Transcript Lecture 8 (May 29, June 5)

Ch12 Analysis of Variance
Outline
Completely randomized designs
Randomized-block designs
Analysis of Variance
Single Factor Analysis of Variance
Single Factor ANOVA
One Way Analysis of Variance
One Way ANOVA
Background
If we have, say, 3 treatments to compare (A, B, C) then
we would need 3 separate t-tests (comparing A with B, A
with C, and B with C).
If we had 7 treatments we would need 21 separate ttests. This would be time-consuming but, more important,
it would be inherently flawed because in each t-test we
accept a 5% chance of our conclusion being wrong
(when we test for p = 0.05). So, in 21 tests we would
expect (by probability) that one test would give us a false
result.
ANalysis Of Variance (ANOVA) overcomes this problem
by enabling us to detect significant differences between
the treatments as a whole.
We do a single test to see if there are differences
between the means at our chosen probability level.
Assumption: equal variances, independent
populations, random sampling
The scheme of one-way
classification
Observations Means Sum of Squares
n1
Sample 1 : y11 , y12 ,
, y1 j ,
, y1n1
y1
2
(
y

y
)
 1j 1
i 1
n2
Sample 2 : y21 , y22 ,
, y2 j ,
, y2 n2
y2
2
(
y

y
)
 2j 2
i 1
ni
Sample i : yi1 , yi 2 ,
, yij ,
, yini
yi
2
(
y

y
)
 ij i
i 1
nk
Sample k : yk1 , yk 2 ,
, ykj ,
, yknk
yk
2
(
y

y
)
 kj k
i 1
Simplify
k
k
ni
T .   yij
i 1 j 1
k
N   ni
i 1
y
n y
i 1
k
n
i 1
y
yi
i
i
T.

N
i
is the overall mean or grand mean of all
observations.
is the mean of the measurements obtained by
the i-th laboratory.
The statistical analysis leading to a comparison of the k different
population means consists essentially of splitting the sum of squares
about the overall grand mean into a component due to treatment
difference, and a component due to error or variation within a sample.
EX
Suppose 3 drying formulas for curing a
glue are studied and the following times
observed.
Formula A: 13 10 8 11
Formula B: 13 11 14 14
Formula C: 4 1 3 4
8
2
4
Each observation can be decomposed as
yij
observation
 y

( yi  y )
grand
deviation due
mean
to treatment
 ( yij  yi )
error
Repeating the decomposition for each observation,
we obtain the arrays
observation
yij
1310 8 11 8
131114 14

 4 1 3 4 2
grand mean treament effects
y
( yi  y )
error
( yij  yi )
 8 8 8 8 8   2 2 2 2 2
 3 0  2 1  2
  8 8 8 8   5 5 5 5
  0  2 1 1
 
 
 
4  8 8 8 8 8 8  5  5  5  5  5  5 1  2 0 1  1



1
k
treatment sum of square SS (Tr )   ni ( yi  y )2
i 1
ni
k
error sum of square SSE   ( yij  yi )2
i 1 j 1
Degrees of freedom for treatment: k-1
Degrees of freedom for error: N-k
Theorem.
SST 
k
ni
k
k
2
(
y

y
)

(
y

y
)

n
(
y

y
)
 ij
 ij i  i i
2
i 1 j 1
SST
ni
2
i 1 j 1
SSE
i 1
SS(Tr)

If i denotes the mean of the i-th population and 
denotes the common variance of the k populations.
2
Yij  i  ij
Yij    i   ij
Where  is the mean of the i in the experiment and
is the effect of the i-th treatment; hence
k
n
i 1
i
i
0
The null hypothesis that the k population means are all
equal can be replaced by the null hypothesis
1  2 
 k  0
The alternative hypothesis that at least two of the
population means are unequal.
To test the null hypothesis that the k population means
are all equal, we shall compare two estimates of  2
One based on the variation among the sample means,
and one based on the variation within the samples.
Each sum of squares is first converted to a mean square.
sum of squares
mean square 
degrees of freedom
When the population means are equal, both
k
treatment mean square 
2
n
(
y

y
)
i i
i 1
k -1
k
and
error mean square 
are estimates of  2
ni
2
(
y

y
)
 ij i
i 1 j 1
N k
If the null hypothesis is true, it can be shown that the two
mean squares are independent and that their ratio
k
F
2
n
(
y

y
)
/(k  1)
i i
i 1
k ni
2
(
y

y
)
 ij i /( N  k )
SS (Tr ) /(k  1)

SSE /( N  k )
i 1 j 1
has an F distribution with k-1 and N-k degrees of freedom.
A large value for F indicates large difference between the
sample means. Therefore, the null hypothesis will be
rejected, ifF  F at  level of significance.
One-way ANOVA
Source of
variance
Sum of
squares
Degree of
freedom
Mean
square
Treatments
SS(Tr)
K-1
SS (Tr )
s 
k 1
Error
SSE
K (n - 1)
s2 
Total
SST
nk - k
2
1
SSE
k (n  1)
Computed
f
s12
s2
Solution of EX
Solution
One-way ANOVA: A, B, C
Source
Factor
Error
Total
DF
SS
MS
F
P
2 270.00 135.00 50.63 0.000
12 32.00 2.67
14 302.00
The value of F0.05 (2,12)  3.89
null hypothesis of equal means.
so we reject the
Exercise
Assume that we have recorded the biomass of 3
bacteria in flasks of glucose broth, and we used
3 replicate flasks for each bacterium
Replicate Bacterium
A
1
12
Bacterium
B
20
Bacterium
C
2
15
19
35
3
9
23
42
40
Solution
One-way ANOVA: A, B, C
Source
Factor
Error
Total
DF SS
MS
F
P
2 1140.2 570.1 64.93 0.000
6
52.7 8.78
8 1192.9
The value of F(2,6) = 5.1 in the level of 0.05
so we reject the null hypothesis of equal means.
12.3 Random-Block designs
Two way ANOVA
Blocks
1
1
2
3
2
13 7
6
6
11 5
3
4
9 3
3 1
15 5
RCB Randomized Complete Block
The randomized block design is an extension of the
paired t-test to situations where the factor of interest has
more than two levels.
Example 1:
Suppose we are interested in how weight gain (Y) in
rats is affected by Source of protein (Beef, Cereal,
and Pork) and by Level of Protein (High or Low).
There are a total of t = 32 = 6 treatment
combinations of the two factors (Beef -High Protein,
Cereal-High Protein, Pork-High Protein, Beef -Low
Protein, Cereal-Low Protein, and Pork-Low Protein) .
Suppose we have available to us a total of N = 60
experimental rats to which we are going to apply the
different diets based on the t = 6 treatment
combinations.
Prior to the experimentation the rats were divided
into n = 10 homogeneous groups of size 6.
The grouping was based on factors that had
previously been ignored (Example - Initial weight
size, appetite size etc.)
Within each of the 10 blocks a rat is randomly
assigned a treatment combination (diet).
The weight gain after a fixed period is
measured for each of the test animals and is
tabulated on the next slide:
Randomized Block Design
Block
1
107
(1)
96
(2)
112
(3)
83
(4)
87
(5)
90
(6)
Block
6
128
(1)
89
(2)
104
(3)
85
(4)
84
(5)
89
(6)
2
102
(1)
72
(2)
100
(3)
82
(4)
70
(5)
94
(6)
7
56
(1)
70
(2)
72
(3)
64
(4)
62
(5)
63
(6)
3
102
(1)
76
(2)
102
(3)
85
(4)
95
(5)
86
(6)
8
97
(1)
91
(2)
92
(3)
80
(4)
72
(5)
82
(6)
4
93
(1)
70
(2)
93
(3)
63
(4)
71
(5)
63
(6)
9
80
(1)
63
(2)
87
(3)
82
(4)
81
(5)
63
(6)
5
111
(1)
79
(2)
101
(3)
72
(4)
75
(5)
81
(6)
10
103
(1)
102
(2)
112
(3)
83
(4)
93
(5)
81
(6)
Example 2:
The following experiment is interested in
comparing the effect four different chemicals
(A, B, C and D) in producing water resistance
(y) in textiles.
A strip of material, randomly selected from
each bolt, is cut into four pieces (samples) the
pieces are randomly assigned to receive one of
the four chemical treatments.
This process is replicated three times
producing a Randomized Block (RB) design.
Moisture resistance (y) were measured for
each of the samples. (Low readings indicate
low moisture penetration).
The data is given in the diagram and table on
the next slide.
Diagram: Blocks (Bolt Samples)
9.9
10.1
11.4
12.1
C
A
B
D
13.4
12.9
12.2
12.3
D
B
A
C
12.7
12.9
11.4
11.9
B
D
C
A
Table
Chemical
A
B
C
D
Blocks (Bolt Samples)
1
2
3
10.1
12.2
11.9
11.4
12.9
12.7
9.9
12.3
11.4
12.1
13.4
12.9
The randomized block design (RBD) consists of a twostep procedure:
1. Matched sets of experimental units, called blocks, are
formed, each block consists of a units. The b blocks
should consist of experimental units that are as similar
as possible (to reduce the within-treatments variation) .
2. One experimental unit from each block is randomly
assigned to each treatment, resulting in a total of ab
responses.
3. If every block has responses from all treatments, the
design is complete, randomized complete block design.
RCB
For example, consider the situation where three different
methods were used to predict the shear strength of steel
plate girders. Say we use four girders as the experimental
units.
RCB
1 b
yi .   yij
b j 1
1 a
y. j   yij
a i 1
1 a b
y.. 
yij

ab i 1 j 1
The total number of responses is ab.
RCB
The appropriate linear statistical model:
We assume
• treatments and blocks are initially fixed effects
• blocks do not interact
•
RCB
The hypotheses of interest are:
i.e., there is no
treatments effect
RCB
has a-1 df
has b-1 df
has (a-1)(b-1) df
RCB
The mean squares are:
RCB
The expected values of these mean squares are:
RCB
RCB
Minitab
Two-way ANOVA: response versus row,
col
Source DF SS
MS F
P
row
2 56 28.0000 3.23 0.112
col
3 90 30.0000 3.46 0.091
Error
6 52 8.6667
Total 11 198
The P-value > 0.05 level of significance, we cannot reject
the null hypothesis.
The Anova Table for Diet Experiment
Source
Block
Diet
ERROR
Total
S.S
5992.41667
4572.88333
3147.28333
13712.58
d.f.
9
5
45
59
M.S.
F
665.82407
9.52
914.576666 13.0766586
69.93963
p-value
0.00000
0.00000
The Anova Table forTextile Experiment
SOURCE
Blocks
Chem
ERROR
Total
SUM OF SQUARES
7.17167
5.20000
0.53500
12.90667
D.F.
2
3
6
MEAN SQUARE
3.5858
1.7333
0.0892
11
F
40.21
19.44
TAIL PROB.
0.0003
0.0017