1.

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.

chem1   <- c(73,    68  ,74,    71, 67)
chem2 <-    c(73,   67, 75, 72, 70)
chem3   <- c(75,    68, 78, 73, 68)
chem4 <-    c(73,   71, 75, 75, 69)
dafr <- data.frame(chem1,chem2,chem3,chem4)
dafr <- stack(dafr)
colnames(dafr) <- c('Numbers','Type')
library(GAD)
dafr$Type <- as.fixed(dafr$Type)
dafr$Bolt <- c(rep(seq(1,5),4))
dafr$Bolt <- as.fixed(dafr$Bolt)

Our linear model effects equation is \(X_{ij}=\mu+\tau_i+\beta_j+\epsilon_{ij}\) where we are testing to see if \(\tau_i\) (the effect of different chemicals) is equal to 0 or not.

We will perform a Randomized Complete Block Design test, with bolt type being blocked(since we are considering bolt type to be a significant source of nuisance variability), and testing chemical types. Our official hypothesis is that the means between the grouped chemical types are the same versus they are not.

Ho: \(\mu_1=\mu_2=\mu_3=\mu_4=\mu\)

Ha: \(\mu_i\neq\mu\) for some i$

or

Ho: \(\tau_i=0\)

Ho: \(\tau_i\neq0\)

model <- lm(Numbers~Type+Bolt,dafr)
gad(model)
## Analysis of Variance Table
## 
## Response: Numbers
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## Type      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

Our p-value is much less than our given alpha value of .15 so therefore we reject the null hypothesis and state there is a difference between the average strength of the cloth given the chemical used.

2.

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.

model <- lm(Numbers~Type,dafr)
gad(model)
## Analysis of Variance Table
## 
## Response: Numbers
##          Df Sum Sq Mean Sq F value Pr(>F)
## Type      3  12.95  4.3167  0.3863 0.7644
## Residual 16 178.80 11.1750

We can see that our p-value is now much greater then our alpha value of .15, therefore we will not reject the null hypothesis and state that there is not enough evidence to support the idea that the chemical used affects the strength of the cloth.

3.

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?

We can see by the results of the two tests that bolt does have a significant amount of nuisance variability and should be blocked. Doing this improves our power of the test manifold. To not block the bolt type would produce incorrect results. Bolt is a significant source of variability in the model.

All Code Used:

chem1   <- c(73,    68  ,74,    71, 67)
chem2 <-    c(73,   67, 75, 72, 70)
chem3   <- c(75,    68, 78, 73, 68)
chem4 <-    c(73,   71, 75, 75, 69)
dafr <- data.frame(chem1,chem2,chem3,chem4)
dafr <- stack(dafr)
colnames(dafr) <- c('Numbers','Type')
library(GAD)
dafr$Type <- as.fixed(dafr$Type)
dafr$Bolt <- c(rep(seq(1,5),4))
dafr$Bolt <- as.fixed(dafr$Bolt)

model <- lm(Numbers~Type,dafr)
gad(model)