Q1
Hypothesis test:
\(H_0: \mu_{\text{1}} = \mu_{\text{2}}= \mu_{\text{3}}=\mu_{\text{4}}\)
\(H_a: {\text{At least one differs}}\)
Linear Effect equation: \[ y_{i,j} = \mu + \tau_i + \beta_j + \epsilon_{i,j} \] Where:
library(GAD)
ch1 <- c(73,68,74,71,67)
ch2 <- c(73,67,75,72,70)
ch3 <- c(75,68,78,73,68)
ch4 <- c(73,71,75,75,69)
blts <-c(1,2,3,4,5)
obs <- c(ch1,ch2,ch3,ch4)
chem <-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
chem <-as.fixed(chem)
bolts <- c(rep(blts,4))
bolts <-as.fixed(bolts)
model <- lm(obs ~ chem + bolts)
gad(model)
## $anova
## Analysis of Variance Table
##
## Response: obs
## Df Sum Sq Mean Sq F value Pr(>F)
## chem 3 12.95 4.317 2.3761 0.1211
## bolts 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 our p-value is less than 0.15, we reject the H0 which means our sample populations have different means.
Q2
Hypothesis test:
\(H_0: \mu_{\text{1}} = \mu_{\text{2}}= \mu_{\text{3}}=\mu_{\text{4}}\)
\(H_a: {\text{At least one differs}}\)
Linear Effect equation: \[ y_{i,j} = \mu + \tau_i + \epsilon_{i,j} \] Where:
model <- lm(obs ~ chem)
gad(model)
## $anova
## Analysis of Variance Table
##
## Response: obs
## Df Sum Sq Mean Sq F value Pr(>F)
## chem 3 12.95 4.3167 0.3863 0.7644
## Residuals 16 178.80 11.1750
Since p-value is more than our alpha we fail to reject H0.
Q3
We compared models in Q1 and Q2 and we came to this conclusion : 1- Our p-value increased significantly. 2- when we consider the bolt types in our model, nuisance variablity is reduced which leads to rejecting the H0.
code:
library(GAD)
ch1 <- c(73,68,74,71,67)
ch2 <- c(73,67,75,72,70)
ch3 <- c(75,68,78,73,68)
ch4 <- c(73,71,75,75,69)
blts <-c(1,2,3,4,5)
obs <- c(ch1,ch2,ch3,ch4)
chem <-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
chem <-as.fixed(chem)
bolts <- c(rep(blts,4))
bolts <-as.fixed(bolts)
model <- lm(obs ~ chem + bolts)
gad(model)
model <- lm(obs ~ chem)
gad(model)