Example of Completely Randomized Design

Examples of Applying a Completely Randomized Design:

Case example 1 : Completely Randomized Design with Same Test

In this case example, the same case example is reused as the example on the decomposition of total variance. It's just that, using a slightly different calculation step. (The following are the results of estrogen testing of some solutions that have undergone certain treatments. The weight of uterine in rats is used as a measure of estrogen activity. The weight of uterine in milligrams of four mice for each control and six different solutions are listed in the following table)

Table 4.    Uterine Weight Data (mg) of 7 Treatments Against Four Mice

Analysis of variance

Hypothesis Testing Steps:

1. Since there are only 7 treatments available, the suitable model is a fixed model. Such models are:
   Y_ij = μ + τ_i + ε_ij ;  i =1, 2, ..., 7 and j = 1,2,3,4

with:

Y_ij    = uterine weight of the j-th rat that obtained the i-th treatment
μ     = uterine heavy population mean
τ_i     = i-th treatment effect
ε_ij     = random effect on j-th mice obtaining i-th treatment.

2. Assumptions: see assumptions for fixed models
3. Hypothesis to be tested:
   • H₀ : All τ_j = 0 (or no effect of treatment on uterine weight of rats)
   • H₁ : Not all τ_j = 0; or there is at least one treatment that affects the uterine weight of rats.

Analysis of variance calculation steps:

Step 1: Calculate the Correction Factor

 $$\begin{matrix}CF=\frac{Y..^2}{rt}=\frac{2249^2}{28}=180642.89\\\\\end{matrix}$$

Step 2: Calculate the Sum of The Total Squares

 $$\begin{matrix}SSTOT=\sum_{i=1}^{t}\sum_{j=1}^{r}{Y_{ij}}^2-CF\\=(89.8^2+93.8^2+....+65.6^2+70.2^2)-180642.89\ \\=5478.51\\\end{matrix}$$

Step 3: Calculate the Sum of Squares of Treatment

 $$\begin{matrix}SST=\sum_{i=1}^{t}\frac{{Y_{i.}}^2}{r}-CF\\=\frac{(384.6^2+353^2+301.6^2+\ 273.8^2+339.6^2+315.6^2+280.8^2)}{4}-180642.89\ \\=2415.94\\\end{matrix}$$

Step 4: Calculate the Sum of Squares of Errors

 $$ SSE=SSTOT-\ SST=3062.57$$

Step 5: Create an Analysis of variance Table with its F-Values table

Table 5.    Analysis of variance of Uterine Weight of Rats

F_(0.05,6,21) = 2,573

F_(0.01,6,21) = 3,812

Step 6: Make a Conclusion

Since Fstat (2.76) > 2,573 then we reject H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅ = μ₆ at a confidence level of 95%

Since Fstat (2.76) ≤ 3,812 then we fail to reject H₀:μ₁ = μ₂ = μ₃ = μ₄ = μ₅ = μ₆ at a confidence level of 99%

This means that at the level of 95% confidence, there is at least one treatment that is different from the others. But at the level of 99% confidence, all the average treatment is no different from others.

Information:

Usually, a one-asterisk (*) is given, when the  F-stat value is greater than F(0.05) and a two-asterisk (**) is given when the F-stat value is greater than F(0.01)

Step 7: Calculate the Coefficient of Variance (CV)

$$\begin{matrix}CV=\frac{\sqrt{MSE}}{\bar{Y}..}\times100\%\\ =\frac{\sqrt{145.84}}{80.32}\times100\%\\ =15.03\%\\ \end{matrix}$$

1. From the table above we can guess some parameters of the experiment:
   • E(MSE) = σ² is suspected with MSE = 145.84
   • E(MST) =  $\sigma^2+\left [\frac{r}{(t-1)}  \right ]\sum_{i=1}^{t}\tau_i^2$  allegedly with MST =  402.66
2. So that if F_stat greater than 1 then the conclusion will be more inclined to reject the null hypothesis and vice versa.
   • Estimators of variance of treatment  $\left [\frac{r}{(t-1)}  \right ]\sum_{i=1}^{t}\tau_i^2$ effects are suspected through  $\frac{E(MST)-\sigma^2}{r}=\frac{402.66-145.84}{4}=64.20$

Average Comparison (by using the LSD Test)

In this example, testing the difference in average pairs between treatments was carried out using one of the post hoc tests, namely LSD. In this case, it is not actually appropriate to use LSD as a further testing procedure, why? (See discussion on treatment mean difference testing)

Table 6.    Table of Average Uterine Weight of Rats

1. Calculate LSD Value_0.05
   • $\begin{matrix}LSD=t_{0.05/2;21}\sqrt{\frac{2MSE}{r}}\\=2.08\times\sqrt{\frac{2(145.84)}{4}}\\=17.76\ \ \\\end{matrix}$
2. Sort Average Treatment (in this example the average is sorted from small to large)
3. Test criteria:
   • Compare the absolute value of the difference between the two averages that we will see the difference with the LSD value with the following test criteria:
   • $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>LSD_{0.05}\ reject\ H_0,\ two\ means\ are\ significantly\ different\\\le LSD_{0.05}\ accept\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$
4. The results of the test of the difference in average pairs (pair wise comparisons) at a significant level of 5%

Calculation with SmartstatXL Excel Add-In

Case example 2 : Completely Randomized Design with Unequal Test

In a biological experiment 4 concentrations of chemicals were used to stimulate the growth of a certain type of plant over a period of time. The following growth data, in centimeters, are recorded from living plants.

Table 7.    Plant growth data (cm)

Analysis of Variance

The hypothesis testing steps for the above case are as follows:

1. The model for the above case is:
   Y_ij = μ + τ_i + ε_ij
   i =1,2,3,4 and j = 1, 2, ..., ri_; where r_i is the multiplicity of replications for the i-th treatment
   with
   Y_ij    = jth plant growth (cm) that obtained the i-th treatment
   μ     = population mean
   τ_i     = i-th treatment effect
   ε_ij       = random effect on the j-th plant that obtained the i-th treatment.
2. Assumptions: see assumptions for fixed models
3. Hypothesis to be tested :     
   • H₀ : All τ_j = 0 or no effect of treatment on plant growth.
   • H₁ : Not all τ_j = 0;  or there is at least one treatment that affects plant growth.

Analysis of variance calculation steps:

Step 1: Calculate the Correction Factor

 $$ CF=\frac{Y..^2}{\sum_{i=1}^{t}r_i}=\frac{181.6^2}{24}=1374.11$$

Step 2: Calculate the Sum of The Total Squares

 $$\begin{matrix}SSTOT=\sum_{i=1}^{t}\sum_{j=1}^{r_i}{Y_{ij}}^2-CF\\=8.2^2+8.8^2+...+7.0^2+6.5^2-1374.11\\=24.673\\\end{matrix}$$

Step 3: Calculate the Sum of Squares of Treatment

 $$\begin{matrix}SST=\sum_{i=1}^{t}\frac{{Y_{i.}}^2}{r_i}-CF\\=\frac{44.8^2}{5}+\frac{49.2^2}{6}+\frac{46.9^2}{7}+\frac{40.7^2}{6}-1374.11\\=21.053\\\\\end{matrix}$$

Step 4: Calculate the Sum of Squares of Errors

 $$\begin{matrix}SSE=SSTOT-SST=24.673-21.053\\=3.620\\\end{matrix}$$

Step 5: Create an Anova Table

Table 8.    Anova Table of Plant Growth

F_{(0.05,3,2 0)} = 3,098

F_{(0.01,3,2 0)} = 4,938

Step 6: Make a Conclusion

Since Fstat (38,768) > 3,098 then we reject H₀: μ₁ = μ₂ = μ₃ = μ₄ at a confidence level of 95%

Since Fstat (38,768) > 4,938 then we reject H₀: μ₁ = μ₂ = μ₃ = μ₄ at a confidence level of 99%

This means that at the level of 99% confidence, there is at least one treatment that is different from the others.

Information:

Usually, a one-asterisk (*) is given, when the  F-stat value is greater than F(0.05) and a two-asterisk (**) is given when the F-stat value is greater than F(0.01)

Step 7: Calculate the Coefficient of Variance (CV)

$$\begin{matrix} CV=\frac{\sqrt{MSE}}{\bar{Y}..}\times100\% =\frac{\sqrt{0.181}}{7.567}\times100\%\\ =5.62\%\\ \end{matrix}$$

Average Comparison (by using the LSD Test)

In this example, testing the difference in average pairs between treatments was carried out using one of the post-hoc tests, namely LSD. In this case, it is not actually appropriate to use LSD as a further testing procedure, why? (See discussion on treatment mean difference testing)

Table 9.    Anova Table of Plant Growth

Table 10.  Table of Average Plant Growth

1. Calculate LSD Value (α = 0.05)
   • $\begin{matrix}LSD=t_{0.05/2;20}\sqrt{MSE(\frac{1}{r_i}+\frac{1}{r_j})}\\LSD\ \ 1:\ \ (k_1\ \ vs\ \ k_2\ or \ k_4)=2.09\times\sqrt{0.181(\frac{1}{5}+\frac{1}{6})}\\=0.538\\LSD\ \ 2:\ \ (k_1\ \ vs\ \ k_3)=2.09\times\sqrt{0.181(\frac{1}{5}+\frac{1}{7})}\\=0.521\\LSD\ \ 3:\ \ (k_2\ or\ k_4\ \ vs\ \ k_3)=2.09\times\sqrt{0.181(\frac{1}{6}+\frac{1}{7})}\\=0.495\\LSD\ \ 4:\ \ (k_2\ vs\ k_4)=2.09\times\sqrt{0.181(\frac{1}{6}+\frac{1}{6})}\\=0.513\\\end{matrix}$
2. Sort Average Treatment (in this example the average is sorted from small to large)
3. Test criteria:
   • Compare the absolute value of the difference between the two averages that we will see the difference with the LSD value with the following test criteria:
   • $ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>LSD_{0.05}\ reject\ H_0,\ two\ means\ are\ significantly\ different\\\le LSD_{0.05}\ accept\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$
4. The results of the test of the difference in average pairs (pair wise comparisons) at a significant level of 5%