3.4 Experiments
3.4.2 Cross-City Prediction Model
Treatment 4 121.7 30.43 0.1925
Error 15 2371.5 158.10
Total 19 2493.2
F0.05, (4, 15) = 3.06
Decision Rule: If Ftab Fcal , reject Ho. Otherwise accept Ho Conclusion:
Since Fcal= (0.1925) < F, (v1, v2) = 3.06, we accept the null hypothesis that there is no significant difference in the examiners’ standards of awards.
Co-efficient of Variation
the co-efficient of variation, the higher the precision.
The co-efficient of variation (C.V) is given by C.V. =
M G
MSE .
x 100%;
Overall (Grand Mean) = N
T..
Overall Mean
The co-efficient of variation for example 1 is
100 58
. 5
00 . 5
x
= 40%
For example 2,
C.V = 100
2 . 17
10 .
158 X
= 73%
Advantages of the Completely Randomized Design (CRD)
1) The CRD is flexible in that the number of treatment and replicate is limited only to the number of experiments available.
2) The statistical analysis is simple even if the number of replicates vary with the treatment.
3) The number of replicates can vary from treatment to treatment, though it is generally desirable to have the same number per treatment.
4) Loss of information due to missing data is small relative to loss with other designs.
5) The number of degrees of freedom for estimating experimental error is maximum. This improves the precision of the experiment.
Disadvantages of the CRD
1) The CRD is not normally used for field experiments, other
2) The major disadvantage of this design is that it is normally appropriate for homogeneous experimental materials and when the number of treatment is small.
The Randomized Complete Block Design (RCBD)
The RCBD is the most widely used design and it was originated in agriculture where land was divided into blocks and blocks were divided into plots that received the treatments under investigation. It is a design in which the units called experimental unit to which the treatments are applied are sub-divided into homogeneous groups called plots, so that the number of experimental units in a plot is equal to the number of treatments being considered. The treatments are then assigned at random to the experimental units within each block. It should be noted that each treatment appears in every block and each receives every treatment.
However, the objective of using the Randomized Complete Block Design is to reduce as small as possible or isolate variability among the experimental units within a block while assuring that treatment means will be free of block effect. The effectiveness of the design depends on the ability to achieve homogeneous blocks of experimental units. The ability to form homogeneous blocks depends on the researcher’s knowledge of the experimental material.
The following are examples of areas where RCBD is applicable:
1) In experiments involving human beings, age and sex may be used as blocks.
2) In animal experiments where different breeds of animal will respond differently to the same treatment. The breed type may be used as a blocking factor.
3) The design can also be used when an experiment must be carried out in more than one laboratory or when several days are required for completion.
Table of sample values for the RCBD Two-Way Classification
Treatment
Blocks 1 2 3 - - - k Total Mean 1. y11 y12 y13 - - - y1k B1
Y1
2. y21 y22 y23 - - - y2k B2
Y2
3. y31 y32 y33 - - - y3k B3 Y3
. . . . . . .
m ym1 ym2 ym3 - - - ymk Bm
Ym
Total T1 T2 T3 - - - Tk R…
Mean Y.1 Y.2 Y.3 - - - Y.k Y..
The model is given as:
Yij = μ + i + βj + eij
Where μ = Grand mean
i = ith row effect βj = jth column effect eij = random error term
Hypothesis: We can consider two different null hypotheses. They are Ho1: There is no difference in treatment means
Ho2: There is no difference in block means
Then we can compute the sum of squares for the three sources of variations as follows:
SSTotal = N
R Y
m
i k
j ij
2
1 1
2 ..
with (mk – 1) d.f Where N = mk; R.. = Grand Total
m Ti2 R..2
SSBlocks =
k
j j
N R K
B
1 2 2
..
with (k-1) d.f.
SSError = SSTotal – SSTreatment – SSBlocks
With (m-1) (k-1) d.f.
ANOVA Table for RCBD Source of
Variation
Degree of Freedom
Sum of
Squares (SS)
Mean Square F-ratio or Fo
Treatment m-1 SSTreatment SSTr=MSTr
m-1
MSTr
MSE
Blocks k-1 SSBlock SSBlock = MSE
k-1
MSB MSE Error (m-1)(k-1) SSError SSError = MSE
(m-1)(k-1)
Total Mk-1 SSTotal
Decision Rule: Reject Ho if Ftab < F0. Otherwise, reject H1
Example 3
The director of a tuberculosis control section of a general hospital wishes to study the effectiveness of 3 newly developed drugs to cure the disease.
The drugs are administered on patients in different hospitals and the number of recovery from the disease per 50 patients is recorded as follows:
Drugs
Hospitals 1 2 3 Total
H1 8 13 10 31
H2 11 15 16 42
N Y R
m
i
ij k
j
2
1
2
1
..
Is there any difference in the number of recovery cases recorded in the 3 hospitals?
Test, using = 5%
Solution
Ho: µ1 = µ2 = µ3 Hi: µ1 ≠ µ2 ≠ µ3 SSTotal =
N = mk
= 3 x 3 = 9
SST = 82 + 132 + … + 172 + 172 -
9 ) 121
( 2
= 1709 – 1626.78
= 82.22
SSTr =
N R m
m Ti
i
2
1
2 ..
= 9
) 121 ( 3
43 45
332 2 2 2
= 1654.33 - 1626.78
= 27.55 SSBlock =
N R k
k Bj
j
2
1
2 ..
= 9
) 121 ( 3
48 42
312 2 2 2
= 1676.33 - 1626.78
= 49.55
SSError = SSTotal - SSTreatment - SSBlock
= 82.22 - 27.55 - 49.55
= 5.12
ANOVA Table Source of Variation
Degree of Freedom
Sum of Squares
Mean Squa re
F-ratio
Treatment 2 27.55 13.78 10.77
Blocks 2 49.55 24.78 19.36
Error 4 5.12 1.28
Total 8 82.22
Fo.o5, (2,4) = 6.94 in both cases.
Decision: Since Fcal> Ftab in both cases, the null hypothesis H0, is rejected.
We therefore conclude that not all the treatment means are equal.
Example 4
Four groups of physical therapy patients were subjected to different diets in 6 clinics. At the end of a specified period of time, each was given a test to measure treatment effectiveness. The following scores were obtained:
Treatments (Diet)
Clinic 1 2 3 4 Total
1. 76 70 90 80 316
2. 75 82 64 88 309
3. 72 80 79 71 302
4. 38 95 74 90 317
5. 66 80 87 60 293
6. 82 88 85 75 330
Total 429 495 479 464 1867
Do these data provided sufficient evidence to indicate a difference among the treatments?
Solution
Ho: µ1 = µ2 = µ3 = µ4
Hi: at least one µi ≠ µj
SSTotal = 762 + 702 + 902 + … + 852 + 752 -
24 ) 1867
( 2
= 147439 - 145237.04
= 2201.96 SSTreatment =
24 ) 1867 ( 6
464 479
495
429 2 2 2 2 2
= 145633.83 - 145237.04
= 396.79 SSBlock =
= 145444.75 - 145237.04
= 207.71
SSError = 2201.96 - 396.79 - 207.71
= 1597.46 ANOVA Table
Source of Variation
Degree of Freedom
Sum of Squares
Mean Square Fo
Treatment 3 396.79 132.26 1.24
Blocks 5 207.71 41.54 0.39
Error 15 1597.46 106.50
Total 23 2201.96
F2, 15 (0.05) = 3.29 F5, 15 (0.05) = 2.90
Decision: Since the value of Fcal < Ftab in both cases, we accept the null hypothesis and conclude that the experiment does not show any significant
24 ) 1867 ( 4
330 293
317 302
309
316 2 2 2 2 2 2 2
Advantages of the RCBD
1) It is easy to understand with simple computation.
2) Grouping experimental units into blocks gives better precision.
3) Missing value can be easily estimated so that arithmetic convenience is not loss.
4) Certain complications that may arise in the cause of experiment are easily handled with this design.
Disadvantage
When the variation among the experimental units between a block is large, a large error term results. This frequently occurs when the number of treatment is large. Thus, it may not be possible to secure sufficiently uniform groups or units for blocks. This is the major disadvantage of the RCBD.
Summary
This lecture discussed a vivid description of the analysis of experiment.
Various designs have been identified. Models are being specified and tests have been conducted to test significance.
Post-Test
1) Give the definition of the following terms:
(a) Treatment (b) Response (c) Experimental units (d) Factors (e) Replication
2) What are the components of the design of an experiment?
3) Mention 3 functions of replication.
4) What is randomization?
5) State and explain briefly the steps involved in planning and execution of an experiment.
6) State three assumptions underlying the use of ANOVA.
7) The design of an experiment can be classified into how many categories and mention them.
8) Give 2 examples each of the areas where the completely
9) State the mathematical model of the RCBD and give the meaning of each term.
10) Give the sample layout for the CRD.
11) Mention 2 advantages and disadvantages of the CRD.
12) Four different groups of children admitted in a large hospital for the treatment of chicken pox were subjected to different treatments. The following scores were obtained from the test given to each of the children to measure treatment effectiveness:
Do these data provide sufficient evidence to indicate a difference among the treatments? Use = 0.05.
Treatment
1 2 3 4
58 74 66 60
95 82 75 90
80 87 88 85
88 66 60 75
75 79 90 82
At = 0.05, can we conclude that the data provide sufficient evidence to indicate a difference among the treatments?
14) In a laboratory test to determine the uric acid values in a 5 groups of pregnant women, the following results were obtained:
5.8 6.0 6.3 5.4 5.7
5.9 6.1 6.3 6.2 5.5
5.1 5.2 5.0 6.7 6.5
5.0 5.4 6.4 6.0 5.3
Can we conclude that there is no significant difference in the mean
15) A medical practitioner in a teaching hospital, wishes to study the effectiveness of three newly manufactured drugs by different pharmaceutical companies for the treatment of malaria. These drugs were administered on patients in four different wards and the number of hours of recovery recorded as follows:
Drugs (Treatment)
Wards 1 2 3
1. 18 24 19
2. 20 21 22
3. 24 27 30
4. 28 22 29
References
Edward, R. Dougherty (1990): Probability and Statistics for the Engineering, Compound and Physical Sciences
James, T. McClave, Terry Sincich (1995, 5th Ed): A first course in Statistics
Lindgen, McElrath, Berry (1978, 4th Ed): Introduction to Probability and Statistics
Murray, R. Spiegel (1972, 1st Ed): Theory and Problems of Statistics Royal Statistical Society (2004): Statistical tables for use in examinations.
Rustagi J.S. (1985): Introduction to Statistical Methods, Volume II.
LECTURE FIVE