Flipped Assignment - 11

Blocked Observation

Hypothesis:

\(H_o: \tau_i = 0\) for all i

\(H_a: \tau_i \neq 0\) for some i

i = the chemical agent used in the experiment

Model Equation:

\(y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij}\)

obs<-c(73,68,74,71,67,
       73,67,75,72,70,
       75,68,78,73,68,
       73,71,75,75,69)

qqnorm(obs[1:5])

qqnorm(obs[6:10])

qqnorm(obs[11:15])

qqnorm(obs[16:20])

chem1<-obs[1:5]
chem2<-obs[6:10]
chem3<-obs[11:15]
chem4<-obs[16:20]
boxplot(chem1,chem2,chem3,chem4, main="Boxplot for 4 Chemical Agents", xlab="Chemical",ylab="tensile strength",names =c("1","2","3","4"))

• The data appears to be normal. We are assuming constant variance because the data set is very few.

chemical<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
bolt<-c(seq(1,5),seq(1,5),seq(1,5),seq(1,5))
bolt<-as.fixed(bolt)
chemical<-as.fixed(chemical)
model<-lm(obs~chemical+bolt) #lm = linear model
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

• The p-value is 0.1211 which is less than our alpha, 0.15. Therefore we reject the null hypothesis. We say that there is a difference in at least one of the chemicals.

Unblocked observation

Hypothesis:

\(H_o: \tau_i = 0\) for all i

\(H_a: \tau_i \neq 0\) for some i

i = the chemical agent used in the experiment

Model Equation:

\(y_{ij} = \mu + \tau_i + \epsilon_{ij}\)

obs<-c(73,68,74,71,67,
       73,67,75,72,70,
       75,68,78,73,68,
       73,71,75,75,69)



chem1<-obs[1:5]
chem2<-obs[6:10]
chem3<-obs[11:15]
chem4<-obs[16:20]




chemical<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
bolt<-c(seq(1,5),seq(1,5),seq(1,5),seq(1,5))
bolt<-as.fixed(bolt)
chemical<-as.fixed(chemical)
model<-lm(obs~chemical) #lm = linear model
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 11.1750

• The p-value, 0.7644 is significantly larger than our alpha, 0.15. Therefore, we fail to reject the null hypotheses that there is no difference in mean tensile strength in chemicals 1-4.

Comparitive Differences:

• You can tell from the p values that the bolts represent a significant amount of nuisances. This allows us to be more confidence in our initial conclusion. By using the blocking method, we lose a degree of freedom (J-1). However, this decreases the sum squared error.