The Randomized Block Design(RBD)

 The Randomized Block Design

When introducing ANOVA , we mentioned that this model will allow us to include more than one categorical factor(explanatory) or confounding variables in the model. 

In a first step we will now include a block variable (factor). This is usually considered a variable that is a confounding variable, i.e. not of interest by itself but has an influence on the response variable and should for this reason be included.

Sometimes a study is designed to include such a variable in order to reduce the variability in the response variable and therefore to require a smaller sample size. Generally each treatment is used exactly once within each block, in conclusion:

if we have k treatments and b block, then the total sample size is n = b · k. 

The concept origins from agricultural studies, when studying yields of certain grain, e.g. grain under different conditions.

Example 1

(Yield and Early Growth Responses to Starter Fertilizer in No-Till Corn Assessed with Precision Agriculture Technologies, Manuel Bermudez and Antonio P. Mallarino (2002)) Several trials were conducted in the 1990’s to evaluate corn yield and early growth responses to starter fertilizer in Iowa farmers’ fields that had 8 to 14 yr of no-till management. Soil series represented in the experimental areas varied across fields and were among typical agricultural soil series of Iowa and neighboring states.

To illustrate, assume three different starters were used. In order to limit the number of field that needed to be planted, 10 locations were chosen, where each field was divided into three parts, then each part was treated with a different starter (randomly assigning the treatment to each part). Beside the fertilizer all other agricultural practises were to be the same. 

For this example:

 response variable=yield kg/m² 

treatment(factor) variable= starter (A, B, C)

block(confounding) variable= location (1,2,3,4,5,6,7,8,9,10)

 Example 2

The cutting speeds of four types of tools are being compared in an experiment. Five cutting materials of varying degree of hardness are to be used as experimental blocks. The data giving the measurement

	Block
Treatment	1	2	3	4	5	Treatment Means
1	12	2	8	1	7	x̄T1=6
2	20	14	17	12	17	x̄T2=16
3	13	7	13	8	14	x̄T3=11
4	11	5	10	3	6	x̄T4=7
Block means	x̄B1=14	x̄B2=7	x̄B3=12	x̄B4=6	x̄B5=11	x̄=10

 The Randomized Block Design Model

x_ij= the measurement for treatment i in block j (remember there is precisely one such measurement) then
x_ij = µ + α_i + β_j + e_ij
where: µ = overall mean
α_i = effect of treatment i (difference with µ)
β_j  = effect of block j (difference with µ)
 e_ij = error in measurement for treatment i and block j.

A positive value for αi indicates that the mean of the response variable is greater than the overall mean for treatment i.

Assume  e_ij ∼ N (0, σ) for all measurements.

In the analysis of variance instead of only explaining the variance through error and treatment, we also include the block as a possible source for variance in the data. For that reason we now also include

Sum of Squares for Block (SSB) with our analysis:

• Sum of Squares for treatment: 

\[SST=\sum\limits_{i=1}^{k}{b(\overline{{{x}_{Ti}}}}-\overline{x}{{)}^{2}},d{{f}_{T}}=k-1\]

• Sum of Squares for block:

\[SSB=\sum\limits_{j=1}^{b}{k(\overline{{{x}_{Bj}}}}-\overline{x}{{)}^{2}},d{{f}_{B}}=b-1\]

• Total Sum of Squares:

\[TotalSS=\sum\limits_{i,j}^{{}}{k({{x}_{ij}}}-\bar{x}{{)}^{2}},d{{f}_{Total}}=n-1\]

• Sum of Squares for error: 

SSE = TotalSS − SST − SSB ,  df_E = n = b − k + 1

ANOVA Table for a Randomized Block Design
Source	df	SS	MS	F
Treatments	k − 1	SST	MST = SST/(k − 1)	MST/MSE
Blocks	b − 1	SSB	MSB = SSB/(b − 1)	MSB/MSE
Error	n − k − b + 1	SSE	MSE = SSE/(n − k − b + 1)
Total	n − 1	TotalSS

 Example 3

Let us find the ANOVA table for the cutting example: 

•  Sum of Squares for treatment:

  \[SST=\sum\limits_{i=1}^{k}{b{{({{{\bar{x}}}_{Ti}}-\bar{x})}^{2}}}\]= 5(6−10)²+5(16−10)²+5(11−10)²+5(7−10)² = 310,df_T = k−1 = 3 

• Sum of Squares for block:

  \[SSB=\sum\limits_{j=1}^{b}{k{{({{{\bar{x}}}_{Bj}}-\bar{x})}^{2}}}\] = 4(14 − 10)² + . . . + 4(11 − 10)² = 184, df_B = b − 1 = 4

• Total Sum of Squares:

  \[TotalSS=\sum\limits_{i,j}^{{}}{k{{({{{\bar{x}}}_{ij}}-\bar{x})}^{2}}}\] = (12−10)²+(2−10)²+. . .+(6−10)² = 518, df_Total = n−1 = 5(4)−1 = 19 

• Sum of Squares for error:

  SSE = TotalSS − SST − SSB = 518 − 310 − 184 = 24, df_E = n − b − k + 1 = 12 

  
  

  ANOVA Table for a Randomized Block Design
  Source	df	SS	MS	F
  Tool	3	310	MST = 103.3	51.7
  Meterial	4	184	MSB = 46	23
  Error	12	24	MSE = 2
  Total	19	TotalSS=518

  
  

Based on the statistics in the ANOVA table, we can now test for a treatment effect and for a block effect.

The ANOVA F-Test(Randomized Block Design) 

1. The Hypotheses are

   H₀ : α₁ = α₂ = . . . = α_k = 0 versus

   H_a : at least one of the values differs from the others 

2.  Assumption: The population follows a normal distribution with means µ₁, µ₂, . . . , µ_k and equal variance σ² for all bk combinations of treatments and blocks. The samples are independent random samples in b independent blocks from each population. 
3. Test statistic:

   F₀ =MST/MSE

   based on df₁ = (k − 1) and df₂ = (n − k − b + 1)
4. P-value: P(F > F₀ ), where F follows an F-distribution with df₁ = (k − 1) and df₂ = (n − k − b + 1). 
5.  Decision:

   If P-value ≤ α, then reject H₀

   If P-value> α, then do not reject H₀

7. Put into context. In order to test for a difference in the blocks,  
   1. The Hypotheses are

      H₀ : β₁ = β₂ = . . . = β_b = 0 versus

      H_a : at least one of the values differs from the others

   2. Assumption: same as above
   3. Test statistic:

      F₀ =MSB/MSE

      based on df₁ = (b − 1) and df₂ = (n − k − b + 1). 

   4. P-value: P(F > F₀), where F follows an F-distribution with df₁ = (b − 1) and df₂ = (n − k − b + 1).
   5. Decision: same as above 
   6. Put into context.

Once we determined with the F test that a difference between the treatment means exist, we will use a multiple comparison analysis, to determine where the differences occur. Use Bonferroni The CIs for the pairwise comparisons are the same like for the one-way ANOVA, only change are the degrees of freedom 

Let γ = C₁µ_T1+C₂µ_T2+. . .+C_kµ_Tk a contrast with sample contrast c = C₁x_T1+C₂x_T2+. . .+C_k ̄x_Tk (1 − α)100% Confidence Interval for γ 

c ± t^df_1−α/2SE(c)

 t^df_1−α/2 is the 1 − α/2 percentile of the t-distribution with df = n − k − b + 1.

Continue Example

We test that the mean cutting speed is independent from the tool used, that is

1. The Hypotheses are

   H₀ : α₁ = α₂ = α₃ = α4 = 0 versus

   H_a : at least one of the values differs from the others α = 0.05

2. Assumption: The cutting speed follows a normal distribution and equal variance σ² for all bk combinations of treatments and blocks.
   It seems to be reasonable that the cutting speed follows a normal distribution under for a given tool cutting a certain material.
   It also seems to be reasonable to assume that the standard deviations are the same, the variation will be caused by the same factors.
   The samples were independent cuts, so the samples are independent.
3. Test statistic:

   F₀ =MST/MSE = 51.7

   based on df₁ = 3 and df₂ = 20 − 4 − 5 + 1 = 12.
   
4. P-value: P(F > F₀) < 0.005, using table IX (since 51.7>7.23).
5. Decision: Since P-value< α = 0.05 reject H₀.
6. The mean cutting speeds are not the same for the four tools

Use Bonferroni, to determine where the differences are. m = k(k − 1)/2 = 6 use α ∗ = α/m = 0.05/6 = 0.0083 To find the critical value, we need α ∗/2 = 0.0041 ≈ 0.005(closest in the table), for df = 12, we get a critical value of t¹²_0.005 = 3.055. Since the sample size is the same for all treatments the Margin of Error will be the same for all pairwise comparisons
ME = t¹²_0.005 * √(MSE * √(1/b + 1//b) = 3.055 * √2 * 2/5 = 2.73
Now rank the sample means and underline those, which differ less than 2.73
     
    means         6   7        11   16
    tools          T1 T4     T2   T3

With an experiment wise error rate of 0.05 the data provide sufficient evidence that the mean cutting speed for tools 1 and 4 are significantly different from the mean cutting speed for tool 2 as well as tool 3, further the mean cutting speed for tools 2 and 3 are significantly different. Mean cutting speeds for tools 1 and 4 are not significantly different.
Also, the mean cutting speeds for tools 1 and 4 are significantly shorter than for the two other tools.
 A test for a difference in the means of the blocks is of no interest, because the materials tested are not relevant, they were only helpful for testing the tools for a variety of materials.