Introduction to ANOVA ANalysis Of VAriance

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Introduction to ANOVA (ANalysis Of VAriance)
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Why ANOVA?

• Effects of CHO loading
• how much?
• 1 gm/kg? 2 gm/kg? 5 gm/kg?
• Effects of bracing on GRF
• which brace
• taping? Swed O? ActiveAnkle?
• Effeks uf alcohol on spelin
• what blood/alcohol level
• 0.04? 0.08? 0.10?

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ANalysis Of VAriance

• 1-way ANOVA
• Grouping variable factor independent variable
• The variable will consist of a number of levels
• If 1-way ANOVA is being used, there will be gt2
  levels of one IV.
• E.G. What type of program has the greatest impact
  on aggression?
• Violent movies, soap operas, or infomercials?
• Type of program is the independent variable or
  factor
• Violent movies is one level of the factor
• soap operas is one level of the factor
• Infomercials is one level of the factor
• Aggression is the dependent variable

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Example of Oneway ANOVA (single factor)

• No reason to assume correlation between the
  cases in the k groups
• (k number of groups)
• Question does CHO affect time to fatigue??
• IV diet (3 levels of IV or factor)
• DV Endurance time on bike

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How to compare more than 2 means?

• ? refers to risk of making a Type 1 error
• with each comparison, we have ? chances of
  making a Type 1 error
• ? 0.05
• 5 times in 100 we will reject a true null
  hypothesis when running each comparison

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Type 1 error rate is exponentially cumulative

• Family Wise error rate
• ? FW 1- (1 - ? )c
• where c is the number of comparisons to be made
• ie if ? 0.05 and three means

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Type 1 error rate is exponentially cumulative

• Family Wise error rate with three means to
  compare
• ? FW 1- (1 - 0.05)3
• 0.143

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Type 1 error rate is exponentially cumulative

• Estimating Family Wise error rate
• ? FW ? c
• where c is the number of comparisons to be made
• Note always overestimates the error rate ie if ?
  0.05 k 3 k 4?????

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ANOVA is an attempt to maintain the FW error
rate at a known (acceptable) level
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Example of ANOVAReturn to our original question

• Question does amount of CHO injected affect time
  to fatigue??
• IV diet (3 levels of IV)
• DV Endurance time on bike
• 1-way ANOVA (0ne IV)
• IV Grouping variable factor
• The IV consists of a number of levels

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Steps to Oneway ANOVA

• set ? (0.05)
• set sample size
• Thirty randomly selected subjects
• Three randomly assigned groups
• n 10 in each group
• Grp 1 Regular Diet
• Grp 2 CHO supp diet (0.5 g/kg)
• Gpr 3 CHO supp diet (1.0 g/kg)
• set HO

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Set statistical hypotheses I

• HO
• Null hypothesis
• Any observed difference between the 3 groups will
  be attributable to random sampling errors

• H1 (HA)
• Alternative hypothesis
• If HO is rejected, the difference is not
  attributable to random sampling errors (perhaps
  diet)?

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Set statistical hypotheses II

• H1
• Alternative hypothesis
• The population means of the three groups differ
  in some wayNote no directional hypothesis
  Null may be false in many different ways

• HO
• Null hypothesis
• The population means of the 3 groups are equal

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Steps

• Set ? (0.05)
• set sample size (n 10/grp)
• set Ho
• test all subjects with a standardized protocol
  (bike)

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Subject Data file ANOVA1.sav
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Steps

• Set ? (0.05)
• set sample size (n 10/grp)
• set Ho
• test all subjects with a standardized protocol
  (bike)
• get descriptive statistics of each group
• histograms
• mean, SD, n
• compare the group means

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How to compare the groups?

• With k 3, ? 0.05,

? FW
???
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Concept of ANOVA

• Evaluate the effect of treatment (the IV)

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Concept of ANOVA

• Evaluate the effect of treatment (the IV) by
  analyzing the amount of variation among the
  subgroup sample means (DV)

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Concept of ANOVA

• Evaluate the effect of treatment (the IV) by
  analyzing the amount of variation among the
  subgroup sample means (DV)

But how much variation is expected if the
subgroup population means are equal?
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Some Nomenclature

• Grand Mean mean of ALL scores, regardless of
  group
• ie all 30 scores
• Group Mean mean of all scores from subjects
  treated the same
• groups of 10

X
X
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3 Sources of Variability(Deviation Scores!!!!)
X - X X - X X - X
Total Variability (individual scores around
Grand Mean)
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3 Sources of Variability
X - X X - X X - X
will sum to 0, so square it for each
subject, then sum. Gives us The Total Sum of
Squares
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3 Sources of Variability
X - X X - X X - X
Within Group Variability (individual scores
around Group Mean)
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3 Sources of Variability
X - X X - X X - X
Within Group Variability (scores around Group
Mean)
Reflects INHERENT variability (all treated the
same)
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Within-group

• Variation between people that is not due to the
  grouping factor
• Example
• You might assign people to three different
  tanning beds to see which has the greatest
  tanning effect
• But folk within each type of bed would still vary
  greatly in the degree of tanning they achieved
• Within group variance is the pooled variance from
  all levels of the grouping factor (similar to
  pooled SD in t-test)

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3 Sources of Variability
X - X X - X X - X
will sum to 0, so square it for each subject,
then sum. Gives us Within Group Sum of Squares
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3 Sources of Variability
X - X X - X X - X
Between Group Variability (Groups around Grand
Mean)
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Between-group variation

• Is the variation normally expected between people
  (within-group variation), plus variation due to
  the grouping factor

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3 Sources of Variability
X - X X - X X - X
Reflects inherent and TREATMENT EFFECT
Between Group Variability (Groups around Grand
Mean)
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3 Sources of Variability
X - X X - X X - X
will sum to 0, so square it for each group,
then sum. Gives us Between Group Sum of
Squares
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Recall

• Size of the Sum of Squares is affected by
• size of each deviation score
• number of cases that are summed

Calculate the MEAN SQUARE of a sum of squares by
dividing through by the degrees of freedom
contributing to the sum.
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3 Sources of Variability
X - X X - X X - X
df for EACH group n-1
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Statistics Humour
Two unbiased estimators were sitting in a bar.
The first says So how do you like married
life? The other replies, "It's pretty good if
you don't mind giving up that one degree of
freedom!"
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3 Sources of Variability
X - X X - X X - X
df for EACH group n-1 df for TOTAL groups k
(n-1)
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3 Sources of Variability
X - X X - X X - X
df for EACH group n-1 df for TOTAL groups k
(n-1) For our Diet study df Within 3 (10 - 1)
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3 Sources of Variability
X - X X - X X - X
df k -1
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3 Sources of Variability
X - X X - X X - X
For our Diet study df Between 3 - 1 2
df k -1
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A new ratio between variabilities for us to
consider
Inherent Variability Treatment Effect Inherent
Variability
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A new ratio between variabilities for us to
consider
Inherent Treatment Inherent
Between Within

Between between group variability Within
within group variability
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A new ratio between variabilities for us to
consider
Inherent Treatment Inherent
MSBetween MSWithin

By using Mean Square, account for different
number of cases contributing to each estimate
of error (random SE).
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A new ratio between variabilities for us to
consider
Inherent Treatment Inherent
MSBetween MSWithin

Note if Treatment effect 0 (ie no effect)
the ratio will be equal to ????
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A new ratio between variabilities for us to
consider
MSBetween MSWithin
F

Note if Treatment effect 0 (ie no effect)
the ratio will be equal to 1.00
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Evaluating Fobserved with the F distribution

• A distribution of F ratios is not normally
  distributed
• follows an F distribution
• positively skewed
• depends on the number of degrees of freedom in
  the numerator (MS between) and the denominator
  (MS within)

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The F distribution (hypothetical)
0 1 2 3 4 5 6 7 8
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Fcritical the F value that must be equaled or
exceeded to classify a difference among group
means as statistically significant (identify a
main effect)
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Fcritical depends on df of MSbetween and MS
within, and chosen ?
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Fcritical depends on df of MSbetween and MS
within, and chosen ?
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The F distribution (hypothetical)
Region of rejection
0 1 2 3 4 5 6 7 8
F.05 ???
For our Diet study, with ? 0.05 and df 2 and
27, Fcritical ???
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The F distribution (hypothetical)
F distribution for df 2, 27
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Concept of evaluating Fobs against Fcrit
F distribution for df 2, 27
Area 0.05 (5)
Fcrit 3.35
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Concept of evaluating Fobs against Fcrit
F distribution for df 2, 27
Area 0.05 (5)
Fcrit 3.35
Fobs lt Fcrit, Decision ?????
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Concept of evaluating Fobs against Fcrit
F distribution for df 2, 27
Area 0.05 (5)
Fcrit 3.35
Fobs ? Fcrit, Decision ?????
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Running Oneway ANOVA (single factor ANOVA) Using
SPSS
Demonstrate with anova1.sav
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1-way ANOVA in SPSS
Procedure Choose the appropriate procedure, and
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1-way ANOVA in SPSS
Dialog box slide the variables
into the appropriate places
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ANOVA in SPSS
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Decision

• Since Fobs 11.13 ? Fcrit of 3.35, our decision
  is to ...

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Decision

• Since Fobs 11.13 ? Fcrit of 3.35, our decision
  is to reject Ho stating that...

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Decision

• Since Fobs 11.13 ? Fcrit of 3.35, our decision
  is to reject Ho stating that the difference among
  the means is not more than would be expected by
  chance and accept HA stating that...

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Decision

• Since Fobs 11.13 ? Fcrit of 3.35, our decision
  is to reject Ho stating that the difference among
  the means is not more than would be expected by
  chance and accept HA stating that the means
  differ in some way.

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Decision

• Since Fobs 11.13 ? Fcrit of 3.35, our decision
  is to reject Ho stating that the difference among
  the means is not more than would be expected by
  chance and accept HA stating that the means
  differ in some way.

Omnibus F identify a significant main effect
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Decision

• Since Fobs 11.13 ? Fcrit of 3.35, our decision
  is to reject Ho stating that the difference among
  the means is not more than would be expected by
  chance and accept HA stating that the means
  differ in some way.

How to determine which means differ?
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Is Normal different from o.5 g/kg? From 1.0
g/kg?Is 0.5 g/kg different from 1.0 g/ kg?
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ANOVA in SPSS
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Significant resultnow what?
There is a significant difference among the means
There are more than 2 means
Among all means, or just two?
Dont know
Better do a follow-up, mate
Rats
Ok then
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Why not use 3 unpairedt-tests?

• Normal vs 0.5 g/kg
• Normal vs 1.0 g/kg
• 0.5 g/kg vs 1.0 g/kg

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Why not use 3 unpairedt-tests?

• Normal vs 0.5 g/kg
• Normal vs 1.0 g/kg
• 0.5 g/kg vs 1.0 g/kg

Because we will be operating with an inflated
?Family Wise .
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Post Hoc tests

• After the Fact comparisons of means used to
  identify which specific pairs of means are
  significantly different
• Designed to maintain a specified ?Family Wise
  level regardless of how many pairs of means are
  compared

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Post Hoc tests

• Follow-up tests
• ONLY compute after a significant ANOVA
• Like a collection of little t-tests
• But they control overall type 1 error
  comparatively well
• They do not have as much power as the omnibus
  test (the ANOVA) so you might get a significant
  ANOVA no sig. Follow-up
• Purpose is to identify the locus of the effect
  (what means are different, exactly?)

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Significant resultnow what?

• Follow-up tests most common
• Tukeys HSD (honestly sig. diff.)
• Formula
• But its easier to use SPSS

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Follow-ups to ANOVA in SPSS
Choose post-hoc test (meaning after this)
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Follow-ups to ANOVA in SPSS
Check the appropriate box for the HSD (Tukey, not
Tukeys b)
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Run Tukeys HSD test Oneway in SPSS

• Use our diet data (ANOVA1.sav)

And one that does
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Assumptions to test in One-Way

• Samples should be independent (as with
  independent t-test does not mean perfectly
  uncorrelated)
• Each of the k populations should be normal
  (important only when samples are smallif theres
  a problem, can use Kruskal-Wallis test)
• The k samples should have equal variances (this
  is the homogeneity of variance assumption, and
  well look at it shortlyviolations are important
  mostly with small samples and unequal ns)

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Homogeneity of variance - SPSS
1. Click on the options button
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Homogeneity of variance - SPSS
3. Click continue
2. Choose homogeneity of variance (Ive also
chosen descriptives here)
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Homogeneity of variance - SPSS
4. SPSS output The test has to be significant for
there to be a violation
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Reporting ANOVA results
Table 1. Descriptive statistics of mean time to
exhaustion (minutes) by diet group (n 10). A
solid line joins pairs of means that are not
significantly different (Tukeys HSD, ?0.05)
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Figure 1. Descriptive statistics of time to
exhaustion with different diets. An asterisk
indicates group means that are not significantly
different (?0.05)
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Reporting ANOVA results
Table 2. ANOVA summary table for the effects of
diet on time to exhaustion.
Optional include in appendix if not in body of
thesis
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Reporting ANOVA results
Descriptive statistics for the mean time to
exhaustion for the three diet groups are
presented in table 1 and graphically in Figure 1.
A oneway ANOVA at ? 0.05 revealed a significant
difference among the diet groups for mean time to
exhaustion (F 2,27 11.13, p 0.0003). Tukeys
HSD was used to identify the source of the
significant omnibus F, and indicated that the
mean time to exhaustion for the regular diet
group (38.9 ? 3.5 minutes) was significantly
shorter than the time for the groups receiving
0.5 grams CHO per kilogram body weight (g/kg) or
1.0 g/kg. These two groups , with means of 44.2
(? 2.9) and 44.7 (? 2.7) minutes respectively,
were not significantly different.
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Reporting ANOVA results
These results suggest that CHO supplements of at
least 0.5 g/kg of CHO will increase time to
exhaustion on the bicycle by about 5.5 minutes
or 14. The data also suggest a plateau effect of
CHO supplementation, with no additional increase
in time to exhaustion seen with 1.0 g/kg
compared to 0.5 g/kg.
In discussion, address whether the
observed increase is physiologically meaningful,
and elaborate on the concept of a plateau
effect with CHO supplements.
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Calculating Tukey-b (HSD) test
Honestly Significant Difference the magnitude
of mean difference that must exist to claim
levels are Significantly Different
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Tukey-b (HSD) test
The studentized range statistic (table E, p.
470) depends on the number of levels to be
compared and df within and ?
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Tukey-b (HSD) test
For our diet study k 3 ( of levels) and
df within 27, ? 0.05 From Table F, q ???
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Tukey-b (HSD) test
For our diet study k 3 ( of levels) and df
within 27, ? 0.05 From Table 8, q 3.51
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Tukey-b (HSD) test
Mean SquareWithin, taken from ANOVA Summary Table
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Tukey-b (HSD) test
For our diet study, MSwithin 9.2815
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Tukey-b (HSD) test
Number of Subjects in EACH group
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Tukey-b (HSD) test
For our diet study, n 10
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Tukey-b (HSD) test
3.382
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Apply Tukeys HSD test value of 3.4 to the diet
data

• Normal vs 0.5 g/kg
• 38.9 vs 44.2 minutes
• difference -5.3 minutes
• Normal vs 1.0 g/kg
• 38.9 vs 44.7 minutes
• difference -5.8 minutes
• 0.5 g/kg vs 1.0 g/kg
• 44.2 vs 44.7 minutes
• difference -0.5

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