0.1 Question 1

  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 alpha=0.15) and draw appropriate conclusions. Be sure to state the linear effects model and hypotheses being tested.

0.1.1 Solution 1

We are testing the Null hypothesis:

\[ H_{0}: \mu _{1} = \mu _{2} = \mu _{3} = \mu _{4} \]

Or \[\tau _{i}=0\forall "i"\]

Compared to the alternative hypothesis:

\(H_{1}:\) One of the \(\mu _{1}\) differs

Compared to \[\tau _{i}\neq 0\exists "i"\]

Linear Effect Equation: \[X_{ij}= \mu + \tau _{i}+\beta _{j}+\epsilon _{ijk}\]

For the experiment when alpha is 0.5

μ is the grand mean value

τi is the fixed effects for treatment i

Where βj is the block effect for j

Where ϵij is the random error

Getting our our data we have

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(rep(seq(1,5),4))
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 a P value (0.1211) less than 0.15, we reject the Null hypothesis

0.2 Question 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 alpha=0.15) and draw appropriate conclusions.  Be sure to state the linear effects model and hypotheses being tested. 

0.2.1 Solution 2

We are testing the Null hypothesis:

\[ H_{0}: \mu _{1} = \mu _{2} = \mu _{3} = \mu _{4} \]

Or \[\tau _{i}=0\forall "i"\]

Compared to the alternative hypothesis:

\(H_{1}:\) One of the \(\mu _{1}\) differs

Compared to \[\tau _{i}\neq 0\exists "i"\]

Linear Effect Equation: \[Y_{ij}= \mu + \tau _{i}+\epsilon _{ij}\]

for the experiment when alpha is 0.15 (Unblocked Observation)

observation2 <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
model2 <- lm(observation2~chemical)
gad(model2)
## Analysis of Variance Table
## 
## Response: observation2
##          Df Sum Sq Mean Sq F value Pr(>F)
## chemical  3  12.95  4.3167  0.3863 0.7644
## Residual 16 178.80 11.1750

Our P value (0.7644) is greater than 0.15 so we do not reject the null hypothesis.

0.3 Question 3

Here are some info and conclusion about the two questions:

  1. There is a P value of 0.1211 in the first question because we blocked bolts. As a result, we reject the Null hypothesis.
  2. For the second question, there was no blocking of bolts so we have a P value of 0.7644, more than the 0.15 P value so we did not reject the hypothesis.
  3. Finally, blocking increases the MSE as a result, getting rid of the block gets rid of the nuisance making a huge difference.

We can conclude from this that there are nuisance variability when we block so we reject the hypothesis.

#Loading the experiment for question 1

chemical<- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
bolt<- c(rep(seq(1,5),4))
obs<- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)

library(GAD)
chemical <- as.fixed(chemical)
bolt<- as.fixed(bolt)
model <- lm(obs~chemical+bolt)
gad(model)

#Question 2

observation <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
model2 <- lm(observation~chemical)
gad(model2)