1 1 Slide Analysis of Variance Chapter 13 BA 303.

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Analysis of Variance

Chapter 13BA 303

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An Introduction to Analysis of Variance

A factor is a variable that the experimenter has selected for investigation.

A treatment is a level of a factor.

Example: A company president is interested in the

number of hours worked by managers at three different plants: Gulfport, Long Beach, and Biloxi.

Factor: Plant Treatment: Gulfport, Long Beach, and

Biloxi

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Analysis of Variance: A Conceptual Overview

Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means.

Data obtained from observational or experimental studies can be used for the analysis.

We want to use the sample results to test the following hypotheses:

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

Ha: Not all population means are equal

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


If H0 is rejected, we cannot conclude that all population means are different.

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


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For each population, the response (dependent) variable is normally distributed.

The variance of the response variable, denoted 2, is the same for all of the populations.

The observations must be independent.

Assumptions for Analysis of Variance


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Sampling Distribution of Given H0 is Truex

1x 3x2x

Sample means are close together because there is only one sampling distribution when H0 is true.


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Sampling Distribution of Given H0 is Falsex

33 1x 2x3x 11 22

Sample means come from different sampling distributions and are not as close together when H0

is false.


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Analysis of Variance

Between-Treatments Estimate of Population Variance Within-Treatments Estimate of Population Variance Comparing the Variance Estimates: The F Test ANOVA Table

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2

1

( )

MSTR1

k

j jj

n x x

k

Between-Treatments Estimateof Population Variance s 2

Denominator is thedegrees of freedom

associated with SSTR

Numerator is calledthe sum of squares

dueto treatments (SSTR)

The estimate of 2 based on the variation of the sample means is called the mean square due to treatments and is denoted by MSTR.

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The estimate of 2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE.

Within-Treatments Estimateof Population Variance s 2

Denominator is thedegrees of freedom

associated with SSE

Numerator is called

the sum of squares

due to error (SSE)

MSE

( )n s

n k

j jj

k

T

1 2

1MSE

( )n s

n k

j jj

k

T

1 2

1

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F Test Statistic

MSE

MSTRF

1 MSTR d.f k

knT MSE d.f

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Comparing the Variance Estimates: 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 MSTR d.f. equal to k - 1 and MSE d.f. equal to nT - k.

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

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F Distribution

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Sampling Distribution of MSTR/MSE

Do Not Reject H0Do Not Reject H0

Reject H0Reject H0

MSTR/MSEMSTR/MSE

Critical ValueCritical ValueFF

Sampling Distributionof MSTR/MSE

a

Comparing the Variance Estimates: The F Test

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MSTRSSTR

-

k 1MSTR

SSTR-

k 1

MSESSE

-

n kT

MSESSE

-

n kT

MSTRMSE

MSTRMSE

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquare F

Treatments

Error

Total

k - 1

nT - 1

SSTR

SSE

SST

nT - k

SST is partitionedinto SSTR and

SSE.

SST’s degrees of freedom

(d.f.) are partitioned into

SSTR’s d.f. and SSE’s d.f.

ANOVA Tablefor a Completely Randomized Design

p-Value

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SST divided by its degrees of freedom nT – 1 is the overall sample variance that would be obtained if we treated the entire set of observations as one data set.

With the entire data set as one sample, the formula for computing the total sum of squares, SST, is:

2

1 1

SST ( ) SSTR SSEjnk

ijj i

x x


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ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error.

Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal population means.


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Test for the Equality of k Population Means

F = MSTR/MSE

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


Hypotheses

Test Statistic

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Test for the Equality of k Population Means

Rejection Rule

where the value of F is based on anF distribution with k - 1 numerator d.f.and nT - k denominator d.f.

Reject H0 if p-value < ap-value Approach:

Critical Value Approach: Reject H0 if F > Fa

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Reed Manufacturing

Janet Reed would like to know if there is anysignificant difference in the mean number of hoursworked per week for the department managers at herthree manufacturing plants (in Buffalo, Pittsburgh,and Detroit). An F test will be conducted using a = .05.

Testing for the Equality of k Population Means

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Reed Manufacturing

A simple random sample of five managers fromeach of the three plants was taken and the number ofhours worked by each manager in the previous weekis shown on the next slide.


Factor . . . Manufacturing plantTreatments . . . Buffalo, Pittsburgh, DetroitExperimental units . . . Managers

Response variable . . . Number of hours worked

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12345

4854575462

7363666474

5163615456

Plant 1Buffalo

Plant 2Pittsburgh

Plant 3DetroitObservation

Sample MeanSample Variance

55 68 5726.0 26.5 24.5


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H0: 1= 2= 3

Ha: Not all the means are equalwhere: 1 = mean number of hours worked per

week by the managers at Plant 1 2 = mean number of hours worked per week by the managers at Plant 2 3 = mean number of hours worked per week by the managers at Plant 3

1. Develop the hypotheses.

p -Value and Critical Value Approaches


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2. Specify the level of significance. a = .05


3. Compute the value of the test statistic.

MSTR = 490/(3 - 1) = 245SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490

(Sample sizes are all equal.)Mean Square Due to Treatments


= (55 + 68 + 57)/3 = 60x

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3. Compute the value of the test statistic.

MSE = 308/(15 - 3) = 25.667

SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308Mean Square Due to Error

(con’t.)

F = MSTR/MSE = 245/25.667 = 9.55



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TreatmentError

Total

490308

798

212

14

24525.667

Source ofVariation

Sum ofSquares

Degrees ofFreedom

MeanSquare

9.55

F

ANOVA Table


p-Value

.0033

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5. Determine whether to reject H0.

We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.

The p-value < .05, so we reject H0.

With 2 numerator d.f. and 12 denominator d.f.,the p-value is .01 for F = 6.93. Therefore, thep-value is less than .01 for F = 9.55.

p –Value Approach

4. Compute the p –value.


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5. Determine whether to reject H0.

Because F = 9.55 > 3.89, we reject H0.

Critical Value Approach

4. Determine the critical value and rejection rule.

Reject H0 if F > 3.89

We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.

Based on an F distribution with 2 numeratord.f. and 12 denominator d.f., F.05 = 3.89.


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ANOVA PRACTICE

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Perform an ANOVA

Using the data in the table on the next slide, compute the mean and standard deviation for the Direct treatment.

Compute the sum of the squares due to treatments (SSTR) and the mean square due to treatments (MSTR).

Compute the sum of the squares due to the error (SSE) and the mean square due to the error (MSE).

Compute the F test statistic.

Using a=0.05, what do you conclude about the null hypothesis that the means of the treatments are equal?

Determine the approximate p-value.

Present the results in an ANOVA table.

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Direct Indirect Combination

1 17.0 16.6 14.44 25.2 0.042 18.5 22.2 3.24 24.0 1.003 15.8 20.5 0.01 21.5 12.254 18.2 18.3 4.41 26.8 3.245 20.2 24.2 14.44 27.5 6.256 16.0 19.8 0.36 25.8 0.647 13.3 21.2 0.64 24.2 0.64

20.4 25.06.26 4.01

x

2)( xxi 2)( xxi ixixix

2s

2)( xxi

1 1 Slide Analysis of Variance Chapter 13 BA 303.

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