Question 1


\(H_0\) : \(u_1\) = \(u_2\) = \(u_3\) = \(u_4\)
\(H_a\) : at least one \(u_I\) differs from the others
alpha = 0.05

Part a


dat<-read.csv(file.choose())
aov.model<-aov(Life~Fluid.Type,data=dat) #weight: measurement group: divided by 
summary(aov.model)
##             Df Sum Sq Mean Sq F value Pr(>F)
## Fluid.Type   1   3.67   3.675   0.874   0.36
## Residuals   22  92.48   4.204


Conclusion


Since p value = 0.874 the test fails to reject H0, which means that the life of fluids is the same with a confidence level of 95%


Part b


plot(aov.model)


Conclusion


There are two clues given by the previous plots that confirm that the model is adequate:
a. Normality: the second plot shows the probability plot for our data. The different points fit an almost perfect straight line. Therefore, our data can be considered as normal.


b. Constant variance: The residuals vs fitted plot shows that the variances for the four populations are very similar. They do not form an irregular shape nor they have significant differences between them.

Question 2


\(u_1\) = 50
\(u_2\) = 60
\(u_3\) = 50
\(u_4\) =60
alpha =0.05
power =0.9
\(sigma^2\) =25 

k=4
u1=50
u2=60
u3=50
u4=60
uI=55
f = sqrt((10)^2/25)
library(pwr)
pwr.anova.test(k=4,n=NULL,f,sig.level=0.05, power=0.90)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 2.170367
##               f = 2
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

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Conclusion


A minimum of 3 samples should be taken from each population with a 90% of probability of rejecting the null hypothesis of equal population means. In other words, 3 samples are required to determine a difference of 10 between the means.

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CODE USE

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dat<-read.csv(file.choose())
aov.model<-aov(Life~Fluid.Type,data=dat)  
summary(aov.model)
plot(aov.model)
k=4
u1=50
u2=60
u3=50
u4=60
uI=55
f = sqrt((10)^2/25)
library(pwr)
?pwr.anova.test
pwr.anova.test(k=4,n=NULL,f,sig.level=0.05, power=0.90)