Answer 4.16
We know that \(\tau_i\) = \(\mu_i\) - \(\mu\) ; and \(\beta_i\) = \(\mu_j\) - \(\mu\)
Tau1 <- mean(TensileStrengths[1:5])-mean(TensileStrengths)
Tau1
## [1] -1.15
Tau2 <- mean(TensileStrengths[6:10])-mean(TensileStrengths)
Tau2
## [1] -0.35
Tau3 <- mean(TensileStrengths[11:15])-mean(TensileStrengths)
Tau3
## [1] 0.65
Tau4 <- mean(TensileStrengths[16:20])-mean(TensileStrengths)
Tau4
## [1] 0.85
Beta1 <- mean(c(73,73,75,73))-mean(TensileStrengths)
Beta1
## [1] 1.75
Beta2 <- mean(c(68,67,68,71))-mean(TensileStrengths)
Beta2
## [1] -3.25
Beta3 <- mean(c(74,75,78,75))-mean(TensileStrengths)
Beta3
## [1] 3.75
Beta4 <- mean(c(71,72,73,75))-mean(TensileStrengths)
Beta4
## [1] 1
Beta5 <- mean(c(67,70,68,69))-mean(TensileStrengths)
Beta5
## [1] -3.25
\(\tau_i\) Values:
\(\tau_1\) = -1.15
\(\tau_2\) = -0.35
\(\tau_3\) = 0.65
\(\tau_4\) = 0.85
\(\beta_i\) Values:
\(\beta_1\) = 1.75
\(\beta_2\) = -3.25
\(\beta_3\) = 3.75
\(\beta_4\) = 1
\(\beta_5\) = -3.25
Answer 4.22
Null Hypothesis: The ingredients have no effect on the reaction time of a chemical process; \(\tau_i\) = 0
Alternative Hypothesis: The ingredients have an effect on the reaction time of a chemical process; \(\tau_i \neq 0\)
Observations <- 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)
Ingredients <- 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")
Batch <- c(rep(1,5), rep(2,5), rep(3,5), rep(4,5), rep(5,5))
Days <- c(rep(seq(1,5),5))
dat <- cbind(Observations, Ingredients, Batch, Days)
dat <- as.data.frame(dat)
dat$Ingredients <- as.fixed(dat$Ingredients)
dat$Batch <- as.fixed(dat$Batch)
dat$Days <- as.fixed(dat$Days)
model2 <- aov(Observations~Batch+Days+Ingredients, data = dat)
summary(model2)
## Df Sum Sq Mean Sq F value Pr(>F)
## Batch 4 15.44 3.86 1.235 0.347618
## Days 4 12.24 3.06 0.979 0.455014
## Ingredients 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
As we can see from the anova test on the Latin Square model, since the P-value = 0.000488 < alpha = 0.05, we reject the null hypothesis and conclude that at least one of the means differs. The factor of interest here is ingredients. The P-values for Batch & Days is much greater than alpha, and they are recognized as blocks as they are a nuisance and not the subjects we are studying on. Hence, the ingredients do have an effect on the reaction time.
All R Code:
TensileStrengths <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
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))
Bolt<-as.fixed(Bolt)
Chemical<-as.fixed(Chemical)
model1 <- lm(TensileStrengths~Chemical+Bolt)
gad(model1)
Tau1 <- mean(TensileStrengths[1:5])-mean(TensileStrengths)
Tau1
Tau2 <- mean(TensileStrengths[6:10])-mean(TensileStrengths)
Tau2
Tau3 <- mean(TensileStrengths[11:15])-mean(TensileStrengths)
Tau3
Tau4 <- mean(TensileStrengths[16:20])-mean(TensileStrengths)
Tau4
Beta1 <- mean(c(73,73,75,73))-mean(TensileStrengths)
Beta1
Beta2 <- mean(c(68,67,68,71))-mean(TensileStrengths)
Beta2
Beta3 <- mean(c(74,75,78,75))-mean(TensileStrengths)
Beta3
Beta4 <- mean(c(71,72,73,75))-mean(TensileStrengths)
Beta4
Beta5 <- mean(c(67,70,68,69))-mean(TensileStrengths)
Beta5
Observations <- 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)
Ingredients <- 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")
Batch <- c(rep(1,5), rep(2,5), rep(3,5), rep(4,5), rep(5,5))
Days <- c(rep(seq(1,5),5))
dat <- cbind(Observations, Ingredients, Batch, Days)
dat <- as.data.frame(dat)
dat$Ingredients <- as.fixed(dat$Ingredients)
dat$Batch <- as.fixed(dat$Batch)
dat$Days <- as.fixed(dat$Days)
model2 <- aov(Observations~Batch+Days+Ingredients, data = dat)
summary(model2)