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Part I – MULTIVARIATE ANALYSIS
C4 ANOVA & MANOVA
© Angel A. Juan & Carles Serrat - UPC 2007/2008
1.4.1: The F-Distribution (quick review)

The F-distribution is basic to regression and analysis
of variance.

The F-distribution has two parameters: the numerator,
m, and denominator, n, degrees of freedom  F(m,n)

Like the chi-square distribution, the F-distribution is
skewed.
1.4.2: Overview of ANOVA & MANOVA

The main purpose of Analysis of Variance (ANOVA) is to test for significant
differences among three or more means. This is accomplished by analyzing the
variance, that is, by partitioning the total variance into the component that is due
to true random error (within-group) and the components that are due to
differences between means (among-group). If we are only comparing two means,
then ANOVA will give the same results as the t test.

The variables that are measured (e.g., a test score) are called dependent or
response variables. The variables that are manipulated or controlled (e.g., a
teaching method or some other criterion used to divide observations into groups
or levels that are compared) are called factors or independent variables.

Multivariate analysis of variance, MANOVA, is an extension of ANOVA methods
to cover cases where there is more than one dependent variable.
(# responses)
(Levels in Factor/s)
1 response
2 levels
1 factor  One-way ANOVA
>=2 levels
2 factors  Two-way ANOVA
>=2 responses
1.4.3: Intro to ANOVA

A t-test can be used to compare two sample means.
What if we want to compare more than two means?

Assume that we have k populations and that a
random sample of size nj (j = 1, 2, …, k) has been
selected from each of these populations. Analysis of
Variance (ANOVA) can be used to test for the
equality of the k population means:
 H 0 : 1  2  ...  k

 H1 : Not all population means are equal


Assumptions for ANOVA:
1.
For each population, the response variable is
normally distributed
2.
The variance of the response variable, σ2, is the
same for all the populations
3.
The observations must be independent
ANOVA is a statistical procedure that can be used to
determine whether the observed differences in the k
sample means are large enough to reject H0
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.
Idea: If the means for the k populations are equal,
we would expect the k sample means to be close
together.
i.e.: if the variability among the sample means is
“small”, it supports H0; if the variability among the
sample means is “large”, it supports H1.
1.4.4: The Logic Behind ANOVA

A1 = the assumptions for ANOVA are satisfied

A2 = the null hypothesis is true
1.
On the one hand, under A1 & A2, each sample will
have come from the same normal probability
distribution with mean μ and variance σ2. Therefore,
we can use the variability among the sample means*
to develop the between-factors estimate of σ2 or
MSF.
2.
3.
On the other hand, when a simple random sample is
selected from each population, each of the sample
variances provides an unbiased estimate of σ2.
Hence, we can combine or pool the individual
estimates of σ2 into one overall estimate, called the
pooled or within-factors estimate of σ2, MSE.
Under A1 & A2, the sampling distribution of
MSF/MSE is an F( k – 1, nT – k ) distribution. Hence,
we will reject H0 if the resulting value of MSF/MSE
appears to be too large to have been selected at
random from the corresponding F distribution.
  
(*) Remember: X  N   ,    X n  N   ,

n

SSF
k 1
MSF  mean square due to factors
SSF  sum of squares due to factors
k  number of factors (one per sample)
k -1 = degrees of freedom associated with SSF
MSF 
If H0 is true, MSF provides an unbiased estimate
of σ2. However, if the means of the k populations
are not equal, MSF is not unbiased.
MSE 
SSE
nT  k
MSE  mean square due to error
SSE  sum of squares due to error
nT  total number of observations
nT  degrees of fredom associated with SSE
The MSE estimate is not affected by whether or
not the population means are equal. It always
provides an unbiased estimate of σ2.
If H0 is true, the two estimates will be similar and
MSF / MSE ≈ 1. If it is false, then MSF >> MSE.
1.4.5: One-way vs. Two-way ANOVA


One-way (one-factor) ANOVA compares
several groups corresponding to a single
categorical (independent) variable or factor,
i.e.: one-way analysis of variance involves a
single factor, which usually has three or more
levels. Each level represents the treatment
applied.
Two-way (two-factors) ANOVA performs an
analysis of variance for testing the equality of
populations means when classification of
treatments is by two categorical (independent)
variables or factors. Within each combination
of the two factors or cell, you might have
multiple observations called replicates. In twoway ANOVA, the data must be balanced, i.e.:
all cells must have the same number of
observations. In two-way ANOVA, possible
interaction between the two factors (the effect
that the two factors have on each other) must
be considered.
Examples of one-way ANOVA applications:
• In agriculture you might be interested in the effects of
potassium (factor) on the growth of potatoes (response)
• In medicine you might want to study the effects of
medication (factor) on the duration of headaches (response)
• In education you might want to study the effects of grade
level (factor) on the time required to learn a skill (response)
• A marketing experiment might consider the effects of
advertising euros (factor) on sales (response)
Examples of two-way ANOVA applications:
• In agriculture you might be interested in the effects of both
potassium and nitrogen (factors) on the growth of potatoes
(response)
• In medicine you might want to study the effects of
medication and dose (factors) on the duration of headaches
(response)
• In education you might want to study the effects of grade
level and gender (factors) on the time required to learn a
skill (response)
• A marketing experiment might consider the effects of
advertising euros and advertising medium (factors) on
sales (response)
1.4.6: One-way ANOVA: Table (Minitab)

File: SCORES.MTW

Stat > ANOVA > One-Way
(Unstacked)...
Boxplot of Group 1; Group 2; Group 3
In performing an ANOVA, you determine what
part of the variance you should attribute to
randomness and what part you can attribute
to other factors. ANOVA does this by splitting
the total sum of squares SST into two parts: a
part due to random error SSE and a part
attributed to differences between groups SSF.
Remember: Var[ y ] 
85
SST    yi  y     yi  yˆi     yˆi  y   SSE  SSF
2
80
75
Data
1
2
 yi  y 

n 1
2
2
The CIs plot also shows that the 95%CI for the means of groups 1 and 3
are clearly disjoint. That suggest that groups 1 and 3 have different means.
70
65
The Boxplots can help to visually detect differences
among
60 the populations being considered.
Group 1
Group 2
Group 3
The ANOVA table shows the following calculations:
• SS Factor, SS Error and SS Total
• MS Factor and MS Error
• Degrees of freedom (DF) associated with SSF and SSE
• F statistic
• p-value for the hypothesis test (reject H0 if p-value <= α)
1.4.7: One-way ANOVA: Checking assumpt.

File: SCORES.MTW
When performing an ANOVA test, you assume
that all the observations come from normally
distributed populations. Obtaining a normal
probability plot for each sample group can
be helpful in order to determine if it is
reasonable to assume normality.
Probability Plot of Group 2
Normal
99
Mean
StDev
N
AD
P-Value
95
90
74
4,472
6
0,325
0,392
In this case, it seems reasonable to assume
that observations from Group 2 are normally
distributed. For each sample group, a similar
plots should be analyzed.
Percent
80
70
60
50
40
Scatterplot of RESI vs GROUP
30
10
20
10
5
5
65
70
75
Group 2
To check for constant variance, use the graphical
method of plotting the residuals versus the
level or group. If the residuals (observed value –
sample mean) have roughly the same range or
vertical spread in each shift, then they satisfy the
constant variance assumption.
80
85
RESI
1
0
-5
-10
In this case, the constant variance assumption
does not appear to be violated.
1,0
1,5
2,0
GROUP
2,5
3,0
1.4.8: One-way ANOVA: Fisher’s LSD

Suppose that one-way ANOVA has provided statistical
evidence to reject the null hypothesis of equal means.
In this case, Fisher’s Least Significant Difference
(LSD) is a multiple comparison procedure (i.e., it
conducts statistical comparisons between pairs of
population means) that can be used to determine
where the differences occur:
Fisher’s LSD procedure provides 95% CI
estimates of the differences between each
pair of means.
In this example, the Fisher’s 95% CI for the
difference between Group 1 and Group 3
does not contain the zero value. Therefore,
both groups seem to have different means.
The same can be said regarding the difference
between Group 2 and Group 3.
Fisher’s LSD procedure is based on the t
test statistic for the two-population case.
1.4.9: One-way ANOVA: Bonferroni adjust.

Fisher’s LSD procedure is used to make several
pairwise comparisons of the form H0: μi = μj vs. H1: μi ≠
μj. In each case, a significance level of α = 0.05 is
used. Therefore, for each test, if the null hypothesis is
true, the probability that we will make a Type I error is
α = 0.05 (particular Type I error rate), and the
probability that we will not make a Type I error on each
test is 1 – 0.05 = 0.95.

If we make C pairwise comparisons, the probability
that we will not make a Type I error on any of the C
tests is (0.95)(0.95)…(0.95) = 0.95C. Therefore, the
probability of making at least one Type I error is 1 –
0.95C (overall Type I error rate). Note that the larger C
the more close to 1 will be this error rate.

One alternative for controlling the overall Type I error
rate, referred as the Bonferroni adjustment, involves
using a smaller particular error rate for each test, i.e.:
take particular = overall / C. Another alternative is to
use Tukey’s procedure instead of Fisher’s LSD.
Using Bonferroni adjustment: For
instance, if we want to use Fisher’s LSD
procedure test C = 3 pairwise comparisons
with a a maximum overall error rate of
0.05, we set the particular error rate to
be 0.05/3 = 0.017.
1.4.10: One-way ANOVA: Tukey’s procedure

Many practitioners are reluctant to use Bonferroni
adjustment since performing individual tests with a
very low particular Type I error rate increases the risk
of making a Type II error.

Tukey’s procedure can be used as an alternative to
Fisher’s LSD. Tukey’s procedure is more conservative
than Fisher’s LSD procedure, since it makes it harder
to declare any particular pair of means to be
significantly different:
Tukey’s procedure provides 95% CI
estimates of the differences between each
pair of means.
In this example, the Tukey’s 95% CI for the
difference between Group 1 and Group 3
does not contain the zero value. Therefore,
both groups seem to have different means.
Note that conclusions for the difference
between Group 2 and Group 3 are different
from those obtained using Fisher’s LSD.
Recall that, for a fixed sample size, any
decrease in the probability of making a
Type I error will result in an increase in
the probability of making a Type II error,
which corresponds to accepting the
hypothesis that the two population means
are equal when in fact they are not.
1.4.11: Two-way ANOVA: Table (Minitab)

File: RATIO2.MTW

Stat > ANOVA > Two-Way...
We want to test three sets of hypothesis:
A marketing department wants to
determine which one of two
magazines (Factor A, 2 levels) has
the lowest ratio of full-page adds to
the number of pages in the magazine
(Response). It also wants to
determine whether there has been a
change in this ratio during the last
three years (Factor B, 3 levels).
H0: There is no interaction between factors
H1: There is an interaction between factors
H0: There is no difference in the mean Response for different levels of Factor A
H1: There is a difference in the mean Response for different values of Factor A
H0: There is no difference in the mean Response for different levels of Factor B
H1: There is a difference in the mean Response for different levels of Factor B
In Minitab, use the Two-way ANOVA when you have exactly two
factors. If you have three or more factors, use the Balanced
ANOVA. Both the Two-way and Balanced ANOVA require that all
combinations of factor levels (cells) have an equal number of
observations, i.e., the data must be balanced.
For a significance level of 0.05, the results indicate a
nonsignificant interaction between year and magazine (p-value
= 0.338). You conclude that the mean ad ratios are probably the
same for the three years (p-value = 0.494). In addition, you
conclude that the mean ad ratios for the two magazines are not
the same (p-value = 0.012).