As mentioned at the end of the previous section, you shall have a clearer understanding of the randomized block design with its application in the illustration below.
Nonso requests Femi to carry assess the efficiency of three types of brushes in cleaning his whole house. The following values which refer to efficiency rate were obtained from Femi's study afterward.
| Brush 1 | Brush 2 | Brush 3 |
Sitting room | \(65\) | \(63\) | \(71\) |
Bedroom | \(67\) | \(66\) | \(72\) |
Kitchen | \(68\) | \(70\) | \(75\) |
Bathroom | \(62\) | \(57\) | \(69\) |
Table 1. Example of Randomized block design.
Would Femi's conclusion indicate variability in the efficiency between the brushes?
Solution:
Note that Femi had carried out blocking by grouping his assessment of the whole house into four such as bedroom, kitchen, sitting room, and bathroom.
First step: Make your hypotheses.
\[ \begin{align} &H_0: \; \text{There is no variability in the efficiency of the brushes.} \\ &H_a: \; \text{There is variability in the efficiency of the brushes.} \end{align} \]
Don't forget that \(H_0\) implies the null hypothesis, and \(H_a\) implies the alternate hypothesis.
Second step: Find the means for the treatments (columns), blocks (row), and the grand mean.
The mean of Treatment 1 is:
\[\bar{y}_{.1}=\frac{262}{4}=65.5\]
The mean of Treatment 2 is:
\[\bar{y}_{.2}=\frac{256}{4}=64\]
The mean of Treatment 3 is:
\[\bar{y}_{.3}=\frac{287}{4}=71.75\]
The mean of Block 1 is:
\[\bar{y}_{1.}=\frac{199}{3}=66.33\]
The mean of Block 2 is:
\[\bar{y}_{2.}=\frac{205}{3}=68.33\]
The mean of Block 3 is:
\[\bar{y}_{3.}=\frac{213}{3}=71\]
The mean of Block 4 is:
\[\bar{y}_{4.}=\frac{188}{3}=62.67\]
The grand mean is:
\[\mu=\frac{805}{12}=67.08\]
Update your table as follows:
| Brush 1(Treatment 1) | Brush 2(Treatment 2) | Brush 3(Treatment 3) | Block total(row summation)& mean |
Sitting room(1st block) | \(65\) | \(63\) | \(71\) | \(199\) | \(63.3\) |
Bedroom(2nd block) | \(67\) | \(66\) | \(72\) | \(205\) | \(68.3\) |
Kitchen(3rd block) | \(68\) | \(70\) | \(75\) | \(213\) | \(71\) |
Bathroom(4th block) | \(62\) | \(57\) | \(69\) | \(188\) | \(62.67\) |
Treatment total(Columnsummation) | \(262\) | \(256\) | \(287\) | \(805\) | \(67.08\) |
Mean ofTreatment | \(65.5\) | \(64\) | \(71.75\) | |
Table 2. Example of Randomized block design.
Third step: Find the sum of squares for total, treatment, blocking, and error.
The total sum of squares, \(SS_T\), is:
Recall that
\[SS_T=\sum_{i=1}^{\alpha} \sum_{j=1}^{\beta}(y_{ij}-\mu)^2\]
\[\begin{align} SS_T& =(65-67.08)^2+(63-67.08)^2 \\ & \quad + \dots+(57-67.08)^2+(69-67.08)^2 \\ &=264.96 \end{align}\]
The sum of squares from treatments, \(SS_t\), is:
Recall that:
\[SS_t=\beta \sum_{j=1}^{\alpha}(\bar{y}_{.j}-\mu)^2\]
and \(beta\) is \(3\).
\[\begin{align} SS_t &=3((65.5-67.08)^2+(64-67.08)^2+(71.75-67.08)^2)\\ &=101.37 \end{align}\]
The sum of squares from blocking, \(SS_b\), is:
Recall that:
\[SS_b=\alpha \sum_{i=1}^{\beta}(\bar{y}_{i.}-\mu)^2\]
and \(\alpha\) is \(4\)
\[\begin{align} SS_b &=4((66.33-67.08)^2+(68.33-67.08)^2+(71-67.08)^2+(62.67-67.08)^2)\\ &=147.76 \end{align}\]
Therefore, you can find the sum of squares of error:
Recall that:
\[SS_e=SS_T-SS_t-SS_b\]
\[\begin{align} SS_e&=264.96-101.37-147.76 \\ &=15.83 \end{align}\]
Fourth step: Find the mean square values for treatment and error.
The mean square value for treatment, \(M_t\), is:
Recall that:
\[M_t=\frac{SS_t}{\alpha -1}\]
\[M_t=\frac{101.37}{4-1}=33.79\]
Recall that \(\alpha\) is the number of blocks which is \(4\) in this case.
The mean square value for error, \(M_e\), is:
Recall that:
[M_e=\frac{SS_e}{(\alpha -1)(\beta -1)}\]
\[M_e=\frac{15.83}{(4-1)(3-1)}=2.64\]
Fifth strep: Find the value of test static.
The test static value, \(F\), is:
Recall that:
\[F=\frac{M_t}{M_e}\]
\[F=\frac{33.79}{2.64} \approx 12.8\]
Sixth Step: Use statistical tables to determine the conclusion.
Here, you have to take some care. You need your numerator degrees of freedom, \(df_n\), and your denominator degrees of freedom \(df_d\).
Note that:
\[df_n=\alpha -1\]
and
\[df_d=(\alpha-1)(\beta-1)\]
Hence,
\[df_n=4-1=3\]
and
\[df_d=(4-1)(3-1)=6\]
You could use a level of significance \(a=0.05\) to carry out your hypothesis test. Find the \(P\)-value at this significant level (\(a=0.05\)) with a \(df_n\) of \(3\) and \(df_d\) of \(6\) which is \(4.76\). It appears that the solved \(F\) value falls very close to a significant level of \(a=0.005\) which has a \(P\)-value of \(12.9\).
You must be able to refer to the table on "Percentiles of F Distribution" to conduct your analysis or use some other statistical software to determine the exact \(P\)-value.
Final step: Communicate your finding.
The \(F\)-value determined from the experiment, \(12.8\) is found in between \(F_{0.01}=9.78\) and \(F_{0.005}=12.9\), and by using statistical software the exact \(P\)-value is \(0.00512\). Since the experiment \(P\)-value (\(0.00512\)) is less than said the chosen level of significance \(a=0.05\), then, you can reject the null hypothesis, \(H_0\): There is no variability in the efficiency of the brushes.
This means that Femi's conclusion indicates variability in the brushes.
Well, I guess that supported my excuse as to why I got tired of cleaning since some brushes weren't that efficient.
Try out more examples on your own, while keeping in mind that randomized blocking is essentially ridding off the nuisance factors through blocking (grouping) before randomization. The goal is to create groups that are similar with less variability as compared to the whole samples. Moreover, if variability is more observable within blocks, this is an indication that blocking is not done properly or the nuisance factor is not very good a variable to block. Hoping you'll start blocking afterwards!