1
252anova 1/26/07 (Open this document in 'Outline' view!)
Roger Even Bove
F. ANALYSIS OF VARIANCE
1. 1-Way Analysis of Variance
a. The ANOVA model - relation to regression
The one-way ANOVA model is used to compare the means of more than two samples, taken from
populations that are all assumed to have the same variance. Each sample (called a treatment) is usually
represented as a column, but there is no requirement that each column have the same number of items in it.
We will assume the model xij    a j  eij , where i  1 through n j , and n 
n j .We thus have m

j 1,m
treatments, n j items in each column and a total of
n observations. (Thus
x ij should be the number in
column i and row j )
b. An ANOVA problem
The following data describes monthly expenses for energy in three random samples of essentially
identical homes. Each column represents expenses on one fuel.   .05  .
Fuel
1
2
3
89
104
86
101
120
98
87
98
100
87
110
96
Sum
364
432
380
 H 0 : 1   2   3
Our hypotheses are 
In the notation used here, i is replaced by a dot to indicate that
H 1 : Not all  equal
a mean has been taken, that is x j is the mean of column j , in particular, the mean of column 1 is
364
432
380
 91 , x2 
 108 and x3 
 95 . The overall or grand mean, is the mean of all the
4
4
4
numbers in the problem, and is often indicated by x  , but x seems to be a more appealing notation.
x1 
364  432  380
 98 .
12
We compute three sums of squares.
(i) The total sum of squares is the same thing as the
 89  98 2

  101  98 2
2
x ij  x  
the problem. SST 
2
j
i
  87  98 
  87  98 2

x
 

numerator of the sample variance of the numbers in
 104  98 2  86  98 2 
 120  98 2  98  98 2 
  1148
 98  98 2  100  98 2 
 110  98 2  96  98 2 
2
(ii) The sum of squares within treatments has the same number of terms, but highlights the contribution to
the total sum of squares generated by the difference between the individual numbers and the column
 89  912
 104  108 2  86  95 2 

2
  101  91  120  108 2  98  95 2 
(treatment) means. SSW 
x ij  x j 2  
  516 (iii)
2
 98  108 2  100  95 2 
  87  91
j
i
  87  912  110  108 2  96  95 2 


The sum of squares between treatments also has the same number of terms, but it highlights the contribution
to the total sum of squares generated by the difference between the column (treatment) means and the
overall mean.
 91  98 2  108  98 2  95  98 2 


  91  98 2  108  98 2  95  98 2 
2
SSB 
x. j  x  
 632
2
2
2
j
i
  91  98   108  98   95  98  
  91  98 2  108  98 2  95  98 2 


But, because of the repetition of the column mean, this can be simplified to
 n x  x 2   491  98 2  4108  98 2  495  98 2  632 .
SSB 
 j .j

 
 

 


j
But note that SSB  SSW  SST , so that the computation of one of the three sums of squares is
unnecessary. The material is summarized in a table like the one below.
Source
SS
DF
MS
F
SSB
MSB
MSB 
F
m 1
SSB
Between
m 1
MSW
SSW
MSW 
nm
Within
SSW
nm
SST
n 1
Total
We fill in the table with the numbers we have computed and compare the F that we have computed with an
F with the appropriate significance level and degrees of freedom shown in the DF column. If the F that
we have computed is larger than the table F , reject the null hypothesis.
Source
SS
DF
MS
F.05
H0
F
Between
632
2
316
5.51
F 2,9  4.26 s
Column means equal
Within
516
9
57.333
Total
1148
11
The ‘s’ for ‘significant difference’ indicates that the null hypothesis of equality of means has been rejected.
‘ns’ for ‘no significant difference’ would indicate that the null hypothesis has not been rejected.
3
c. A format for ANOVA
If we use the same simplifications that we use in calculating a sample variance, we can get the tableau
below.
Fuel
2
104
120
98
110
3
86
98
100
96
Sum
Sum
364 +
432 +
380
 1176 
nj
4+
4+
4
 12  n
x j
91.00
108.00
95.00
SS
33260 +
46920 +
36216
x 2j
8281 +
11664 +
9025
 x
1176
 98  x
12
 116396 
ij
 x
 28970   x
2
ij
2
j
2   xij2  nx 2  116396 1298 2  1148
2
2
2
2
2
2
2
. j  x    n j x. j  nx  491  4108   495   12 98 
 x
SSB   x
SST 
1
89
101
87
87
ij
x
 428970   12 98 2  632
Source
SS
Between
DF
632
2
MS
316
F
F.05
5.51
F 2,9  4.26 s
H0
Column means equal
Within
516
9
57.333
Total
1148
11
Explanation: Since the Sum of Squares (SS) column must add up, 516 is found by subtracting 632
from 1148. Since n  12 , the total degrees of freedom are n  1  11 . Since there are 3 random samples or
columns, the degrees of freedom for Between is 3 – 1 = 2. Since the Degrees of Freedom (DF) column must
add up, 9 = 11 – 2. The Mean Square (MS) column is found by dividing the SS column by the DF column.
MSB
316 is MSB and 57.333 is MSW . F 
, and is compared with F.05 from the F table
MSW
df1  2, df 2  9 . To see this as Minitab output go to 252anovaex1.
d. Confidence Intervals
i. A single Confidence Interval
If we desire a single interval, we use the formula for the difference between two means when the variance is
known. For example, if we want the difference between means of column 1 and column 2.
1   2  x1  x2   t n  m  s
2
1
1

, where s  MSW .
n1 n 2
4
ii. Scheffé Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible intervals

1
1 
between column means, use 1   2  x1  x2   m  1Fm1,n m   s
.

 n
n 2 
1

iii. Bonferroni Confidence Interval
If we only need k different intervals, use 1   2  x1  x2   t n  m  s
1
1

n1 n 2
2k
iv. Tukey Confidence Interval
This also applies to all possible differences.
1   2  x1  x2   q m,n  m 
s
2
1
1
This gives rise to Tukey’s HSD (Honestly Significant

n1 n 2
Difference) procedure. Two sample means x .1 and x .2 are significantly different if x.1  x.2 is greater
than 1  2  qm, n  m 
s
2
1
1

n1 n2
2. 2 -Way Analysis of Variance
a. The 2-way model
We will assume R rows, C columns and P observations per cell. Thus our model reads
xijk     i   j   ij   ijk , where i  1 through R, j  1 through C, and k  1 through P . We will be
testing three pairs of Hypotheses - (i) H 01 : All row means equal (All  i zero); H 11 : Not all row means
equal , (ii) H 02 : All column means equal (All  j zero); H 12 : Not all column means equal, (iii) H 03 :
No interaction (All  ij zero) ; H 13 : Interaction.
This is similar to one-way ANOVA with RC groups , but the ‘between’ variation is itemized as to whether
it is due to variation between row means, variation between column means or interaction. If we remember
that n  RCP and m  RC , we can rewrite the one way ANOVA table diagram on the previous table as
below. As previously, we get the items in the MS column by dividing the numbers in the SS column, by the
numbers in the DF column. The F is then found by dividing MSB by MSW .
Source
SS
DF
MS
F
F
H0
RC  1
SSB
MSB
Between
___
___
Treatment means equal
MSW
Within
SSW
RC P  1
RCP  1
SST
Total
We can now rewrite the same table with the ‘between’ items split up.
Source
SS
DF
MS
SSR
MSR
Rows
R 1
C 1
SSC
MSC
Columns
SSI
Interaction
R  1C  1 MSI
Within
SSW
RC P  1
Total
SST
RCP  1
MSW
F
___
___
___
F
___
___
___
H0
Row means equal
Column means equal
No Interaction
5
b. An example
Insulation 1
Insulation 2
(Factor B1 )
(Factor B 2 )
89
87
Fuel 1 (Factor A1 )
101
87
120
98
Fuel 2 (Factor A2 )
110
104
100
86
Fuel 3 (Factor A3 )
98
96
This problem has R  3 rows, C  2 columns and, within each cell P  2 measurements. We can compute
a table of means which shows means for each cell, row and column, as well as an overall mean.
Insulation 1
Insulation 2
Row means
(Factor B1 )
(Factor B 2 )
Fuel 1 (Factor A1 )
x11   95
x12   87
x1    91
Fuel 2 (Factor A2 )
x 21   115
x 22   101
x 2    108
Fuel 3 (Factor A3 )
x31   99
x32   91
x3    95
Column Means
x1   103
x2   93
x  x    98
Now we do the computation of sums of squares, using the same simplification that we use in computing a
sample variance.
SST 
 x
i
j
x
ijk
2
k
 89  98   101  98 2  87  98 2  87  98 2  120  98 2        96  98 2
2
 89 2  101 2  87 2  87 2  120 2        96 2  1298 2  1148
SSW 
 x
i
j
 xij 
ijk
2
k
 89  95   101  95 2  87  87 2  87  87 2  120  115 2        96  912
2
 89 2  101 2  295 2  87 2  87 2  287 2  120 2  110 2  2115 2      
 86 2  96 2  2912  192
SSR  CP
 x
i 
x

2



 22 91  98 2  108  98 2  95  98 2  22912  108 2  105 2  398 2
i
 632
SSC  RP
 x
j 
x

2

i
SSI  P
 x
i
ij 



 32103  98 2  93  98 2  32103 2  93 2  298 2  300
 xi    x  j   x
2 , but we do not compute this because
j
SST  SSR  SSC  SSI  SSW , so that SSI  SST  SSR  SSC  SSW  1148  632  300  192  24

6
Out ANOVA table is thus:
Source
SS
DF
MS
F
Rows  A
632
2
316
Columns B 
300
1
300
9.36s
24
2
12
0.38ns
Interaction  AB 
F
2,6   5.14
F.05
F 1,6   5.99
9.88s
.05
2,6 
F.05
 5.14
H0
Row means equal
Column means equal
No Interaction
Within
192
6
32
Total
1148
11
So we reject H 01 and H 02 ,but do not reject H 03 .
To explain further, We get the degrees of freedom for rows by taking the number of rows minus 1,. We do
the same for columns. Then the interaction degrees of freedom are the product of row and column degrees
of freedom. The total degrees of freedom comes from subtracting 1 from the total number of items in the
problem. The ‘within’ degrees of freedom comes from subtracting the other degrees of freedom from the
‘total’ degrees of freedom. The ‘MS’ column comes from dividing the SS column by the DF column. The
‘F’ column is calculated by dividing the items in the ‘MS’ column by s  MSW  32 .
To see this as Minitab output go to 252anovaex2. An example of 2-way ANOVA with one
measurement per cell is in 252anovaex3.
c. Confidence Intervals
i. A Single Confidence Interval
If we desire a single interval we use the formula for a Bonferroni Confidence Interval below with m  1 .
ii. Scheffé Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible intervals
between means, use the following formulas.
For cell means, use 11   21  x11  x 21  
For row means, use 1   2  x1  x 2  
RC  1FRC 1, RCP 1 2MSW
P
R  1FR 1, RC P 1 2MSW
PC
For column means, use
 1   2  x1  x2  
C  1FC 1, RCP 1 2MSW
Note that if P  1 , replace RC P  1 with
PR
R  1C  1 .
.
.
7
iii. Bonferroni Confidence Interval
If we only need m different intervals, use for cell means 11   21  x11  x 21   t RC P 1
2m
Use for row means 1   2  x1  x 2   t RC P 1
2m
2MSW
.
P
2MSW
.
PC
Use for column means  1   2  x1  x2   t RC P 1
2MSW
.
PR
2m
iv. Tukey Confidence Interval
For cell means, use 11   21  x11  x 21   qRC, RC P 1
For row means, use 1   2  x1  x 2   qR , RC P 1
MSW
.
PC
For column means, use  1   2  x1  x2   qC , RC P 1
Note that if P  1 , replace RC P  1 with
MSW
.
P
MSW
PR
R  1C  1 .
3. More than 2-Way analysis of Variance See 252anovaex4.
4. Kruskal-Wallis Test
Equivalent to one-way ANOVA when the underlying distribution is non-normal.
H 0 : Columns come from same distribution or medians equal.
Example: Use same example as for one-way ANOVA, but assume that data comes from non-normal source.
Assume that   .05 . There are n  12 data items, so rank them from 1 to 12. Let n i be the number of
items in column i and SRi be the rank sum of column i . n 
Treatment
1
89
101
87
87
n
Original Data
Treatment Treatment
2
3
104
86
120
98
98
100
110
96
SRi
i
.
Treatment
1
4
9
2.5
2.5
18.0
Ranked Data
Treatment Treatment
2
3
10
1
12
6.5
6.5
8
11 .
5 .
39.5
20.5
4
4
4
ni
To check the ranking, note that the sum of the three rank sums is 18.0 + 39.5 + 20.5 = 78.0, and that the
nn  1 12 13 

 78 .
sum of the first n numbers is
2
2
8
 12
 SRi 2 

  3n  1
Now, compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 
 12  18 .0 2 39 .52 20 .52 

  313   1 576 .125   39  5.3173 . If we look up this result in the (4,



4
4 
13
12 13   4
4, 4) section of the Kruskal-Wallis table (Table 9) , we find that the p-value for H  5.6538 is .054 and that
the p-value for H  4.6539 is .097, so the p-value for H  5.3173 must lie between these two. Since both
are above   .05 , do not reject H 0 .

If the size of the problem is larger than those shown in Table 9, use the  2 distribution, with df  m  1 ,
where m is the number of columns. For example, if each of m  3 columns contains 6 items,   .05 and
H  5.3173 , compare H with  .2052   5.9915 . Since H is smaller than  .205 , do not reject the null
hypothesis.
5. Friedman Test
Equivalent to two-way ANOVA with one observation per cell when the underlying distribution is nonnormal.
H 0 : Columns come from same distribution or medians equal. Note that the only difference between this
and the Kruskal-Wallis test is that the data is cross-classified in the Friedman test.
Example: Three groups of 4 matched workers are trained to do a task by four different methods. When each
worker is observed later, he or she is given a grade of 1 through 10 on performance of the task. Note that
because this data is ordinal, ANOVA is not appropriate. Assume that   .05 . In the data below, the
methods are represented by c  4 columns, and the groups by r  3 rows.. In each row the numbers are
ranked from 1 to c  4 . For each column, compute SRi , the rank sum of column i .
Group 1
Group 2
Group 3
Method
1
9
6
9
Original Data
Method Method
2
3
4
1
5
2
1
2
Method
4
7
8
6
Method
1
4
3
4
11
Ranked Data
Method Method
2
3
2
1
2
1
1
2
5
4
Method
4
3
4
3
10
SRi
To check the ranking, note that the sum of the four rank sums is 11 + 5 + 4 + 10 = 30, and that the sum of
cc  1
the c numbers in a row is
. However, there are r rows, so we must multiply the expression by r .
2
rcc  1 345
SRi 

 30 .
So we have
2
2

9
 12
Now compute the Friedman statistic  F2  
 rc c  1

 12
112  52  42  10 2





3
4
5


 SR   3r c  1
2
i
i


1

  335   262   45  7.4 . If we find the place on the
5



Friedman Table (Table 8) for 4 columns and 3 rows, we find that the p-value for  F2  7.4 is .033. Since
the p-value is below   .05 , reject the null hypothesis.
If the size of the problem is larger than those shown in Table 10, use the  2 distribution, with df  c  1 ,
where c is the number of columns. For example, if each of c  5 columns contains 6 items,   .05 and
 F2  7.4 , compare  F2 with  .2054   9.4877 . Since  F2 is not larger than  .205 , do not reject the null
hypothesis.
6. Tests for Equality of Variances – See 252mvar.