CH 10: ANALYSIS OF VARIANCE
WHAT IS ANOVA?
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Moving from chapter 8 where we covered hypothesis testing, we
now need to understand how to compare more than two means. So
3 or 4 or more means will be tested at the same time.
We will follow the same 4 step process as in chapter 8.
The null hypothesis will now be H0: µ1 = µ2 = µ3 = … = µk
The alternative hypothesis will always be H1: at least two
means differ. We do not know which two population means
differ, so we are simply testing if they differ. You will learn how to
find out which means differ in more advanced Statistics courses.
So why is this hypothesis test called the analysis of variance
(ANOVA)? Even though we are testing whether more than 2
population means differ or not, we do not actually compare the
means in the test. We have to compare the variances of the
different populations and then decide whether the means are
equal or not.
1
CONCEPTS AND DEFINITIONS
Independent
Variable
This is the factor/variable of interest. It is the variable that
is divided into the different groups/ treatments. Eg. We
want to compare 3 types of learning materials for
educational research (Type A, Type B, Type C). Each
learning material would be a different group/treatment.
Dependent
Variable
This is the response variable. So after the different types of
learning materials were administered, what were the average test
scores for those groups/treatments?
ni
Number of observations in each group/treatments
k
Number of groups/treatments
n
Total number of observations in ALL the groups/treatments (n =
n1 + n2 + … + nk)
yij
The ith observation in the jth group/treatment (y11 ; y35 ; y42 ; y73 )
𝒚𝒋
The mean of the observations in the jth group/treatment
𝒚
The grand mean = the mean of ALL the observations
2
ASSUMPTIONS FOR ANOVA
1.
observations are assumed to be normally distributed within
each group or treatment
2.
observations are from a random sample and are independent
3.
assumption that all variances are equal
(𝜎12 = 𝜎22 = 𝜎32 = ⋯ = 𝜎𝑘1 )
3
HYPOTHESIS TESTING FOR MORE THAN 2 POPULATION
MEANS (ANOVA)
Step 1: Hypotheses
H0: µ1 = µ2 = µ3 = … = µk
H1: at least two means differ.
Step 2: CV and RR
𝛼 can be 10% or 5% or 2.5% or 1% or 0.5%
d.f. (numerator) = ν1 = k – 1
d.f. (denominator) = ν2 = n – k
We use the F-distribution to find the critical value for an
ANOVA test. It is smooth curve that is skewed to the right and
is always positive. We will thus always have a one tail test with
the rejection region in the right hand tail (Page A10-A19).
CV: Fcrit = Fα, ν1, ν2 = F0.05, k-1, n-k
RR: {F│F > F0.05, k-1, n-k}
4
HYPOTHESIS TESTING FOR MORE THAN 2 POPULATION
MEANS (ANOVA)
Step 3: Test Statistic
In order to test whether or not differences exist between more the 2
populations, we have to calculate 2 types of deviations.
o We need to firstly see what is the deviation or difference between the
groups/treatments and then we need to see what is the deviation or
difference within the groups/treatments. But how do we do this?
o We will use sums of squares in order to minimize these deviations or
differences.
o This involves 6 steps:
1) TSS = Total Sum of Squares = SST + SSE
Measures the total deviation between treatments and within
treatments.
𝟐
𝒌
𝒊=𝟏
TSS = 𝒌𝒋=𝟏 𝒏𝒊=𝟏 𝒚𝒊𝒋 − 𝒚
TSS = 𝒌𝒋=𝟏 𝒏𝒊=𝟏 𝒚𝟐𝒊𝒋 −
2) SST = Sum of Squares for Treatments
Measures the deviation between each group or treatment
mean and the grand mean.
SST =
𝒌
𝒋=𝟏 𝒏𝒋
× 𝒚𝒋 − 𝒚
𝟐
𝟐
𝒏
𝒚
𝒊=𝟏 𝒊𝒋
𝒏
5
STEP 3: TEST STATISTIC
3) SSE = Sum of Squares for Error = TSS - SST
Measures the deviation between each observation
and each group or treatment mean.
SSE =
𝒌
𝒋=𝟏
𝒏
𝒊=𝟏
𝒚𝒊𝒋 − 𝒚𝒋
𝟐
4)MST = Mean Square for Treatment
Measures the average or Mean of the Sum of
Squares for Treatment.
MST =
𝑆𝑆𝑇
𝑘−1
5)MSE = Mean Square for Error
Measures the average or Mean of the Sum of
Squares for Error.
MSE =
𝑆𝑆𝐸
𝑛−𝑘
6) F-Statistic = F = MST/MSE
6
7
ANOVA OUTPUT TABLE
8
HYPOTHESIS TESTING FOR MORE THAN 2
POPULATION MEANS (ANOVA)
Step 4: Conclusion
Option 1: Reject the null hypothesis if Fstat > Fcrit which
means there is enough statistical evidence at the 5%
significance level to conclude that population means are not
equal and at least 2 means differ.
Option 2: Do not reject the null hypothesis if Fstat < Fcrit
which means that there is sufficient evidence at the 5%
significance level to conclude that the population means are
all equal.
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Exercise 1
A university president collects data showing the number of
absences over the past academic year for a random sample
of six professors in the Faculty of Science. She does the
same for a random sample of nine professors in the
Economic and Management Science Faculty and for a
random sample of eight professors in the Faculty of Arts.
Faculty of Science
Absences
Management and Economic Sciences Faculty
Arts Faculty
8
5
9
10
7
10
6
6
10
8
7
9
4
7
7
8
6
5
8
13
8
7
1
Test at the 5% significance level if there is sufficient
evidence to infer whether the mean absences for the 3
faculties are the same or not. Use the 4 step process.
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Solution:
Step 1: Hypotheses
H0: 𝜇1 = 𝜇2 = 𝜇3
H1: at least two means differ
Step 2: CV and RR
𝛼 = 5% = 0.05
ν1 = d.f. (numerator) = k – 1 = 3 – 1 = 2
So we look in the top row for numerator df = 2 (somewhere down that column)
ν2 = d.f. (denominator) = n – k = 23 – 3 = 20
So we look for 20 down the first column (on page A14)
We look in the block of values that intersect at numerator df = 2 and denominator df = 20 and 𝛼 = 0.05
Fcrit = 𝐹0.05,2,20 = 3.49
RR: {F│F > F0.05, 2, 20} = {F│F > 3.49}
11
Step 3: Test Statistic
Group (or treatment)
Science
EMS
Arts
8
5
9
10
7
10
6
6
10
8
7
9
4
7
7
8
6
5
8
13
8
7
Observations
1
Sample size = nj
(n =
23
)
6
9
8
Total
= ∑yj
44
55
70
Total
= ∑yj2
344
373
654
Mean
= 𝑦𝑗
7.3333
6.1111
8.75
Grand Total
= ∑∑yij = 44 + 55 + 70 = 169
Grand Mean
= 𝑦 = 169/23 = 7.3478
12
Step 3: Test Statistic
Source
of Sum of squares
D.f
Mean square
F-statistic
Step 3.4
Step 3.6
variation
Step 3.2
SST   n j  y j  y 
k
2
j 1
Treatments (T)
𝟐
Numerator
𝟐
3–1=2
SST = 6 x (7.3333 – 7.3478) +
9 x (6.1111 – 7.3478) +
8 x (8.75 – 7.3478)
1  k  1
(done in step2)
𝟐
= 29.4954
2  n  k
SSE    yij  y j 
Step 3.3
k
nj
2
j 1 i 1
Sampling error
(E)
Denominator
SSE = TSS – SST
23 – 3 = 20
= 129.2174 – 29.4954
= 99.722
(done in step2)
MST 
MST =
SST
k 1
𝟐𝟗.𝟒𝟗𝟓𝟒
𝟐
= 14.7477
Step 3.5
F
F=
MST
MSE
𝟏𝟒.𝟕𝟒𝟕𝟕
𝟒.𝟗𝟖𝟔𝟏
= 2.9578
SSE
MSE 
nk
MSE =
𝟗𝟗.𝟕𝟐𝟐
𝟐𝟎
= 4.9861
Step 3.1
Total (T)
2
 k nj 
  yij 
 j 1 i 1 
k nj
2

TSS   yij  
n
j 1 i 1
(𝟒𝟒+𝟓𝟓+𝟕𝟎)𝟐
TSS = (344+373+654) –
𝟐𝟑
= 1371
-
𝟐𝟖𝟓𝟔𝟏
𝟐𝟑
= 1371 – 1241.782609
= 129.2174
  n  1 Step 4: Conclusion
Total
23 – 1 = 22
F(critical) = 3.49 > F(statistic) = 2.9578
Therefore, the F statistic falls in the
acceptance region and we do not reject
the null hypothesis. We can infer13
that the
mean absences of the three faculties are
the same.
Exercise 2
In a collaborative trial, four laboratories were sent samples
from a reservoir and requested to perform ten assays and
report the results based on percentage of a labelled amount
of the drug (see Table 1). Were there any significant
differences based on the laboratory performing the
analysis? Table 2 shows the partial results of the ANOVA
test output. Answer this question using the 4 step process.
Table 1: Descriptive data for four different laboratories (rounded off to four decimals)
Groups
Lab (A)
Lab (B)
Lab (C)
Lab (D)
Count
10
10
10
10
Sum
999
996.9
995.1
1000
Average
99.9
99.69
99.51
100
Variance
0.0622
0.1721
0.0477
0.1156
Table 2: ANOVA table for four different laboratories (rounded off to four decimals)
Source of Variation
SS
df
MS
F
F crit
Between Groups (T)
1.4370
___
0.4790
___
___
Within Groups (E)
3.5780
36
Total
5.0150
___
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Step 1: Hypotheses
H0: 𝜇1 = 𝜇2 = 𝜇3 = 𝜇4
H1: at least two means differ
Step 2: CV and RR
𝛼 = not given so use default of 5% = 0.05 (can be different)
ν1 = d.f. (numerator) = k – 1 = 4 – 1 = 3
So we look in the top row for numerator df = 3 (somewhere down that column)
ν2 = d.f. (denominator) = n – k = 40 – 4 = 36
So we look for 36 down the first column (on page A18) but 36 is not there, so we round off to the nearest
ten. That means we round up to 40.
We look in the block of values that intersect at numerator df = 3 and denominator df = 40 and 𝛼 = 0.05
Fcrit = 𝐹0.05,3,40 = 2.84
RR: {F│F > F0.05, 3, 40} = {F│F > 2.84}
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Step 3: Test Statistic
Source of
Variation
Between Groups
(T)
SS
df
MS
1.4370
Done in step 2
0.4790
Df(numerator)
=k–1
=4–1
=3
Check this is correct:
Within Groups
(E)
3.5780
Total
5.0150
36
MST =
F stat
𝟏.𝟒𝟑𝟕𝟎
𝟎.𝟒𝟕𝟗
𝟎.𝟎𝟗𝟗𝟒
𝟑
= 0.479
MSE =
F=
𝑺𝑺𝑬
= 4.8189
𝒏 −𝒌
𝟑.𝟓𝟕𝟖𝟎
=
𝟑𝟔
= 0.0994
Df(total)
=n–1
= 40 – 1
= 39
Also = 3 + 36
Step 4: Conclusion
Compare F(crit) to F(stat). F(crit) = 2.84 < F(stat) = 4.8189
and therefore the F statistic falls in the rejection region.
There is sufficient statistical evidence to infer that at least
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two means differ.