# Randomised Block Design

Suppose the response of three different treatments, $$A$$, $$B$$ and $$C$$ are measured in two different hospitals of a country. The data are given below.

My question is: so far I understand it is a randomised block design if I want to compare the treatments. But whatever I search on the internet, I see there is only one observation per each cell. And the model is :

$$Y_{ij}=\beta_0+\beta_i+\gamma_j+\epsilon_{ij},$$

where $$\epsilon_{ij}\sim N(0,\sigma^2)$$, $$\beta_i$$ is the effect of the $$i$$th treatment, and $$\gamma_j$$ is the effect of the $$j$$th block.

But in my particular example, there are multiple observations per cell. It makes me doubt if I am adopting the correct experimental design. Is it possible to have more than one observation per cell in a randomised block design? If so, will the model become

$$Y_{ijk}=\beta_0+\beta_{ik}+\gamma_{jk}+\epsilon_{ijk},\quad k=1,\ldots, n~?$$

• It should be something like $Y_{ijk}=\beta_0+\beta_i+\gamma_j+\varepsilon_{ijk}$, where $k=1,2,\ldots,n_{ij}$. Commented Apr 28, 2020 at 7:20

But in my particular example, there are multiple observations per cell. It makes me doubt if I am adopting the correct experimental design. Is it possible to have more than one observation per cell in a randomised block design?

It is certainly possible! With unequal number of observations per cell, as you have, you loose some properties, such as balance and orthogonality. As noted in a comment, you can model as $$Y_{ijk}=\beta_0+\beta_i+\gamma_j+\varepsilon_{ijk}$$ where $$k=1,2,\ldots,n_{ij}$$.

With R we can analyze as follows:

mod <- lm(response ~ T + Hospital, data=yourdata)
summary(mod)

Call:
lm(formula = response ~ T + Hospital, data = yourdata)

Residuals:
Min       1Q   Median       3Q      Max
-0.65714 -0.18714  0.03714  0.27286  0.41143

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   2.0571     0.1927  10.678 8.68e-07 ***
TB           -0.1400     0.2593  -0.540 0.601089
TC           -1.2600     0.2593  -4.859 0.000662 ***
Hospital2     0.1714     0.2090   0.820 0.431116
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.3709 on 10 degrees of freedom
Multiple R-squared:  0.7618,    Adjusted R-squared:  0.6903
F-statistic: 10.66 on 3 and 10 DF,  p-value: 0.001858

Data was constructed:

response <- c(1.4, 2.4, 2.2, 2.4, 2.1, 1.7, 2.5, 1.8, 2.0, 0.7, 1.1, 0.5,
0.9, 1.3)
T <- c(rep("A", 4), rep("B", 5), rep("C", 5))
Hospital <- c(1,1,1,2,1,1,2,2,2,1,1,2,2,2)
yourdata <- data.frame(response=response, T=factor(T),
Hospital =factor(Hospital))
• In the case of a binary treatment, you would need to add an interaction to get a consistent estimate for the average treatment effect. I imagine at least this is necessary here. Commented Apr 6 at 13:19