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One Way Analysis of Variance

– Designed experiments usually involve comparisons among more than two means.

– The use of Z or t tests with more than two means is not efficient.

– A more efficient approach is Analysis of Variance (Anova)

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One Way Analysis of Variance

Three approaches in this course:1. One variable completely randomized

• analogous to independent groups Z and t test

2. One variable randomized block• analogous to dependent pairs Z and t test

3. Two variable completely randomized

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µ2

µ1

µ4

µ3

4

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X2

X1

X4

X3

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Completely randomized design

• Compares the variability among the treatment means (X1, X2, X3, … XP) to error variability.

• Is the variability among the treatment means so large that it could not be due to sampling error?

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Completely randomized design

• To answer that question, we need to measure two things:

1. Variability among the treatment means• how different from each other are the means?

2. Error variability• how different could the treatment means be just on the basis of chance?

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Variability among treatment means

• Recall the formula for a variance:

S2 = Σ(Xi – X)2 (1)

(n-1)

• The numerator measures the variability of individual scores around the sample mean.

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Variability among treatment means

S2 = Σ(Xi – X)2 (1)

(n-1)

SST = Σni(Xi – X)2 (2)

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Individual treatment means

Grand mean

X = XG

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Variability among treatment means

SST = Σni(Xi – X)2 (2)

Here, we’re finding the difference between each element in a set of scores and the mean of that set – but this time, the elements are themselves sample means

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Variability among treatment means

• In Eqn. 2, the sum (Σ) measures the variability among the sample means.

– SST is the Sum of Squared deviations of the Treatment means from the grand mean.

• In other words, SST is the Sum of Squares for Treatments. The more different the treatment means are from each other, the bigger SST is.

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X1

X3

X2

X4

X

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Error variability

• SSE =

Σ(X1j – X1)2 + Σ(X2j – X2)2 + … + Σ(XPj – XP)2

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Mean for Sample 1One individual score

in Sample 1

Once again, we’re finding the difference between each element in a set of scores and the mean of that set – but this time we’re working at the level of individual samples

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Error variability

SSE =Σ(X1j – X1)2 + Σ(X2j – X2)2 + … + Σ(XPj – XP)2

– SSE = the Sum of Squares for Error. This measures the total variability of individual scores around their respective treatment means.

– Key point : people who get the same treatment should all have the same score. Any deviation from that state (from X) reflects sampling error.

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2. Sampling variability

Important note:

(S1)2 = Σ(X1j – X1)2 (n1 – 1)

Therefore, (n1 – 1) (S1)2 = Σ(X1j – X1)2

Same is true for S22, S3

2, … SP2.

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2. Sampling variability

Thus:

SSE = (n1 – 1) S12 + (n2 – 1) S2

2 + … + (nP – 1) SP2

Now, we’re almost ready. One last issue:

SST is the sum of P terms (that is, P deviations) SSE is the sum of N = Σni terms

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2. Sampling variability

How can we compare SST to SSE? To make SST and SSE commensurable, we divide each by their degrees of freedom.

SST = MST (Mean Square Treatment) P-1

SSE = MSE (Mean Square Error) N-P

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The Analysis of Variance – F-test

When there is a treatment effect, MST will be much larger than MSE. Therefore, the ratio of MST to MSE will be much larger than 1.0

F = MSTMSE

How much larger than 1 must F be for us to reject H0? Check F table for α and d.f.

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The Analysis of Variance – F-test

d.f. numerator = p – 1

d.f. denominator = n – p

Important note: for Anova, F test is always one-tailed. You’re asking “is the treatment variance larger than the error variance?”

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Computational Formulae

CM = (ΣXi)2

NSSTotal = ΣXi

2 – CM

SST = T12 + T2

2 + … + TP2 – CM

n1 n2 nP

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Example 1

You want to know whether different situations produce different amounts of stress. The amount of the hormone corticosterone circulating in the blood is a good measure of how stressed a person is. You randomly assign 15 students to three groups of five each. Subjects in Group 1 have their corticosterone levels measured immediately after returning from vacation (low stress). Group 2 subjects are measured after one week of classes (moderate stress). Group 3 is measured immediately before final exam week (high stress).

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Example 1All measurements are made at the same time of day. Scores in milligrams of corticosterone per 100 milliliters of blood are (α = .05):Vacation Class Final ExamX X2 X X2 X X2

2 4 8 64 10 1003 9 10 100 13 1697 49 7 49 14 1962 4 5 25 13 1696 36 10 100 15 22520 102 40 338 65 859

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Example 1

n1 = 5 n2 = 5 n3 = 5

X1 = 4.0 X2 = 8.0 X3 = 13.0

∑X = 125∑X2 = 1299

X = XG = 125/15 = 8.33

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Example 1

SSTotal = ∑X2 – CM = 1299 – 1252 = 257.333

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SSTreat = T12 + T2

2 + … + TP2 – CM

n1 n2 nP

= 202 +402 + 652 – 1252 = 203.333 5 5 5 15

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Example 1

SSError = SSTotal – SSTreat

SSError = 257.333 – 203.333 = 54

Now we are ready for our hypothesis test…

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Example 1 – Hypothesis Test

HO: μ1 = μ2 = μ3

HA: At least one pair of means differ

Test statistic:F = MSTreat

MSError

Rejection region: Fobt > F(2, 12, α = .05) = 3.88

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Example 1 – Summary Table

Source df SS MS F

Treat 2 203.333 101.67 22.59Error 12 54 4.5Total 14 257.33

Decision: Reject HO. At least two of the situations differ in how much stress they produce.

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Example 2

The first question we have to answer is, what is n1 (n for the Rain group)?

We are told that d.f. = 27. That means that n-1 = 27, so n = 28.n2 = 10 and n3 = 8 and 10 + 8 = 18

so n1 = 28 – 18 = 10.

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Example 2

Since we don’t have raw scores, we cannot use computational formulae. Therefore, we use the conceptual formulae.

X = 10 (12.1) + 10 (22.7) + 8 (19.5) = 18.0 28

SSTreat = 10 (12.1 – 18.0)2 + 10 (22.7 – 18.0)2 + 8 (19.5 – 18.0)2

= 587.0

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Example 2

SSError = 9 (4.0)2 + 9 (5.1)2 + 7 (6.9)2

= 711.36

SSTotal = 587 + 711.36 = 1298.36

Now we are ready for our hypothesis test.

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Example 2 – Hypothesis Test

HO: μ1 = μ2 = μ3

HA: At least one pair of means differ

Test statistic:F = MSTreat

MSError

Rejection region: Fobt > F(2, 25, α = .05) = 3.39

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Example 2 – Summary Table

• Source df SS MS F

• Treat2 587 293.5 10.32• Error 25 711.36 28.45• Total 27 1298.36

• Decision: Reject HO. At least two of the CDs differ significantly in length.

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Example 3Simple reaction times to green, red, and yellow instrument panel lights were compared. The three colors were randomly assigned to 31 different subjects who were instructed to press a button in response to the light. Shown below are average RTs in milliseconds for these subjects.

Green Red YellowX 201 215 218S 2.9 3.5 3.4ni 10 11 10

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Example 3

Is there an overall significant difference? (α = .05)

SSTreat = ∑ni (Xi – X)2

X = [10 (201) + 11 (215) + 10 (218)] 3

= 211.45

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Example 3

SSTreat = 10 (201 – 211.45)2 + 11 (215 – 211.45)2 + 10 (218 – 211.45) 2

= 1659.67

SSError = 9 (2.9)2 + 10 (3.5)2 + 9 (3.4)2

Now we’re ready for our hypothesis test.

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Why?

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Example 3 – Hypothesis Test

HO: μ1 = μ2 = μ3

HA: At least one pair of means differ

Test statistic: F = MSTreat

MSError

Rejection region: Fobt > F(2, 28, α = .05) = 3.34

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Example 3 – Summary Table

Source df SS MS F

Treat 2 1659.67 829.84 76.91Error 28 302.23 10.79Total 30 1961.9

Decision: Reject HO. At least two of the treatments differ in average response time.

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