Question 7.12

From the exercise 6.21, the data is as follows. But, since we want to block the replications, let’s add the variable “blocking” in order to do that:

length <- rep(c(rep("-",7),rep("+",7)),8)
type <- rep(c(rep("-",14),rep("+",14)),4)
brk <- rep(c(rep("-",28),rep("+",28)),2)
slope <- c(rep("-",56),rep("+",56))

blocking <- rep(seq(1,7),16)

obs <- c(10.0, 18.0, 14.0, 12.5, 19.0, 16.0, 18.5,
         0.0, 16.5, 4.5, 17.5, 20.5, 17.5, 33.0,
         4.0, 6.0, 1.0, 14.5, 12.0, 14.0, 5.0,
         0.0, 10.0, 34.0, 11.0, 25.5, 21.5, 0.0,
         0.0, 0.0, 18.5, 19.5, 16.0, 15.0, 11.0,
         5.0, 20.5, 18.0, 20.0, 29.5, 19.0, 10.0,
         6.5, 18.5, 7.5, 6.0, 0.0, 10.0, 0.0,
         16.5, 4.5, 0.0, 23.5, 8.0, 8.0, 8.0,
         4.5, 18.0, 14.5, 10.0, 0.0, 17.5, 6.0,
         19.5, 18.0, 16.0, 5.5, 10.0, 7.0, 36.0,
         15.0, 16.0, 8.5, 0.0, 0.5, 9.0, 3.0,
         41.5, 39.0, 6.5, 3.5, 7.0, 8.5, 36.0,
         8.0, 4.5, 6.5, 10.0, 13.0, 41.0, 14.0,
         21.5, 10.5, 6.5, 0.0, 15.5, 24.0, 16.0,
         0.0, 0.0, 0.0, 4.5, 1.0, 4.0, 6.5,
         18.0, 5.0, 7.0, 10.0, 32.5, 18.5, 8.0)

data1 <- data.frame(length,type,brk,slope,blocking,obs)

Now, to analyze the data, differently from the question 6.21, we will a term \(\zeta_z\) in the linear equation that represents the source of nuisance that we are blocking. And, since this is a known source of nuisance, it is correct to assume that this variable does not interact with the levels.

So, the statistical test will be as following:

model.aov.1 <- aov(obs~length*type*brk*slope+blocking,data=data1)
summary(model.aov.1)

##                       Df Sum Sq Mean Sq F value  Pr(>F)   
## length                 1    917   917.1  10.673 0.00151 **
## type                   1    388   388.1   4.517 0.03616 * 
## brk                    1    145   145.1   1.689 0.19687   
## slope                  1      1     1.4   0.016 0.89888   
## blocking               1    152   152.1   1.769 0.18663   
## length:type            1    219   218.7   2.545 0.11398   
## length:brk             1     12    11.9   0.138 0.71068   
## type:brk               1    115   115.0   1.338 0.25020   
## length:slope           1     94    93.8   1.092 0.29877   
## type:slope             1     56    56.4   0.657 0.41976   
## brk:slope              1      2     1.6   0.019 0.89084   
## length:type:brk        1      7     7.3   0.084 0.77206   
## length:type:slope      1    113   113.0   1.315 0.25437   
## length:brk:slope       1     39    39.5   0.459 0.49952   
## type:brk:slope         1     34    33.8   0.393 0.53224   
## length:type:brk:slope  1     96    95.6   1.113 0.29411   
## Residuals             95   8164    85.9                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Just like the question 6.21, it is possible to assume that only the levels tyoe and brk are significant for the data, since the repectivelly p-values are below \(\alpha=0.05\).

However, since we want, not only test the levels significance, test the effect of blocking the replicates, let’s compare the \(MSE_{6.21}\) with \(MSE_{7.12}\).

From the previous question, we have a \(MSE_{6.21}=86.6\) and from the analysis of this data, we have a \(MSE_{7.12}= 85.9\). Therefore we can conclude that blocking is in fact a source of nuisance, because the error decreased. But since the error decreased only 0.8% of the MSE, blocking the replicates is not really effective.

Question 7.20

To propose the design of our experiment, let’s use the following function in R.

library(agricolae)
levels <- c(2,2,2,2,2,2)
experiment <- design.ab(levels, r=1,design="crd",randomization=FALSE)

After that, according to table 7.9, the most effective blocks for a \(2^6\) design factorial with 4 blocks are the effects ABCF and CDEF.

The full design is in the attached excel spreadsheet and it is possible to see in the image below:

Homework 12

2022-11-21

Question 7.12

Question 7.20

Question 7.21