1. CHAPTER 3 Analysis of Variance (ANOVA)

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.
A treatment is a level of a factor.
Experimental units are the objects of interest in the experiment. [effect in experiment]
Variation between treatment groups captures the effect of the treatment.
Variation within treatment groups represents random error not explained by the experimental treatments.

3. EXAMPLE:
Mazlan studied the effect of three learning skills; peer group discussion, extra exercises, and additional reference books towards student’s score. Define which are factor, treatment and experimental units.
Answer:
Factor – Learning skill
Treatment - peer group discussion, extra exercises, and additional reference books
Experimental units - student’s score

4. Analysis of Variance: A Conceptual Overview
Assumptions for Analysis of Variance
The populations from which the samples were obtained
must be normally or approximately normal distributed
The variance of the response variable, denoted  2,
is the same for all of the populations.
The observations (samples) must be independent of
each other

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 or not.
Effects model for CRD:

6. 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

7. * Treatments are not equal * All treatments are equal
CONCLUSION
Fail to Reject H0
Reject H0
No difference in mean
Difference in mean
Between- group variance estimate approximately equal to the within-group variance
Between- group variance estimate will be larger than within-group variance
* Treatments are not equal
* All treatments are equal

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 .

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 [ F test ] = MSTR/MSE,
where numerator degrees of freedom are t – 1 and denominator degrees of freedom are N – t.
- F, t-1,N-t refer table ms 30
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. EXAMPLE ONE- WAY ANOVA

17. One-Way ANOVA - An Example
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.
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?

18. One-Way ANOVA - An Example
Factor Car wax
Treatments Type I, Type 2, Type 3
Experimental units Cars
Response variable Number of washes

19. 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

20. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
1. 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

21. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
2. Test Statistic - 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

22. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
2. 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

23. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
3. F value
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
4. Rejection Rule (Draw picture)
Critical Value Approach: Do not Reject H0 if F < 3.89
p-Value Approach: Do not Reject H0 if p-value > .05

24. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
5. Conclusion
F test < F alfa (3.89), do not fall in rejection region
so we do not reject H0.
2. There is insufficient evidence to conclude that
the mean number of washes for the three wax
types are not all the same.

25. EXAMPLE 2 Step 2 Reject H0 5 IMPORTANT STEP: HYPOTHESIS TESTING
TEST STATISTIC – F TEST
F – VALUE (CRITICAL VALUE)
REJECTION REGION
CONCLUSION
SSTR - MSTR
SSE - MSE
SST = SSTR +SSE
F TEST = MSTR/MSE
BUILD ANOVA TABLE
Reject H0