Analysis of Variance (ANOVA)

• ANOVA can be used to test for the equality of three or more population means

• We want to use the sample results to test the following hypotheses.

H0: 1=2=3=. . . = k

Ha: Not all population means are equal

• Rejecting H0 means that at least two population means have different values.

Assumptions of ANOVA

• The values from each population are normally distributed.

• The variance, 2, is the same for all of the populations.

• The observations must be independent.

Impact of Assumptions

• Given these assumptions, what is the source of variation among the observed variables?

• Two values could come from different populations, or distributions, which could have different means.

• Two values could come from the same distribution.

ANOVA:Testing for the Equality of k Population Means

• The ANOVA table

• Between-treatments estimate of 2

• Within-treatments estimate of 2

• Comparing these two estimates: The F Test

Preliminaries

• Compute the jth sample mean and variance, j = 1,…,k

• Compute the overall sample mean

1

)(1

2

21

j

n

i

jij

jj

n

iij

jn

xxsn

xx

jj

)( 21

11

k

n

iij

k

j

nnn

x

x

j

Note: the book calls this nT

Between-Treatment Estimate of 2

• Sum of squares due to treatments

• A between-treatments estimate of 2, MSTR, can be computed by

k

j

jj xxnSSTR1

2)(

1

k

SSTRMSTR

Within-Treatment Estimate of 2

• The within-treatments estimate of 2

kn

sn

kn

SSEMSE

T

k

jjj

T

1

2)1(

The ANOVA Table

Source of Sum of Degrees of Mean

Variation Squares Freedom Squares F

Between SSTR k - 1 MSTR MSTR/MSE

Within SSE nT - k MSE

Total SST nT - 1

jn

iij

k

j

xxSST1

2

1

)(

Example: Kelli Home Products, Inc.

• Kelli Home Products, 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 is shown on the next slide. Kelli Home Products, Inc. must decide which wax to market. Are the three waxes equally effective?

Example continued

Observation Type 1 Type 2 Type 3

1 48 73 51

2 54 63 63

3 57 66 61

4 54 64 54

5 62 74 56

Sample Mean 55 68 57

Sample Variance 26.0 26.5 24.5

The F-Test

• If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with k – 1 numerator and nT – k denominator df

• If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates 2.

Sampling Distribution of MSTR/MSE

• The figure below shows the rejection region associated with a level of significance equal to where F denotes the critical value.

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSEFF

Critical ValueCritical Value

The F-Test

• We will reject H0 if the resulting value of MSTR/MSE appears to be too large to have been selected at random from the appropriate F distribution.

• Our decision rule becomes

.HReject 0,1, knk TFFIf

• Assuming = .05, F.05 = 3.89 (2 d.f. numerator, 12 d.f. denominator). Therefore, reject H0 if F > 3.89

• Test Statistic:

F = MSTR/MSE = 245/25.667 = 9.55

• Decision: Reject H0

• There appears to be a difference in the types of wax at a 5% level of significance.

Example continued

Multiple Comparison Procedures

• If ANOVA provides statistical evidence to reject the null hypothesis of equal population means, then Fisher’s least significance difference (LSD) procedure can be used to determine where the differences occur.

Fisher’s LSD Procedure

• Pairwise hypothesis tests of the form

• Test Statistic

ji

ji

H

H

:

:

1

0

)/11( ji

ji

nnMSE

xxt

Fisher’s LSD Procedure

• Rejection rule:

.Hreject then ,t- or t t t If 0n/2,n/2, TT kk

Alternative Fisher’s LSD Procedure

• Hypotheses

• Rejection Rule

where

ji

ji

H

H

:

:

1

0

.Hreject then LSD,x If 0i jx

)11( t LSDTn/2,

jik nnMSE

Using LSD in our Example

• Assuming = .05, t.025,12 = 2.179.

– Test Statistic

– Conclusion

The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

98.6)51

5125.67(2.179 LSD

13685521 xx

Using LSD in our Example

• Note that

• Since 13 > 6.98, and 11 > 6.98, we conclude that types 1 and 2 have different means, and types 2 and 3 have different means, but there is no significant difference between the means of types 1 and 3.

115768

25755

136855

32

31

21

xx

xx

xx

Getting it Done in EXCEL

• Use Tools/Data Analysis/ANOVA: Single Factor

• Data should be in columns which represent each treatment

Experimental Design

• Statistical studies can be classified as being either experimental or observational.

• In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest.

• In an observational study, no attempt is made to control the factors.

• Cause-and-effect relationships are easier to establish in experimental studies than in observational studies.