1. CHAPTER 3 Analysis of Variance (ANOVA) PART 1
MADAM SITI AISYAH ZAKARIA
EQT 271
SEM /2016

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 Sum of squares: Treatment Error Block Interaction Total
Response variable / dependent variable
Factor level /Independent variable
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 is a statistical inference method used to test the equality of three or more population means.
Response variable/dependent variable is the result that your experiment is measuring.
A factor / independent is a variable that affects the outcome of the response variable.
A treatment is a level of a factor.
Experimental units are the objects or individual that experiment uses.
Variation between treatment groups captures the effect of the treatment. A : B : C [Between group-variance]
Variation within treatment groups represents random error not explained by the experimental treatments. A ONLY [Within group variance]

5. 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 response variable, factor, treatment and experimental units.
Answer:
Response variable – student’s score
Factor – Learning skill
Treatment - peer group discussion, extra exercises, and additional reference books [level of factor]
Experimental units – score from student at random.

6. EXAMPLE:
You are interested to conduct a study in attempt to determine whether or not the means of statistics test score are not the same for the different sleeping times ( 4,6,8 hours). State the response variable, factor, treatment and experimental unit.
Answer:
Response variable – statistics test scores
Factor – sleeping times
Treatment – the number of hours of sleep 4, 6, 8 hours [level of factor]
Experimental units – students who are selected in random

7. Analysis of Variance: A Conceptual Overview
Assumptions for Analysis of Variance
There must be k random samples drawn from each of k population
The populations from which the samples are drawn must be normally or approximately normal distributed.
The population variances must be equal ;
The k samples are independent of each other; that is the subjects in one group are not related to subjects in a second group in any way.

8. Variability in response variable
Comparison of population mean
ANOVA
Study the total variation in the response variable/ dependent variable, y
Total variation
The variation of the group means
( )around the grand mean
The variation of observations within each group around the group mean
VARIATION BETWEEN GROUPS
(reflects the effect of the factor levels on the response variable); A: B: C
VARIATION WITHIN GROUPS
(represents random error from sampling); A
Note: large value between group variation higher chance to reject null hypothesis the groups’ population means are significantly different.

9. F test value approximately equal to 1 F test value = greater than 1
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
Variability of response variable
F test value approximately equal to 1
F test value = greater than 1

10. One-Way ANOVA (Completely Randomized Design)
A completely randomized design (CRD) is often used in scientific experiments with the purpose of comparing treatments, processes, materials or products.
Purpose: used when there is only one independent variable or also called as factor which gives influence to the response variable, y
Effects model for CRD:

11. One-Way ANOVA, cont.
Example: you would want to compare the effect of three different types of vehicle (small, sedan, multi-purpose-vehicle (MPV) on fuel consumption.
Response variable – fuel consumption
Factor – three types of vehicle
Hypothesis: k= 3 population means of fuel consumption for the three different types of vehicle.
H0: µ1 = µ2 = µ3 (all three population means are equal)
H1: At least one mean is different from the other.
Note: Reject H0 – does not require that ALL means of fuel consumption are significantly differ from each other. It rejected if at least one of the mean is significantly different from the other.

12. 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).

13. 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

14. 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

15. One-Way ANOVA, cont. Calculations, cont.:
Mean square treatment (MSTR) = SSTR/(k– 1), where k is the number of treatment groups... between-groups variance.
Mean square error (MSE) = SSE/(N – k),
where N is the number of elements sampled and k is the number of treatment groups... within-groups variance.
F-Ratio [ F test ] = MSTR/MSE,
where numerator degrees of freedom are k – 1 and denominator degrees of freedom are N – k.
- F, k-1,N-k refer table ms 30
If F-Ratio > F or p-value <  , reject H0 at the  level.

16. 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, k-1,N-k
Critical Value

17. Characteristics of F-distribution
The distribution is not a symmetric distribution
F is calculated on ratio of variance so it could not be negative
Shape of F-dist is determined by two degree of freedom, d.f
df1 = k – 1; df2 = N - k
* k = number of group ;
* N = Total number of observation

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

19. 5 important step 1. HYPOTHESIS TESTING 2. TEST STATISTIC – F TEST
H0: µ1 = µ2 = µ3 (all three population means are equal)
H1: At least one mean is different from the other.
2. TEST STATISTIC – F TEST
3. F (alfa) – VALUE (CRITICAL VALUE)
4. REJECTION REGION
5. CONCLUSION/DECISION
Reject H0 – There is sufficient evidence to show that at least one population mean is different from the others.
Fail to Reject H0 – There is insufficient evidence to show that at least one population mean is different from the others.

20. EXAMPLE ONE- WAY ANOVA

21. One-Way ANOVA - An Example
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? Use

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

23. 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

24. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
1. Hypothesis
H0: 1=2=3
H1: at least one population mean number of washes for all types is not equal from the other
(population mean number of washes for all types 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

25. One-Way ANOVA - An Example
Testing for the Equality of k Population Means: CRD
2. Test Statistic 1 5 2 6 4 3

26. 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

27. 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: Fail to Reject H0 if F < 3.89
p-Value Approach: Fail to Reject H0 if p-value > .05

28. One-Way ANOVA - An Example
Testing for the Equality of k Population Means:
A Completely Randomized Design
5. Conclusion/ Decision
F test < F alfa (3.89), do not fall in rejection region
so we fail to reject H0.
2. There is insufficient evidence to conclude that at least one population mean is different from the other or at least one of the mean number of washes for the three wax types is different from the other. So, this three waxes equally effective. ( same effect)

29. EXAMPLE 2 2 step: Reject H0 Factor:….. Treatments: …..
Experimental Unit:..
Response variable:…
2 step:
5 IMPORTANT STEP:
HYPOTHESIS TESTING
TEST STATISTIC – F TEST
F (alfa) – VALUE (CRITICAL VALUE)
REJECTION REGION
CONCLUSION
SST
SSTR MSTR
SSE = SST -SSTR MSE
F TEST = MSTR/MSE
BUILD ANOVA TABLE
Reject H0

30. answer
SSTR =
SSE =
SST =
F TEST = 4.83

31. EXERCISE NOTE PAGE 33-36 (ALL)
Take out all the information from the exercise:
1. Factor:…..
2. Treatments: …..
3. Experimental Unit:..
4. Response variable:…
5 IMPORTANT STEP:
HYPOTHESIS TESTING
TEST STATISTIC – F TEST
F (alfa) – VALUE (CRITICAL VALUE)
REJECTION REGION
CONCLUSION

32. THE END