1. CHAPTER 4 Analysis of Variance (ANOVA)

2. Learning Objectives
Describe the relationship between analysis of variance, the design of experiments, and the types of applications to which the experiments are applied.
Differentiate one-way and two-way ANOVA techniques.
Arrange data into a format that facilitates their analysis by the appropriate ANOVA technique.
Use the appropriate methods in testing hypothesis relative to the experimental data.

3. Key Terms Factor level, treatment, block, interaction
Experimental units
Replication
Within-group variation
Between-group variation
Completely randomized design
Randomized block design
Factorial experiment
Sum of squares:
Treatment
Error
Block
Interaction
Total

4. Key Concepts
ANOVA can be used to analyze the data obtained from experimental or observational studies.
A factor is a variable that the experimenter has selected for investigation.
A treatment is a level of a factor.
Experimental units are the objects of interest in the experiment.
Variation between treatment groups captures the effect of the treatment.
Variation within treatment groups represents random error not explained by the experimental treatments.

5. Analysis of Variance: A Conceptual Overview
Assumptions for Analysis of Variance
For each population, the response (dependent)
variable is normally distributed.
The variance of the response variable, denoted  2,
is the same for all of the populations.
The observations must be independent.

6. One-Way ANOVA (Completely Randomized Design)
A completely randomized design (CRD) is an experimental design in which the treatments are randomly assigned to the experimental units.
Purpose: Examines two or more levels of an independent variable to determine if their population means could be equal.
Effects model for CRD:

7. One-Way ANOVA, cont. Hypothesis: H0: µ1 = µ2 = ... = µt *
H1: µi  µj for at least one pair (i,j)
(At least one of the treatment group means differs from the rest. OR At least two of the population means are not equal) @
* where t = number of treatment groups or levels

8. One-Way ANOVA, cont.
Format for data: Data appear in separate columns or rows, organized by treatment groups. Sample size of each group may differ.
Calculations:
Sum of squares total (SST) = sum of squared differences between each individual data value (regardless of group membership) minus the grand mean, , across all data... total variation in the data (not variance).

9. One-Way ANOVA, cont.
Calculations, cont.:
Sum of squares treatment (SSTR) = sum of squared differences between each group mean and the grand mean, balanced by sample size... between-groups variation (not variance).
Sum of squares error (SSE) = sum of squared differences between the individual data values and the mean for the group to which each belongs... within-group variation (not variance).

10. One-Way ANOVA, cont. Calculations, cont.:
Mean square treatment (MSTR) = SSTR/(t – 1), where t is the number of treatment groups... between-groups variance.
Mean square error (MSE) = SSE/(N – t),
where N is the number of elements sampled and t is the number of treatment groups... within-groups variance.
F-Ratio = MSTR/MSE,
where numerator degrees of freedom are t – 1 and denominator degrees of freedom are N – t.
If F-Ratio > F or p-value <  , reject H0 at the  level.

11. One-Way ANOVA, cont. Comparing the Variance Estimates: The F Test a
Sampling Distribution of MSTR/MSE
Sampling Distribution
of MSTR/MSE
Reject H0
Do Not Reject H0 a
MSTR/MSE F
Critical Value

12. One-Way ANOVA, cont. ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F
p-Value
Treatments
SSTR
t-1
Error
SSE
N-t
Total
SST
N-1

13. One-Way ANOVA - An Example
Example 4.1: Safety researchers, interested in determining if occupancy of a vehicle might be related to the speed at which the vehicle is driven, have checked the following speed (MPH) measurements for two random samples of vehicles:
Driver alone:
1+ rider(s):
a. What are the null and alternative hypothesis?
H0: µ1 = µ2 where Group 1 = driver alone
H1: µ1 ¹ µ2 Group 2 = with rider(s)

14. One-Way ANOVA - An Example
b. Use ANOVA and the level of significance in testing the appropriate null hypothesis.
SSTR = 10(63.7 – 60)2 + 12( – 60)2 =
SSE = (64 – 63.7 )2 + (50 – 63.7 ) (74 – 63.7 )2
+ (44 – ) 2 + (52 – ) (67 – ) 2 =
SST = (64 – 60 )2 + (50 – 60 ) (74 – 60 )2
+ (44 – 60) 2 + (52 – 60) (67 – 60) 2
= 1738

15. One-Way ANOVA - An Example
Compare calculated values to those in the Excel output:
The test statistic The p-value The critical bound

16. One-Way ANOVA - An Example
Source of Sum of Degrees of Mean
Variation Squares Freedom Square F-Ratio
Treatments
Error
Total
I. H0: µ1 = µ H1: µ1 ¹ µ2
II. Rejection Region:
a = 0.05
dfnum = 1 If F > 4.35, reject H0.
dfdenom = 20
0.95
0.05
F=4.35

17. One-Way ANOVA - An Example
III. Test Statistic: F = / = 3.38
IV. Conclusion: Since the test statistic of F = 3.38 falls below the critical value of F0.05,1,20 = 4.35, we do not reject H0 with at most 5% error.
V. Implications: There is not enough evidence to conclude that the speed at which a vehicle is driven changes depending on whether the driver is alone or has at least one passenger.
p-value:
To calculate the p-value, in a cell within a Microsoft Excel spreadsheet, type: =FDIST(3.38,1,20)
The answer is: p-value =

18. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Example 4.2 : AutoShine, Inc.
AutoShine, Inc. is considering marketing a long-
lasting car wax. Three different waxes (Type 1, Type 2,
and Type 3) have been developed.
In order to test the durability of these waxes, 5 new
cars were waxed with Type 1, 5 with Type 2, and 5
with Type 3. Each car was then repeatedly run
through an automatic carwash until the wax coating
showed signs of deterioration.

19. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Example: AutoShine, Inc.
The number of times each car went through the
carwash before its wax deteriorated is shown on the
next slide. AutoShine, Inc. must decide which wax
to market. Are the three waxes equally effective?
Factor Car wax
Treatments Type I, Type 2, Type 3
Experimental units Cars
Response variable Number of washes

20. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Wax
Type 1
Wax
Type 2
Wax
Type 3
Observation 1 2 3 4 5 27 30 29 28 31 33 28 31 30 29 28 30 32 31
Sample Mean
Sample Variance

21. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Hypothesis
H0: 1=2=3
H1: Not all the means are equal
where:
1 = mean number of washes using Type 1 wax
2 = mean number of washes using Type 2 wax
3 = mean number of washes using Type 3 wax

22. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Mean Square Between Treatments
Because the sample sizes are all equal:
= ( )/3 = 29.8
SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2
MSTR = 5.2/(3 - 1) = 2.6
Mean Square Error
SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2
MSE = 33.2/(15 - 3) = 2.77

23. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Rejection Rule
p-Value Approach: Reject H0 if p-value < .05
Critical Value Approach: Reject H0 if F > 3.89
where F0.05,2,12 = 3.89 is based on an F distribution
with 2 numerator degrees of freedom and 12
denominator degrees of freedom

24. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
Test Statistic
F = MSTR/MSE = 2.60/2.77 =0 .939
Conclusion
The p-value is greater than .10, where F0.10,2,12 = 2.81.
(Excel provides a p-value of .42.)
Therefore, we cannot reject H0.
There is insufficient evidence to conclude that
the mean number of washes for the three wax
types are not all the same.

25. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Squares F
p-Value
Treatments
5.2 2
2.60
0.939
0.42
Error
33.2 12
2.77
Total
38.4 14

26. Two-Way ANOVA without Replication (Randomized Block Design)
If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.
Purpose: Reduces variance within treatment groups by removing known fluctuation among different levels of a second dimension, called a “block.”

27. Randomized Block Design, cont.
Effects model for RBD:
Two Sets of Hypothesis:
Treatment Effect:
H0: 1 =  2 = ... =  t =0
H1:  j  0 at least one j
Block Effect:
H0:  i = 0 for each value of i through n
H1:  i ≠ 0 at least one i

28. Randomized Block Design, cont.
Format for data: Data appear in a table, where location in a specific row and a specific column is important.
Calculations:
Sum of squares total (SST) = sum of squared differences between each individual data value (regardless of group membership) minus the grand mean, , across all data... total variation in the data (not variance).

29. Randomized Block Design, cont.
Calculations, cont.:
Sum of squares treatment (SSTR) = sum of squared differences between each treatment group mean and the grand mean, balanced by sample size... between-treatment-groups variation (not variance).
Sum of squares block (SSBL) = sum of squared differences between each block group mean and the grand mean, balanced by sample size... between-block-groups variation (not variance).
Sum of squares error (SSE):
SSE = SST – SSTR – SSBL

30. Randomized Block Design, cont.
Calculations, cont.:
Mean square treatment (MSTR) = SSTR/(t – 1),
where t is the number of treatment groups.
Mean square block (MSBL) = SSBL/(n – 1),
where n is the number of block groups. Controls the size of SSE by removing variation that is explained by the blocking categories.
Mean square error (MSE):

31. Randomized Block Design, cont.
Calculations, cont.:
Test Statistics, F-Ratios:
F-Ratio, Treatment = MSTR/MSE,
where numerator degrees of freedom are t – 1 and denominator degrees of freedom are (t – 1)(n – 1) . This F-ratio is the test statistic for the hypothesis that the treatment group means are equal. To reject the null hypothesis means that at least one treatment group had a different effect than the rest.
F-Ratio, Block = MSBL/MSE,
where numerator degrees of freedom are n – 1 and denominator degrees of freedom are (t – 1)(n – 1). This F-ratio is the test statistic for the hypothesis that the block group means are equal. To reject the null hypothesis means that at least one block group had a different effect on the dependent variable than the rest.
If F-Ratio > F or p-value <  , reject H0 at the  level.

32. Randomized Block Design, cont. ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F
p-Value
Treatments
SSTR
t-1
Blocks
SSBL
n-1
Error
SSE
(t-1)(n-1)
Total
SST
tn-1

33. Randomized Block Design, cont.
Example 4.3: Crescent Oil Co.
Crescent Oil has developed three new blends of
gasoline and must decide which blend or blends to
produce and distribute. A study of the miles per
gallon ratings of the three blends is being conducted
to determine if the mean ratings are the same for the
three blends.

34. Randomized Block Design, cont.
Example: Crescent Oil Co.
Five automobiles have been tested using each of
the three gasoline blends and the miles per gallon
ratings are shown on the next slide.
Factor Gasoline blend
Treatments Blend X, Blend Y, Blend Z
Blocks Automobiles
Response variable Miles per gallon

35. Randomized Block Design, cont.
Type of Gasoline (Treatment)
Automobile
(Block)
Block
Means
Blend X
Blend Y
Blend Z 1 2 3 4 5 31 30 29 33 26 30 29 31 25 30 29 28 26
30.333
29.333
28.667
31.000
25.667
Treatment
Means

36. Randomized Block Design, cont.
Hypothesis: H0: 1 =  2 =  3=0
H1:  j  0 at least one j
Mean Square Due to Treatments:
The overall sample mean is 29. Thus,
SSTR = 5[( )2 + ( )2 + ( )2] = 5.2
MSTR = 5.2/(3 - 1) = 2.6
Mean Square Due to Blocks:
SSBL = 3[( ) ( )2] = 51.33
MSBL = 51.33/(5 - 1) = 12.8
Mean Square Due to Error:
SSE = = 5.47
MSE = 5.47/[(3 - 1)(5 - 1)] = 0 .68

37. Randomized Block Design, cont.
ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F
p-Value
Treatments
5.20 2
2.60
3.82
0.07
Blocks
51.33 4
12.80
***
Error
5.47 8
0.68
Total
62.00 14

38. Randomized Block Design, cont.
Rejection Rule:
p-Value Approach: Reject H0 if p-value < 0.05
Critical Value Approach: Reject H0 if F > 4.46
For  = 0.05, F0.05,2,8 = 4.46
(2 d.f. numerator and 8 d.f. denominator)

39. Randomized Block Design, cont.
Test Statistic:
F = MSTR/MSE = 2.6/.68 = 3.82
Conclusion:
The p-value is greater than .05 (where F0.05,2,8 = 4.46) and less than (where F0.10,2,8 = 3.11). (Excel provides a p-value of 0.07). Therefore, we cannot reject H0.
There is insufficient evidence to conclude that
the miles per gallon ratings differ for the three
gasoline blends.

40. Two-Way ANOVA with Replication (Factorial Experiment)
Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required.
Replication means an independent repeat of each factor combination.
Purpose: Examines (1) the effect of Factor A on the dependent variable, y; (2) the effect of Factor B on the dependent variable, y; along with (3) the effects of the interactions between different levels of the two factors on the dependent variable , y.

41. Factorial Experiment, cont.
Effects model for factorial experiment:

42. Two-Factor Factorial Experiment
Three Sets of Hypothesis:
Factor A Effect:
H0: 1 =  2 = ... =  a =0
H1: at least one i  0
Factor B Effect:
H0: 1 = 2 = ... = b =0
H1: at least one j ≠ 0
Interaction Effect:
H0: ( )ij = 0 for all i,j
H1: at least one ( )ij  0

43. Two-Factor Factorial Experiment, cont.
Format for data: Data appear in a grid, each cell having two or more entries. The number of values in each cell is constant across the grid and represents r, the number of replications within each cell.
Calculations:
Sum of squares total (SST) = sum of squared differences between each individual data value (regardless of group membership) minus the grand mean, , across all data... total variation in the data (not variance).

44. Two-Factor Factorial Experiment, cont.
Calculations, cont.:
Sum of squares Factor A (SSA) = sum of squared differences between each group mean for Factor A and the grand mean, balanced by sample size... between-factor-groups variation (not variance).
Sum of squares Factor B (SSB) = sum of squared differences between each group mean for Factor B and the grand mean, balanced by sample size... between-factor-groups variation (not variance).

45. Two-Factor Factorial Experiment, cont.
Calculations, cont.:
Sum of squares Error (SSE) = sum of squared differences between individual values and their cell mean... within-groups variation (not variance).
Sum of squares Interaction:
SSAB = SST – SSA – SSB – SSE

46. Two-Factor Factorial Experiment, cont.
Calculations, cont.:
Mean Square Factor A (MSA) = SSA/(a – 1),
where a = the number of levels of Factor A .
Mean Square Factor B (MSB) = SSB/(b – 1),
where b = the number of levels of Factor B .
Mean Square Interaction (MSAB) = SSAB/(a – 1)(b – 1).
Mean Square Error (MSE) = SSE/ab(r – 1),
where ab(r – 1) = the degrees of freedom on error .

47. Two-Factor Factorial Experiment, cont.
Calculations - F-Ratios:
F-Ratio, Factor A = MSA/MSE ,
where numerator degrees of freedom are a – 1 and denominator degrees of freedom are ab(r – 1). This F-ratio is the test statistic for the hypothesis that the Factor A group means are equal. To reject the null hypothesis means that at least one Factor A group had a different effect on the dependent variable than the rest.
F-Ratio, Factor B = MSB/MSE,
where numerator degrees of freedom are b – 1 and denominator degrees of freedom are ab(r – 1). This F-ratio is the test statistic for the hypothesis that the Factor B group means are equal. To reject the null hypothesis means that at least one Factor B group had a different effect on the dependent variable than the rest.

48. Two-Factor Factorial Experiment, cont.
Calculations - F-Ratios:
F-Ratio, Interaction = MSAB/MSE,
where numerator degrees of freedom are (a – 1)( b – 1) and denominator degrees of freedom are ab(r – 1). This F-ratio is the test statistic for the hypothesis that Factors A and B operate independently. To reject the null hypothesis means that there is some relationship where levels of Factor A operate differently with different levels of Factor B.
If F-Ratio>F or p-value< , reject H0 at the  level.

49. Two-Factor Factorial Experiment, cont. Two-Factor ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F
p-Value
Factor A
SSA
a-1
Factor B
SSB
b-1
Interaction
SSAB
(a-1)(b-1)
Error
SSE
ab(r-1)
Total
SST
abr-1

50. Two-Factor Factorial Experiment,
Example 4.4: State of Ohio Wage Survey
A survey was conducted of hourly wages for a
sample of workers in two industries at three locations
in Ohio. Part of the purpose of the survey was to
determine if differences exist in both industry type and location. The sample data are shown on the next slide.

51. Two-Factor Factorial Experiment- An Example
Industry
Cincinnati
Cleveland
Columbus I
$12.10
$11.80
$12.90
11.80
11.20
12.70
12.10
12.00
12.20 II
12.40
12.60
13.00
12.50
Factors:
Factor A: Industry Type (2 levels)
Factor B: Location (3 levels)
Replications:
Each experimental condition is repeated 3 times

52. Two-Factor Factorial Experiment- An Example
Hypothesis for this analysis: Factor A Effect: H0: 1 =  2 =0 H1: at least one i  0 Factor B Effect: H0: 1 = 2 =3 =0 H1: at least one j ≠ 0 Interaction Effect: H0: ( )ij = 0 for all i,j H1: at least one ( )ij  0

53. Two-Factor Factorial Experiment- An Example Result: Two-Way ANOVA Table
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F
p-Value
Factor A
0.50 1
4.19
0.06
Factor B
1.12 2
0.56
4.69
0.03
Interaction
0.37
0.19
1.55
0.25
Error
1.43 12
0.12
Total
3.42 17

54. Two-Factor Factorial Experiment-An Example
Conclusions Using the Critical Value Approach
Industries: F = 4.19 < Fa = 4.75
Mean wages do not differ by industry type.
Locations: F = 4.69 > Fa = 3.89
Mean wages differ by location.
Interaction: F = 1.55 < Fa = 3.89
Interaction is not significant.

55. Two-Factor Factorial Experiment-An Example
Conclusions Using the p-Value Approach
Industries: p-value = .06 > a = .05
Mean wages do not differ by industry type.
Locations: p-value = .03 < a = .05
Mean wages differ by location.
Interaction: p-value = .25 > a = .05
Interaction is not significant.