4.3

We want to test the hypotheses of equal means against the hypotheses of at least one being different:

\[ H_0: \tau_i = 0 \]

\[ H_a: \tau_i \neq 0 \]

The linear effects model is shown by the equation below:

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

where \(\mu_i, \tau_i, \beta_j, \epsilon_{ij}\), represents, respectively the average of the ith treatment, the interaction between treatments, additive effect of the blocks and normal random 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.
chemical<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
bolt <- c(seq(1,5),seq(1,5),seq(1,5),seq(1,5))
obs<- c(73,68,74,71,67,
        73,67,75,72,70,
        75,68,78,73,68,
        73,71,75,75,69)

chemical<-as.fixed(chemical)
bolt <- as.fixed(bolt)
model<-lm(obs~chemical+bolt)
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    
## bolt      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

Conclusion:

With an \(\alpha\) = 0.05, the null hypothesis is not rejected (P-value = 0.1211).

4.16

4.22

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

\[ i = j = k = (1,2,3,4,5) \]

\(\mu\) = Baseline mean

\(\tau\) = Treatments (Ingredients)

\(\beta\) = Batch of materials

\(\alpha\) = Day of production

\(\epsilon\) = Random error

library(GAD)
batch<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
day<-c(seq(1,5),seq(1,5),seq(1,5),seq(1,5),seq(1,5))
variable<-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")
value<-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)
dat<-data.frame(batch,day,variable,value)

dat$batch<-as.fixed(dat$batch)
dat$day<-as.fixed(dat$day)
dat$variable<-as.fixed(dat$variable)
model<-lm(value~batch+day+variable,data = dat)
anova(model)
## Analysis of Variance Table
## 
## Response: value
##           Df Sum Sq Mean Sq F value    Pr(>F)    
## batch      4  15.44   3.860  1.2345 0.3476182    
## day        4  12.24   3.060  0.9787 0.4550143    
## variable   4 141.44  35.360 11.3092 0.0004877 ***
## Residuals 12  37.52   3.127                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Conclusion:

Since the p-value for treatments is 0.0004877 which is very less than 0.05 therefore we can successfully reject the null hypothesis that the mean are equal with 95% confidence.