Assignment11

Q1.

Linear effects equation \[Y_{ij} = \mu + \tau_{i} + \beta_{j}+\epsilon_{ij}\]where

\(\mu\)=Grand Mean

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

\(\beta_{j}\)= Block Effect for “j”

\(\epsilon_{ij}\)= Random Error for “j” from treatment “i”

Hypothesis: \[H_0:\ \mu_1 = \mu_2 = \mu_3 = \mu_4 \]\[ H_1:\ \text{At least one } \mu_i \text{ differs from the others}\]

library(GAD)
strength <- 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))
bolt <- c(rep(seq(1,5),4))
chemical <- as.fixed(chemical)
bolt <- as.fixed(bolt)
model1 <- lm(strength ~ chemical + bolt)
gad(model1)

## $anova
## Analysis of Variance Table
## 
## Response: strength
##           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 ***
## Residuals 12  21.80   1.817                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Since the p value of 0.1211 is less than 0.15, therefore reject H0. There is significant evidence to conclude that at least one chemical agent has a different effect on the strength of a particular type of cloth.

Q2.

Linear effects equation \[Y_{ij} = \mu + \tau_{i} +\epsilon_{ij}\]where

\(\mu\)=Grand Mean

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

\(\epsilon_{ij}\)= Random Error for “j” from treatment “i”

Hypothesis: \[H_0:\ \mu_1 = \mu_2 = \mu_3 = \mu_4 \]\[ H_1:\ \text{At least one } \mu_i \text{ differs from the others}\]

strength2 <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
chemical2 <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
chemical2 <- as.fixed(chemical2)
model2 <- lm(strength2 ~ chemical2)
gad(model2)

## $anova
## Analysis of Variance Table
## 
## Response: strength2
##           Df Sum Sq Mean Sq F value Pr(>F)
## chemical2  3  12.95  4.3167  0.3863 0.7644
## Residuals 16 178.80 11.1750

Since the p value of 0.7644 is greater than 0.15, therefore fail to reject H0. There is insignificant evidence to conclude that at least one chemical agent has a different effect on the strength of a particular type of cloth.

Q3.

The results from Questions 1 and 2 are different. With blocking, the p-value is 0.1211 < 0.15, so we reject H0. Without blocking, the p-value is 0.7644 > 0.15, so we fail to reject H0. Blocking on Bolt reduces the error variability, showing that Bolt is an important source of nuisance variability and improves our ability to detect differences among the chemicals.