Analysis of Variance

1
Chapter 15

• Analysis of Variance
• (ANOVA)

2
Single-Factor ANOVA

• In Chapter 11 we compared the means of two
  populations or treatments.
• To decide whether the means for more than two
  populations or treatments are identical, we use a
  method called single-factor analysis of variance.
• Example The rating of luxury cars

Does the data support that the true average
(population mean) rating of all three types of
cars are identical, or at least two of them are
different from one another?
3
15.1 Single-Factor ANOVA and the F-Test

• A single-factor analysis of variance (ANOVA)
  problem involves a comparison of k population or
  treatment means.
• H0 µ1µ2 µk, against
• Ha At least two of the µs are different.
• The analysis is based on k independently selected
  samples, one from each population or for each
  treatment.

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ANOVA Notation

• k number of populations or treatments being
  compared
• Population or treatment 1 2 k
• Population or treatment mean µ1 µ2 µk
• Population or treatment variance s12 s22
  sk2
• Sample size n1 n2 nk
• Sample mean
• Sample variance s12 s22 sk2
• The total number of observations in the data set
  is
• N n1 n2 nk
• The sum of all N observations-the grand total-is
  denoted by
• T n1 n2 nk
• The grand mean is

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Example Strength of Cardboard

• The following table presents the results of a
  single-factor experiment involving k 4 types of
  cardboard boxes with respect to compression
  strength (in pounds).

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Assumptions for ANOVA

• Each of the k population or treatment response
  distributions is normal.
• s1 s2 sk (i.e. the k normal
  distributions have identical standard
  deviations).
• The observations in the sample from any
  particular one of the k populations or treatments
  are independent of one another.
• When comparing population means, k random samples
  are selected independently of one another. When
  comparing treatment means, treatments are
  assigned at random to subjects or objects (or
  subjects are assigned at random to treatments).

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Definition of Treatment Sum of Square and
Error Sum of Squares

• A measure of disparity among the sample means is
  the treatment sum of squares, denoted by SSTr
• A measure of variation within the k samples,
  called the error sum of squares and denoted by
  SSE, is
• Each sum of squares has an associated number of
  degrees of freedom
• treatment df k - 1
• error df N k

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Mean Squares

• A mean square is a sum of squares divided by its
  number of degrees of freedom.
• mean square for treatments MSTr
• mean square for error MSE

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Example Cardboard Strength
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Example Cardboard Strength Calculations
The number of degrees of freedom are treatment df
k - 1 3 error df N - k 6 6 6 6 - 4
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11
Relationship between MSTr and MSE

• When H0 is true (µ1 µ2 µk),
• µMSTR µMSE
• When H0 is false
• µMSTR gt µMSE
• And the greater the differences among the µs,
  the larger µMSTr will be relative to µMSE

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The Single-Factor ANOVA F Test

• Null hypothesis H0 µ1µ2 µk
• Test statistics F
• An F distribution always arises in a ratio with a
  sum of squares on the numerator and another sum
  of squares on the denominator.
• When H0 is true and the ANOVA assumptions are
  reasonable, F has an F distribution with df1 k
  - 1 and df2 N - k.
• All F test in this book are upper-tailed. P-value
  is the area captured in the upper tail of the
  corresponding F curve.
• Reject H0 when P-value lt a.

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F Curve and Upper-Tailed Test
Exercise Find P-value if the calculated F value
is 9.20 with df14 and df26.
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Single-Factor ANOVA Table
15
Exercise Cardboard Strength Revisited
(Conclusion)

• H0 µ1µ2µ3µ4
• Ha At least two among µ1, µ2, µ3, and µ4 are
  different.
• In Example 15.2 MSTr 42,455.86 and MSE
  1691.92
• Conduct an F test.

• df1 k - 1 3 and df2 N k 24 4 20.
• Table of Upper-Tail F Curve Areas shows 8.10
  captures tail
• area lt .001. Because 25.09 gt 8.10, it follows
  that P-value lt
• .001.
• Reject H0. True average compression strength
  appears
• different on the types of box.

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• Example Hormones and Body Fat
• To investigate the effect of four treatments on
  various body characteristics, each of 57 female
  subjects who were over age 65 was assigned at
  random to one of the following four treatments
  (1) placebo growth hormone and placebo
  steroid (P P), (2) placebo growth hormone
  and the steroid estradiol (P S), (3) growth
  hormone and placebo steroid (G P), and (4)
  growth hormone and the steroid estradiol (G S).
• The table on the next slide lists data on change
  in body fat mass over the 26-week period
  following the treatments.
• Carry out an F test to see whether true mean
  change in body fat mass differs for the four
  treatments. Use a significance level a 0.01.

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Let µ1, µ2, µ3, and µ4 denote the true mean
change in body fat for the treatments P P, P
S, G P, and G S, respectively. H0 µ1 µ2
µ3 µ4 Ha At least two among µ1, µ2, µ3, and µ4
are different.

• Use Excel to create ANOVA table and calculate
  the test statistics F (See next slides)
• Enter the data.
• Click Data
• Choose Data Analysis
• Choose ANOVA Single Factor
• Select Input and Output Range, and click OK.

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Choose Anova Single Factor in Data Analysis
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The Excel generates an ANOVA table after you
enter the input range.

• Conclusion Since P-value 0.0000162 lt a (0.01),
  we reject H0. We conclude that the mean change in
  body fat mass is not the same for all four
  treatments.

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15.2 Multiple Comparison

• When H0 µ1 µ2 . . . µk is rejected by the
  F test, we believe that there are differences
  among the k population or treatment means. A
  natural question is Which means differ?
• A multiple comparisons procedure is a method for
  identifying differences among the µs once the
  hypothesis of overall equality has been rejected.
• One such method is the Tukey-Kramer (T-K)
  multiple comparison procedure.
• The T-K procedure is based on computing
  confidence intervals for difference between each
  possible pair of µs. For example, if k 3,
  there are three differences to consider
• µ1 - µ2 , µ1 - µ3 and µ2 - µ3

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The Tukey-Kramer (T-K) Procedure

• Construct all confidence intervals for the
  difference between each pairs of µs

• q is the critical value for the Studentized
  Range Distribution available in Appendix Table 7
  based on error df ( N k ) and confidence
  level.
• Recall that we may use Excel to find MSE and
  error df.
• When there are k populations or treatments
  being compared, you need to compute ½ k (k - 1)
  confidence intervals.
• Two means are judged to be significantly
  different if the corresponding interval does not
  include 0.

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Example Hormones and Body Fat Revisited

• Recall that the data on the right list the change
  in body fat mass resulting from a double-blind
  experiment designed to compare the four
  treatments P P, P S, G P and G S.
• Does the true mean change in body fat mass differ
  for the four treatments?

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Recall The Single-Factor ANOVA F test concluded
that the mean change in body fat mass is not the
same for all four treatments since P-value 0.

• Use the ANOVA table to find MSE 1.92 and error
  df 53.
• Appendix Table 7 gives the 95 Studentized range
  critical value q 3.74 (using the error df 60,
  the closet table value of error degree N k
  53.)

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includes 0. We conclude that µ1 is not
significantly different from µ2.
does not include 0. We conclude that µ1 is
significantly different from µ3.
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Exercise Find the remaining T-K intervals
.
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• Summarizing the Results of the Tukey-Kramer
  Procedure
• List the sample means in increasing order.
• Use the T-K intervals to determine the group of
  means that do not differ significantly from the
  first in the list. Draw a horizontal line
  extending from the smallest mean to the last mean
  in the group.
• Repeat the above procedure from the second
  smallest. Continue considering the means on the
  order list.

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Exercise Sleep Time

• As biologist wished to study the effect of
  ethanol on sleep time. A sample of 20 rats,
  matched for age and other characteristics, was
  selected , and each rat was given an oral
  injection having a particular concentration of
  ethanol per body weigh. The rapid eye movement
  (REM) sleep time for each rat was then recorded
  for a 24-hr period, with the results shown on the
  right.

Does true average REM sleep time depend on the
treatment used?
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• Answers to Exercise Sleep Time
• Single-Factor ANOVA F Test
• H0 µ1 µ2 µ3 µ4
• Ha At least two among µ1, µ2, µ3, and µ4 are
  different.
• Based on the ANOVA table, we reject H0 since
  P 0. The true REM sleep time is not the same
  for all four treatments.
• The T-K intervals are list in the table on the
  right.
• The corresponding underscore pattern is

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Appendix Using SPSS for ANOVA F test and T-K
procedure.

• Enter data in two columns
• Column 1 groups
• group 1 P P
• group 2 P S
• group 3 G P
• group 4 G S
• Column 2 change
• (in body fat mass)

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• Click Analyze, click General Linear Model, then
  click Univariate. (see the figure on right)
• You now see the Univariate dialog box below.
  Click group, and move it to Fix Factor(s) box.
• Also in the Univariate dialog, click change, and
  move it to Dependent Variable box. (See the
  figure on the right below. )

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• Click Options. The Univariate Options dialog
  box appears.
• Click group and then click ?to move group to the
  Display Means For box.
• Click Descriptive statistics in the Display box.
• Click Continue.

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• Click Post Hoc. The Post Hoc Multiple Comparison
  for Observed Means dialog box appears.
• In the Factor(s) box, click group, and click
  ?to make it appear in the Post Hoc Tests for
  box.
• In the Equal Variances Assumed box, click
  Tukey.
• Click Continue.
• Click OK.

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• The first half of the output is shown in the
  right figure.
• The Descriptive Statistics provide the mean,
  standard deviation and sample size for each
  group.
• The Tests of Between-Subjects Effects shows the
  ANOVA table.
• The second half of the output continues on next
  slide.

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• The group table lists the 95 confidence
  interval for each group.
• The T-K intervals are listed in Multiple
  Comparison table. Notice that the T-K interval
  for µ1-µ2 is (-1.0395, 1.7395), while the T-K
  interval for µ2-µ1 is (-1.7395, 1.0395), etc.