Q 4.3, Q 4.16

bolts<- c(rep(seq(1,5),4))
chemical <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
obs <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
data<-data.frame(bolts,chemical,obs)

Hypothesis:

Null Hypothesis:

$H_o : \mu_1 = \mu_2 =\mu_i$

Alternative Hypothesis:

$Ha$ : Atleast one $\mu_i$ differs

As we have the Fixed Effect, our hypothesis comes to be.

Null Hypothesis:

$H_o : \tau_i=0$ for all i

Alternative Hypothesis:

$H_a$ : $\tau_i \neq 0$ for some i

Linear Effects Equation:

$y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij}$

library(GAD)

## Warning: package 'GAD' was built under R version 4.1.3

## Loading required package: matrixStats

## Warning: package 'matrixStats' was built under R version 4.1.3

## Loading required package: R.methodsS3

## Warning: package 'R.methodsS3' was built under R version 4.1.3

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

chemical<- as.fixed(chemical)
bolts<- as.fixed(bolts)

model<-lm(obs~bolts+chemical)
gad(model)

## Analysis of Variance Table
## 
## Response: obs
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## bolts     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

P-value is 0.1211 > $\alpha$ = $0.05, we fail to reject Null Hypothesis.

We could say that the chemical agents doesn’t have effect on the strength of the cloth for given level of significance.

The Model Parameters: $\tau_i$ is of the chemicals and $\beta_j$ is of bolts. From the ANOVA table above.

Source Code (Q 4.3, Q 4.16).

bolts<- c(rep(seq(1,5),4))
chemical <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
obs <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
data<-data.frame(bolts,chemical,obs)

library(GAD)
chemical<- as.fixed(chemical)
bolts<- as.fixed(bolts)

model<-lm(obs~bolts+chemical)
gad(model)

4.22

Days<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
Batch<-c(rep(seq(1,5),5))
Ingredients<-c("A","C","B","D","E","B","E","A","C","D","D","A","C","E","B", "C","D","E","B","A","E","B","D","A","C")
obser<- c(8,11,4,6,4,7,2,9,8,2,1,7,10,6,3,7,3,1,6,8,3,8,5,10,8)

Days <-as.factor(Days)
Ingredients <-as.factor(Ingredients)
Batch <-as.factor(Batch)

Hypothesis (Fixed Effects)

Null Hupothesis:

$\tau_1 = \tau_2 = \tau_i$ = 0$

Alternative Hypothesis:

$\tau_i \neq 0$

Model Analysis.

Model <- aov(obser ~ Batch+Days+Ingredients)
anova(Model)

## Analysis of Variance Table
## 
## Response: obser
##             Df Sum Sq Mean Sq F value    Pr(>F)    
## Batch        4  15.44   3.860  1.2345 0.3476182    
## Days         4  12.24   3.060  0.9787 0.4550143    
## Ingredients  4 141.44  35.360 11.3092 0.0004877 ***
## Residuals   12  37.52   3.127                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

HW-7

Pavan,

2022-10-16

Q 4.3, Q 4.16

Hypothesis:

Null Hypothesis:

\(H_o : \mu_1 = \mu_2 =\mu_i\)

Alternative Hypothesis:

\(Ha\) : Atleast one \(\mu_i\) differs

As we have the Fixed Effect, our hypothesis comes to be.

Null Hypothesis:

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

Alternative Hypothesis:

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

Linear Effects Equation:

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

P-value is 0.1211 > \(\alpha\) = $0.05, we fail to reject Null Hypothesis.

We could say that the chemical agents doesn’t have effect on the strength of the cloth for given level of significance.

The Model Parameters: \(\tau_i\) is of the chemicals and \(\beta_j\) is of bolts. From the ANOVA table above.

Source Code (Q 4.3, Q 4.16).

4.22

Hypothesis (Fixed Effects)

Null Hupothesis:

\(\tau_1 = \tau_2 = \tau_i\) = 0$

Alternative Hypothesis:

\(\tau_i \neq 0\)

Model Analysis.

P value of the ingredients is 0.000218 < \(\alpha\) = $0.05, we have sufficient evidence to reject Null Hypothesis.

We could conclude as the different ingredients (A, B, C, D, E) has effects on reaction time of a chemical process.