COMPLETELY RANDOM DESIGN (CRD) -Design can be used when experimental units are essentially homogeneous.

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1 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? D 1 11 C D 1 B B 3 13 A C 4 14 B D 5 15 C C 6 16 B A 7 17 C A 8 18 D B 9 19 A D 10 0 A -Note that there is no blocking of experimental units into replicates. -Every experimental unit has the same probability of receiving any treatment.

2 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).. 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.. 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, y y y y i. i..... = = n j= 1 = y = y i... a y ij Treatment total, where n n = Treatment mean i= 1 = n j= 1 N = y ij = Experiment total, where a Experiment mean, where N = number of observations in a treatment = = number of treatments total number of observations in the experiment. Linear Additive Model for the CRD Y = µ + τ + ε ij i ij where: Y ij is the j th observation of the i th treatment, µ is the population mean,

3 τ i is the treatment effect of the i th treatment, and ε is the random error. ij -Using this model we can estimate τ i or ε ij for any observation if we are given Y ij and µ. Example Treatment 1 Treatment Treatment Y i Y.. = 7 Y i. Y.. = 8 Y i. Y We can now write the linear model for each observation ( Y ). -Write in µ for each observation. ij Treatment 1 Treatment Treatment 3 4 = 8 9 = 8 8 = 8 5 = 8 10 = 8 11 = 8 6 = 8 11 = 8 8 = 8 Y i Y.. = 7 Y i. Y.. = 8 Y i. Y Write in the respective τ i for each observation where τ = Y i. Y.. i Treatment 1 Treatment Treatment 3 4 = = = = = = = = = Y i Y.. = 7 Y i. Y.. = 8 Y i. Y

4 -Write in theε ij for each observation. Treatment 1 Treatment Treatment 3 4 = = = = = = = = = Y i Y.. = 7 Y i. Y.. = 8 Y i. Y Note for each treatment ε ij = 0. -If you are asked to solve for τ 3, what is the answer? -If you are asked to solve for ε 3, what is the answer? -Question: If you are given just the treatment totals ( Y i. s), how would you fill in the values for each of the observations such that the Error SS = 0. Example 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. Given the following information, fill in the values for all Error SS = 0. Treatment 1 Treatment Treatment 3 Y ij s such that the Experimental Y i Y.. = 7 Y i. Y.. = 8

5 Answer Treatment 1 Treatment Treatment Y i Y.. = 7 Y i. Y.. = 8 Note in the previous two examples that τ i = 0. This is true for all situations. Given H : µ = µ =... = µ H t 0 A i= 1 t 1 : µ µ for at least one pair of treatments (i,i') µ i i i' = µ (i.e., t the sum of the treatment means divided by the number of equals the experiment mean). This definition implies that τ i = 0 = ( Y i. Y.. ). t i= 1 treatments The hypothesis written above can be rewritten in terms of the treatment effects τ i as: H 0 : 1 τ = τ =... = τ = 0 a H A : τ 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, τ i, are zero. ANOVA for Any Number of Treatments with Equal Replication Given the following data: Treatment Replicate A B C Y i Y... =444 Y ij 3,875 5,805 7,48

6 Step 1. Write the hypotheses to be tested. H H or 1 1 : µ = µ = µ : µ = µ µ µ µ = µ or o A µ µ µ 3 H o : All three means are equal. H A : At least one of the means is different from the other means. Step. Calculate the Correction Factor. CF Y = rt = 4*3 = 16,48.0 Step 3. Calculate the Total SS TotalSS = Y ij CF = ( ) CF = 17,108 16,48 = Step 4. Calculate the Treatment SS (TRT SS) Yi. TRTSS = CF r 13 = ,48 = = 300.5

7 Step 5. Calculate the Error SS Error SS = Total SS Treatment SS = = Step 6. Complete the ANOVA table Sources of variation Df SS MS F Treatment t-1 = NS Error t(r-1) = Total rt-1 = Step 7. Look up Table F-values. F 0.05;,9 = 4.6 F 0.01;,9 = 8.0 Step 8. Make conclusions Since F-calc (3.563) < 4.6 we fail to reject Ho: at the 95% level of confidence. -Since F-calc (3.563) < 8.0 we fail to reject Ho: at the 99% level of confidence.

8 Step 9. Calculate Coefficient of Variation (CV). s % CV = *100 Y Remember that the Error MS = s. % CV = * *3 = ( 6.494/ 37) = 17.6% *100 ANOVA for Any Number of Treatments with Unequal Replication Given the following data: Treatment Replicate A B C D Y i Y.. =35.5 Y ij Step 1. Write the hypotheses to be tested. H o : µ 1 = µ = µ 3 = µ 4 H A : At least one of the means is different from one of the other means. Step. Calculate the Correction Factor. Y 35.5 CF = = r i 17.. = 74.13

9 Step 3. Calculate the Total SS TotalSS = Y ij CF = ( ) CF = = Step 4. Calculate the Treatment SS (TRT SS) Yi. TRTSS = CF r i 10 = = = Step 5. Calculate the Error SS Error SS = Total SS Treatment SS = = Step 6. Complete the ANOVA table Sources of variation Df SS MS F Treatment t-1 = ** Error By subtraction = Total Total number of observations -1 =

10 Step 7. Look up Table F-values. F 0.05;3,13 = 3.41 F 0.01;3,13 = 5.74 Step 8. Make conclusions. Since F-calc (6.39) > 3.41 we reject Ho: at the 95% level of confidence. Since F-calc (6.39) > 5.74 we reject Ho: at the 99% level of confidence Step 9. Calculate Coefficient of Variation (CV). s % CV = *100 Y Remember that the Error MS = s. % CV = * = ( 0.59 /.088) = 10.8% *100

11 ANOVA with Sampling (Equal Number of Samples Per Experimental Unit) Linear Model Yijk = µ + τ + ε + δ i ij ijk Where: Y ijk is the k th sample of the j th observation of the i th treatment, µ is the population mean, τ is the treatment effect of the i th treatment, i ε is the random error, and ij δ ijk is the sampling error. ANOVA table SOV Df F Treatment t-1 Treatment MS/Experimental Error MS Experimental error (tr-1) - (t-1) Sampling Error (trs-1) - (tr-1) Total trs-1 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 ( σ ).. Experimental Error = variation among experimental units treated alike ( σ + ). The Experimental Error MS is expected to be larger than the Sampling Error MS. s s rσ E 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.

12 Example (without using a pooled error mean square) Temperature 8 o 1 o 16 o Pot number Pot number Pot number Plant Y ij Y i Y =156.0 Note i = treatment, j = replicate, and k = sample. Step 1. Calculate correction factor: Y... rts = 156 3(3)(4) = 676 Step. Calculate the Total SS: TotalSS = Y ijk CF = ( ) = CF = 36.5 Step 3. Calculate the Treatment SS: Yi.. TreatmentSS = CF rs = + 3(4) (3(4) (4) = = 15.04

13 Step 4. Calculate the SS Among Experimental Units Total (SSAEUT) Yij. SSAEUT = CF s 15 = = = 3.5 Step 5. Calculate the Experimental Error SS: Experimental Error SS = SSAEUT SS TRT = = 8.08 Step 6. Calculate the Sampling Error SS: Sampling Error SS = Total SS SSAEUT = = 13.5 Step 7. Complete the ANOVA Table: SOV Df SS MS F Treatment t-1 = * Experimental Error (tr-1) - (t-1) = Sampling Error (trs-1) - (tr-1) = Total trs-1 = Step 8 Look up Table F-values. F-value = Trt MS/Expt Error MS F 0.05;,6 = 5.14 F 0.01;,6 = 10.9

14 Step 8. Make conclusions. Since F-calc (5.498) > 5.14 we reject Ho: at the 95% level of confidence. Since F-calc (5.498) < 10.9 we fail to reject Ho: at the 99% level of confidence ANOVA When the Number of Subsamples are Not Equal Y... TotalSS = Yijk df = #observations 1 total# ofobservations Yi.. Y... TreatmentSS = r s total# ofobs. j k df = # treatments 1 SSAEUT Yij. Y... = s total# ofobs. k df = # Experimental units 1 SS Experimental Error = SSAEUT SS TRT SS Sampling Error = Total SS SSAEUT df = SSAEUT df TRT df df = Total df SSAEUT Assumptions Underlying ANOVA Experimental errors are random, independently, and normally distributed about a mean of zero and with a common variance (i.e. treatment variances are homogenous). The above assumption can be express as NID(0, σ ). Departure from this assumption can affect both the level of significance and the sensitivity of F- or t-tests to real departures from H o :

15 This results in the rejection of Ho when it is true (i.e. a Type I Error) more often than α calls for. The power of the test also is reduced if the assumption of NID(0, σ ) is violated. Violation of the assumption NID(0, σ ) with the fixed model is usually of little consequence because ANOVA is a very robust technique. Violation of the basic assumptions of ANOVA can be investigated by observing plots of the residuals. Residuals will be discussed in more detail when Transformations are discussed later in the semester.

16 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 Df SS MS F Treatment t-1 = Experimental Error (tr-1) - (t-1) = Sampling Error (trs-1) - (tr-1) = Total trs-1 = Pooled Error Step 7.1 Calculate the Sampling Error Mean Square. Step 7. 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 = and the expected mean square for the Experimental Error = σ s + rσ E. 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 Square will approach the value zero. σ s rσ in the Experimental Error Mean E

17 F = =.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 σ. Error MS ( σ + ) is going to be larger than the Sampling Error MS ( ) S sσ E Thus, if you are testing α = 0.01, then you need to use the F-table for α = S = F ExptErrdf 0.01;6,7 = F 0.01,( )( SampErrdf ) Step 7.4 Make conclusions: Since the calculated value of F (.786) is less than the Table-F value (3.558), we fail to reject H o : 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 = (6+7) = 33 Pooled Error MS = Sampling Error SS + Experimental Error SS Sampling Error df + Experimental Error df = = Step 8. Complete the ANOVA by calculating the F-value using the Pooled Error MS as the denominator of the F-test. SOV Df SS MS F Treatment t-1 = ** Experimental Error (tr-1) - (t-1) = Sampling Error (trs-1) - (tr-1) = Total trs-1 = Pooled Error F-value = Trt MS/Pooled Error MS

18 Step 9 Look up Table F-values. F 0.05;,33 = 3.30 F 0.01;,33 = 5.37 Step 10. Make conclusions. Since F-calc (11.57) > 3.30 we reject H o : at the 95% level of confidence. Since F-calc (11.57) > 5.37 we reject H o : at the 99% level of confidence

19 SAS commands for a CRD with no sampling; options pageno=1; data crd; input Trt $ yield; datalines; a 3 a 36 a 31 a 33 b 4 b 6 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;

20 ANOVA for a CRD with no Sampling The ANOVA Procedure Class Level Information Class Levels Values Trt 3 a b c Number of Observations Read 1 Number of Observations Used 1

21 ANOVA for a CRD with no Sampling The ANOVA Procedure Dependent Variable: yield Source DF Sum of Squares Mean Square F Value Pr > F Model Error Corrected Total R-Square Coeff Var Root MSE yield Mean Source DF Anova SS Mean Square F Value Pr > F Trt

22 ANOVA for a CRD with no Sampling The ANOVA Procedure NoteThis test controls the Type I comparisonwise error rate, not the : experimentwise error rate. Alpha 0.05 Error Degrees of Freedom 9 Error Mean Square Critical Value of t.616 Least Significant Difference Means with the same letter are not significantly different. t Grouping Mean N Trt A c A B A b B B a

23 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; ;; 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/lsd e=rep*trt; title 'CRD with Sampling - Using Experimental Error as the Denominator';

24 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/lsd; title 'CRD with Sampling - Using Pooled Error as the Denominator'; run; ods rtf close;

25 CRD with Sampling - Using Experimental Error as the Denominator The ANOVA Procedure Class Level Information Class Levels Values trt rep sample Number of Observations Read 36 Number of Observations Used 36

26 CRD with Sampling - Using Experimental Error as the Denominator Dependent Variable: kwt The ANOVA Procedure Source DF Sum of Squares Mean Square F Value Pr > F Model Error Corrected Total This MS is the Sampling Error MS R-Square Coeff Var Root MSE kwt Mean Source DF Anova SS Mean Square F Value Pr > F trt <.0001 trt*rep This MS is the Experimental Error MS Tests of Hypotheses Using the Anova MS for trt*rep as an Error Term Source DF Anova SS Mean Square F Value Pr > F trt Correct result of the F- test using the Expt. Error MS as the denominator of the F- test.

27 CRD with Sampling - Using Experimental Error as the Denominator The ANOVA Procedure t Tests (LSD) for kwt Note: This test controls the Type I comparisonwise error rate, not the experimentwise error rate. Alpha 0.05 Error Degrees of Freedom 6 Error Mean Square Critical Value of t Least Significant Difference Means with the same letter are not significantly different. t Grouping Mean N trt A A B A B B

28 CRD with Sampling - Using Pooled Error as the Denominator The ANOVA Procedure Class Level Information Class Levels Values trt rep sample Number of Observations Read 36 Number of Observations Used 36

29 CRD with Sampling - Using Pooled Error as the Denominator Dependent Variable: kwt The ANOVA Procedure Source DF Sum of Squares Mean Square F Value Pr > F Model Error Corrected Total R-Square Coeff Var Root MSE kwt Mean Pooled Error MS Source DF Anova SS Mean Square F Value Pr > F trt Correct value for F-test using the Pooled Error MS as the denominator of the F-test.

30 CRD with Sampling - Using Pooled Error as the Denominator The ANOVA Procedure t Tests (LSD) for kwt Note: This test controls the Type I comparisonwise error rate, not the experimentwise error rate. Alpha 0.05 Error Degrees of Freedom 33 Error Mean Square Critical Value of t.0345 Least Significant Difference Means with the same letter are not significantly different. t Grouping Mean N trt A B B B