Assignment 11

Question 1

Test Hypothesis

H_o: \(\mu_1 = \mu_2 = \mu_3 = \mu_4\) - Null Hypothesis

H_a: At least 1 differs - Alternative Hypothesis

\(\alpha\) = 0.15

Linear Effects

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

#install.packages("GAD")
library(GAD)

## Loading required package: matrixStats

## Loading required package: R.methodsS3

## R.methodsS3 v1.8.1 (2020-08-26 16:20:06 UTC) successfully loaded. See ?R.methodsS3 for help.

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))
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

From the result f_o is 2.3761 with a corresponding p-value of 0.1211 is significantly lesser than \(\alpha\) = 0.15. Therefore we reject H_o that the means are equal, and conclude that none of the means are different.

Question 2

Test Hypothesis

H_o: \(\mu_1 = \mu_2 = \mu_3 = \mu_4\) - Null Hypothesis

H_a: At least 1 differs - Alternative Hypothesis

\(\alpha\) = 0.15

Linear Effects

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

chemical2<-c(rep(1,5), rep(2,5), rep(3,5), rep(4,5))
replications<-c(73, 68, 74, 71, 67, 73, 67, 75, 72, 70, 75, 68, 78, 73, 68, 73, 71, 75, 75, 69)

chemical2<-as.fixed(chemical2)

model2<-lm(replications~chemical2)
gad(model2)

## Analysis of Variance Table
## 
## Response: replications
##           Df Sum Sq Mean Sq F value Pr(>F)
## chemical2  3  12.95  4.3167  0.3863 0.7644
## Residual  16 178.80 11.1750

From the result f_o is 0.3863 with a corresponding p-value of 0.7644 is significantly greater than \(\alpha\) = 0.15. Therefore we fail to reject H_o that the means are equal, and conclude that none of the means are different.

Question 3

After running both experiments, we can conclude that the nuisance factor is significant in experiment 1 because we used blocking which has an effect on the mean square error by reducing it and increasing the F statistic value. However in experiment 2 when there was no blocking effect the mean square error is high with a low F statistic value

library(GAD)

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))
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)

chemical2<-c(rep(1,5), rep(2,5), rep(3,5), rep(4,5))
replications<-c(73, 68, 74, 71, 67, 73, 67, 75, 72, 70, 75, 68, 78, 73, 68, 73, 71, 75, 75, 69)

chemical2<-as.fixed(chemical2)

model2<-lm(replications~chemical2)
gad(model2)