Flipped Assignment 9
Author 1 “Jai Sai Teja Reddy Devalampeta”
Author 2 “Rahul Vithalani”
Last compiled on September 28, 2023 at 1:59 PM - CDT
1 Assignment on Sample size determination for ANOVA
1.1 Question 1
Given Data
#Question1
k<-4 #no. of populations
Var<- 4.5 #Variance of data
alpha<- 0.05 #Significance level
p <- 0.8 #power1.2 Question 1a
Assuming all three cases of variability with alpha 0.05, performed the Power analysis of ANOVA
#1a
diff1<- 1
library(pwr)
?pwr.anova.test
#Assuming Min Variability
pwr.anova.test(k = 4, n = NULL, f = diff1*sqrt(1/(2*k)),
sig.level = 0.05, power = p)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 22.806
## f = 0.3535534
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group#Assuming Intermediate variability
pwr.anova.test(k = 4, n = NULL, f = (diff1/2)*sqrt((k+1)/(3*(k-1))),
sig.level = 0.05, power = p)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 20.62807
## f = 0.372678
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group#Assuming Max variability and our no. of pop are 4 which is even
pwr.anova.test(k = 4, n = NULL, f = (diff1/2),
sig.level = 0.05, power = p)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 11.92611
## f = 0.5
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group- Comment: Observing the results from the analysis,
- No of samples required in min variability = 23
- No of samples required in Intermediate variability = 21
- No of samples required in max variability = 12
1.3 Question 1b
Assuming all three cases of variability with alpha 0.1
#1b
diff2<- 0.5
#Assuming Min Variability
pwr.anova.test(k = 4, n = NULL, f = diff2*sqrt(1/(2*k)),
sig.level = 0.05, power = p)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 88.20348
## f = 0.1767767
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group#Assuming Intermediate variability
pwr.anova.test(k = 4, n = NULL, f = (diff2/2)*sqrt((k+1)/(3*(k-1))),
sig.level = 0.05, power = p)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 79.4821
## f = 0.186339
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group#Assuming Max variability and our no. of pop are 4 which is even
pwr.anova.test(k = 4, n = NULL, f = (diff2/2),
sig.level = 0.05, power = p)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 44.59927
## f = 0.25
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group- Comment: Observing the results from the analysis,
- No of samples required in min variability = 89
- No of samples required in Intermediate variability = 80
- No of samples required in max variability = 45
1.4 Question 2
Given Data
#Question2
pop1<- c(17.6,18.9,16.3,17.4,20.1,21.6)
pop2<- c(16.9,15.3,18.6,17.1,19.5,20.3)
pop3<- c(21.4,23.6,19.4,18.5,20.5,22.3)
pop4<- c(19.3,21.1,16.9,17.5,18.3,19.8)
data <- data.frame(
Fluid_Type = rep(1:4, each=6),
Life = c(pop1, pop2, pop3,pop4))
data$Fluid_Type<- as.factor(data$Fluid_Type)##Question 2a
#2a
pwr.anova.test(k = 4, n = 6, f = 1/sd(data$Life), sig.level = 0.1, power=NULL)##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 6
## f = 0.4890694
## sig.level = 0.1
## power = 0.5618141
##
## NOTE: n is number in each group- Comment: Power is 56%
##Question 2b
#2b
?aov
anova_result <- aov(Life ~ Fluid_Type, data = data)
summary(anova_result)## Df Sum Sq Mean Sq F value Pr(>F)
## Fluid_Type 3 30.16 10.05 3.047 0.0525 .
## Residuals 20 65.99 3.30
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1- Comment: our p-value is 0.0525, we will reject the null hypothesis at a significance level of 0.10, and we would conclude that there’s a significant difference between mean lives of the tested fluids.
##Quetion 2c
#2c
plot(anova_result)
- Comment: Observed Normality of data and Almost equal variance.
- most value fitted in.
1.5 Question 2d
#2d
?TukeyHSD
TukeyHSD(anova_result, conf.level = 0.9)## Tukey multiple comparisons of means
## 90% family-wise confidence level
##
## Fit: aov(formula = Life ~ Fluid_Type, data = data)
##
## $Fluid_Type
## diff lwr upr p adj
## 2-1 -0.7000000 -3.2670196 1.8670196 0.9080815
## 3-1 2.3000000 -0.2670196 4.8670196 0.1593262
## 4-1 0.1666667 -2.4003529 2.7336862 0.9985213
## 3-2 3.0000000 0.4329804 5.5670196 0.0440578
## 4-2 0.8666667 -1.7003529 3.4336862 0.8413288
## 4-3 -2.1333333 -4.7003529 0.4336862 0.2090635plot(TukeyHSD(anova_result, conf.level = 0.9))
- Comment: only 2-3 fluid type lies outside. and significantly
differ.
2 Complete R Code
#Question1
k<-4 #no. of populations
Var<- 4.5 #Variance of data
alpha<- 0.05 #Significance level
p <- 0.8 #power
#1a
diff1<- 1
library(pwr)
?pwr.anova.test
#Assuming Min Variability
pwr.anova.test(k = 4, n = NULL, f = diff1*sqrt(1/(2*k)),
sig.level = 0.05, power = p)
#Assuming Intermediate variability
pwr.anova.test(k = 4, n = NULL, f = (diff1/2)*sqrt((k+1)/(3*(k-1))),
sig.level = 0.05, power = p)
#Assuming Max variability and our no. of pop are 4 which is even
pwr.anova.test(k = 4, n = NULL, f = (diff1/2),
sig.level = 0.05, power = p)
#1b
diff2<- 0.5
#Assuming Min Variability
pwr.anova.test(k = 4, n = NULL, f = diff2*sqrt(1/(2*k)),
sig.level = 0.05, power = p)
#Assuming Intermediate variability
pwr.anova.test(k = 4, n = NULL, f = (diff2/2)*sqrt((k+1)/(3*(k-1))),
sig.level = 0.05, power = p)
#Assuming Max variability and our no. of pop are 4 which is even
pwr.anova.test(k = 4, n = NULL, f = (diff2/2),
sig.level = 0.05, power = p)
#Question2
pop1<- c(17.6,18.9,16.3,17.4,20.1,21.6)
pop2<- c(16.9,15.3,18.6,17.1,19.5,20.3)
pop3<- c(21.4,23.6,19.4,18.5,20.5,22.3)
pop4<- c(19.3,21.1,16.9,17.5,18.3,19.8)
data <- data.frame(
Fluid_Type = rep(1:4, each=6),
Life = c(pop1, pop2, pop3,pop4))
data$Fluid_Type<- as.factor(data$Fluid_Type)
#2a
pwr.anova.test(k = 4, n = 6, f = 1/sd(data$Life), sig.level = 0.1, power=NULL)
#2b
?aov
anova_result <- aov(Life ~ Fluid_Type, data = data)
summary(anova_result)
#2c
plot(anova_result)
#2d
?TukeyHSD
TukeyHSD(anova_result, conf.level = 0.9)
plot(TukeyHSD(anova_result, conf.level = 0.9))