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

1.1 Solution 1: (Blocked Case)

Stating the Hypothesis:

Null Hypothesis: \[H_0: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\]

OR \[\tau_{i}=0 \space \forall \space "i"\]

Alternative Hypothesis:

\[ H_a: Atleast \space one \space \mu_{i} \space differs \]

OR \[\tau_{i}\neq 0 \space\space\exists \space "i"\]

Writing Linear Effects Equation:

\[ y_{i,j}=\mu +\tau_{i}+\beta_{j}+\epsilon_{i,j} \]

where,

\(\mu\)=Grand Mean

\(\tau_{i}\)= Fixed Effects for treatment “i”

\(\beta_{j}\)= Block Effect for “j”

\(\epsilon_{i,j}\)= Random Error

Now analyzing the experiment at \(\alpha\)=0.15

library(GAD)
observation <- 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))
bolts <- c(rep(seq(1,5),4))
chemical <- as.fixed(chemical)
bolts <- as.fixed(bolts)

Running the Model with Blocked Observations:

model <- lm(observation~chemical+bolts)
gad(model)
## Analysis of Variance Table
## 
## Response: observation
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## chemical  3  12.95   4.317  2.3761    0.1211    
## bolts     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:

---> P value is 0.1211 < 0.15 hence we reject Null Hypothesis

2 Question 2:

1.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. 

Stating the Hypothesis: (Un-Blocked Case)

Null Hypothesis: \[H_0: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\]

OR \[\tau_{i}=0\space\space \forall \space "i"\]

Alternative Hypothesis:

\[ H_a: Atleast \space one \space \mu_{i} \space differs \]

OR \[\tau_{i}\neq 0 \space \space\exists \space "i"\]

Writing Linear Effects Equation: (Un-Blocked)

\[ y_{i,j}=\mu +\tau_{i}+\epsilon_{i,j} \]

where,

\(\mu\)=Grand Mean

\(\tau_{i}\)= Fixed Effects for treatment “i”

\(\epsilon_{i,j}\)= Random Error

Now analyzing the experiment at \(\alpha\)=0.15

And Running the Model with Un-Blocked Observations:

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

Conclusion:

---> P value is 0.7644 > 0.15 hence we fail to reject Null Hypothesis

3 Problem 3:

Conclusion:

---> We can see in first question that we blocked bolts , and we got our p value as 0.1211 which was less than 0.15 ,hence we reject Null Hypothesis. But on the other hand in Question 2 we did not consider blocking on bolts(replication) which gave us a p value of 0.7644 which is greater than 0.15, hence we fail to reject Null Hypothesis. This tells us that when we block we included nuisance variability and it led us to reject the Null hypothesis. Therefore, bolt of cloth did represented a significant amount of nuisance variability as it changed our decision of rejecting Null hypothesis.