Chapter 13 Analysis of Variance and Experimental Design

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Lesson 8
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Chapter 13 Analysis of Variance and Experimental
Design

• An Introduction to Analysis of Variance
• Analysis of Variance Testing for the Equality
  of
• k Population Means

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Introduction to Analysis of Variance
Analysis of Variance (ANOVA) can be used to
test for the equality of three or more
population means.
Data obtained from observational or
experimental studies can be used for the
analysis.
We want to use the sample results to test the
following hypotheses
H0 ?1???2???3??. . . ?k
Ha Not all population means are equal
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Introduction to Analysis of Variance
H0 ?1???2???3??. . . ?k
Ha Not all population means are equal
If H0 is rejected, we cannot conclude that all
population means are different.
Rejecting H0 means that at least two population
means have different values.
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Introduction to Analysis of Variance
Sample means are close together because there is
only one sampling distribution when H0 is true.
?
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Introduction to Analysis of Variance
Sample means come from different sampling
distributions and are not as close together when
H0 is false.
?3
?1
?2
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Assumptions for Analysis of Variance
For each population, the response variable is
normally distributed.
The variance of the response variable, denoted ?
2, is the same for all of the populations.
The observations must be independent.
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Analysis of VarianceTesting for the Equality of
K Population Means

• Between-Samples Estimate of Population Variance
• Within-Samples Estimate of Population Variance
• Comparing the Variance Estimates The F Test
• The ANOVA Table

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Between-Treatments Estimateof Population Variance

• A between-treatment estimate of ? 2 is called
  the mean square treatment and is denoted MSTR.

Numerator is the sum of squares due to
treatments and is denoted SSTR
Denominator represents the degrees of freedom
associated with SSTR
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Within-Samples Estimateof Population Variance

• The estimate of ? 2 based on the variation of the
  sample observations within each sample is called
  the mean square error and is denoted by MSE.

Numerator is the sum of squares due to error and
is denoted SSE
Denominator represents the degrees of freedom
associated with SSE
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Comparing the Variance Estimates The F Test

• If the null hypothesis is true and the ANOVA
• assumptions are valid, the sampling
  distribution of
• MSTR/MSE is an F distribution with MSTR
  d.f.
• equal to k - 1 and MSE d.f. equal to nT - k.

• If the means of the k populations are not
  equal, the
• value of MSTR/MSE will be inflated because
  MSTR
• overestimates ? 2.

• Hence, we will reject H0 if the resulting
  value of
• MSTR/MSE appears to be too large to have
  been
• selected at random from the appropriate F
• distribution.

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Sampling Distribution of MSTR/MSE

• The figure below shows the rejection region
  associated with a level of significance equal to
  ? where F? denotes the critical value.

Do Not Reject H0
Reject H0
MSTR/MSE
F?
Critical Value
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Test for the Equality of k Population Means

• Hypotheses

H0 ?1???2???3??. . . ?k? Ha Not all
population means are equal

• Test Statistic

F MSTR/MSE
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Test for the Equality of k Population Means

• Rejection Rule

Reject H0 if p-value lt a
p-value Approach
Critical Value Approach
Reject H0 if F gt Fa
where the value of F?? is based on an F
distribution with k - 1 numerator d.f. and nT - k
denominator d.f.
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The ANOVA Table

• Source of Sum of Degrees of Mean
• Variation Squares Freedom Squares
  F
• Treatment SSTR k - 1 MSTR MSTR/MSE
• Error SSE nT - k MSE
• Total SST nT - 1
• SST divided by its degrees of freedom nT - 1 is
  simply the overall sample variance that would be
  obtained if we treated the entire nT observations
  as one data set.

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ANOVA Table

• Source of Sum of Degrees of
  Mean
• Variation Squares Freedom
  Squares F
• Treatments SSTR k - 1
• Error SSE nT - k
• Total SST nT - 1

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Test for the Equality of k Population Means

• Example Reed Manufacturing

A simple random sample of five managers
from each of the three plants was taken and the
number of hours worked by each manager for
the previous week is shown on the next slide.
Conduct an F test using a .05.
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Test for the Equality of k Population Means
Plant 3 Detroit
Plant 2 Pittsburgh
Plant 1 Buffalo
Observation
1 2 3 4 5
48 54 57 54 62
51 63 61 54 56
73 63 66 64 74
Sample Mean
55 68 57
Sample Variance
26.0 26.5 24.5
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Example Reed Manufacturing

• Analysis of Variance
• Hypotheses
• H0 ? 1??? 2??? 3?
• Ha Not all the means are equal
• where
• ???? 1 mean number of hours worked per
  week by the managers at Plant 1
• ? 2 mean number of hours worked per
  week by the managers at Plant 2
• ???? 3 mean number of hours worked per
  week by the managers at Plant 3

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Example Reed

• ANOVA
• Mean Square Between Treatments
• Since the sample sizes are all equal
• x (x1 x2 x3)/3 (55 68 57)/3
  60
• SSTR 5(55 - 60)2 5(68 - 60)2 5(57 - 60)2
  490
• MSTR 490/(3 - 1) 245
• Mean Square Error
• SSE 4(26.0) 4(26.5) 4(24.5) 308
• MSE 308/(15 - 3) 25.667

_
_
_

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Example Reed.

• ANOVA
• Rejection Rule
• Assuming ? .05, F.05 3.89 (2 d.f. numerator
• and 12 d.f. denominator). Reject H0 if F gt
  3.89.
• Test Statistic
• F MSTR/MSE 245/25.667 9.55
• Conclusion
• Since F 9.55 gt F.05 3.89, we reject H0. The
• mean number of carwashes are not the same for
• all three waxes.

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Example BWM.

• ANOVA Table
• Source of Sum of Degrees of
  Mean
• Variation Squares Freedom
  Squares F
• Treatments 490 2 245
  9.55
• Error 308 12 25.667
• Total 798 14