One Way Analysis of Variance

1
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)

2
One Way Analysis of Variance

• Three approaches in this course
• One Independent Variable
• Completely randomized design (today)
• analogous to independent groups Z and t test
• Randomized block design
• analogous to dependent pairs Z and t test
• Two Independent Variables
• Two factor completely randomized Anova

3
Completely randomized design
X2
Draw a sample
Make an inference
4
Completely randomized design

• The approach is to compare the differences among
  the treatment means ( , , , ) to
  the amount of error variability.
• Differences among treatment means vs. error
  variability
• Our question is, are the differences among the
  treatment means so large that they could not be
  due to sampling error?

5
Completely randomized design

• In order to answer our question, we need
  numerical measures of two things
• differences between the treatment means
• how different from each other are the means?
• sampling variability within each treatment
• how different could the treatment means be just
  on the basis of chance?

6
1. Differences among treatment means

• Recall the formula for a variance
• S2 S(Xi - )2 (Eqn. 1)
• (n-1)
• The numerator of Eqn. 1 measures variability of
  individual scores around the sample mean. Compare
  with Eqn. 2 (next slide)

7
1. Differences among treatment means

• S2 S(Xi - )2 (Eqn. 1)
• (n-1)
• SST Sni( - )2 (Eqn. 2)

Grand mean
Individual treatment means
8
1. Differences among treatment means

• In Eqn. 2, the sum (S) measures the variability
  among the treatment (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 will be.

9
2. Error variability

• SSE S(X1j X1)2 S(X2j X2)2 S(XPj
  XP)2

One individual score In Treatment 1 sample
Mean for Treatment 1
10
2. Error variability

• SSE S(X1j X1)2 S(X2j X2)2 S(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 S(X1j X1)2
• (n1 1)
• Therefore, (n1 1) (S1)2 S(X1j X1)2
• Same is true for S2, S3, SP.

2
2
2
12
2. Sampling variability

• Thus
• SSE (n1 1) S1 (n2 1) S2 (nP 1) SP
• Now, were almost ready. One last issue
• SST is the sum of P terms (deviations)
• SSE is the sum of N Sni terms

2
2
2
13
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 MST
• MSE
• How much larger than 1 must F be for us to reject
  H0? Check F table for a and d.f.

15
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.
• Youre asking is the treatment variance larger
  than the error variance?

16
Computational Formulae

• CM (SXi)2
• N
• SSTotal SXi2 CM
• SST T1 T2 TP CM
• n1 n2 nP
• SSE SSTotal - SST

2
2
2
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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 1

• All measurements are made at the same time of
  day. Scores in milligrams of corticosterone per
  100 millilitres of blood are (a .05)
• Vacation Class Final Exam
• X X2 X X2 X X2
• 2 4 8 64 10 100
• 3 9 10 100 13 169
• 7 49 7 49 14 196
• 2 4 5 25 13 169
• 6 36 10 100 15 225
• ? 20 102 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
• 15
• SSTreat T1 T2 TP CM
• n1 n2 nP
• 202 402 652 1252 203.333
• 5 5 5 15

2
2
2
21
Example 1

• SSError SSTotal SSTreat
• SSError 257.333 203.333 54
• Now we are ready for our hypothesis test

22
Example 1 Hypothesis Test

• HO µ1 µ2 µ3
• HA At least one pair of means differ
• Test statistic F MSTreat
• MSError
• Rejection region Fobt gt F(2, 12, a .05)
  3.88

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Example 1 Summary Table

• Source df SS MS F
• Treat 2 203.333 101.67 22.59
• Error 12 54 4.5
• Total 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, therefore, n 28.
• n2 10 and n3 8,
• so n1 28 18 10.

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Conceptual formulas - review

• Well need these conceptual formulas
• SST Sni( - )2
• SSE S(X1j X1)2 S(X2j X2)2 S(XPj
  XP)2
• (See Slides 7 and 9 above)

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Example 2

• Since we dont 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.

28
Example 2 Hypothesis Test

• HO µ1 µ2 µ3
• HA At least one pair of means differ
• Test statistic F MSTreat
• MSError
• Rejection region Fobt gt F(2, 25, a .05)
  3.39

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Example 2 Summary Table

• Source df SS MS F
• Treat 2 587 293.5 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 3

• Simple 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 Yellow
• X 201 215 218
• S 2.9 3.5 3.4
• ni 10 11 10

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Example 3

• Is there an overall significant difference? (a
  .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
• 1659.67
• SSError 9 (2.9)2 10 (3.5)2 9 (3.4)2
• Now were ready for our hypothesis test.

33
Example 3 Hypothesis Test

• HO µ1 µ2 µ3
• HA At least one pair of means differ
• Test statistic F MSTreat
• MSError
• Rejection region Fobt gt F(2, 28, a .05)
  3.34

34
Example 3 Summary Table

• Source df SS MS F
• Treat 2 1659.67 829.84 76.91
• Error 28 302.23 10.79
• Total 30 1961.9
• Decision Reject HO. At least two of the
  treatments differ in average response time.