Question 4.3

For the hypothesis that we are trying to test is:

\[ H_o: \mu_1=\mu_2=\mu_3=\mu_4=\mu \\ Ha: \mu_k \neq \mu \;for \; any\;k \]

Given the following linear equation:

\[ y_{ij}=\mu+\tau_i+\beta_j+\epsilon_{ij} \]

Where \(\tau_i\) is the treatment effect error and \(\beta_j\) is the block error.

library(GAD)
## Loading required package: matrixStats
## Loading required package: R.methodsS3
## R.methodsS3 v1.8.2 (2022-06-13 22:00:14 UTC) successfully loaded. See ?R.methodsS3 for help.
bolts <- c(seq(1,5),seq(1,5),seq(1,5),seq(1,5))
chemical <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))

obs <- c(73, 68, 74, 71, 67, 73, 67, 75, 72, 70, 75, 68, 78, 73, 68, 73, 71, 75, 75, 69)

bolts <- as.fixed(bolts)
chemical <- as.fixed(chemical)

model <- lm(obs~chemical+bolts)
gad(model)
## Analysis of Variance Table
## 
## Response: obs
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## chemical  3  12.95   4.317  2.3761    0.1211    
## bolts     4 157.00  39.250 21.6055 2.059e-05 ***
## Residual 12  21.80   1.817                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

From the result above, since the p-value of the chemicals is higher than \(\alpha\) (\(p-value=0.1211 > \alpha=0.05\)) we cannot successfully reject Ho and consider that there is at least one mean that differ between chemicals

Question 4.16

For this question, the results were calculated in a excel spreadsheet. These results are presented in the image below:

Results for the treatments and blocks errors

Question 4.22

For this test, the hypothesis that we are going to test is:

\[ H_o: \mu_1=\mu_2=\mu_3=\mu_3=\mu \\ H_a: \mu_k \neq \mu \; for \; any \; k \]

And the linear equation is considering two sources of nuisance variability:

\[ y_{ijk}=\mu+\tau_i+\beta_k+\alpha_k+\epsilon_{ijk} \]

library(GAD)

day <- c(seq(1,5),seq(1,5),seq(1,5),seq(1,5),seq(1,5))
batch <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
ing <- c("A","B","D","C","E","C","E","A","D","B","B","A","C","E","D","D","C","E","B","A","E","D","B","A","C")

obs <- c(8,7,1,7,3,11,2,7,3,8,4,9,10,1,5,6,8,6,6,10,4,2,3,8,8)

day <- as.fixed(day)
batch <- as.fixed(batch)
ing <- as.fixed(ing)

dat <- data.frame(batch,day,ing,obs)

mod <- aov(obs~batch+day+ing,data=dat)
summary(mod)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## batch        4  15.44    3.86   1.235 0.347618    
## day          4  12.24    3.06   0.979 0.455014    
## ing          4 141.44   35.36  11.309 0.000488 ***
## Residuals   12  37.52    3.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

From the result above, since the p-value of the chemicals is higher than \(\alpha\) (\(p-value=0.000488 < \alpha=0.05\)) we can successfully reject Ho and consider that the ingredients impact directly into the reaction time of the chemical process