Homework week 6

Problem 3.23

library(agricolae)
Fluidtype<-c(17.6,18.9,16.3,17.4,20.1,21.6,16.9,15.3,18.6,17.1,19.5,20.3,21.4,23.6,19.4,18.5,20.5,22.3,19.3,21.1,16.9,17.5,18.3,19.8)
Type <- c(rep(1,6), rep(2,6), rep(3,6), rep(4,6))
Data <- data.frame(Fluidtype, Type)

Data$Type <- as.factor(Data$Type)
str(Data)

## 'data.frame':    24 obs. of  2 variables:
##  $ Fluidtype: num  17.6 18.9 16.3 17.4 20.1 21.6 16.9 15.3 18.6 17.1 ...
##  $ Type     : Factor w/ 4 levels "1","2","3","4": 1 1 1 1 1 1 2 2 2 2 ...

b<-aov(Fluidtype~Type, data = Data)
summary(b)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## Type         3  30.17   10.05   3.047 0.0525 .
## Residuals   20  65.99    3.30                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(b)

Fluid1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
mean(Fluid1)

## [1] 18.65

Fluid2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
mean(Fluid2)

## [1] 17.95

Fluid3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
mean(Fluid3)

## [1] 20.95

Fluid4<-c(19.3,21.1,16.9,17.5,18.3,19.8)
mean(Fluid4)

## [1] 18.81667

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4 differs

Here the p value is slightly more than 0.05, so we fail to reject null hypothesis.

a)There is no indication that the fluids differ.

b)We would you select fluid 3 , given that the objective is long life.

c)After analyzing the residuals from this experiments, we can say that basic analysis of varience assumptions is satisfied.

Problem 3.28

library(agricolae)
material<-c(110,157,194,178,1,2,4,18,880,1256,5276,4355,495,7040,5307,10050,7,5,29,2)
Type<-c(rep(1,4),rep(2,4),rep(3,4),rep(4,4),rep(5,4))
Data<-data.frame(material,Type)
Data$Type<- as.factor(Data$Type)
str(Data)

## 'data.frame':    20 obs. of  2 variables:
##  $ material: num  110 157 194 178 1 ...
##  $ Type    : Factor w/ 5 levels "1","2","3","4",..: 1 1 1 1 2 2 2 2 3 3 ...

av<-aov(material~Type,data = Data)
summary(av)

##             Df    Sum Sq  Mean Sq F value  Pr(>F)   
## Type         4 103191489 25797872   6.191 0.00379 **
## Residuals   15  62505657  4167044                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(av)

library(MASS)
boxplot(Data$material~Data$Type)

boxcox(material~Type)

m <- log(material)
qqnorm(m)

plot(m~Type,xlab="Materials",ylab="Failure Time",main="Variance check")

av <- aov(m~Type, data = Data)
summary(av)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Type         4 165.06   41.26   37.66 1.18e-07 ***
## Residuals   15  16.44    1.10                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu5=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4,mu5 differs

a)Here p value is less than 0.05, so we reject null hypotheis. So we can say that materials have different effects on mean failure time.

b)Here from the plots we can say that the variences are not stable and data is not normally distributed, so we need to do a boxcox transfromation.

c)As the variences are not stable and data is not normal, we need to do a log transformation because the maximum value is close to zero.

Probelm 3.29

library(agricolae)
method<-c(31,10,21,4,1,62,40,24,30,35,53,27,120,97,68)
Type<-c(rep(1,5),rep(2,5),rep(3,5))
Data<-data.frame(method,Type)
Data$Type<-as.factor(Data$Type)
str(Data)

## 'data.frame':    15 obs. of  2 variables:
##  $ method: num  31 10 21 4 1 62 40 24 30 35 ...
##  $ Type  : Factor w/ 3 levels "1","2","3": 1 1 1 1 1 2 2 2 2 2 ...

av1<-aov(method~Type,data = Data)
summary(av1)

##             Df Sum Sq Mean Sq F value  Pr(>F)   
## Type         2   8964    4482   7.914 0.00643 **
## Residuals   12   6796     566                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(av1)

library(MASS)
boxplot(Data$method~Data$Type)

boxcox(method~Type)

lambda<-0.5
Tmethod<-method^(lambda)
boxplot(Tmethod~Data$Type)

DataT <- cbind(Tmethod,Type)
DataT <- data.frame(DataT)
DataT$Type <- as.factor(DataT$Type)
str(DataT)

## 'data.frame':    15 obs. of  2 variables:
##  $ Tmethod: num  5.57 3.16 4.58 2 1 ...
##  $ Type   : Factor w/ 3 levels "1","2","3": 1 1 1 1 1 2 2 2 2 2 ...

aov.modelT<-aov(Tmethod~Type,data=DataT)
summary(aov.modelT)

##             Df Sum Sq Mean Sq F value  Pr(>F)   
## Type         2  63.90   31.95    9.84 0.00295 **
## Residuals   12  38.96    3.25                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(aov.modelT)

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4 differs

a)Here p value is less than 0.05, so we reject the null hypotheis. No, all methods do not have the same effect on mean particle count.

b)From the graphs we can say that there are potential concerns as the variences are not stable and data is not normally distributed.

c)As variences are not stable and data is not normally distributed, we need to do a boxcox transformation.

Probelm 3.51

library(agricolae)
Fluidtype<-c(17.6,18.9,16.3,17.4,20.1,21.6,16.9,15.3,18.6,17.1,19.5,20.3,21.4,23.6,19.4,18.5,20.5,22.3,19.3,21.1,16.9,17.5,18.3,19.8)
Type <- c(rep(1,6), rep(2,6), rep(3,6), rep(4,6))
Data <- data.frame(Fluidtype, Type)

Data$Type <- as.factor(Data$Type)
str(Data)

## 'data.frame':    24 obs. of  2 variables:
##  $ Fluidtype: num  17.6 18.9 16.3 17.4 20.1 21.6 16.9 15.3 18.6 17.1 ...
##  $ Type     : Factor w/ 4 levels "1","2","3","4": 1 1 1 1 1 1 2 2 2 2 ...

kruskal.test(Fluidtype~Type,data = Data)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  Fluidtype by Type
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4 differs

From the test,the p value is greater than 0.05,so we can say that we failed to reject null hypothesis and the conclusions obtained with those from the usual analysis of variance and Kruskal–Wallis test are the same as we fail to reject the null hypothesis in both cases.

Homework week 6

Nitish Kulkarni

10/10/2021

Problem 3.23

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4 differs

Here the p value is slightly more than 0.05, so we fail to reject null hypothesis.

a)There is no indication that the fluids differ.

b)We would you select fluid 3 , given that the objective is long life.

c)After analyzing the residuals from this experiments, we can say that basic analysis of varience assumptions is satisfied.

Problem 3.28

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu5=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4,mu5 differs

a)Here p value is less than 0.05, so we reject null hypotheis. So we can say that materials have different effects on mean failure time.

b)Here from the plots we can say that the variences are not stable and data is not normally distributed, so we need to do a boxcox transfromation.

c)As the variences are not stable and data is not normal, we need to do a log transformation because the maximum value is close to zero.

Probelm 3.29

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4 differs

a)Here p value is less than 0.05, so we reject the null hypotheis. No, all methods do not have the same effect on mean particle count.

b)From the graphs we can say that there are potential concerns as the variences are not stable and data is not normally distributed.

c)As variences are not stable and data is not normally distributed, we need to do a boxcox transformation.

Probelm 3.51

Null Hypothesis(H0): mu1=mu2=mu3=mu4=mu

Alternate Hypothesis(H1): at least one of mu1,mu2,mu3,mu4 differs

From the test,the p value is greater than 0.05,so we can say that we failed to reject null hypothesis and the conclusions obtained with those from the usual analysis of variance and Kruskal–Wallis test are the same as we fail to reject the null hypothesis in both cases.