Q1)Problem 3.23
Ftypes<-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)
z<-c(rep(1,6), rep(2,6), rep(3,6), rep(4,6))
dat<-data.frame(Ftypes,z)
z<-as.factor(dat$z)
boxplot(Ftypes~z)
anv<-aov(Ftypes~z, data = dat)
summary(anv)
## Df Sum Sq Mean Sq F value Pr(>F)
## z 1 3.67 3.675 0.874 0.36
## Residuals 22 92.48 4.204
plot(anv)
Ftype1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
Ftype2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
Ftype3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
Ftype4<-c(19.3,21.1,16.9,17.5,18.3,19.8)
mean(Ftype1)
## [1] 18.65
mean(Ftype2)
## [1] 17.95
mean(Ftype3)
## [1] 20.95
mean(Ftype4)
## [1] 18.81667
Here we can see that the value of P is greater than alpha hence we failed to reject the null hypothesis
Here we can see that the mean of fluid 3 is the highest hence it has the longest life
From the boxplot we can see that the basic assumption that the variances are equal is unsatisfied
Q2)Problem 3.51
kruskal.test(Ftypes~z, data = dat)
##
## Kruskal-Wallis rank sum test
##
## data: Ftypes by z
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015
As the value of P is 0.1015 which is greater than alpha hence we fail to reject the null hypothesis
Q2)Problem 3.28
materials<-c(110,157,194,178,1,2,4,18,880,1256,5276,4355,495,7040,5307,10050,7,5,29,2)
y<-c(rep(1,4), rep(2,4), rep(3,4), rep(4,4),rep(5,4))
dat1<-data.frame(materials,y)
y<-as.factor(dat1$y)
avn<-aov(materials~y,data = dat1)
plot(avn)
boxplot(materials~y)
##we can see that the variances are not approximately the same hence we need to equate the variance using boxcox
library(MASS)
boxcox(materials~y)
### here we can see that the lambda is near 0 hence we take lambda as 0.1
lambda<-0.1
material<-materials^(lambda)
boxplot(material~y)
nav<-aov(material~y, data = dat1)
summary(nav)
## Df Sum Sq Mean Sq F value Pr(>F)
## y 1 0.028 0.02821 0.103 0.752
## Residuals 18 4.940 0.27446
plot(nav)
library(agricolae)
b<-LSD.test(avn, "material", console = TRUE)
##
## Study: avn ~ "material"
##
## LSD t Test for materials
##
## Mean Square Error: 8552889
##
## materials, means and individual ( 95 %) CI
##
## materials std r LCL UCL Min Max
## 1 1 NA 1 -6143.2144 6145.214 1 1
## 2 2 0 2 -4342.6156 4346.616 2 2
## 4 4 NA 1 -6140.2144 6148.214 4 4
## 5 5 NA 1 -6139.2144 6149.214 5 5
## 7 7 NA 1 -6137.2144 6151.214 7 7
## 18 18 NA 1 -6126.2144 6162.214 18 18
## 29 29 NA 1 -6115.2144 6173.214 29 29
## 110 110 NA 1 -6034.2144 6254.214 110 110
## 157 157 NA 1 -5987.2144 6301.214 157 157
## 178 178 NA 1 -5966.2144 6322.214 178 178
## 194 194 NA 1 -5950.2144 6338.214 194 194
## 495 495 NA 1 -5649.2144 6639.214 495 495
## 880 880 NA 1 -5264.2144 7024.214 880 880
## 1256 1256 NA 1 -4888.2144 7400.214 1256 1256
## 4355 4355 NA 1 -1789.2144 10499.214 4355 4355
## 5276 5276 NA 1 -868.2144 11420.214 5276 5276
## 5307 5307 NA 1 -837.2144 11451.214 5307 5307
## 7040 7040 NA 1 895.7856 13184.214 7040 7040
## 10050 10050 NA 1 3905.7856 16194.214 10050 10050
##
## Alpha: 0.05 ; DF Error: 18
## Critical Value of t: 2.100922
##
## Groups according to probability of means differences and alpha level( 0.05 )
##
## Treatments with the same letter are not significantly different.
##
## materials groups
## 10050 10050 a
## 7040 7040 ab
## 5307 5307 ab
## 5276 5276 ab
## 4355 4355 ab
## 1256 1256 b
## 880 880 b
## 495 495 b
## 194 194 b
## 178 178 b
## 157 157 b
## 110 110 b
## 29 29 b
## 18 18 b
## 7 7 b
## 5 5 b
## 4 4 b
## 2 2 b
## 1 1 b
a)all five material doesnt have the same mean effect failure time as in boxplot we can see that material 4 differs a lot from the others
b) from Above table of LSD we can see that the treatments with the same letters are the same
Q3) Problem 3.29
method<-c(31,10,21,4,1,62,40,24,30,35,53,27,120,97,68)
dat2<-c(rep(1,5), rep(2,5), rep(3,5))
cmb<-data.frame(method,dat2)
dat2<-as.factor(cmb$dat2)
library(MASS)
boxplot(method~dat2)
### here we can see that the variances are unequal hence we need to do the boxcox
boxcox(method~dat2)
## we can see that lambda is less than 0.5 hence we take lambda as 0.4
lambda<-0.4
methods<-method^lambda
boxplot(methods~dat2)
methods1<-log(method)
boxplot<-(methods1~dat2)
anva<-aov(methods~dat2, data = cmb)
library(agricolae)
LSD.test(anva, "methods", console = TRUE)
##
## Study: anva ~ "methods"
##
## LSD t Test for methods
##
## Mean Square Error: 1.010105
##
## methods, means and individual ( 95 %) CI
##
## methods std r LCL UCL Min Max
## 1 1.000000 NA 1 -1.1712564 3.171256 1.000000 1.000000
## 1.74110112659225 1.741101 NA 1 -0.4301553 3.912358 1.741101 1.741101
## 2.51188643150958 2.511886 NA 1 0.3406301 4.683143 2.511886 2.511886
## 3.37977444523543 3.379774 NA 1 1.2085181 5.551031 3.379774 3.379774
## 3.56520491593201 3.565205 NA 1 1.3939485 5.736461 3.565205 3.565205
## 3.73719281884655 3.737193 NA 1 1.5659364 5.908449 3.737193 3.737193
## 3.89805984091619 3.898060 NA 1 1.7268035 6.069316 3.898060 3.898060
## 3.9495232751503 3.949523 NA 1 1.7782669 6.120780 3.949523 3.949523
## 4.14598014312126 4.145980 NA 1 1.9747238 6.317237 4.145980 4.145980
## 4.37344829577311 4.373448 NA 1 2.2021919 6.544705 4.373448 4.373448
## 4.89452270907168 4.894523 NA 1 2.7232663 7.065779 4.894523 4.894523
## 5.21142720534249 5.211427 NA 1 3.0401708 7.382684 5.211427 5.211427
## 5.40758761965103 5.407588 NA 1 3.2363312 7.578844 5.407588 5.407588
## 6.23316600928143 6.233166 NA 1 4.0619096 8.404422 6.233166 6.233166
## 6.78691638054318 6.786916 NA 1 4.6156600 8.958173 6.786916 6.786916
##
## Alpha: 0.05 ; DF Error: 13
## Critical Value of t: 2.160369
##
## least Significant Difference: 3.07062
##
## Treatments with the same letter are not significantly different.
##
## methods groups
## 6.78691638054318 6.786916 a
## 6.23316600928143 6.233166 ab
## 5.40758761965103 5.407588 abc
## 5.21142720534249 5.211427 abc
## 4.89452270907168 4.894523 abc
## 4.37344829577311 4.373448 abcd
## 4.14598014312126 4.145980 abcd
## 3.9495232751503 3.949523 abcde
## 3.89805984091619 3.898060 abcde
## 3.73719281884655 3.737193 abcde
## 3.56520491593201 3.565205 bcde
## 3.37977444523543 3.379774 bcde
## 2.51188643150958 2.511886 cde
## 1.74110112659225 1.741101 de
## 1 1.000000 e
qqnorm(method)
qqline(method)
a)No All methods does not have the same effect on the mean particle count
from above LSD table we can see that treatments with the same letters are same
There are no pottential concerns as we can see from normal probability plot the data seems to be normally distributed