Flip Assignment 9 Group 4
Sams Zubayer Mridha, Ken, Venkata
2025-09-30
Answer to Question No 1
#For Minimum Variability
power.anova.test(groups = 4, n = NULL,
between.var =var(c(18,19,19,20)), within.var = 3.5,
sig.level = 0.05, power = .8)##
## Balanced one-way analysis of variance power calculation
##
## groups = 4
## n = 20.08368
## between.var = 0.6666667
## within.var = 3.5
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group#For Intermediate Variability
power.anova.test(groups = 4, n = NULL,
between.var =var(c(18,18.66,19.33,20)), within.var = 3.5,
sig.level = 0.05, power = .8)##
## Balanced one-way analysis of variance power calculation
##
## groups = 4
## n = 18.16131
## between.var = 0.7414917
## within.var = 3.5
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group#For Maximum Variability
power.anova.test(groups = 4, n = NULL,
between.var =var(c(18,18,20,20)), within.var = 3.5,
sig.level = 0.05, power = .8)##
## Balanced one-way analysis of variance power calculation
##
## groups = 4
## n = 10.56952
## between.var = 1.333333
## within.var = 3.5
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group##We can consider minimum variability number of Sample needs to collect 21 ##We can consider intermidiate variability number of Sample needs to collect 19 ##We can consider maximum variability number of Sample needs to collect 11
Answer to the Question No:2
library(tidyr)
fluid1<- c(17.6,18.9,16.3,17.4,20.1,21.60)
fluid2<- c(16.9,15.3,18.6,17.1,19.5,20.3)
fluid3<- c(21.4, 23.6, 19.4, 18.5, 20.5,22.3)
fluid4<-c(19.3, 21.1, 16.9, 17.5, 18.3, 19.8)
fluid<- c(fluid1, fluid2, fluid3, fluid4)
fluid.type<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6))
df <- data.frame(fluid.type, fluid)
df$fluid.type <- as.factor(df$fluid.type)
str(df)## 'data.frame': 24 obs. of 2 variables:
## $ fluid.type: Factor w/ 4 levels "1","2","3","4": 1 1 1 1 1 1 2 2 2 2 ...
## $ fluid : num 17.6 18.9 16.3 17.4 20.1 21.6 16.9 15.3 18.6 17.1 ...(a) Test The Hypothesis
model_anova <- aov(fluid~fluid.type, data=df)
summary(model_anova)## Df Sum Sq Mean Sq F value Pr(>F)
## fluid.type 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 ' ' 1##From the summary we can see that P value 0.0525 is less than the alpha value, so we can reject the null hypothesis, life time of the fuilds are differents.
#(b) Model Adequecy
plot(model_anova)
##From the model we can see that the normality and constant variance assumtion holds.
##(C)
Tukey.test <- TukeyHSD(model_anova, conf.level = .9)
plot(Tukey.test)
#from the tukey test we can observe model 2 and 3 differs from the mean
value
#For Minimum Variability
power.anova.test(groups = 4, n = NULL,
between.var =var(c(18,19,19,20)), within.var = 3.5,
sig.level = 0.05, power = .8)
#For Intermediate Variability
power.anova.test(groups = 4, n = NULL,
between.var =var(c(18,18.66,19.33,20)), within.var = 3.5,
sig.level = 0.05, power = .8)
#For Maximum Variability
power.anova.test(groups = 4, n = NULL,
between.var =var(c(18,18,20,20)), within.var = 3.5,
sig.level = 0.05, power = .8)
## Answer to the Question No:2
library(tidyr)
fluid1<- c(17.6,18.9,16.3,17.4,20.1,21.60)
fluid2<- c(16.9,15.3,18.6,17.1,19.5,20.3)
fluid3<- c(21.4, 23.6, 19.4, 18.5, 20.5,22.3)
fluid4<-c(19.3, 21.1, 16.9, 17.5, 18.3, 19.8)
fluid<- c(fluid1, fluid2, fluid3, fluid4)
fluid.type<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6))
df <- data.frame(fluid.type, fluid)
df$fluid.type <- as.factor(df$fluid.type)
str(df)
## 2(a) Test The Hypothesis
model_anova <- aov(fluid~fluid.type, data=df)
summary(model_anova)
#(b) Model Adequecy
plot(model_anova)
#(c) Tukeys Test
Tukey.test <- TukeyHSD(model_anova, conf.level = .9)
plot(Tukey.test)