1. Statistics for Business and Economics
Chapter 8
Design of Experiments and Analysis of Variance

2. Learning Objectives Describe Analysis of Variance (ANOVA)
Explain the Rationale of ANOVA
Compare Experimental Designs
Test the Equality of 2 or More Means
Completely Randomized Design
Factorial Design
As a result of this class, you will be able to ...

3. Experiments

4. Experiment Investigator controls one or more independent variables
Called treatment variables or factors
Contain two or more levels (subcategories)
Observes effect on dependent variable
Response to levels of independent variable
Experimental design: plan used to test hypotheses

5. Examples of Experiments
Thirty stores are randomly assigned 1 of 4 (levels) store displays (independent variable) to see the effect on sales (dependent variable).
Two hundred consumers are randomly assigned 1 of 3 (levels) brands of juice (independent variable) to study reaction (dependent variable).

6. Experimental Designs Experimental Designs Completely Randomized
Factorial
One-Way
ANOVA
Two-Way
ANOVA

7. Completely Randomized Design

8. Experimental Designs Experimental Designs Completely Randomized
Factorial
One-Way
ANOVA
Two-Way
ANOVA

9. Completely Randomized Design
Experimental units (subjects) are assigned randomly to treatments
Subjects are assumed homogeneous
One factor or independent variable
Two or more treatment levels or classifications
Analyzed by one-way ANOVA

10. Randomized Design Example
Factor (Training Method)
Factor levels
Level 1
Level 2
Level 3
(Treatments)
Are the mean training times the same for 3 different methods?
9 subjects
3 methods (factor levels)
Experimental



units
21 hrs.
17 hrs.
31 hrs.
Dependent
variable
27 hrs.
25 hrs.
28 hrs.
(Response)
29 hrs.
20 hrs.
22 hrs. 78

11. One-Way ANOVA F-Test

12. Experimental Designs Experimental Designs Completely Randomized
Factorial
One-Way
ANOVA
Two-Way
ANOVA

13. One-Way ANOVA F-Test
Tests the equality of two or more (k) population means
Variables
One nominal scaled independent variable
Two or more (k) treatment levels or classifications
One interval or ratio scaled dependent variable
Used to analyze completely randomized experimental designs
Note:
There is one dependent variable in the ANOVA model. MANOVA has more than one dependent variable.
Ask, what are nominal & interval scales?

14. Conditions Required for a Valid ANOVA F-test: Completely Randomized Design
Randomness and independence of errors
Independent random samples are drawn
Normality
Populations are approximately normally distributed
Homogeneity of variance
Populations have equal variances

15. One-Way ANOVA F-Test Hypotheses
H0: 1 = 2 = 3 = ... = k
All population means are equal
No treatment effect
Ha: Not All i Are Equal
At least 2 pop. means are different
Treatment effect
1  2  ...  k is Wrong X
f(X)  1 = 2 3 1 2 3 X
f(X)  =

16. Variances AMONG differ Variances WITHIN differ
Why Variances?
Same treatment variation
Different random variation
Pop 1
Pop 2
Pop 3
Pop 4
Pop 6
Pop 5
Variances AMONG differ B
Different treatment variation
Same random variation
Pop 1
Pop 2
Pop 3
Pop 4
Pop 6
Pop 5
Variances WITHIN differ A
In all cases, the population means are DIFFERENT! Should reject Ho.
Panel A:
Same treatment effect - means are equal.
Different random variation - standard deviations are different.
As variances WITHIN get bigger, we are more likely to conclude equal means.
In Populations 4, 5, & 6, it is possible to draw a sample and falsely conclude population means are equal.
Panel B:
Different treatment effect - means are different.
Same random variation - standard deviations are equal.
As variances AMONG get bigger, we are more likely to conclude population means are different.
In Populations 4,5, & 6, it is possible to draw a sample and falsely conclude population means are equal.
Possible to conclude means are equal! 82

17. One-Way ANOVA Basic Idea
Compares two types of variation to test equality of means
Comparison basis is ratio of variances
If treatment variation is significantly greater than random variation then means are not equal
Variation measures are obtained by ‘partitioning’ total variation

18. One-Way ANOVA Partitions Total Variation
Variation due to treatment
Total variation
Variation due to random sampling
Variation due to Random Sampling are due to Individual Differences Within Groups.
Sum of Squares Among
Sum of Squares Between
Sum of Squares Treatment
Among Groups Variation
Sum of Squares Within
Sum of Squares Error
Within Groups Variation 85

19. Total Variation
Response, X X
Group 1
Group 2
Group 3

20. Treatment Variation
Response, X X3 X X2 X1
Group 1
Group 2
Group 3

21. Random (Error) Variation
Response, X X3 X2 X1
Group 1
Group 2
Group 3

22. One-Way ANOVA F-Test Test Statistic
F = MST / MSE
MST is Mean Square for Treatment
MSE is Mean Square for Error
Degrees of Freedom
1 = k - 1
2 = n - k
k = Number of groups
n = Total sample size

23. One-Way ANOVA Summary Table
Degrees of Freedom
Mean Square (Variance)
Source of Variation
Sum of Squares F
Treatment
k - 1
SST
MST = SST/(k - 1)
MST
MSE
n = sum of sample sizes of all populations.
k = number of factor levels
All values are positive. Why? (squared terms)
Degrees of Freedom & Sum of Squares are additive; Mean Square is NOT.
Error
n - k
SSE
MSE = SSE/(n - k)
Total
n - 1
SS(Total) = SST+SSE

24. One-Way ANOVA F-Test Critical Value
If means are equal, F = MST / MSE  1. Only reject large F!
Reject H 
Do Not
Reject H F
F(α; k – 1, n – k)
Always One-Tail!
© T/Maker Co.

25. One-Way ANOVA F-Test Example
As production manager, you want to see if three filling machines have different mean filling times. You assign 15 similarly trained and experienced workers, 5 per machine, to the machines. At the .05 level of significance, is there a difference in mean filling times?
Mach1 Mach2 Mach

26. One-Way ANOVA F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
1 = 2 = 3
Not all equal
Test Statistic:
Decision:
Conclusion:
.05 F
3.89
 = .05

27. Summary Table Solution
From Computer
Degrees of Freedom
Mean Square (Variance)
Source of Variation
Sum of Squares F
Treatment (Machines)
3 - 1 = 2
25.60
Error
= 12
.9211
Total
= 14

28. One-Way ANOVA F-Test Solution
H0: 1 = 2 = 3
Ha: Not All Equal
 = .05
1 = 2 2 = 12
Critical Value(s):
Test Statistic:
Decision:
Conclusion: F
MST
MSE  23
5820
9211
25.6 .
Reject at  = .05
 = .05
There is evidence population means are different F
3.89

29. One-Way ANOVA F-Test Thinking Challenge
You’re a trainer for Microsoft Corp. Is there a difference in mean learning times of 12 people using 4 different training methods ( =.05)?
M1 M2 M3 M
Use the following table.
You assign randomly 3 people to each method, making sure that they are similar in intelligence etc.
© T/Maker Co.

30. Summary Table (Partially Completed)
Degrees of Freedom
Mean Square (Variance)
Source of Variation
Sum of Squares F
Treatment
348
(Methods)
Error 80
Total

31. One-Way ANOVA F-Test Solution*
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
1 = 2 = 3 = 4
Not all equal
Test Statistic:
Decision:
Conclusion:
.05 F
4.07
 = .05

32. Summary Table Solution*
Degrees of Freedom
Mean Square (Variance)
Source of Variation
Sum of Squares F
Treatment (Methods)
4 - 1 = 3
348
116
11.6
Error
= 8 80 10
Total
= 11
428

33. One-Way ANOVA F-Test Solution*
H0: 1 = 2 = 3 = 4
Ha: Not All Equal
 = .05
1 = 2 = 8
Critical Value(s):
Test Statistic:
Decision:
Conclusion: F
MST
MSE 
116 10 11 6 .
Reject at  = .05
 = .05
There is evidence population means are different F
4.07

34. Tukey Procedure
Tells which population means are significantly different
Example: μ1 = μ 2 ≠ μ 3
Post hoc procedure
Done after rejection of equal means in ANOVA
Output from many statistical computer programs X
f(X) m 1 = 2 3
2 Groupings
105

35. Randomized Block Design
Reduces sampling variability (MSE)
Matched sets of experimental units (blocks)
One experimental unit from each block is randomly assigned to each treatment

36. Randomized Block Design Total Variation Partitioning
Variation Due to Random Sampling
SS(Total)
SSB
Total Variation
Variation Due to Blocks
Variation Due to Treatment
SSE
SST

37. Conditions Required for a Valid ANOVA F-test: Randomized Block Design
The blocks are randomly selected, and all treatments are applied (in random order) to each block
The distributions of observations corresponding to all block-treatment combinations are approximately normal
All block-treatment distributions have equal variances

38. Randomized Block Design F-Test Test Statistic
F = MST / MSE
MST is Mean Square for Treatment
MSE is Mean Square for Error
Degrees of Freedom
1 = k – 1 2 = n – k – b + 1
k = Number of groups
n = Total sample size
b = Number of blocks

39. Randomized Block Design Example
A production manager wants to see if three assembly methods have different mean assembly times (in minutes). Five employees were selected at random and assigned to use each assembly method. At the .05 level of significance, is there a difference in mean assembly times?
Employee Method 1 Method 2 Method 3

40. Random Block Design F-Test Solution*
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
1 = 2 = 3
Not all equal
Test Statistic:
Decision:
Conclusion:
.05 F
4.46
 = .05

41. Summary Table Solution*
Degrees of Freedom
Mean Square (Variance)
Source of Variation
Sum of Squares F
Treatment (Methods)
3 - 1 = 2
5.43
2.71
12.9
Block (Employee)
5 - 1 = 4
10.69
2.67
12.7
Error
= 8
1.68
.21
Total
= 14
17.8

42. Random Block Design F-Test Solution*
H0: 1 = 2 = 3
Ha: Not all equal
 = .05
1 = 2 = 8
Critical Value(s):
Test Statistic:
Decision:
Conclusion: F
MST
MSE 
2.71
.21
12.9
Reject at  = .05
 = .05
There is evidence population means are different F
4.46

43. Factorial Experiments

44. Experimental Designs Experimental Designs Completely Randomized
Factorial
One-Way
ANOVA
Two-Way
ANOVA

45. Factorial Design
Experimental units (subjects) are assigned randomly to treatments
Subjects are assumed homogeneous
Two or more factors or independent variables
Each has two or more treatments (levels)
Analyzed by two-way ANOVA

46. Two-Way ANOVA Data Table
Factor
Factor B A 1 2
... b
Observation k 1
X111
X121
...
X1b1
Xijk
X112
X122
...
X1b2 2
X211
X221
...
X2b1
X212
X222
...
X2b2
Level i Factor A
Level j Factor B : : : : :
Treatment a
Xa11
Xa21
...
Xab1
Xa12
Xa22
...
Xab2

47. Factorial Design Example
Factor 2 (Training Method)
Factor Levels
Level 1
Level 2
Level 3   
Level 1
15 hr.
10 hr.
22 hr.
(High)   
Factor 1 (Motivation)
11 hr.
12 hr.
17 hr.  
Treatment 
Level 2
27 hr.
15 hr.
31 hr.
(Low)   
29 hr.
17 hr.
49 hr.

48. Advantages of Factorial Designs
Saves time and effort
e.g., Could use separate completely randomized designs for each variable
Controls confounding effects by putting other variables into model
Can explore interaction between variables

49. Two-Way ANOVA

50. Experimental Designs Experimental Designs Completely Randomized
Factorial
One-Way
ANOVA
Two-Way
ANOVA

51. Two-Way ANOVA
Tests the equality of two or more population means when several independent variables are used
Same results as separate one-way ANOVA on each variable
No interaction can be tested
Used to analyze factorial designs

52. Interaction
Occurs when effects of one factor vary according to levels of other factor
When significant, interpretation of main effects (A and B) is complicated
Can be detected
In data table, pattern of cell means in one row differs from another row
In graph of cell means, lines cross

53. Graphs of Interaction
Effects of motivation (high or low) and training method (A, B, C) on mean learning time
Interaction
No Interaction
Average
Average
Response
High
Response
High
Low
Low A B C A B C

54. Two-Way ANOVA Total Variation Partitioning
Variation Due to Random Sampling
Variation Due to Interaction
SS(AB)
SS(Total)
Total Variation
Variation Due to Treatment A
Variation Due to Treatment B
SSA
SSB
SSE

55. Conditions Required for Valid F-Tests in Factorial Experiments
Normality
Populations are approximately normally distributed
Homogeneity of variance
Populations have equal variances
Independence of errors
Independent random samples are drawn

56. Two-Way ANOVA Hypotheses
Test for Treatment Means
H0: The ab treatment means are equal
Ha: At least two of the treatment means differ
Test Statistic
F = MST / MSE
Degrees of Freedom
1 = ab – 1 2 = n – ab

57. Two-Way ANOVA Hypotheses
Test for Factor Interaction
H0: The factors do not interact
Ha: The factors do interact
Test Statistic
F = MS(AB) / MSE
Degrees of Freedom
1 = (a – 1)(b – 1) 2 = n – ab

58. Two-Way ANOVA Hypotheses
Test for Main Effect of Factor A
H0: No difference among mean levels of factor A
Ha: At least two factor A mean levels differ
Test Statistic
F = MS(A) / MSE
Degrees of Freedom
1 = (a – 1) 2 = n – ab

59. Two-Way ANOVA Hypotheses
Test for Main Effect of Factor B
H0: No difference among mean levels of factor B
Ha: At least two factor B mean levels differ
Test Statistic
F = MS(B) / MSE
Degrees of Freedom
1 = (b – 1) 2 = n – ab

60. Two-Way ANOVA Summary Table
Source of Variation
Degrees of Freedom
Sum of Squares
Mean Square F
A (Row)
a - 1
SS(A)
MS(A)
MS(A)
MSE
B (Column)
b - 1
SS(B)
MS(B)
MS(B)
MSE
AB (Interaction)
(a - 1)(b - 1)
SS(AB)
MS(AB)
MS(AB)
MSE
Error
n - ab
SSE
MSE
Same as other designs
Total
n - 1
SS(Total)

61. Factorial Design Example
Human Resources wants to determine if training time is different based on motivation level and training method. Conduct the appropriate ANOVA tests. Use α = .05 for each test.
Training Method
Factor Levels
Self– paced
Classroom
Computer
15 hr.
10 hr.
22 hr.
Motivation
High
11 hr.
12 hr.
17 hr.
27 hr.
31 hr.
Low
29 hr.
49 hr.

62. Treatment Means F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
The 6 treatment means are equal
At least 2 differ
Test Statistic:
Decision:
Conclusion:
.05 F
4.39
 = .05

63. Two-Way ANOVA Summary Table
Source of Variation
Degrees of Freedom
Sum of Squares
Mean Square F
Model 5
546.75
240.35
7.65
Error 6
188.5
31.42
Corrected Total 11
735.25

64. Treatment Means F-Test Solution
H0: The 6 treatment means are equal
Ha: At least 2 differ
 = .05
1 = 2 = 6
Critical Value(s):
Test Statistic:
Decision:
Conclusion: F
4.39
 = .05
Reject at  = .05
There is evidence population means are different

65. Interaction F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
The factors do not interact
The factors interact
Test Statistic:
Decision:
Conclusion:
.05 F
5.14
 = .05

66. Two-Way ANOVA Summary Table
Source of Variation
Degrees of Freedom
Sum of Squares
Mean Square F
A (Row) 1
546.75
546.75
17.40
B (Column) 2
531.5
265.75
8.46
AB (Interaction) 2
123.5
61.76
1.97
Error 6
188.5
31.42
Same as other designs
Total 11
SS(Total)

67. Interaction F-Test Solution
H0: The factors do not interact
Ha: The factors interact
 = .05
1 = 2 = 6
Critical Value(s):
Test Statistic:
Decision:
Conclusion: F
5.14
 = .05
Do not reject at  = .05
There is no evidence the factors interact

68. Main Factor A F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
No difference between motivation levels
Motivation levels differ
Test Statistic:
Decision:
Conclusion:
.05 F
5.99
 = .05

69. Two-Way ANOVA Summary Table
Source of Variation
Degrees of Freedom
Sum of Squares
Mean Square F
A (Row) 1
546.75
546.75
17.40
B (Column) 2
531.5
265.75
8.46
AB (Interaction) 2
123.5
61.76
1.97
Error 6
188.5
31.42
Same as other designs
Total 11
SS(Total)

70. Main Factor A F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
No difference between motivation levels
Motivation levels differ
Test Statistic:
Decision:
Conclusion:
.05
Reject at  = .05 F
5.99
 = .05
There is evidence motivation levels differ

71. Main Factor B F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
No difference between training methods
Training methods differ
Test Statistic:
Decision:
Conclusion:
.05 F
5.14
 = .05

72. Two-Way ANOVA Summary Table
Source of Variation
Degrees of Freedom
Sum of Squares
Mean Square F
A (Row) 1
546.75
546.75
17.40
B (Column) 2
531.5
265.75
8.46
AB (Interaction) 2
123.5
61.76
1.97
Error 6
188.5
31.42
Same as other designs
Total 11
SS(Total)

73. Main Factor B F-Test Solution
H0:
Ha:
 =
1 = 2 =
Critical Value(s):
No difference between training methods
Training methods differ
Test Statistic:
Decision:
Conclusion:
.05
Reject at  = .05 F
5.14
 = .05
There is evidence training methods differ

74. Conclusion Described Analysis of Variance (ANOVA)
Explained the Rationale of ANOVA
Compared Experimental Designs
Tested the Equality of 2 or More Means
Completely Randomized Design
Factorial Design