chapter 3 analysis of variance (anova) part 1 madam siti aisyah zakaria eqt 271 sem 2 2014/2015

CHAPTER 3Analysis of Variance

(ANOVA)PART 1

MADAM SITI AISYAH ZAKARIAEQT 271

SEM 2 2014/2015

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.

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

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. A : B : C [Between group-variance]

• Variation within treatment groups represents random error not explained by the experimental treatments. A ONLY [Within group variance]

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 skillTreatment - peer group discussion, extra

exercises, and additional reference books

Experimental units - student’s score

The populations from which the samples were obtained must be normally or approximately normal distributed 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 variance of the response variable, denoted 2, is the same for all of the populations.

The observations (samples) must be independent of each otherThe observations (samples) must be independent of each other

Assumptions for Analysis of Variance

Analysis of Variance: A Conceptual Overview

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:

1,2,...,

1, 2,...,

j

ij j ij

i nx

j t

where,

: overall mean effect

: the effect of treatment

: random error

j

ij

j

• 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

0 1 2

0

: .... 0

: 0 for at least one t

j

H

H j

One-Way ANOVA, cont.

CONCLUSION

Fail to Reject H0

No difference in mean

Between- group variance estimate

approximately equal to the within-

group variance

F test value approximately equal to 1

Reject H0

Difference in mean

Between- group variance estimate will be larger than

within-group variance

F test value = greater than 1

* All treatments are equal

* Treatments are not equal

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).

2

1 1

1 1

1

SST SSTR + SSE

where, ,

j

j

nt

ijj i

nt

ij tj i

jj

x x

x

x N nN

x


• 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 .

2

.1

SSTRt

j jj

n x x

2

2.

1 1 1

.12

SSE = ( 1)

where, 1

j

j

nt k

ij j j jj i j

n

ij jj

jj

x x n s

x x

sn


• 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.

Sampling Distribution of MSTR/MSE

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSE

Critical ValueCritical ValueFF

Sampling DistributionSampling Distributionof MSTR/MSEof MSTR/MSE

One-Way ANOVA, cont.Comparing the Variance Estimates: The F Test

One-Way ANOVA, cont.ANOVA Table

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquare

F p-Value

Treatments

SSTR t-1

Error SSE N-t

Total SST N-1

-1

SSTRMSTR

t

SSEMSE

N t

MSTR

MSE

One-Way ANOVA - An ExampleExample 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: 64 50 71 55 67 61 80 56 59 74

1+ rider(s): 44 52 54 48 69 67 54 57 58 51 62 67

a. What are the null and alternative hypothesis?H0: µ1 = µ2 where Group 1 = driver alone

H1: µ1 µ2 Group 2 = with rider(s)

One-Way ANOVA - An Exampleb. Use ANOVA and the 0.025 level of significance in testing

the appropriate null hypothesis.

SSTR = 10(63.7 – 60)2 + 12(56.917 – 60)2 = 250.983SSE = (64 – 63.7 )2 + (50 – 63.7 )2 + ... + (74 – 63.7 )2

+ (44 – 56.917) 2 + (52 – 56.917) 2 + ... + (67 – 56.917) 2

= 1487.017SST = (64 – 60 )2 + (50 – 60 )2 + ... + (74 – 60 )2

+ (44 – 60) 2 + (52 – 60) 2 + ... + (67 – 60) 2 = 1738

1 1 1

2 2 2

63.7, 9.3577, 10

56.916, 7.9711, 12

60.0

x s n

x s n

x

One-Way ANOVA - An ExampleCompare calculated values to those in the Excel output:

Anova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Alone 10 637 63.7 87.56666667

WithPass 12 683 56.91666667 63.53787879

ANOVA

Source of Variation SS df MS F P-value F crit

Between Groups 250.9833333 1 250.9833333 3.37566268 0.081071382 4.351250027

Within Groups 1487.016667 20 74.35083333

Total 1738 21

The test statistic The p-value The critical bound

EXAMPLEONE- WAY ANOVA

AutoShine, Inc. is considering marketing a long-lasting car wax. Three different waxes (Type 1, Type 2,and Type 3) have been developed.

Example 4.2 : AutoShine, Inc.

In order to test the durability of these waxes, 5 newcars were waxed with Type 1, 5 with Type 2, and 5with Type 3. Each car was then repeatedly runthrough an automatic carwash until the wax coatingshowed signs of deterioration.

One-Way ANOVA - An Example

The number of times each car went through thecarwash before its wax deteriorated is shown on thenext slide. AutoShine, Inc. must decide which waxto 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


12345

2730292831

3328313030

2928303231

Sample MeanSample Variance

ObservationWaxType 1

WaxType 2

WaxType 3

2.5 3.3 2.529.0 30.4 30.0

Testing for the Equality of k Population Means:A Completely Randomized Design


1. Hypothesis

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

H0: 1=2=3

H1: Not all the means are equal



Because the sample sizes are all equal:

MSE = 33.2/(15 - 3) = 2.77

MSTR = 5.2/(3 - 1) = 2.6

SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2

SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2

Mean Square Error

2. Test Statistic - Mean Square Between Treatments

= (29 + 30.4 + 30)/3 = 29.8



1 2 3 / 3x x x x

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquares F

Treatments

Error

Total

2

14

5.2

33.2

38.4

12

2.60

2.77

0.939

2. ANOVA Table


p-Value

0.42


3. F value



4. Rejection Rule (Draw picture)

p-Value Approach: Do not Reject H0 if p-value > .05

Critical Value Approach: Do not Reject H0 if F < 3.89

where F0.05,2,12 = 3.89 is based on an F distributionwith 2 numerator degrees of freedom and 12denominator degrees of freedom

2. There is insufficient evidence to conclude thatthe mean number of washes for the three waxtypes are not all the same.

5. Conclusion

1. F test < F alfa (3.89), do not fall in rejection region so we do not reject H0.



EXAMPLE 2

Reject H0

5 IMPORTANT STEP:1.HYPOTHESIS TESTING2.TEST STATISTIC – F TEST3.F – VALUE (CRITICAL VALUE)4.REJECTION REGION5.CONCLUSION

Step 21. SSTR - MSTR

2. SSE - MSE3. SST = SSTR +SSE4. F TEST = MSTR/MSE5. BUILD ANOVA TABLE

EXERCISETEXT BOOK PAGE 143 ( 1 &

2)5 IMPORTANT STEP:1.HYPOTHESIS TESTING2.TEST STATISTIC – F TEST3.F – VALUE (CRITICAL VALUE4.REJECTION REGION5.CONCLUSION

chapter 3 analysis of variance (anova) part 1 madam siti aisyah zakaria eqt 271 sem 2 2014/2015

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