#Question 3.7
#NULL HYPOTHESES: All means are same
#ALTERNATE HYPOTHESES: Atleast one mean is different
#install.packages("agricolae")
library(agricolae)
#importing the data
#answer c
t1<-c(3129, 3000, 2865, 2890)
t2<-c(3200, 3300, 2975, 3150)
t3<-c(2800, 2900, 2985, 3050)
t4<-c(2600, 2700, 2600, 2765)
s<-c(t1,t2,t3,t4)
tec<-c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
d<-cbind(tec,s)
df<-as.data.frame(d)
df$tec<-as.factor(df$tec)
aov.model<-aov(s~tec,data = df)
summary(aov.model)
## Df Sum Sq Mean Sq F value Pr(>F)
## tec 3 489740 163247 12.73 0.000489 ***
## Residuals 12 153908 12826
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#As P value is less than alpha we reject Null Hypotheses
#question 3.44
#LSD
lsd<-LSD.test(s,tec,12,12836,console = TRUE)
##
## Study: s ~ tec
##
## LSD t Test for s
##
## Mean Square Error: 12836
##
## tec, means and individual ( 95 %) CI
##
## s std r LCL UCL Min Max
## 1 2971.00 120.55704 4 2847.575 3094.425 2865 3129
## 2 3156.25 135.97641 4 3032.825 3279.675 2975 3300
## 3 2933.75 108.27242 4 2810.325 3057.175 2800 3050
## 4 2666.25 80.97067 4 2542.825 2789.675 2600 2765
##
## Alpha: 0.05 ; DF Error: 12
## Critical Value of t: 2.178813
##
## least Significant Difference: 174.55
##
## Treatments with the same letter are not significantly different.
##
## s groups
## 2 3156.25 a
## 1 2971.00 b
## 3 2933.75 b
## 4 2666.25 c
plot(lsd)
#answer d
meanstr<-c(rep(mean(t1),4),rep(mean(t2),4),rep(mean(t3),4),rep(mean(t4),4))
residual<-s-meanstr
qqnorm(residual)
#from normal probability we can say that given data is normal
plot(meanstr,residual, xlab="tensile strength",ylab="residual",main="variance")
#answer e
#from the plot we can conclude that there is not much variation in data. so we can say that variance for different technique is equal
#answer f
#install.packages("car",dependencies = TRUE)
library(car)
## Loading required package: carData
scatterplot(s~tec)
#Question 3.10
#NULL HYPOTHESIS:H0:MU1=MU2=MU3=MU4
#ALTERNATE HYPOTHESIS:H1:atleast one of the mean is different
#Question 3.10 b,c
P1 <- c(7, 7, 15, 11, 9)
P2 <- c(12, 17, 12, 18, 18)
P3 <- c(14, 19, 19, 18, 18)
P4 <- c(19, 25, 22, 19, 23)
P5 <- c(7, 10, 11, 15, 11)
s<-c(P1,P2,P3,P4,P5)
g<-c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5)
d<-cbind(g,s)
df<-as.data.frame(d)
df$g<-as.factor(df$g)
aov.model<-aov(s~g,data = df)
summary(aov.model)
## Df Sum Sq Mean Sq F value Pr(>F)
## g 4 475.8 118.94 14.76 9.13e-06 ***
## Residuals 20 161.2 8.06
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#Answer b
#the fisher LSDmethod states that pair mean of 2 and 3 are slightly equal as wellas pair mean of 1 and 5
library(agricolae)
lsd<-LSD.test(s,g,20,8.06,console = TRUE)
##
## Study: s ~ g
##
## LSD t Test for s
##
## Mean Square Error: 8.06
##
## g, means and individual ( 95 %) CI
##
## s std r LCL UCL Min Max
## 1 9.8 3.346640 5 7.151566 12.44843 7 15
## 2 15.4 3.130495 5 12.751566 18.04843 12 18
## 3 17.6 2.073644 5 14.951566 20.24843 14 19
## 4 21.6 2.607681 5 18.951566 24.24843 19 25
## 5 10.8 2.863564 5 8.151566 13.44843 7 15
##
## Alpha: 0.05 ; DF Error: 20
## Critical Value of t: 2.085963
##
## least Significant Difference: 3.745452
##
## Treatments with the same letter are not significantly different.
##
## s groups
## 4 21.6 a
## 3 17.6 b
## 2 15.4 b
## 5 10.8 c
## 1 9.8 c
plot(lsd)
#answer c
#Normal probability plot of residuals is almost very close to normality as the data lies in single line
plot(aov.model)
#Question 3.44
#how many observations must be taken
#given data
p1=50
p2=60
p3=50
p4=60
v=25
library(pwr)
pwr.anova.test(k=4,n=NULL, f=sqrt((5)^2/25),sig.level = 0.05,power=0.90)
##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 4.658119
## f = 1
## sig.level = 0.05
## power = 0.9
##
## NOTE: n is number in each group
#Therefore by the power test number of observations to be selected from each populations is approximately equal to 5
#Question 3.45
#given data
p1=50
p2=60
p3=50
p4=60
v=36
library(pwr)
pwr.anova.test(k=4, n=NULL,f=sqrt((5)^2/36),sig.level = 0.05,power=0.90)
##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 6.180857
## f = 0.8333333
## sig.level = 0.05
## power = 0.9
##
## NOTE: n is number in each group
#Answer a
#Therefore by the power test number of observations to be selected from each populations is approximately equal to 7
#Answer b
#given data
p1-50
## [1] 0
p2=60
p3=50
p4=60
v=49
library(pwr)
pwr.anova.test(k=4,n=NULL,f=sqrt((5)^2/49),sig.level=0.05,power=0.90)
##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 7.998751
## f = 0.7142857
## sig.level = 0.05
## power = 0.9
##
## NOTE: n is number in each group
#Therefore by the power test number of observations to be selected from each populations is approximately equal to 8
#Answer c
#As the variance increases number of samples to be collected from each population also increases inorder to obtain the results with power level
#Answer d
# In General, when an experiment has to be set up and samples have to be collected, it would be a better approach to determine the upper and lower limits of variance and then make the best possible estimate of number of samples to be collected.