Question 4.3 & 4.16

Linear Effects Equation:

Xij = mu + τi + βj + ϵij

Null Hypothesis(H0): Ti = 0 for all i

Alternate Hypothesis(Ha): Ti != 0 for some i

library(GAD)

## Loading required package: matrixStats

## Loading required package: R.methodsS3

## R.methodsS3 v1.8.2 (2022-06-13 22:00:14 UTC) successfully loaded. See ?R.methodsS3 for help.

observations<-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(seq(1,5),seq(1,5),seq(1,5),seq(1,5))

data<-data.frame(observations,chemical,Bolt)
chemical <- as.fixed(chemical)
Bolt <- as.fixed(Bolt)


model1 <- lm(observations~Bolt+chemical)
gad(model1)

## Analysis of Variance Table
## 
## Response: observations
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## Bolt      4 157.00  39.250 21.6055 2.059e-05 ***
## chemical  3  12.95   4.317  2.3761    0.1211    
## Residual 12  21.80   1.817                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p-value is 0.1211 > 0.05 level of signifance .so we fail to reject the null hypothesis(H0).

we could say that the chemical agents doesn’t signifcantly effect the strength of the cloth for the given level of significance.

Question 4.22

Observation <- c(8,7,1,7,3,11,2,7,3,8,4,9,10,1,5,6,8,6,6,10,4,2,3,8,8)
batch <- c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5)
Ingredient <- c(1,2,4,3,5,3,5,1,4,2,2,1,3,5,4,4,3,5,2,1,5,4,2,1,3)
Day <- c(rep(seq(1,5),5)) 
Ingredient <- as.factor(Ingredient)
batch <- as.factor(batch)
Day <- as.factor(Day)
Data1<- data.frame(Observation, batch, Day, Ingredient)

Xij = mu + τi + βj + ϵij

Fixed Effects

Assuming τi is the ingredient effect, and thus we define our null and alternative hypotheses as

Null Hypothesis H0: τi = 0

Alternative Hypothesis Ha: τi != 0

Analysis

model2<-aov(Observation~Ingredient+batch+Day,data=Data1)
summary(model2)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Ingredient   4 141.44   35.36  11.309 0.000488 ***
## batch        4  15.44    3.86   1.235 0.347618    
## Day          4  12.24    3.06   0.979 0.455014    
## Residuals   12  37.52    3.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p value of the ingredients is less than the level of significance (0.05) so we reject null Hypothesis

we conclude there’s a significant effect of ingredients on the reaction time of chemical process.

Homework 7

Rishidhar

2022-10-16

Question 4.3 & 4.16

Linear Effects Equation:

Xij = mu + τi + βj + ϵij

Null Hypothesis(H0): Ti = 0 for all i

Alternate Hypothesis(Ha): Ti != 0 for some i

The p-value is 0.1211 > 0.05 level of signifance .so we fail to reject the null hypothesis(H0).

we could say that the chemical agents doesn’t signifcantly effect the strength of the cloth for the given level of significance.

Question 4.22

Xij = mu + τi + βj + ϵij

Fixed Effects

Assuming τi is the ingredient effect, and thus we define our null and alternative hypotheses as

Null Hypothesis H0: τi = 0

Alternative Hypothesis Ha: τi != 0

Analysis

The p value of the ingredients is less than the level of significance (0.05) so we reject null Hypothesis

we conclude there’s a significant effect of ingredients on the reaction time of chemical process.