-Design can be used when experimental units are essentially homogeneous. ... Very flexible design (i.e. number of treatments and replicates is only limited by.
COMPLETELY RANDOM DESIGN (CRD)
Description of the Design -Simplest design to use. -Design can be used when experimental units are essentially homogeneous. -Because of the homogeneity requirement, it may be difficult to use this design for field experiments. -The CRD is best suited for experiments with a small number of treatments.
Randomization Procedure -Treatments are assigned to experimental units completely at random. -Every experimental unit has the same probability of receiving any treatment. -Randomization is performed using a random number table, computer, program, etc. Example of Randomization -Given you have 4 treatments (A, B, C, and D) and 5 replicates, how many experimental units would you have?
1
D 11
C
2
D 12
B
3
B 13
A
4
C 14
B
5
D 15
C
6
C 16
B
7
A 17
C
8
A
9
B
18
D
19
A
-Note that there is no “blocking” of experimental units into replicates. -Every experimental unit has the same probability of receiving any treatment.
10
D 20
A
Advantages of a CRD 1. Very flexible design (i.e. number of treatments and replicates is only limited by the available number of experimental units). 2. Statistical analysis is simple compared to other designs. 3. Loss of information due to missing data is small compared to other designs due to the larger number of degrees of freedom for the error source of variation. Disadvantages 1. If experimental units are not homogeneous and you fail to minimize this variation using blocking, there may be a loss of precision. 2. Usually the least efficient design unless experimental units are homogeneous. 3. Not suited for a large number of treatments.
Analysis of the Fixed Effects Model Notation Statistical notation can be confusing, but use of the Y-dot notation can help simplify things.
The dot in the Y-dot notation implies summation across over the subscript it replaces.
For example, n
yi. yij Treatment total, where n number of observations in a treatment j 1
yi. yi. n Treatment mean y.. i 1 a
n
y j 1
ij
Experiment total, where a number of treatments
y.. y.. N Experiment mean, where N total number of observations in the experiment .
Linear Additive Model for the CRD
Y ij i ij where: Yij is the jth observation of the ith treatment, is the population mean,
i is the treatment effect of the ith treatment, and ij is the random error. -Using this model we can estimate i or ij for any observation if we are given Yij and . Example
Yi. Y i. Y i. Y ..
Treatment 1 4 5 6 15 5 -3
Treatment 2 9 10 11 30 10 2
Treatment 3 8 11 8 27 9 1
Y.. = 72 Y .. = 8
-We can now write the linear model for each observation ( Y ij ). -Write in for each observation.
Yi. Y i. Y i. Y ..
Treatment 1 4=8 5=8 6=8 15 5 -3
Treatment 2 9=8 10 = 8 11 = 8 30 10 2
Treatment 3 8=8 11 = 8 8=8 27 9 1
Y.. = 72 Y .. = 8
-Write in the respective i for each observation where i Y i. Y ..
Yi. Y i. Y i. Y ..
Treatment 1 4=8–3 5=8–3 6=8–3 15 5 -3
Treatment 2 9=8+2 10 = 8 + 2 11 = 8 + 2 30 10 2
Treatment 3 8=8+1 11 = 8 + 1 8=8+1 27 9 1
Y.. = 72 Y .. = 8
-Write in the ij for each observation.
Yi. Y i. Y i. Y ..
Treatment 1 4=8–3-1 5=8–3 +0 6=8–3+1 15 5 -3
-Note for each treatment
Treatment 2 9=8+2-1 10 = 8 + 2 + 0 11 = 8 + 2 + 1 30 10 2
ij
Treatment 3 8=8+1-1 11 = 8 + 1 + 2 8=8+1-1 27 9 1
Y.. = 72 Y .. = 8
0.
-If you are asked to solve for 3 , what is the answer? -If you are asked to solve for 23 , what is the answer? -Question: If you are given just the treatment totals ( Yi. ’s), how would you fill in the values for each of the observations such that the Error SS = 0. Answer: Remember that the Experimental Error is the failure of observations treated alike to be the same. Therefore, if all treatments have the same value in each replicate, the Experimental Error SS =0. Example Given the following information, fill in the values for all Y ij ’s such that the Experimental Error SS = 0.
Yi. Y i.
Treatment 1
Treatment 2
Treatment 3
15 5
30 10
27 9
Y.. = 72 Y .. = 8
Answer Treatment 1 5 5 5 15 5
Yi. Y i.
Treatment 2 10 10 10 30 10
Note in the previous two examples that
Given H 0 : 1 2 ... t
Treatment 3 9 9 9 27 9
i
Y.. = 72 Y .. = 8
0. This is true for all situations.
H A : i i ' for at least one pair of treatments (i, i' ) t
i 1
i
t
(i.e., the sum of the treatment means divided by thenumber of treatments equals the experiment mean).
t
This definition implies that i 0 (Yi. Y.. ). i 1
The hypothesis written above can be rewritten in terms of the treatment effects τ i as:
H 0 : 1 2 ... a 0
H A : i 0 for at least i. Thus, when we are testing the null hypothesis that all treatments means are the same, we are testing at the same time the null hypothesis that all treatment effects, τi, are zero.
ANOVA for Any Number of Treatments with Equal Replication Given the following data:
Replicate 1 2 3 4 Yi.
Y
2 ij
Treatment B 23 42 36 26 31 47 33 34 123 149
A
3,875
5,805
C 47 43 43 39 172 Y...=444 7,428
Step 1. Write the hypotheses to be tested.
H o : 1 2 3 H A : 1 2 3
Ho: All three means are equal.
or
HA: At least one of the means is different from the other means.
1 2 3 or
1 2 3 Step 2. Calculate the Correction Factor. CF
Y..2 444 2 16,428.0 rt 4*3
Step 3. Calculate the Total SS TotalSS Yij2 CF (232 36 2 312 ... 39 2 ) CF 17,108 16,428 680.0
Step 4. Calculate the Treatment SS (TRT SS) Yi.2 TRTSS CF r 123 2 149 2 172 2 16,428 4 4 4 16728.5 16428.0 300.5
Step 5. Calculate the Error SS Error SS = Total SS – Treatment SS = 680 – 300.5 = 379.5 Step 6. Complete the ANOVA table
Sources of variation Treatment Error Total
Df t-1 = 2 t(r-1) = 9 rt-1 = 11
SS 300.5 379.5 680.0
MS 150.25 42.167
F 3.563NS
Step 7. Look up Table F-values. F0.05;2,9 = 4.26 F0.01;2,9 = 8.02 Step 8. Make conclusions.
-Since F-calc (3.563) < 4.26 we fail to reject Ho: at the 95% level of confidence. -Since F-calc (3.563) < 8.02 we fail to reject Ho: at the 99% level of confidence. 0
3.563
4.26
8.02
Step 9. Calculate Coefficient of Variation (CV). %CV
s *100 Y
Remember that the Error MS = s2.
%CV
42.167 *100 444 4*3
6.494 / 37 *100 17.6%
SAS commands for a CRD with no sampling; options pagenot=1;
data crd; input Trt $ yield; datalines; a 23 a 36 a 31 a 33 b 42 b 26 b 47 b 34 c 47 c 43 c 43 c 39 ;; ods rtf file='crd.rtf'; proc anova; class trt; model yield=trt; means trt/lsd; title 'ANOVA for a CRD with no Sampling'; run; ods rtf close;
ANOVA for a CRD with no Sampling The ANOVA Procedure Class Level Information Class Levels Values Trt
3 abc
Number of Observations Read
12
Number of Observations Used
12
ANOVA for a CRD with no Sampling The ANOVA Procedure Dependent Variable: yield
Source
DF
Sum of Squares Mean Square F Value
Model
2 300.5000000 150.2500000
Error
9 379.5000000
Corrected Total
Pr > F
3.56 0.0725
42.1666667
11 680.0000000
R-Square Coeff Var Root MSE yield Mean
0.441912 17.55023 6.493587
Source DF Trt
37.00000
Anova SS Mean Square F Value
2 300.5000000 150.2500000
Pr > F
3.56 0.0725
ANOVA for a CRD with no Sampling The ANOVA Procedure
Note This test controls the Type I comparisonwise error rate, not the experimentwise error rate. :
Alpha
0.05
Error Degrees of Freedom
9
Error Mean Square
42.1666 7
Critical Value of t
2.26216
Least Significant Difference
10.387
Means with the same letter are not significantly different. t Grouping
A
Mean N Trt
43.000
4 c
37.250
4 b
30.750
4 a
A B
A
B B
ANOVA for Any Number of Treatments with Unequal Replication Given the following data:
Replicate 1 2 3 4 5 Yi.
Y
2 ij
A 2.0 2.2 1.8 2.3 1.7 10 20.26
Treatment B C 1.7 2.0 1.9 2.4 1.5 2.7 2.5 2.4 5.1 12 8.75
29.06
D 2.1 2.2 2.2 1.9 8.4 Y..=35.5 17.7
Step 1. Write the hypotheses to be tested.
H o : 1 2 3 4 HA: At least one of the means is different from one of the other means. Step 2. Calculate the Correction Factor.
Y..2 35.52 CF 74.132 ri 17
Step 3. Calculate the Total SS TotalSS Yij2 CF (2.0 2 2.2 2 1.8 2 ... 1.9 2 ) CF 75.77 74.132 1.638
Step 4. Calculate the Treatment SS (TRT SS) Y2 TRTSS i. CF ri
10 2 5.12 12 2 8.4 2 74.132 3 5 4 5 75.110 74.132 0.978
Step 5. Calculate the Error SS Error SS = Total SS – Treatment SS = 1.638 – 0.978 = 0.660
Step 6. Complete the ANOVA table Sources of variation Treatment Error Total
Df t-1 = 3 By subtraction = 13 Total number of observations -1 = 16
SS 0.978 0.660 1.638
MS 0.326 0.051
F 6.392**
Step 7. Look up Table F-values. F0.05;3,13 = 3.41 F0.01;3,13 = 5.74 Step 8. Make conclusions.
0
3.41 5.74 6.39
Step 9. Calculate Coefficient of Variation (CV). %CV
s *100 Y
Remember that the Error MS = s2.
%CV
0.051 *100 35.5 17
0.2259 / 2.088 *100 10.82%
Since F-calc (6.392) > 3.41 we reject Ho: at the 95% level of confidence. Since F-calc (6.392) > 5.74 we reject Ho: at the 99% level of confidence.
ANOVA with Sampling (Equal Number of Samples Per Experimental Unit) Linear Model Yijk i ij ijk Where: Yijk is the kth sample of the jth observation of the ith treatment, is the population mean, i is the treatment effect of the ith treatment,
ij is the random error, and
ijk is the sampling error.
ANOVA table SOV Treatment Experimental error Sampling Error Total
Df t-1 (tr-1) - (t-1) (trs-1) - (tr-1) trs-1
F Treatment MS/Experimental Error MS
Facts about ANOVA with Sampling
There are two sources of variation that contribute to the variance appropriate to comparisons among treatment means. 1. Sampling Error = variation among sampling units treated alike ( s2 ). 2. Experimental Error = variation among experimental units treated alike ( s2 r E2 ).
The Experimental Error MS is expected to be larger than the Sampling Error MS.
If the Experimental Error variance component is not important, the Sampling Error MS and the Experimental Error MS will be of the same order of magnitude.
If the Experimental Error variance component is important, the Experimental Error MS will be much larger than the Sampling Error MS.
Example (without using a pooled error mean square) Temperature 12o Pot number 2 3 3.5 4.5 3.5 4.0 3.0 4.0 4.0 5.0 14.0 17.5 49.5
o
8 Pot number Plant 1 2 3 1 1 3.5 2.5 3.0 5.0 2 4.0 4.5 3.0 5.5 3 3.0 5.5 2.5 4.0 4 4.5 5.0 3.0 3.5 Yij. 15.0 17.5 11.5 18.0 Yi.. 44.0 Y…=156.0 Note i = treatment, j = replicate, and k = sample. Step 1. Calculate correction factor: Y...2 156 2 676 rts 3(3)(4)
Step 2. Calculate the Total SS: TotalSS Yijk2 CF
3.5 2 4.0 2 3.0 2 ... 5.5 2 CF 712.5 676.0 36.5
Step 3. Calculate the Treatment SS: TreatmentSS
Yi..2 CF rs
44 2 49.5 2 62.5 2 676.0 3 ( 4 ) ( 3 ( 4 ) 3 ( 4 ) 691.04 676.0 15.042
1 5.0 4.5 5.0 4.5 19.0
16o Pot number 2 5.5 6.0 5.0 5.0 21.5 62.5
3 5.5 4.5 6.5 5.5 22.0
Step 4. Calculate the SS Among Experimental Units Total (SSAEUT)
SSAEUT
Yij2. s
CF
15 2 17.5 2 11.5 2 22.0 2 676.0 ... 4 4 4 4 699.25 676.0 23.25 Step 5. Calculate the Experimental Error SS: Experimental Error SS = SSAEUT – SS TRT = 23.25 – 15.042 = 8.208 Step 6. Calculate the Sampling Error SS: Sampling Error SS = Total SS – SSAEUT = 36.5 – 23.25 = 13.25 Step 7. Complete the ANOVA Table: SOV Treatment Experimental Error Sampling Error Total
Df t-1 = 2 (tr-1) - (t-1) = 6 (trs-1) - (tr-1) = 27 trs-1 = 35
Step 8 Look up Table F-values. F0.05;2,6 = 5.14 F0.01;2,6 = 10.92
SS 15.042 8.208 13.25 36.5
MS 7.521 1.368
F 5.498*
F-value = Trt MS/Expt Error MS
Step 8. Make conclusions.
0
5.14
Since F-calc (5.498) > 5.14 we reject Ho: at the 95% level of confidence. Since F-calc (5.498) < 10.92 we fail to reject Ho: at the 99% level of confidence.
10.92
5.498
OPTIONAL MATERIAL: ANOVA of CRD With Sampling (Use of the Pooled Error MS)
We will use the data from the previous example for performing the F-test on treatments using a Pooled Error Mean Square.
The advantage of using the Pooled Error Mean Square as the denominator of the F-test is that you will have more degrees of freedom associated with denominator.
This may allow you to detect smaller significant differences between treatment means than you may have by using the Experimental Error Mean Square.
Example Steps 1-6 are the same as in the previous example. Step 7. Complete the ANOVA, but don’t calculate the F-value. SOV Treatment Experimental Error Sampling Error Total Pooled Error
Df t-1 = 2 (tr-1) - (t-1) = 6 (trs-1) - (tr-1) = 27 trs-1 = 35 33
SS 15.042 8.208 13.25 36.5 21.458
MS 7.521 1.368 0.491
F
0.65
Step 7.1 Calculate the Sampling Error Mean Square. Step 7.2 Test the homogeneity of the Sampling and Experimental Error Mean Square using the folded F-test. If they are homogeneous a Pooled Error MS can be calculated and used as the denominator of the F-test on treatments.
We need to remember that the expected mean square for the Sampling Error = s2 and the expected mean square for the Experimental Error = s2 r E2 . Thus the Experimental Error MS is expected to be > the Sampling Error MS.
Remember from the t-test that the Folded F-test involves the larger MS/smaller MS. Thus, in our case the Folded F = Experimental Error MS / Sampling Error MS.
If the variances are similar, the component of r E2 in the Experimental Error Mean Square will approach the value zero.
F
1.368 0.491
2.786
Step 7.3 Look up the table F-value This F-test is a one-tail test because there is the expectation that the Experimental Error MS S2 s E2 is going to be larger than the Sampling Error MS S2 .
Thus, if you are testing = 0.01, then you need to use the F-table for α = 0.01.
F 0.01,( ExptErrdf )( SampErrdf) = F 0.01;6, 27 3.558
Step 7.4 Make conclusions: Since the calculated value of F (2.786) is less than the Table-F value (3.558), we fail to reject Ho: Sampling Error MS = Experimental Error MS at the 99% level of confidence.
Therefore, we can calculate a Pooled Error MS
Step 7.5: Calculate the Pooled Error df and the Pooled Error MS Pooled Error df = Sampling Error df + Experimental Error df = (26+7) = 33 Pooled Error MS = Sampling Error SS + Experimental Error SS Sampling Error df + Experimental Error df =
13.25 8.208 0.650 26 7
Step 8. Complete the ANOVA by calculating the F-value using the Pooled Error MS as the denominator of the F-test. SOV Treatment Experimental Error Sampling Error Total Pooled Error
Df t-1 = 2 (tr-1) - (t-1) = 6 (trs-1) - (tr-1) = 27 trs-1 = 35 33
SS 15.042 8.208 13.25 36.5 21.458
MS 7.521 1.368 0.491
F 11.57**
0.65
Step 9 Look up Table F-values. F-value = Trt MS/Pooled Error MS
F0.05;2,33 = 3.30 F0.01;2,33 = 5.37 Step 10. Make conclusions.
0
Since F-calc (11.57) > 3.30 we reject Ho: at the 95% level of confidence. Since F-calc (11.57) > 5.37 we reject Ho: at the 99% level of confidence.
3.30 5.37
11.57
ANOVA When the Number of Subsamples are Not Equal TotalSS Yijk2
Y...2 total # ofobservations
df = #observations – 1
TreatmentSS
Yi..2 Y...2 r j sk total # ofobs.
df = # treatments –1
SSAEUT
Yij2. sk
Y...2 total # ofobs.
df = # Experimental units – 1
SS Experimental Error = SSAEUT – SS TRT
df = SSAEUT df – TRT df
SS Sampling Error = Total SS – SSAEUT
df = Total df – SSAEUT df
CRD with Sampling (Using the Experimental Error and Pooled Error MS as the denominator of the F-tests) options pageno=1; data crdsamp; input trt rep sample kwt; datalines; 8 1 1 3.5 8 1 2 4 8 1 3 3 8 1 4 4.5 8 2 1 2.5 8 2 2 4.5 8 2 3 5.5 8 2 4 5 8 3 1 3 8 3 2 3 8 3 3 2.5 8 3 4 3 12 1 1 5 12 1 2 5.5 12 1 3 4 12 1 4 3.5 12 2 1 3.5 12 2 2 3.5 12 2 3 3 12 2 4 4 12 3 1 4.5 12 3 2 4 12 3 3 4 12 3 4 5 16 1 1 5 16 1 2 4.5 16 1 3 5 16 1 4 4.5 16 2 1 5.5 16 2 2 6 16 2 3 5 16 2 4 5 16 3 1 5.5 16 3 2 4.5 16 3 3 6.5 16 3 4 5.5 ;; ods rtf file='example.rtf'; run; proc anova; class trt rep sample; model kwt=trt rep*trt;
*Comment The rep*trt commmand = the experimental error; test h=trt e=rep*trt; *The previous statement tells SAS to use the Experimental Error MS as the denominator of the F-test; means trt; title 'CRD with Sampling - Using Experimental Error as the Denominator'; run; proc anova; class trt rep sample; model kwt=trt; *The previous statement does not have the rep*trt statement, which is equivalent to the Expt. Error MS; means trt; title 'CRD with Sampling - Using Pooled Error as the Denominator'; run; ods rtf close;
CRD with Sampling - Using Experimental Error as the Denominator The ANOVA Procedure Class Level Information Class
Levels Values
trt
3 8 12 16
rep
3 123
sample
4 1234
Number of Observations Read
36
Number of Observations Used
36
CRD with Sampling - Using Experimental Error as the Denominator The ANOVA Procedure Dependent Variable: kwt Source
DF
Sum of Squares Mean Square F Value
Model
8
23.25000000
2.90625000
Error
27
13.25000000
0.49074074
Corrected Total
35
36.50000000
5.92
Pr > F
0.0002
This MS is the Sampling Error MS
R-Square Coeff Var Root MSE kwt Mean
0.636986 16.16605 0.700529 4.333333
Source DF trt trt*rep
Anova SS Mean Square F Value
2
15.04166667
7.52083333
6
8.20833333
1.36805556
Pr > F
15.33 F
0.0440
Correct result of the Ftest using the Expt. Error MS as the denominator of the Ftest.
CRD with Sampling - Using Experimental Error as the Denominator The ANOVA Procedure Level of trt
kwt N
Mean
Std Dev
8
12
3.66666667
1.00754728
12
12
4.12500000
0.74238559
16
12
5.20833333
0.62005620
CRD with Sampling - Using Pooled Error as the Denominator The ANOVA Procedure Class Level Information Class
Levels Values
trt
3 8 12 16
rep
3 123
sample
4 1234
Number of Observations Read
36
Number of Observations Used
36
CRD with Sampling - Using Pooled Error as the Denominator The ANOVA Procedure Dependent Variable: kwt Source
DF
Sum of Squares Mean Square F Value
Model
2
15.04166667
7.52083333
Error
33
21.45833333
0.65025253
Corrected Total
35
36.50000000
11.57
Pr > F
0.0002
Pooled Error MS R-Square Coeff Var Root MSE kwt Mean
0.412100 18.60882 0.806382 4.333333
Source DF trt
2
Anova SS Mean Square F Value
15.04166667
7.52083333
11.57
Pr > F
0.0002
Correct value for F-test using the Pooled Error MS as the denominator of the F-test.
CRD with Sampling - Using Pooled Error as the Denominator The ANOVA Procedure Level of trt
kwt N
Mean
Std Dev
8
12
3.66666667
1.00754728
12
12
4.12500000
0.74238559
16
12
5.20833333
0.62005620
