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=0.15) and draw appropriate conclusions. Be sure to state the linear effects model and hypotheses being tested.
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= .15 , we concluded that we have enough evidence to reject the null hypothesis that the chemicals same effect on the cloth.
Assume now that she didn’t block on Bolt and rather ran the experiment at a completely randomized design on random pieces of cloth, resulting in the following data. Analyze the data from this experiment (use α=0.15) and draw appropriate conclusions. Be sure to state the linear effects model and hypotheses being tested.
chemical<- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
chemical<- as.fixed(chemical)
model<- lm(obs~chemical)
gad(model)## Analysis of Variance Table
##
## Response: obs
## Df Sum Sq Mean Sq F value Pr(>F)
## chemical 3 12.95 4.3167 0.3863 0.7644
## Residual 16 178.80 11.1750Linear effects model observations = \(\mu\) + \(\tau_i\) +\(\epsilon_{i,j}\)
where \(\tau_i\) is the chemical effects and \(\mu\) is the grand mean
Since the p- value is greater than 0.15, we concluded that we accept the null hypotheses that the chemicals have the same effect.
Comment on any differences in the findings from questions 1 and 2. Do you believe that the Bolt of cloth represents a significant amount of nuisance variability?
Blocking makes the MSE increase , therefore removing the block takes away the nuisance making a significant difference. The p-value for blocking is 0.1211 and the p- value for not blocking is 0.7644.