Q1a)
library(pwr)
pwr.anova.test(k=4,n=NULL,f=sqrt(((1)^2)/4.5) ,sig.level = 0.05,power = 0.8)
##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 13.28401
## f = 0.4714045
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group
Answer to Q1)a) We will have to take 14 samples from each group
Q1b)
library(pwr)
pwr.anova.test(k=4,n=NULL,f=sqrt(((0.5)^2)/4.5) ,sig.level = 0.05,power = 0.8)
##
## Balanced one-way analysis of variance power calculation
##
## k = 4
## n = 50.04922
## f = 0.2357023
## sig.level = 0.05
## power = 0.8
##
## NOTE: n is number in each group
Answer to Q1)b) We will have to take 51 samples from each group.
Q2)
FType <- c("1","1","1","1","1","1","2","2","2","2","2","2","3","3","3","3","3","3","4","4","4","4","4","4")
Obs <- c(17.6,18.9,16.3,17.4,20.1,21.6,16.9,15.3,18.6,17.1,19.5,20.3,21.4,23.6,19.4,18.5,20.5,22.3,19.3,21.1,16.9,17.5,18.3,19.8)
dat1 <- cbind(FType,Obs)
dat1 <- as.data.frame(dat1)
dat1$FType <- as.factor(dat1$FType)
dat1$Obs <- as.numeric(dat1$Obs)
Q2)a)
library(pwr)
pwr.anova.test(k=4,n=6,f=sqrt(((1)^2)/var(dat1$Obs)) ,sig.level = 0.10,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
Answer to Q2)a) Power is 56.18%
Q2)b)
Null Hypotheses : Ho : \(H_o: \mu_1= \mu_2 = \mu_3 = \mu_4 = \mu\)
Alternative Hypothese : Ha: at least one of the \(\mu_i\) differs
library(pwr)
first.model <- aov(dat1$Obs~dat1$FType,data = dat1)
summary(first.model)
## Df Sum Sq Mean Sq F value Pr(>F)
## dat1$FType 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
Answer to Q2)b) We reject Ho as p value < 0.1 , hence the mean differ
Q2)c)
plot(first.model)
Answer Q2)c) From the plots we can see residual points fall in fairly straight line , hence assumption of normality is adequate .
The Residual Vs Fitted plot shows that we have a constant variance , as the maximum and minimum points of all groups fall as in a rectangular.
As we know for adequacy we have two assumptions i.e. Normality and Constant variance , and in our experiment both are adequate.
Q2)d)
tukey_firstmodel <- TukeyHSD(first.model,conf.level = 0.90)
plot(tukey_firstmodel)
Answer Q2)d) From Tukey HSD and from plot we found out that pair 3-2 differ from other with family wise error of alpha = 0.10 .