A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order to each bolt. The resulting tensile strengths follow. Analyze the data from this experiment (use α=0.05) and draw appropriate conclusions.
Chemicals 1-4 are factors and Bolts 1-5 are blocks.
observations = \(\mu\) + \(\tau_i\) +\(\beta_j\) +\(\epsilon_{i,j}\)
where \(\tau_i\) is the chemical effect, \(\beta_j\) is the bolt effect, and \(\mu\) is the grand mean
Chemical_1 <- c(73,68,74,71,67)
Chemical_2 <- c(73,67,75,72,70)
Chemical_3 <- c(75,68,78,73,68)
Chemical_4 <- c(73,71,75,75,69)
obs <- c(Chemical_1,Chemical_2,Chemical_3,Chemical_4)
Blots <-c(rep(seq(1,5),4))
chemical<- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
chemical<- as.fixed(chemical)
Blots<- as.fixed(Blots)
model<- lm(obs~chemical+Blots)
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
## Blots 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 ' ' 1With α= .05<< Pr=0.1211 ,we fail to reject Ho, we concluded that we do not have enough evidence to reject the null hypothesis that the chemicals same effect on the cloth.
Assuming that chemical types and bolts are fixed, estimate the model parameters \(\tau_i\) and \(\beta_j\) in Problem 4.3
Chemical_1 <- c(73,68,74,71,67)
Chemical_2 <- c(73,67,75,72,70)
Chemical_3 <- c(75,68,78,73,68)
Chemical_4 <- c(73,71,75,75,69)
obs <- c(Chemical_1,Chemical_2,Chemical_3,Chemical_4)
u1<-mean(Chemical_1)
u2<-mean(Chemical_2)
u3<-mean(Chemical_3)
u4<-mean(Chemical_4)
U<-mean(obs)
tau_1= u1-U
tau_2= u2-U
tau_3= u3-U
tau_4= u4-UThe effect of five different ingredients (A, B, C, D, E) on the reaction time of a chemical process is being studied. Each batch of new material is only large enough to permit five runs to be made. Furthermore, each run requires approximately hours, so only five runs can be made in one day. The experimenter decides to run the experiment as a Latin square so that day and batch effects may be systematically controlled. She obtains the data that follow. Analyze the data from this experiment (use α=0.05) and draw conclusions.
statistical model: \(y_{i,j,k}\) = \(\mu\) + \(α_i\) + \(\tau_i\) +\(\beta_j\) +\(\epsilon_{i,j}\) null hypothesis: The ingredients have no affect on the reaction time, τj = 0 alternative hypothesis: The ingredients have an affect on the reaction time, τj != 0
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 )
ingredient <- 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")
ingredient <- as.factor(ingredient)
batch <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
batch <- as.factor(batch)
day <- c(rep(seq(1,5),5))
day <- as.factor(day)
aov.model <-aov(obs~batch+day+ingredient)
summary(aov.model)## 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
## ingredient 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 ' ' 1The p-value of the ingredient is significantly lower than that of alpha, indicating that we reject the null hypothesis. Additionally, the p-values of the two blocks (batch & day) are much greater than alpha, indicating that they are sources of nuisance variability. In conclusion, the ingredients do have an effect on the reaction time.