Homework 6
Rishidhar Rao Avirineni
2022-10-09
#Question 3.23
#Null Hypothesis (H0):μ1=μ2=μ3=μ4
#Alternative Hypothesis (Ha):At least one of the μ1,μ2,μ3,μ4 differs.
Fluid_type1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
Fluid_type2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
Fluid_type3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
Fluid_type4<-c(19.3,21.1,16.9,17.5,18.3,19.8)
Fluid_type<-c(Fluid_type1,Fluid_type2,Fluid_type3,Fluid_type4)
group1 <- as.factor(c(rep(1,6),rep(2,6),rep(3,6),rep(4,6)))
Fluid_life<- as.factor(cbind(Fluid_type,group1))
test1<-aov(Fluid_type~group1)
summary(test1)## Df Sum Sq Mean Sq F value Pr(>F)
## group1 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 ' ' 1μ1<-mean(Fluid_type1)
μ2<-mean(Fluid_type2)
μ3<-mean(Fluid_type3)
μ4<-mean(Fluid_type4)
plot(test1)
#(a) Is there any indication that the fluids differ? Use α = 0.05. #Answer: p-value(0.0525) > 0.05 , so we fail to reject null hypothesis. #and there is no differ in the fluids
#(b) Which fluid would you select, given that the objective is long life? #Answer:The mean life of fluid_type 3 is greater .so i would choose this.
#(c) Analyze the residuals from this experiment. Are the basic analysis of #variance assumptions satisfied? #Answer:By Checking both the assumption of normality and constant variance #therefore basic analysis of variance assumptions satisfied.
#Question 3.28
#(a)
Material1<- c(110,157,194,178)
Material2<- c(1,2,4,18)
Material3<- c(880,1256,5276,4355)
Material4<- c(495,7040,5307,10050)
Material5<- c(7,5,29,2)
Material<- c(Material1,Material2,Material3,Material4,Material5)
group2 <- as.factor(c(rep(1,4),rep(2,4),rep(3,4),rep(4,4),rep(5,4)))
FT <- as.data.frame(cbind(group2,Material))
test2 <- aov(Material~group2)
summary(FT)## group2 Material
## Min. :1 Min. : 1.0
## 1st Qu.:2 1st Qu.: 6.5
## Median :3 Median : 167.5
## Mean :3 Mean : 1768.3
## 3rd Qu.:4 3rd Qu.: 2030.8
## Max. :5 Max. :10050.0#(b)
plot(test2)
library(MASS)
boxcox(Material~group2) ## Lambda(λ)=0
Materialtest<- log(Material) ## Here we are using log transformation
test3 <- aov(Material~group2)
summary(test3)## Df Sum Sq Mean Sq F value Pr(>F)
## group2 4 103191489 25797872 6.191 0.00379 **
## Residuals 15 62505657 4167044
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(test3)
boxplot(Materialtest~group2)
#(a) Do all five materials have the same effect on mean failure time? #Answer: The p-value < 0.05, we reject the Null Hypothesis(H0).
#(b) Plot the residuals versus the predicted response. Construct a normal #probability plot of the residuals.What information is conveyed by these plots? #Answer: By observing plot of Test 2 data is neither normal or have constant variance
#(c) Based on your answer to part (b) conduct another analysis of the failure time data and draw appropriate conclusions. #Answer: p-value < .05, we reject the Null Hypothesis(H0). #By observing plot of Test 3 data is approximately normal and doe not have constant variance #From the box plot is equal to material 1 has least sd of all the materials.
#(a)
Method1<-c(31,10,21,4,1)
Method2<-c(62,40,24,30,35)
Method3<-c(53,27,120,97,68)
Method <-c(Method1,Method2,Method3)
group3 <- as.factor(c(rep(1,5),rep(2,5),rep(3,5)))
Method_count<- as.data.frame(cbind(Method,group3))
Method_count<- aov(Method~group3)
summary(Method_count)## Df Sum Sq Mean Sq F value Pr(>F)
## group3 2 8964 4482 7.914 0.00643 **
## Residuals 12 6796 566
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1#(b)
plot(Method_count)
#(c)
library(MASS)
boxcox(Method~group3)
# λ=0.5
trans_count <- Method^0.5
qqnorm(trans_count)
plot(trans_count~group3)
kruskal.test(Method~group3)##
## Kruskal-Wallis rank sum test
##
## data: Method by group3
## Kruskal-Wallis chi-squared = 8.54, df = 2, p-value = 0.01398#(a) Do all methods have the same effect on mean particle count? #Answer:The p-value <0.05, we reject the Null Hypothesis(H0).
#(b)Plot the residuals versus the predicted response. Construct a normal #probability plot of the residuals. Are there potential concerns about the #validity of the assumptions? #Answer:From observed plot we conclude that data is approximately normal and unequal variance
#(c)Based on your answer to part (b) conduct another analysis of the particle #count data and draw appropriate conclusions. #Answer:After transformation of the data, we have unequal variances. Therefore #it is better to use non parametric test.
#Question 3.51 & 3.52
kruskal.test(Fluid_type~group1)##
## Kruskal-Wallis rank sum test
##
## data: Fluid_type by group1
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015#Question 3.51 & 3.52 #Use the Kruskal–Wallis test for the experiment in Problem 3.23. Compare the #conclusions obtained with those from the usual analysis of variance.