ST 511
Analysis of
Variance
Introduction
Analysis of variance compares two or
more populations of quantitative data.
 Specifically, we are interested in
determining whether differences exist
between the population means.
 The procedure works by analyzing the
sample variance.

§1
One Way Analysis of
Variance

The analysis of variance is a procedure
that tests to determine whether
differences exist between two or more
population means.

To do this, the technique analyzes the
sample variances
One Way Analysis of Variance:
Example

A magazine publisher wants to compare
three different styles of covers for a
magazine that will be offered for sale at
supermarket checkout lines. She
assigns 60 stores at random to the
three styles of covers and records the
number of magazines that are sold in a
one-week period.
One Way Analysis of Variance:
Example

How do five bookstores in the same city
differ in the demographics of their
customers? A market researcher asks
50 customers of each store to respond
to a questionnaire. One variable of
interest is the customer’s age.
Idea Behind ANOVA –
two types of variability
1. Within group variability
2. Between group variability
30
25
x3  20
20
x 2  15
16
15
14
11
10
9
x3  20
20
19
x 2  15
x1  10
12
10
9
x1  10
7
1
A small variability within
The sample means are the same as before,
theTreatment
samples
makes it2easier
Treatment 3 Treatment
Treatment 3
1 Treatment
1 within-sample
Treatment 2 variability
but the larger
to draw a conclusion about the
makes it harder to draw a conclusion
about the population means.
population means.
Idea behind ANOVA: recall the
two-sample t-statistic

Difference between 2 means, pooled variances, sample sizes
both equal to n
n (xy)
t
xy
sp
t 
2



11
n n

2
sp
n ( x  y )2
2
s 2p
Numerator of t2: measures variation between the groups in
terms of the difference between their sample means
Denominator: measures variation within groups by the pooled
estimator of the common variance.
If the within-group variation is small, the same variation
between groups produces a larger statistic and a more
significant result.
One Way Analysis of Variance:
Example

Example 1
– An apple juice manufacturer is planning to
develop a new product -a liquid concentrate.
– The marketing manager has to decide how to
market the new product.
– Three strategies are considered
 Emphasize
convenience of using the product.
 Emphasize the quality of the product.
 Emphasize the product’s low price.
One Way Analysis of Variance

Example 1 - continued
– An experiment was conducted as follows:
 In three cities an advertisement campaign
was launched .
 In each city only one of the three
characteristics (convenience, quality, and
price) was emphasized.
 The weekly sales were recorded for twenty
weeks following the beginning of the
campaigns.
One Way Analysis of Variance
Convnce
Weekly
sales
529
658
793
514
663
719
711
606
461
Weekly
529
sales
498
663
604
495
485
557
353
557
542
614
Quality
Price
804
630
774
717
679
604
620
697
706
615
492
719
787
699
572
Weekly
523
sales
584
634
580
624
672
531
443
596
602
502
659
689
675
512
691
733
698
776
561
572
469
581
679
532
One Way Analysis of Variance

Solution
– The data are quantitative
– The problem objective is to compare
sales in three cities.
– We hypothesize that the three
population means are equal
Defining the Hypotheses
• Solution
H0: 1 = 2 = 3
H1: At least two means differ
To build the statistic needed to test
the hypotheses use the following
notation:
Notation
Independent samples are drawn from k populations (treatment groups).
First
observation,
first sample
Second observation,
second sample
1
2
k
X11
x21
.
.
.
Xn1,1
X12
x22
.
.
.
Xn2,2
X1k
x2k
.
.
.
Xnk,k
n2
nk
x2
xk
n1
x1
Sample size
Sample mean
X is the “response variable”.
The variables’ values are called “responses”.
Terminology

In the context of this problem…
Response variable – weekly sales
Responses – actual sale values
Experimental unit – weeks in the three cities
when we record sales figures.
Factor – the criterion by which we classify the
populations (the treatments). In this problems
the factor is the marketing strategy.
Factor levels – the population (treatment)
names. In this problem factor levels are the 3
marketing strategies: 1) convenience, 2) quality,
3) price
The rationale of the test statistic
Two types of variability are
employed when testing for the
equality of the population means
1. Within sample variability
2.Between sample variability
H0: 1 = 2 = 3
H1: At least two means differ
The rationale behind the test statistic – I
If the null hypothesis is true, we
would expect all the sample means to
be close to one another (and as a
result, close to the grand mean).
 If the alternative hypothesis is true,
at least some of the sample means
would differ.
 Thus, we measure variability between
sample means.

Variability between sample means
• The variability between the sample
means is measured as the sum of
squared distances between each mean
and the grand mean.
This sum is called the
Sum of Squares for Groups
SSG
In our example, treatments are
represented by the different
advertising strategies.
Sum of squares for treatment groups
(SSG)
k
SSG   n j (x j x)
2
j1
There are k treatments
The size of sample j The mean of sample j
Note: When the sample means are close to
one another, their distance from the grand
mean is small, leading to a small SSG.
Thus, large SSG indicates large variation
between sample means, which supports H1.
Sum of squares for treatment groups
(SSG)

Solution – continued
Calculate SSG
x1  577.55 x2  653.00 x3  608.65
k
SSG   n j (x j x) 2
j1
The grand mean is calculated by
n1 x1  n2 x2  ...  nk xk
x
n1  n2  ...  nk
= 20(577.55 - 613.07)2 +
+ 20(653.00 - 613.07)2 +
+ 20(608.65 - 613.07)2 =
= 57,512.23
Sum of squares for treatment
groups (SSG)
Is SSG = 57,512.23 large
enough to reject H0 in favor of
H1?
The rationale behind test statistic – II
Large variability within the samples
weakens the “ability” of the sample
means to represent their
corresponding population means.
 Therefore, even though sample means
may markedly differ from one
another, SSG must be judged relative
to the “within samples variability”.

Within samples variability

The variability within samples is
measured by adding all the squared
distances between observations and
their sample means.
This sum is called the
Sum of Squares for Error
SSE
In our example, this is the sum of
all squared differences between
sales in city j and the sample mean
of city j (over all the three cities).
Sum of squares for errors (SSE)

Solution –
continued
Calculate SSE
s12  10, 775.00 s22  7, 238.11 s32  8, 670.24
k
nj
SSE   (xij  x j )  (n1 - 1)s12 + (n2 -1)s22 +
2
j 1 i 1
(n3 -1)s32
= (20 -1)10,775 + (20 -1)7,238.11+ (20-1)8,670.24
= 506,983.50
Sum of squares for errors (SSE)
Is SSG = 57,512.23 large
enough relative to SSE =
506,983.50 to reject the null
hypothesis that specifies that
all the means are equal?
The mean sum of squares
To perform the test we need
to calculate the mean squares
as follows:
Calculation of MSG Mean Square for
treatment Groups
SSG
MSG 
k 1
57, 512.23

3 1
 28, 756.12
Calculation of MSE
Mean Square for Error
SSE
MSE 
nk
506,983.50

60  3
 8,894.45
Calculation of the test statistic
MSG
F
MSE
28, 756.12

8,894.45
 3.23
with the following degrees of freedom:
v1=k -1 and v2=n-k
Required Conditions:
1. The populations tested
are normally distributed.
2. The variances of all the
populations tested are
equal.
The F test rejection region
the hypothesis test:
And finally
H0: 1 = 2 = …=k
H1: At least two means differ
Test statistic:
MSG
F
MSE
R.R: F>Fa,k-1,n-k
The F test
Ho: 1 = 2= 3
H1: At least two means differ
MSG
MSE
28, 756.12

8, 894.45
 3.23
F
Test statistic F= MSG/ MSE= 3.23
R.R.: F  Fa k 1nk  F0.05,31,603  3.15
Since 3.23 > 3.15, there is sufficient
evidence to reject Ho in favor of H1, and
argue that at least one of the mean sales
is different than the others.
The F test p- value
Use Excel to find the p-value
fx
Statistical
F.DIST.RT(3.23,2,57) =
.0469

0.1
0.08
p Value = P(F>3.23) = .0469
0.06
0.04
0.02
0
-0.02
0
1
2
3
4
Excel single factor ANOVA
Anova: Single Factor
SUMMARY
Groups
Convenience
Quality
Price
Count
20
20
20
ANOVA
Source of Variation
Between Groups
Within Groups
SS
57512
506984
Total
564496
Sum
Average Variance
11551
577.55 10775.00
13060
653.00
7238.11
12173
608.65
8670.24
df
2
57
59
SS(Total) = SSG + SSE
MS
28756
8894
F
P-value
3.23
0.0468
F crit
3.16
Multiple Comparisons


When the null hypothesis is rejected,
it may be desirable to find which
mean(s) is (are) different, and at
what ranking order.
Two statistical inference procedures,
geared at doing this, are presented:
– “regular” confidence interval calculations
– Bonferroni adjustment
Multiple Comparisons
Two means are considered different
if the confidence interval for the
difference between the
corresponding sample means does not
contain 0. In this case the larger
sample mean is believed to be
associated with a larger population
mean.
 How do we calculate the confidence
intervals?

“Regular” Method


This method builds on the equal variances
confidence interval for the difference
between two means.
The CI is improved by using MSE rather than
sp2 (we use ALL the data to estimate the
common variance instead of only the data
from 2 samples)
1 1
( xi  x j )  ta 2, n k  s

ni n j
d . f .  n  k , s  MSE
Experiment-wise Type I error rate
(the effective Type I error)




The preceding “regular” method may result in an
increased probability of committing a type I error.
The experiment-wise Type I error rate is the
probability of committing at least one Type I
error at significance level a. It is calculated by:
experiment-wise Type I error rate = 1-(1 – a)g
where g is the number of pairwise comparisons (i.e.
g = k C 2 = k(k-1)/2.
For example, if a=.05, k=4, then
experiment-wise Type I error rate =1-.735=.265
The Bonferroni adjustment determines the
required Type I error probability per pairwise
comparison (a*) , to secure a pre-determined
overall a.
Bonferroni Adjustment
 The procedure:
– Compute the number of pairwise comparisons
(g)
[g=k(k-1)/2], where k is the number of
populations.
– Set a* = a/g, where a is the true probability
of making at least one Type I error (called
experiment-wise Type I error).
– Calculate the following CI for i – j
1 1
( xi  x j )  ta * 2, n  k  s

ni n j
d . f .  n  k , s  MSE
Bonferroni Method


Example - continued
– Rank the effectiveness of the marketing
strategies (based on mean weekly sales).
– Use the Bonferroni adjustment method
Solution
– The sample mean sales were 577.55, 653.0,
608.65.
– We calculate g=k(k-1)/2 to be 3(2)/2 = 3.
– We set a* = .05/3 = .0167, thus t.0167/2, 60-3 = 2.467
(Excel).
– Note that s = √8894.447 = 94.31
x1  x2  577.55  653  75.45
x1  x3  577.55  608.65  31.10
x2  x3  653  608.65  44.35
ta * 2  s
1 1


ni n j
2.467 *94.31 1/ 20  1/ 20  73.57
Bonferroni Method: The Three
Confidence Intervals
1 1
( xi  x j )  ta * 2, n  k  s

ni n j
d . f .  n  k , s  MSE
x1  x2  577.55  653  75.45
x1  x3  577.55  608.65  31.10
x2  x3  653  608.65  44.35
ta * 2  s
1 1


ni n j
2.467 *94.31 1/ 20  1/ 20  73.57
1  2 :  75.45  73.57 (149.02, 1.88)
1  3 :  31.10  73.57 (104.67, 42.47)
2  3 :44.35  73.57 (29.22,117.92)
There is a significant difference between 1
and 2.
Bonferroni Method: Conclusions
Resulting from Confidence Intervals
Do we have evidence to distinguish two means?
 Group 1 Convenience: sample mean 577.55
 Group 2 Quality: sample mean 653
 Group 3 Price: sample mean 608.65
1  2 :  75.45  73.57 (149.02, 1.88)
 31.10  73.57 (104.67, 42.47)
1  3 : (
1  3 :  31.10 73.57
104.67, 42.47)
2  3 :44.35  73.57 (29.22,117.92)

List the group numbers in increasing order of their sample
means; connecting overhead lines mean no significant
difference
1 3 2
Bonferroni Method: Conclusions
Resulting from Confidence Intervals
Do we have evidence to distinguish two means?
Group 1 Convenience: sample mean 577.55
 Group 2 Quality: sample mean 653
 Group 3 Price: sample mean 608.65

1  2 :  75.45  73.57 (149.02, 1.88)
1  3 :  31.10  73.57 (104.67, 42.47)
2  3 :44.35  73.57 (29.22,117.92)

List the group numbers in increasing order of their sample
means; connecting overhead lines mean no significant
difference
1 3 2