DOE HW WEEK 6
Sahil Gaikwad
10/8/2021
Question 3.23
f1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
f2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
f3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
f4<-c(19.3,21.1,16.9,17.5,18.3,19.8)
readings<-c(f1,f2,f3,f4)
fluidtype<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6))
fluidtype<-as.factor(fluidtype)
d<-cbind(readings,fluidtype)
d<-as.data.frame(d)
str(d)## 'data.frame': 24 obs. of 2 variables:
## $ readings : num 17.6 18.9 16.3 17.4 20.1 21.6 16.9 15.3 18.6 17.1 ...
## $ fluidtype: num 1 1 1 1 1 1 2 2 2 2 ...aov.model<-aov(readings~fluidtype)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## fluidtype 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 ' ' 1plot(aov.model)
library(agricolae)
lsdtest<-LSD.test(aov.model,"fluidtype",20,3.30,console = TRUE)##
## Study: aov.model ~ "fluidtype"
##
## LSD t Test for readings
##
## Mean Square Error: 3.299667
##
## fluidtype, means and individual ( 95 %) CI
##
## readings std r LCL UCL Min Max
## 1 18.65000 1.952178 6 17.10309 20.19691 16.3 21.6
## 2 17.95000 1.854454 6 16.40309 19.49691 15.3 20.3
## 3 20.95000 1.879096 6 19.40309 22.49691 18.5 23.6
## 4 18.81667 1.554885 6 17.26975 20.36358 16.9 21.1
##
## Alpha: 0.05 ; DF Error: 20
## Critical Value of t: 2.085963
##
## least Significant Difference: 2.187666
##
## Treatments with the same letter are not significantly different.
##
## readings groups
## 3 20.95000 a
## 4 18.81667 ab
## 1 18.65000 b
## 2 17.95000 bplot(lsdtest)
a)
Testing Hypothesis:-
Null Hypothesis : \(H_o : \mu_1 = \mu_2 = \mu_3 = .... \mu_i = \mu\)
Alternative Hypothesis : \(H_a\) : At least one of the \(\mu_i\) differs
As p=0.0525 value is slightly greater than 0.05, hence we conclude that we failed to reject the null hypothesis and there is indication that fluid differs from each other
Using the Fisher LSD method with alpha=0.05 to make comparisons between pairs of means
From the LSD method we can conclude that t3 is significantly different from t1 and t2 as it does not lie in the same group at 5% level of significance and the least Significant Difference: 2.187666
mean(f1)## [1] 18.65mean(f2)## [1] 17.95mean(f3)## [1] 20.95mean(f4)## [1] 18.81667print(mean(f1))## [1] 18.65print(mean(f2))## [1] 17.95print(mean(f3))## [1] 20.95print(mean(f4))## [1] 18.81667sd(f1)## [1] 1.952178sd(f2)## [1] 1.854454sd(f3)## [1] 1.879096sd(f4)## [1] 1.554885b)
The mean of f1=18.65, f2=17.95, f3=20.95 and f4=18.81667. Also according to Fisher LSD test results, fluid 3 is different from the others, and it’s average life also exceeds the average lives of the other three fluid types. Hence, we will select fluid type 3 for long life objective
f1<-rnorm(6,mean=18.65,sd=1.952178)
f2<-rnorm(6,mean=17.95,sd=1.854454)
f3<-rnorm(6,mean=20.95,sd=1.879096)
f4<-rnorm(6,mean=18.81667,sd=1.554885)
fluidtype<-c(f1,f2,f3,f4)
x<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6))
meanx<-c(rep(mean(f1),6),rep(mean(f2),6),rep(mean(f3),6),rep(mean(f4),6))
residuals<-fluidtype-meanx
qqnorm(residuals)
qqline(residuals)
plot(meanx,residuals,xlab="Life (in h) at 35 kV Load",ylab="residuals",
main="constant variance check")
c)
According to the normal probability plot, we can conclude that the data is fairly distributed but it has some points that do not lie along the line in the upper region. This might indicate the potential outliers in the data. Hence we can conclude that variances are fairly constant and assumption is satisfied
Question 3.28
minutes<-c(110,157,194,178,1,2,4,18,880,1256,5276,4355,495,7040,5307,10050,7,5,29,2)
types<-c(rep(1,4),rep(2,4),rep(3,4),rep(4,4),rep(5,4))
readings<-data.frame(minutes,types)
readings$types<-as.factor(readings$types)
str(readings)## 'data.frame': 20 obs. of 2 variables:
## $ minutes: num 110 157 194 178 1 ...
## $ types : Factor w/ 5 levels "1","2","3","4",..: 1 1 1 1 2 2 2 2 3 3 ...library(agricolae)
aov.model<-aov(minutes~types,data=readings)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## types 4 103191489 25797872 6.191 0.00379 **
## Residuals 15 62505657 4167044
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(aov.model)
Testing Hypothesis:-
Null Hypothesis : \(H_o : \mu_1 = \mu_2 = \mu_3 = .... \mu_i = \mu\)
Alternative Hypothesis : \(H_a\) : At least one of the \(\mu_i\) differs
a)
As the p-value is 0.00379 which is smaller than the 0.05, hence we reject the null hypothesis.
m1<-rnorm(4,mean=159.75,sd=36.46345)
m2<-rnorm(4,mean=6.25,sd=7.9320026)
m3<-rnorm(4,mean=2941.75,sd=2201.40597)
m4<-rnorm(4,mean=5723,sd=3998.43544)
m5<-rnorm(4,mean=10.75,sd=12.338962)
material<-c(m1,m2,m3,m4,m5)
x<-c(rep(1,4),rep(2,4),rep(3,4),rep(4,4),rep(5,4))
meanx<-c(rep(mean(m1),4),rep(mean(m2),4),rep(mean(m3),4),rep(mean(m4),4),rep(mean(m5),4))
residuals<-material-meanx
qqnorm(residuals)
qqline(residuals)
plot(meanx,residuals,xlab="Failure Time (minutes)",ylab="residuals",
main="constant variance check")
b)
From the plot of residuals versus predicted we can conclude that the variance of the original observations is not constant.
From the residuals plotted in the normal probability plot we can conclude that the normality assumption is invalid as data is not normally distributed and a data transformation is required
library(MASS)
boxplot(minutes~readings$types)
boxcox(minutes~types)
Trans_minutes<-log(minutes)
plot(Trans_minutes~readings$types,xlab="Materials",ylab="Failure Time",main="Variance check")
trans_aov.model<-aov(Trans_minutes~readings$types)
summary(trans_aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## readings$types 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 ' ' 1plot(trans_aov.model)
From boxplot we found that variances were not stable, so we performed the boxcox transformation to stabilize the variances of fluid types.
As value of lambda is almost equal to zero, we will perform the log transformation to stablize the data.
As the p-value is 1.18e-07which is smaller than the 0.05, hence we reject the null hypothesis.
c)
There is nothing unusual about the residual plots when the transformation is applied.
Question 3.29
count<-c(31,10,21,4,1,62,40,24,30,35,53,27,120,97,68)
method<-c(rep(1,5),rep(2,5),rep(3,5))
observations<-data.frame(count,method)
observations$method<-as.factor(observations$method)
str(observations)## 'data.frame': 15 obs. of 2 variables:
## $ count : num 31 10 21 4 1 62 40 24 30 35 ...
## $ method: Factor w/ 3 levels "1","2","3": 1 1 1 1 1 2 2 2 2 2 ...aov.model<-aov(count~method,data=observations)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## method 2 8964 4482 7.914 0.00643 **
## Residuals 12 6796 566
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(aov.model)
lsdtest<-LSD.test(aov.model,"method",12,566,console = TRUE)##
## Study: aov.model ~ "method"
##
## LSD t Test for count
##
## Mean Square Error: 566.3333
##
## method, means and individual ( 95 %) CI
##
## count std r LCL UCL Min Max
## 1 13.4 12.46194 5 -9.788411 36.58841 1 31
## 2 38.2 14.56709 5 15.011589 61.38841 24 62
## 3 73.0 36.48972 5 49.811589 96.18841 27 120
##
## Alpha: 0.05 ; DF Error: 12
## Critical Value of t: 2.178813
##
## least Significant Difference: 32.79336
##
## Treatments with the same letter are not significantly different.
##
## count groups
## 3 73.0 a
## 2 38.2 b
## 1 13.4 bplot(lsdtest)
Testing Hypothesis:-
Null Hypothesis : \(H_o : \mu_1 = \mu_2 = \mu_3 = .... \mu_i = \mu\)
Alternative Hypothesis : \(H_a\) : At least one of the \(\mu_i\) differs
a)
As the p-value is 0.00643 which is smaller than the 0.05, hence we reject the null hypothesis.
Using the Fisher LSD method with alpha=0.05 to make comparisons between pairs of means
From the LSD method we can conclude that m3 is significantly different from m1 and m2 as it does not lie in the same group at 5% level of significance and the least Significant Difference: 32.79336
Hence we can conclude that, at least one method has a different effect on mean particle count.
m1<-rnorm(5,mean=13.40,sd=12.46194)
m2<-rnorm(5,mean=38.2,sd=14.567086)
m3<-rnorm(5,mean=73,sd=36.489724)
method<-c(m1,m2,m3)
x<-c(rep(1,5),rep(2,5),rep(3,5))
meanx<-c(rep(mean(m1),5),rep(mean(m2),5),rep(mean(m3),5))
residuals<-method-meanx
qqnorm(residuals)
qqline(residuals)
plot(meanx,residuals,xlab="Failure Time (minutes)",ylab="residuals",
main="constant variance check")
b)
From the plot of residuals versus predicted plot we can conclude that the variance of the original observations is not constant, which indicates that the variance assumption is not valid and a data transformation is required
library(MASS)
boxplot(count~observations$method)
boxcox(count~method)
lambda<-0.4
Trans_method<-method^(lambda)
boxplot(Trans_method~observations$method)
transdata<-cbind(Trans_method,method)
transdata<-data.frame(transdata)
transdata$method<-as.factor(transdata$method)
str(transdata$method)## Factor w/ 15 levels "-1.53216192975142",..: 1 4 3 5 2 11 6 9 10 8 ...transmodel<-aov(Trans_method~method,data=transdata$method)
summary(transmodel)## Df Sum Sq Mean Sq F value Pr(>F)
## method 1 21.102 21.102 196.2 8.49e-09 ***
## Residuals 12 1.291 0.108
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 1 observation deleted due to missingnessplot(transmodel)
As the p-value is 1.35e-08which is smaller than the 0.05, hence we reject the null hypothesis.
Question 3.51
Kruskal–Wallis test for problem no.3.23
Null Hypothesis : \(H_o : \mu_1 = \mu_2 = \mu_3 = .... \mu_i = \mu\)
Alternative Hypothesis : \(H_a\) : At least one of the \(\mu_i\) differs
f1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
f2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
f3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
f4<-c(19.3,21.1,16.9,17.5,18.3,19.8)
readings<-c(f1,f2,f3,f4)
fluidtype<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6))
fluidtype<-as.factor(fluidtype)
d<-cbind(readings,fluidtype)
d<-as.data.frame(d)
kruskal.test(readings~fluidtype,data=d)##
## Kruskal-Wallis rank sum test
##
## data: readings by fluidtype
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015From the Kruskal-Wallis Rank Sum Test we get, P value is 0.1015which is higher than 0.05 , hence we fail to reject the Null Hypothesis.