Single Factor Analysis of Variance (ANOVA) · logo1 The SituationTest StatisticComputing the Quantities Single Factor Analysis of Variance (ANOVA) Bernd Schroder¨ Bernd Schroder¨

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The Situation Test Statistic Computing the Quantities

Single Factor Analysis of Variance(ANOVA)

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Single Factor Analysis of Variance (ANOVA)

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ANOVA Analyzes Responses from SeveralExperiments or Treatments

1. Data is sampled from multiple populations or fromexperiments with multiple treatments. Multiple means“more than two.” For two, we can use hypothesis tests (theexact tests are not covered in this course).

2. The characteristic that differentiatespopulations/treatments is called the factor. The differenttreatments or populations are the levels of the factor.

3. Examples.I Testing different levels of medication/toxins etc. for effect.I Testing different soil samples for mineral content.I Testing the frequency of a given allele in different

races/ethnic groups.



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1. Data is sampled from multiple populations or fromexperiments with multiple treatments.

Multiple means“more than two.” For two, we can use hypothesis tests (theexact tests are not covered in this course).






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1. Data is sampled from multiple populations or fromexperiments with multiple treatments. Multiple means“more than two.”

For two, we can use hypothesis tests (theexact tests are not covered in this course).






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2. The characteristic that differentiatespopulations/treatments is called the factor.

The differenttreatments or populations are the levels of the factor.





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3. Examples.

I Testing different levels of medication/toxins etc. for effect.I Testing different soil samples for mineral content.I Testing the frequency of a given allele in different




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3. Examples.I Testing different levels of medication/toxins etc. for effect.

I Testing different soil samples for mineral content.I Testing the frequency of a given allele in different




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3. Examples.I Testing different levels of medication/toxins etc. for effect.I Testing different soil samples for mineral content.

I Testing the frequency of a given allele in differentraces/ethnic groups.



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ANOVA Terminology

1. I populations or treatments of equal size J are to becompared.

2. µi denotes the actual mean of the ith population.3. Null hypothesis. H0 : µ1 = µ2 = · · · = µI (no difference,

or, no effect)4. Alternative hypothesis. Ha : At least two means differ.5. For example, if among 10 pain relievers, all have a sample

average time until pain lessens of around 20 minutes andone has a sample average of around 10 minutes, then itpretty much looks like that one is different.When it’s not that obvious, we need a testing procedure(finer analysis).



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ANOVA Terminology1. I populations or treatments of equal size J are to be

compared.

2. µi denotes the actual mean of the ith population.3. Null hypothesis. H0 : µ1 = µ2 = · · · = µI (no difference,





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compared.2. µi denotes the actual mean of the ith population.

3. Null hypothesis. H0 : µ1 = µ2 = · · · = µI (no difference,or, no effect)

4. Alternative hypothesis. Ha : At least two means differ.5. For example, if among 10 pain relievers, all have a sample




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compared.2. µi denotes the actual mean of the ith population.3. Null hypothesis.

H0 : µ1 = µ2 = · · · = µI (no difference,or, no effect)





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compared.2. µi denotes the actual mean of the ith population.3. Null hypothesis. H0 : µ1 = µ2 = · · · = µI

(no difference,or, no effect)





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compared.2. µi denotes the actual mean of the ith population.3. Null hypothesis. H0 : µ1 = µ2 = · · · = µI (no difference,

or, no effect)





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or, no effect)4. Alternative hypothesis.

Ha : At least two means differ.5. For example, if among 10 pain relievers, all have a sample




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or, no effect)4. Alternative hypothesis. Ha : At least two means differ.

5. For example, if among 10 pain relievers, all have a sampleaverage time until pain lessens of around 20 minutes andone has a sample average of around 10 minutes, then itpretty much looks like that one is different.When it’s not that obvious, we need a testing procedure(finer analysis).



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average time until pain lessens of around 20 minutes andone has a sample average of around 10 minutes

, then itpretty much looks like that one is different.When it’s not that obvious, we need a testing procedure(finer analysis).



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average time until pain lessens of around 20 minutes andone has a sample average of around 10 minutes, then itpretty much looks like that one is different.

When it’s not that obvious, we need a testing procedure(finer analysis).



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ANOVA Terminology (cont.)

6. Xi,j is the random variable that denotes the jth measurementfrom the ith population/treatment group.xi,j will be the observed value (“as always”)Data is often displayed in a matrix.

7. Individual sample means: Xi· =∑

Jj=1 Xij

JThe dot says we summed over the second variable.

8. Sample variance: S2i =

∑Jj=1(Xij −Xi·

)2

J−1

9. Grand mean: X·· =∑

Ii=1 ∑

Jj=1 Xij

IJ



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ANOVA Terminology (cont.)6. Xi,j is the random variable that denotes the jth measurement

from the ith population/treatment group.

xi,j will be the observed value (“as always”)Data is often displayed in a matrix.


Jj=1 Xij



∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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from the ith population/treatment group.xi,j will be the observed value (“as always”)

Data is often displayed in a matrix.


Jj=1 Xij



∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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from the ith population/treatment group.xi,j will be the observed value (“as always”)Data is often displayed in a matrix.


Jj=1 Xij



∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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Jj=1 Xij

J

The dot says we summed over the second variable.


∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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Jj=1 Xij



∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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Jj=1 Xij



∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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Jj=1 Xij



∑Jj=1(Xij −Xi·

)2

J−1


Ii=1 ∑

Jj=1 Xij

IJ



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Underlying Assumptions and Their Consequences

1. All populations are assumed to be normally distributedwith the same variance σ2. Hence all Xij are normallydistributed and E(Xij) = µi and V(Xij) = σ2.

2. If the largest sample standard deviation is at most twice thesmallest sample standard deviation, then it is (still)reasonable to assume that the σs are equal.

3. To check normality, use a normal probability plot.4. If the null hypothesis µ1 = µ2 = · · · = µI is true, then all

sample averages should be close to each other.5. To determine if the variation is consistent with the null

hypothesis, we compare a measure of the variance betweenthe samples (“between-samples” variation) to a measure ofthe variation “within” the samples. (Remember that weassume all populations have the same σ ).



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Underlying Assumptions and Their Consequences1. All populations are assumed to be normally distributed

with the same variance σ2.

Hence all Xij are normallydistributed and E(Xij) = µi and V(Xij) = σ2.







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with the same variance σ2. Hence all Xij are normallydistributed and E(Xij) = µi and V(Xij) = σ2.







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2. If the largest sample standard deviation is at most twice thesmallest sample standard deviation

, then it is (still)reasonable to assume that the σs are equal.






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3. To check normality, use a normal probability plot.

4. If the null hypothesis µ1 = µ2 = · · · = µI is true, then allsample averages should be close to each other.

5. To determine if the variation is consistent with the nullhypothesis, we compare a measure of the variance betweenthe samples (“between-samples” variation) to a measure ofthe variation “within” the samples. (Remember that weassume all populations have the same σ ).



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3. To check normality, use a normal probability plot.4. If the null hypothesis µ1 = µ2 = · · · = µI is true

, then allsample averages should be close to each other.




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sample averages should be close to each other.




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hypothesis

, we compare a measure of the variance betweenthe samples (“between-samples” variation) to a measure ofthe variation “within” the samples. (Remember that weassume all populations have the same σ ).



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hypothesis, we compare a measure of the variance betweenthe samples

(“between-samples” variation) to a measure ofthe variation “within” the samples. (Remember that weassume all populations have the same σ ).



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hypothesis, we compare a measure of the variance betweenthe samples (“between-samples” variation)

to a measure ofthe variation “within” the samples. (Remember that weassume all populations have the same σ ).



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hypothesis, we compare a measure of the variance betweenthe samples (“between-samples” variation) to a measure ofthe variation “within” the samples.

(Remember that weassume all populations have the same σ ).



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1. Mean square for treatments.

MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2

2. Mean square for error MSE =S2

1 + · · ·+S2I

I.

3. The test statistic for single factor ANOVA is F =MSTrMSE

.

4. The J in MSTr re-scales the spread of the means back to the spread ofindividual samples.

5. If the null hypothesis is true: E(MSTr) = E(MSE) = σ2. If the nullhypothesis is false: E(MSTr) > E(MSE) = σ2.

6. When the null hypothesis is true, the statistic F =MSTrMSE

has an

F-distribution with ν1 = I−1 and ν2 = I(J−1).

7. A rejection region f > Fα,I−1,I(J−1) gives a test of significance level α .

8. For p-values, use the area to the right of the test statistic.



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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2

2. Mean square for error

MSE =S2

1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.


5. If the null hypothesis is true: E(MSTr) = E(MSE) = σ2.

If the nullhypothesis is false: E(MSTr) > E(MSE) = σ2.


has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an






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MSTr =J

I−1

[(X1·−X··

)2 + · · ·+(XI·−X··

)2]

=J

I−1

I

∑i=1

(Xi·−X··

)2


1 + · · ·+S2I

I.


.




has an



8. For p-values, use the area to the right of the test statistic.Bernd Schroder Louisiana Tech University, College of Engineering and Science


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Example. Perform an ANOVA onthe enclosed test data to see if the“true average performances” can beconsidered equal.



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Keeping Track of the Data

The key to ANOVA (by hand) is orderly bookkeeping.Also remember that all this was done before computers. Soanything that could save a few operations, or help minimizerounding errors, was appreciated.



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Keeping Track of the DataThe key to ANOVA (by hand) is orderly bookkeeping.

Also remember that all this was done before computers. Soanything that could save a few operations, or help minimizerounding errors, was appreciated.



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Keeping Track of the DataThe key to ANOVA (by hand) is orderly bookkeeping.Also remember that all this was done before computers.

Soanything that could save a few operations, or help minimizerounding errors, was appreciated.



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Keeping Track of the DataThe key to ANOVA (by hand) is orderly bookkeeping.Also remember that all this was done before computers. Soanything that could save a few operations

, or help minimizerounding errors, was appreciated.



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Keeping Track of the DataThe key to ANOVA (by hand) is orderly bookkeeping.Also remember that all this was done before computers. Soanything that could save a few operations, or help minimizerounding errors

, was appreciated.



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Keeping Track of the DataThe key to ANOVA (by hand) is orderly bookkeeping.Also remember that all this was done before computers. Soanything that could save a few operations, or help minimizerounding errors, was appreciated.



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1. Grand total: x·· =I

∑i=1

J

∑j=1

xij

2. Total sum of squares: SST =I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··

3. Treatment sum of squares: SSTr =I

∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij

4. Error sum of squares: SSE =I

∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1

(What happened to J? It’s the dummy sum over j!)

6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1

(What happened to J?

It’s the dummy sum over j!)

6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1)

, F =MSTrMSE



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∑i=1

J

∑j=1

xij


∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··


∑i=1

J

∑j=1

(xi·− x··)2 =1J

I

∑i=1

x2i·−

1IJ

x2··,

where xi· =J

∑j=1

xij


∑i=1

J

∑j=1

(xij − xi·)2

5. MSTr =SSTrI−1


6. MSE =SSE

I(J−1), F =

MSTrMSE



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Are the Claimed Formulas Right?

SST =I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1

x2ij −2xijx··+ x2

··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··

Treatment sum of squares: Similar.



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Are the Claimed Formulas Right?SST

=I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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Are the Claimed Formulas Right?SST =

I

∑i=1

J

∑j=1

(xij − x··)2

=I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij

=I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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I

∑i=1

J

∑j=1

(xij − x··)2 =I

∑i=1

J

∑j=1


··

=I

∑i=1

J

∑j=1

x2ij −2x··

I

∑i=1

J

∑j=1

xij +I

∑i=1

J

∑j=1

x2··

=I

∑i=1

J

∑j=1

x2ij −

2IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij + IJ

(1IJ

I

∑i=1

J

∑j=1

xij

)2

=I

∑i=1

J

∑j=1

x2ij −

1IJ

I

∑i=1

J

∑j=1

xij

I

∑i=1

J

∑j=1

xij =I

∑i=1

J

∑j=1

x2ij −

1IJ

x2··




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Fundamental Identity: SST = SSTr +SSE

xij − x·· = (xij − xi·)+(xi·− x··)

(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2

Now sum over i,j. The middle term drops out after summing

over j, becauseJ

∑j=1

(xij − xi·) = 0. Hence

SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1

(xi·− x··)2 = SSE+SSTr

1. SST measures the total variation of the data.2. SSE is the contribution from the variation within the

populations/treatment groups.3. SSTr is the contribution from between the

populations/groups.



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Fundamental Identity: SST = SSTr +SSExij − x··

= (xij − xi·)+(xi·− x··)

(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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Fundamental Identity: SST = SSTr +SSExij − x·· = (xij − xi·)+(xi·− x··)

(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2

= (xij − xi·)2 +2(xij − xi·)(xi·− x··)+(xi·− x··)

2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2

Now sum over i,j.

The middle term drops out after summing

over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j

, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1

(xij − xi·) = 0.

Hence

SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST

=I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2

=I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1

(xi·− x··)2

= SSE+SSTr



populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1


1. SST measures the total variation of the data.

2. SSE is the contribution from the variation within thepopulations/treatment groups.

3. SSTr is the contribution from between thepopulations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1



populations/treatment groups.

3. SSTr is the contribution from between thepopulations/groups.



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(xij − x··)2 = (xij − xi·)

2 +2(xij − xi·)(xi·− x··)+(xi·− x··)2


over j, becauseJ

∑j=1


SST =I

∑i=1

J

∑j=1

(xij − x··)2 =

I

∑i=1

J

∑j=1

(xij − xi·)2 +

I

∑i=1

J

∑j=1




populations/groups.Bernd Schroder Louisiana Tech University, College of Engineering and Science


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Use a significancelevel of α = 0.05 and also computethe p-value.



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Example. Perform an ANOVA onthe enclosed test data to see if the“true average performances” can beconsidered equal. Use a significancelevel of α = 0.05 and also computethe p-value.



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Example. Perform an ANOVA onthe enclosed test data to see if the“true average performances” can beconsidered equal. Use a significancelevel of α = 0.05 and also computethe p-value.



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Documents

Single Factor Analysis of Variance (ANOVA) · logo1 The SituationTest StatisticComputing the Quantities Single Factor Analysis of Variance (ANOVA) Bernd Schroder¨ Bernd Schroder¨