. Two-Factor ANOVA With Replication
Box Machine1 Machine2 Machine
As production manager, you want to see if 3 filling machines have different mean filling times when used with 5 types of boxes. At the .05 level, is there a difference in machines, in boxes? Is there an interaction?
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

29. Summary Table (Boxes) (Machines) Source of Degrees of Sum of Mean F
P-Value
Variation
Freedom
Squares
Square
Sample
5 - 1 = 4
7.4714
1.8678
3.6868
.0277
(Boxes)
Columns
3 - 1 = 2
53.149
1.52E-09
(Machines)
Interaction
(5-1)(3-1) = 8
1.2129
2.3941
.0690
9.7032
Within (Error)
5·3·(2-1)=15
7.5994
.5066
Total
3·5·2 -1 = 29
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

30. The Tukey-Kramer Procedure: With Replication Factor A
No Macro - compute by formula only
MSW (Within) from ANOVA Printout
Q from Table E.9 in the Book page 860 alpha = .05 or .01
r is the number of levels of factor A (across table) rc(n’-1) (down the table) c is the number of levels of factor B n’ is the number of replications
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

31. The Tukey-Kramer Procedure: With Replication Factor B
No Macro compute by formula only
MSW (Within) from ANOVA Printout
Q from Table E.9 in the Book page 860 alpha = .05 or .01
c is the number of levels of factor B (across table) rc(n’-1) (down the table) r is the number of levels of factor A n’ is the number of replications
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.

32. Chapter Summary Described one-factor analysis of variance
ANOVA assumptions
ANOVA test for difference in c means
The Tukey-Kramer procedure for multiple comparisons
Described two-factor analysis of variance
Examined effects of multiple factors
Examined interaction between factors for the model with replicated observations
The Tukey-Kramer procedure for multiple comparisons for both factor A and factor B for the model with replication
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc.