1. CHAPTER 4 Analysis of Variance (ANOVA)
EQT 373

2. 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. (eg: multivitamin)
A treatment is a level of a factor. (eg: different brands of multivitamins)
Experimental units are the objects of interest in the experiment. (eg: students)
EQT 373

3. Overview Analysis of Variance (ANOVA) EQT 373 One-Way ANOVA
(Completely
Randomized Design)
Randomized Block Design
Two-Way ANOVA
(Factorial Experiment)
EQT 373

4. When dealing with two-sample inference, we are actually dealing with a special case of one-factor problem with two levels (k=2). This factor level is also known as treatment.
Example: Let say we want to compare the different brands of multivitamins of Brand A and Brand B, then two samples are needed. In this case the treatment is the multivitamins with two levels (k=2), Brand A and Brand B. Suppose we need to compare four brands of multivitamins, than in this case, there is one factor at four levels (k=4). ANOVA is a very common procedure dealing with comparison of more than two populations means. If we are dealing with only one factor, then it is referred as one-way ANOVA.

5. 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:

6. Hypothesis:
All population means are equal
i.e., no factor effect (no variation in means among groups)
At least one population mean is different
i.e., there is a factor effect
Does not mean that all population means are different (some pairs may be the same)
EQT 373

7. or EQT 373 The Null Hypothesis is True
All Means are the same: No Factor Effect
The Null Hypothesis is Not True
At least one of the means is different: Factor Effect is present or
EQT 373

8. Format for data: Data appear in separate columns or rows, organized by treatment groups. Sample size of each group may differ.
Calculations:
Total variation can be split into two parts:
Sum of squares total (SST) ,
Sum of squares treatment (SSTR) ,
Sum of squares error (SSE) ,

9. a The F Test F Test Reject H0 Do Not Reject H0 MSTR/MSE F
Critical Value
If F Test > F,t-1,N-t or p-value <  , reject H0 at the  level.

10. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean
p-Value
Treatments
SSTR
t-1
Error
SSE
N-t
Total
SST
N-1

11. Example:
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 ≠ µ2 Group 2 = with rider(s)

12. b) Use ANOVA and the 0.05 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

13. Compare calculated values to those in the Excel output:
The test statistic The p-value The critical bound

14. Source of Sum of Degrees of Mean
Variation Squares Freedom Square F-Ratio
Treatments
Error
Total
I. H0: µ1 = µ H1: µ1≠ µ2
II. Rejection Region:
a = 0.05; F0.05,1,20 =4.35
If F > 4.35, reject H0.
0.95
0.05
F=4.35

15. III. Test Statistic: F = 250.983 / 74.351 = 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.
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 =

16. Example
Testing for the Equality of k Population Means:
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.
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?

17. Factor Car wax
Treatments Type I, Type 2, Type 3
Experimental units Cars
Response variable Number of washes
Observation
Wax type 1
Wax Type 2
Wax Type 3 1 27 33 29 2 30 28 3 31 4 32 5
Sample Mean
29.0
30.4
30.0

18. Hypothesis
H0: 1=2=3
H1: Not all the means are equal
where:  = mean number of washes using wax
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

19. 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

20. Rejection Rule
Critical Value Approach: Reject H0 if F > F0.05,2,12
where F0.05,2,12 = 3.89
p-Value Approach : Reject H0 if p-value < .05
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 0.42)
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.

21. 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.”
The block is the nuisance factor (the actual experimental unit).

22. 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

23. Calculations:
Total variation can now be split into three parts:
Sum of squares total (SST) ,
Sum of squares treatment (SSTR) ,
Sum of squares block (SSBL) ,
Sum of squares error (SSE) , SSE = SST – SSTR – SSBL
t is the levels of the treatments, n is number of blocks

24. 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):

25. F-Ratio, Treatment = MSTR/MSE,
where numerator degrees of freedom are t – 1 and denominator degrees of freedom are (t – 1)(n – 1) . If F-Ratio > F or p-value <  , reject H0 at the  level: 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). If F-Ratio > F or p-value <  , reject H0 at the  level: Reject the null hypothesis means that at least one block group had a different effect on the dependent variable than the rest.

26. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean
Treatments
SSTR
t-1
Blocks
SSBL
n-1
Error
SSE
(t-1)(n-1)
Total
SST
tn-1

27. Example
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.
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

28. 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

29. 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

30. p-Value Approach: Reject H0 if p-value < 0.05
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
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)
Test Statistic: F = MSTR/MSE = 2.6/.68 = 3.82

31. 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.

32. Two-Way ANOVA (Factorial Experiment)
Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required.
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.

33. Effects model for factorial experiment:

34. 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

35. 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:
Total variation can now be split into four parts:
Sum of squares total (SST) ,
Sum of squares Factor A (SSA) ,
Sum of squares Factor B (SSB) ,
Sum of squares Error (SSE) ,
Sum of squares Interaction (SSAB) : SSAB = SST – SSA – SSB – SSE

36. 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 .

37. 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.

38. 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.

39. 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

40. Factors Factor A: Industry Type (2 levels)
Example
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 below.
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

41. Hypothesis: 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

42. 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

43. Conclusions Using the Critical Value Approach
Industries: F = < 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.