Single Factor Analysis of Variance (ANOVA) Bernd Schr oder logo1 - - PowerPoint PPT Presentation

SLIDE 1

logo1 The Situation Test Statistic Computing the Quantities

Single Factor Analysis of Variance (ANOVA)

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 2

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 3

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 4

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.”

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 5

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 6

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

• 2. The characteristic that differentiates

populations/treatments is called the factor.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 7

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

• 2. The characteristic that differentiates

populations/treatments is called the factor. The different treatments or populations are the levels of the factor.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 8

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

• 2. The characteristic that differentiates

populations/treatments is called the factor. The different treatments or populations are the levels of the factor.

• 3. Examples.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 9

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

• 2. The characteristic that differentiates

populations/treatments is called the factor. The different treatments or populations are the levels of the factor.

• 3. Examples.

◮ Testing different levels of medication/toxins etc. for effect. Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 10

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

• 2. The characteristic that differentiates

populations/treatments is called the factor. The different treatments or populations are the levels of the factor.

• 3. Examples.

◮ Testing different levels of medication/toxins etc. for effect. ◮ Testing different soil samples for mineral content. Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 11

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Analyzes Responses from Several Experiments or Treatments

• 1. Data is sampled from multiple populations or from

experiments with multiple treatments. Multiple means “more than two.” For two, we can use hypothesis tests (the exact tests are not covered in this course).

• 2. The characteristic that differentiates

populations/treatments is called the factor. The different treatments or populations are the levels of the factor.

• 3. Examples.

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

races/ethnic groups.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 12

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 13

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

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

compared.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 14

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

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

compared.

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 15

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 16

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 17

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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,
• r, no effect)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 18

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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,
• r, no effect)
• 4. Alternative hypothesis.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 19

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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,
• r, no effect)
• 4. Alternative hypothesis. Ha : At least two means differ.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 20

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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,
• r, 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 and

• ne has a sample average of around 10 minutes

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 21

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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,
• r, 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 and

• ne has a sample average of around 10 minutes, then it

pretty much looks like that one is different.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 22

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology

• 1. 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,
• r, 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 and

• ne has a sample average of around 10 minutes, then it

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 23

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology (cont.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 24

logo1 The Situation Test Statistic Computing the Quantities

ANOVA Terminology (cont.)

• 6. Xi,j is the random variable that denotes the jth measurement

from the ith population/treatment group.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 25

logo1 The Situation Test Statistic Computing the Quantities

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”)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 26

logo1 The Situation Test Statistic Computing the Quantities

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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 27

logo1 The Situation Test Statistic Computing the Quantities

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.

• 7. Individual sample means: Xi· = ∑J

j=1 Xij

J

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 28

logo1 The Situation Test Statistic Computing the Quantities

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.

• 7. Individual sample means: Xi· = ∑J

j=1 Xij

J The dot says we summed over the second variable.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 29

logo1 The Situation Test Statistic Computing the Quantities

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.

• 7. Individual sample means: Xi· = ∑J

j=1 Xij

J The dot says we summed over the second variable.

• 8. Sample variance: S2

i = ∑J j=1

• Xij −Xi·

2 J −1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 30

logo1 The Situation Test Statistic Computing the Quantities

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.

• 7. Individual sample means: Xi· = ∑J

j=1 Xij

J The dot says we summed over the second variable.

• 8. Sample variance: S2

i = ∑J j=1

• Xij −Xi·

2 J −1

• 9. Grand mean: X·· = ∑I

i=1 ∑J j=1 Xij

IJ

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 31

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 32

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

with the same variance σ2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 33

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 34

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest sample standard deviation

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 35

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 36

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

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

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 37

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 38

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 39

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 40

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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 between the samples

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 41

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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 between the samples (“between-samples” variation)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 42

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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 between the samples (“between-samples” variation) to a measure of the variation “within” the samples.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 43

logo1 The Situation Test Statistic Computing the Quantities

Underlying Assumptions and Their Consequences

• 1. All populations are assumed to be normally distributed

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

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

smallest 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 between the samples (“between-samples” variation) to a measure of the variation “within” the samples. (Remember that we assume all populations have the same σ).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 44

logo1 The Situation Test Statistic Computing the Quantities Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 45

logo1 The Situation Test Statistic Computing the Quantities

• 1. Mean square for treatments.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 46

logo1 The Situation Test Statistic Computing the Quantities

• 1. Mean square for treatments.

MSTr = J I −1

• X1· −X··

2 +···+

• XI· −X··

2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 47

logo1 The Situation Test Statistic Computing the Quantities

• 1. Mean square for treatments.

MSTr = J I −1

• X1· −X··

2 +···+

• XI· −X··

2 = J I −1

I

∑

i=1

• Xi· −X··

2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 48

logo1 The Situation Test Statistic Computing the Quantities

• 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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 49

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 50

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 51

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

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

individual samples.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 52

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

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

individual samples.

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 53

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

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

individual samples.

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

hypothesis is false: E(MSTr) > E(MSE) = σ2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 54

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

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

individual samples.

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

hypothesis is false: E(MSTr) > E(MSE) = σ2.

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

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 55

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

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

individual samples.

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

hypothesis is false: E(MSTr) > E(MSE) = σ2.

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

MSE 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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 56

logo1 The Situation Test Statistic Computing the Quantities

• 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 +···+S2 I

I .

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

MSE .

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

individual samples.

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

hypothesis is false: E(MSTr) > E(MSE) = σ2.

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

MSE 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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 57

logo1 The Situation Test Statistic Computing the Quantities

Example. Perform an ANOVA on the enclosed test data to see if the “true average performances” can be considered equal.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 58

logo1 The Situation Test Statistic Computing the Quantities

Example. Perform an ANOVA on the enclosed test data to see if the “true average performances” can be considered equal.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 59

logo1 The Situation Test Statistic Computing the Quantities

Keeping Track of the Data

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 60

logo1 The Situation Test Statistic Computing the Quantities

Keeping Track of the Data

The key to ANOVA (by hand) is orderly bookkeeping.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 61

logo1 The Situation Test Statistic Computing the Quantities

Keeping Track of the Data

The key to ANOVA (by hand) is orderly bookkeeping. Also remember that all this was done before computers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 62

logo1 The Situation Test Statistic Computing the Quantities

Keeping Track of the Data

The key to ANOVA (by hand) is orderly bookkeeping. Also remember that all this was done before computers. So anything that could save a few operations

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 63

logo1 The Situation Test Statistic Computing the Quantities

Keeping Track of the Data

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 64

logo1 The Situation Test Statistic Computing the Quantities

Keeping Track of the Data

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 65

logo1 The Situation Test Statistic Computing the Quantities Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 66

logo1 The Situation Test Statistic Computing the Quantities

• 1. Grand total: x·· =

I

∑

i=1 J

∑

j=1

xij

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 67

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 68

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 69

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

• 4. Error sum of squares: SSE =

I

∑

i=1 J

∑

j=1

(xij −xi·)2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 70

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

• 4. Error sum of squares: SSE =

I

∑

i=1 J

∑

j=1

(xij −xi·)2

• 5. MSTr = SSTr

I −1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 71

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

• 4. Error sum of squares: SSE =

I

∑

i=1 J

∑

j=1

(xij −xi·)2

• 5. MSTr = SSTr

I −1 (What happened to J?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 72

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

• 4. Error sum of squares: SSE =

I

∑

i=1 J

∑

j=1

(xij −xi·)2

• 5. MSTr = SSTr

I −1 (What happened to J? It’s the dummy sum over j!)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 73

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

• 4. Error sum of squares: SSE =

I

∑

i=1 J

∑

j=1

(xij −xi·)2

• 5. MSTr = SSTr

I −1 (What happened to J? It’s the dummy sum over j!)

• 6. MSE =

SSE I(J −1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 74

logo1 The Situation Test Statistic Computing the Quantities

• 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

x2

ij − 1

IJ x2

··

• 3. Treatment sum of squares: SSTr =

I

∑

i=1 J

∑

j=1

(xi· −x··)2 = 1 J

I

∑

i=1

x2

i· − 1

IJ x2

··,

where xi· =

J

∑

j=1

xij

• 4. Error sum of squares: SSE =

I

∑

i=1 J

∑

j=1

(xij −xi·)2

• 5. MSTr = SSTr

I −1 (What happened to J? It’s the dummy sum over j!)

• 6. MSE =

SSE I(J −1), F = MSTr MSE

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 75

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 76

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 77

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 78

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2 =

I

∑

i=1 J

∑

j=1

x2

ij −2xijx·· +x2 ··

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 79

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2 =

I

∑

i=1 J

∑

j=1

x2

ij −2xijx·· +x2 ··

=

I

∑

i=1 J

∑

j=1

x2

ij −2x·· I

∑

i=1 J

∑

j=1

xij +

I

∑

i=1 J

∑

j=1

x2

··

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 80

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2 =

I

∑

i=1 J

∑

j=1

x2

ij −2xijx·· +x2 ··

=

I

∑

i=1 J

∑

j=1

x2

ij −2x·· I

∑

i=1 J

∑

j=1

xij +

I

∑

i=1 J

∑

j=1

x2

··

=

I

∑

i=1 J

∑

j=1

x2

ij − 2

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij +IJ

• 1

IJ

I

∑

i=1 J

∑

j=1

xij 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 81

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2 =

I

∑

i=1 J

∑

j=1

x2

ij −2xijx·· +x2 ··

=

I

∑

i=1 J

∑

j=1

x2

ij −2x·· I

∑

i=1 J

∑

j=1

xij +

I

∑

i=1 J

∑

j=1

x2

··

=

I

∑

i=1 J

∑

j=1

x2

ij − 2

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij +IJ

• 1

IJ

I

∑

i=1 J

∑

j=1

xij 2 =

I

∑

i=1 J

∑

j=1

x2

ij − 1

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 82

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2 =

I

∑

i=1 J

∑

j=1

x2

ij −2xijx·· +x2 ··

=

I

∑

i=1 J

∑

j=1

x2

ij −2x·· I

∑

i=1 J

∑

j=1

xij +

I

∑

i=1 J

∑

j=1

x2

··

=

I

∑

i=1 J

∑

j=1

x2

ij − 2

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij +IJ

• 1

IJ

I

∑

i=1 J

∑

j=1

xij 2 =

I

∑

i=1 J

∑

j=1

x2

ij − 1

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij =

I

∑

i=1 J

∑

j=1

x2

ij − 1

IJx2

··

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 83

logo1 The Situation Test Statistic Computing the Quantities

Are the Claimed Formulas Right?

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2 =

I

∑

i=1 J

∑

j=1

x2

ij −2xijx·· +x2 ··

=

I

∑

i=1 J

∑

j=1

x2

ij −2x·· I

∑

i=1 J

∑

j=1

xij +

I

∑

i=1 J

∑

j=1

x2

··

=

I

∑

i=1 J

∑

j=1

x2

ij − 2

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij +IJ

• 1

IJ

I

∑

i=1 J

∑

j=1

xij 2 =

I

∑

i=1 J

∑

j=1

x2

ij − 1

IJ

I

∑

i=1 J

∑

j=1

xij

I

∑

i=1 J

∑

j=1

xij =

I

∑

i=1 J

∑

j=1

x2

ij − 1

IJx2

··

Treatment sum of squares: Similar.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 84

logo1 The Situation Test Statistic Computing the Quantities

Fundamental Identity: SST = SSTr +SSE

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 85

logo1 The Situation Test Statistic Computing the Quantities

Fundamental Identity: SST = SSTr +SSE

xij −x··

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 86

logo1 The Situation Test Statistic Computing the Quantities

Fundamental Identity: SST = SSTr +SSE

xij −x·· = (xij −xi·)+(xi· −x··)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 87

logo1 The Situation Test Statistic Computing the Quantities

Fundamental Identity: SST = SSTr +SSE

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 88

logo1 The Situation Test Statistic Computing the Quantities

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 89

logo1 The Situation Test Statistic Computing the Quantities

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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 90

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 91

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

j=1

(xij −xi·) = 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 92

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

j=1

(xij −xi·) = 0. Hence

SST

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 93

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

j=1

(xij −xi·) = 0. Hence

SST =

I

∑

i=1 J

∑

j=1

(xij −x··)2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 94

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 95

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

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

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 96

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 97

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 98

logo1 The Situation Test Statistic Computing the Quantities

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

• ver j, because

J

∑

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.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 99

logo1 The Situation Test Statistic Computing the Quantities

Example. Perform an ANOVA on the enclosed test data to see if the “true average performances” can be considered equal.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 100

logo1 The Situation Test Statistic Computing the Quantities

Example. Perform an ANOVA on the enclosed test data to see if the “true average performances” can be considered equal. Use a significance level of α = 0.05 and also compute the p-value.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 101

logo1 The Situation Test Statistic Computing the Quantities

Example. Perform an ANOVA on the enclosed test data to see if the “true average performances” can be considered equal. Use a significance level of α = 0.05 and also compute the p-value.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 102

logo1 The Situation Test Statistic Computing the Quantities Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)

SLIDE 103

logo1 The Situation Test Statistic Computing the Quantities Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Single Factor Analysis of Variance (ANOVA)