1
$\begingroup$

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.

enter image description here

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

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

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

1 Answer 1

0
$\begingroup$

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))
$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – num_39
    Commented Apr 6 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.