Lecture & Examples
Topic 2: Completely Randomized Design
The completely randomized design is the simplest form of
experimental designs. In a completely randomized design,
each treatment is applied to each experimental unit
completely by chance. Although the completely
randomized design is very simple, it has many
advantages:
(1) It is very flexible. Any number of treatments and any
number of experimental units can be used.
(2) The statistical analysis is easy even if the number of
experimental units in different treatments is different.
(3) The statistical analysis remains simple in the presence
of missing values. We can actually prove that the relative
loss of information due to missing data is smaller than any
other experimental designs. Completely randomized
design is extremely attractive in the case when
experimental units are homogeneous. For example, it is
the method of choice for many laboratory experiments,
e.g., in physics, chemistry, or cookery, where a quantity
of material, after through mixing, is divided into small
samples or batches to which the treatments are applied.
Steps for Conducting an Analysis of Variance
(ANOVA) for a Completely Randomized Design:
Step 1: To make sure the data come from a completely
randomized design. This means that each treatment is
assigned to an experimental unit completely by chance.
Step 2: Use a graphical procedure such as box-plots or
dot-plots to visualize the equal variance assumption. The
normality assumption is guaranteed if the data truly
comes from a completely randomized design. Suppose
that the equal variance assumption is not satisfied, you
can either use a nonparametric method, discussed in
Chapter 15, or find a statistical consultant to get help. The
method discussed in Chapter 15 is just one alternative
when the equal variance assumption is not satisfied. It is
not the "best" alternative.
Step 3: Create an ANOVA table that is similar to Table
10.1 below with any statistical package such as SAS (used
in this lecture) or SPSS. In Table 10.1, we assume that
there are p treatments and n experimental units.
Table 10.1: ANOVA Table for a Completely Randomized Design
Source
df
SS
MS
F
Treatment
SST
MST(a)
MST/MSE
(p  1)
(b)
Error
SSE
MSE
(n  p)
Total
SS(Total)
(n  1)
(a) MST = SST/(p  1)
(b) MSE = SSE/(n  p)
Note:
(1) The degrees of freedom for treatment is (p  1) in a
completely randomized design with p treatments.
(2) The degrees of freedom for error is (n  p) in a
completely randomized design with p treatments and n
experimental units.
(3) SST + SSE = SS(Total)
(4) We can complete the ANOVA table if we know any
two of these three quantities: SST, SSE, or SS(Total).
(5) We can complete the ANOVA table if we know
both MSE and MST.
(6) Partially complete ANOVA table will always be
available in exams or practice problems.
Step 4: Use the F-test provided in the ANOVA table to
perform the following test:
H0: 1 = 2 = . . . = p
Ha: At least two treatment means are different.
The output from SAS or SPSS always includes the pvalue of the above F-test. We can reject the null
hypothesis at significance level  when "p-value  ."
Step 5: Suppose the F-test suggests that one can reject the
null hypothesis, one can then use the multiple comparison
procedures discussed in Section 10.3 to find the mean
differences. However, all of those multiple comparison
procedures discussed are very conservative and will not
be discussed this semester.
Example 10.4:
A manufacturer of television sets is interested in the effect
on tube conductivity of four different types of coating for
color picture tubes. The following conductivity data are
obtained.
Coating Type
1
2
3
4
143
152
134
129
Conductivity
141
150
149
137
136
132
127
132
146
143
127
129
(a) Complete the following ANOVA table.
General Linear Models Procedure
Dependent Variable: RESP
Sum of
Source
DF
Squares
Model
(1)
844.68750
Error
(2)
(4)
Corrected Total(3)
1080.93750
Mean
Square
(5)
(6)
F Value
(7)
Pr > F
.00029
Solution: n = 16, p = 4, df(Treatment) = (p  1),
df(Error) = (n  p), df(Total) = (n  1),
SSE = SS(Total)  SST, MST = SST/(p  1),
MSE = SSE/(n  p),
F = MST/MSE
Note: In the above SAS printout, Model = Treatment and
Pr > F = p-value.
General Linear Models Procedure
Dependent Variable: RESP
Sum of
Source
DF
Squares
Model
3
844.68750
Error
12
236.25000
Corrected Total 15
1080.93750
Mean
Square
281.56250
19.68750
F Value
14.30
Pr > F
.00029
(b) Test the null hypothesis that 1 = 2 = 3 = 4, against
the alternative that at least two of the means differ. Use 
= 0.05.
Solution:
H0: 1 = 2 = 3 = 4
Ha: At least two means differ
Test Statistic: Fc = 14.30
p-value: p-value = 0.00029
Since the p-value = 0.00029 <  = 0.05, one can reject
the null hypothesis.
Example 10.5:
A manufacturer suspects that the batches of raw material
furnished by her supplier differ significantly in calcium
content. There is a large number of batches currently in
the warehouse. Five of these are randomly selected for
study. A chemist makes five determinations on each batch
and obtains the following data.
Batch 1
23.46
23.48
23.56
23.39
23.40
Batch 2
23.59
23.46
23.42
23.49
23.50
Batch 3
23.51
23.64
23.46
23.52
23.49
Batch 4
23.28
23.40
23.37
23.46
23.39
Batch 5
23.29
23.46
23.37
23.32
23.38
(a) Complete the following ANOVA table.
General Linear Models Procedure
Dependent Variable: RESP
Sum of
Source
DF
Squares
Model
(1)
(4)
Error
(2)
(5)
Corrected Total(3)
(6)
Mean
Square
0.0242440
0.0043800
F Value
(7)
Pr > F
.00363
Solution: n = 25, p = 5, SST = (MST)(p  1),
SSE = (MSE)(n  p), SS(Total) = SST + SSE
General Linear Models Procedure
Dependent Variable: RESP
Sum of
Source
DF
Squares
Model
4
0.0969760
Error
20
0.0876000
Corrected Total 24
0.1845760
Mean
Square
0.0242440
0.0043800
F Value
5.54
Pr > F
.00363
(b) Is there a significant variation in calcium content from
batch to batch? Use
 = 0.05.
Solution:
H0: 1 = 2 = 3 = 4 = 5
Ha: At least two means differ
Test Statistic: Fc = 5.54
p-value: p-value = 0.00363
Since the p-value = 0.00363 <  = 0.05, one can reject
the null hypothesis that all means are equal. Thus, there
is a significant variation in calcium content from batch
to batch, for  = 0.05.