---

question 1

min= 18 , max= 20 , variance = 3.5 , alpha= 0.05 , power= 0.8

min case:

min=18

max =20

( 18+20 ) / 2= 19

\(\mu=( 18, 19, 19, 20)\)

\(\bar{\mu} = \frac{18+19+19+20}{4}=19\)

\[\sigma_{\mu}^{2} = \frac{(18-19)^2 +(19-19)^2 + (19-19)^2 + (20-19)^2}{4}= 0.5\]

variance between (18,19,19,20) = 0.5

power.anova.test(groups=4, n=NULL, between.var=var(c(18,19,19,20)), within.var= 3.5, sig.level=0.05, power= 0.80)

## 
##      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

sample size in min case= 21

Intermediate case :

(20-18) /( 4-1 ) = 0.667

\[\mu = ( 18 , 18.667 , 19.334, 20) \]

\(\bar{\mu} = \frac{18+18.667+19.334+20}{4}=19\)

power.anova.test(groups=4, n=NULL , between.var=var(c(18,18.667,19.334,20)), within.var=3.5, sig.level=0.05, power= 0.8)

## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 4
##               n = 18.17695
##     between.var = 0.7408149
##      within.var = 3.5
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group

sample size in intermediate case= 19

max case :

\(\mu = (18 , 18, 20 , 20 )\)

power.anova.test(groups= 4, n= NULL, between.var = var(c(18,18,20,20)),within.var=3.5, sig.level=0.05, power=0.8)

## 
##      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

sample size in max case = 11

question 2

\(H_0: \mu_1=\mu_2=\mu_3=\mu_4\)

\(H_a:\) At least one mean is different

fluid1<-c ( 17.6,18.9,16.3,17.4,20.1,21.6)
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)
dat<-data.frame(fluid1,fluid2,fluid3,fluid4)
dat

##   fluid1 fluid2 fluid3 fluid4
## 1   17.6   16.9   21.4   19.3
## 2   18.9   15.3   23.6   21.1
## 3   16.3   18.6   19.4   16.9
## 4   17.4   17.1   18.5   17.5
## 5   20.1   19.5   20.5   18.3
## 6   21.6   20.3   22.3   19.8

library(tidyr)
dat<-pivot_longer(dat,c(fluid1,fluid2,fluid3,fluid4))
aov.model<-aov(value~name,data=dat)
summary(aov.model)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## name         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

reject \(H_0\) , p-value= 0.0525 < alpha= 0.10.

plot(aov.model)

yes, the model is adequate because the residual plots show no obvious pattern, the residuals are

approximately normally distributed and the variance appears to be reasonably constant.

TukeyHSD(aov.model,conf.level=0.90)

##   Tukey multiple comparisons of means
##     90% family-wise confidence level
## 
## Fit: aov(formula = value ~ name, data = dat)
## 
## $name
##                     diff        lwr       upr     p adj
## fluid2-fluid1 -0.7000000 -3.2670196 1.8670196 0.9080815
## fluid3-fluid1  2.3000000 -0.2670196 4.8670196 0.1593262
## fluid4-fluid1  0.1666667 -2.4003529 2.7336862 0.9985213
## fluid3-fluid2  3.0000000  0.4329804 5.5670196 0.0440578
## fluid4-fluid2  0.8666667 -1.7003529 3.4336862 0.8413288
## fluid4-fluid3 -2.1333333 -4.7003529 0.4336862 0.2090635

plot(TukeyHSD(aov.model))

fluid 2 and fluid 3 are significantly different because the adjusted p-value = 0.0440578 < alpha=0.10.