1 one way analysis of variance – designed experiments usually involve comparisons among more than...
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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)
Lecture 15
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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
Lecture 15
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Lecture 15
µ2
µ1
µ4
µ3
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Lecture 15
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?
Lecture 15
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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?
Lecture 15
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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.
Lecture 15
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Variability among treatment means
S2 = Σ(Xi – X)2 (1)
(n-1)
SST = Σni(Xi – X)2 (2)
Lecture 15
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
Lecture 15
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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.
Lecture 15
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Lecture 15
X1
X3
X2
X4
X
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Error variability
• SSE =
Σ(X1j – X1)2 + Σ(X2j – X2)2 + … + Σ(XPj – XP)2
Lecture 15
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.
Lecture 15
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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.
Lecture 15
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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
Lecture 15
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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
Lecture 15
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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.
Lecture 15
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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?”
Lecture 15
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Computational Formulae
CM = (ΣXi)2
NSSTotal = ΣXi
2 – CM
SST = T12 + T2
2 + … + TP2 – CM
n1 n2 nP
SSE = SSTotal - SSTLecture 15
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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).
Lecture 15
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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
Lecture 15
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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
Lecture 15
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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
Lecture 15
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Example 1
SSError = SSTotal – SSTreat
SSError = 257.333 – 203.333 = 54
Now we are ready for our hypothesis test…
Lecture 15
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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
Lecture 15
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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.
Lecture 15
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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.
Lecture 15
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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
Lecture 15
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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.
Lecture 15
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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
Lecture 15
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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.
Lecture 15
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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
Lecture 15
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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
Lecture 15
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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.
Lecture 15
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
Lecture 15
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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.
Lecture 15