FlippedAssignment9

1.

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

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=.80)

## 
##      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

power.anova.test(groups=4,n=NULL,between.var=var(c(18,18,20,20)),within.var=3.5,sig.level=0.05,power=.80)

## 
##      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

2a.

FT1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
FT2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
FT3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
FT4<-c(19.3,21.1,16.9,17.5,18.3,19.8)
Life<-c(FT1,FT2,FT3,FT4)
FT<-rep(c("FT1","FT2","FT3","FT4"),each=6)
dat<-data.frame(Life,FT)
model<-aov(Life~FT,data=dat)
summary(model)

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

With the p-value=.0525 being less than \(\alpha=.1\), we have enough evidence to reject the null hypothesis.

2b.

qqnorm(dat$Life)

resmodel<-lm(Life~FT,data=dat)
res<-resid(resmodel)
plot(fitted(resmodel),res)

The normal probability plot appears to be approximately linear with its theoretical and sample percentiles. In addition to the residuals vs predicted life model indicating an even spread of samples across the means, the model appears to be adequate.

2c.

TukeyHSD(model,conf.level=.9)

##   Tukey multiple comparisons of means
##     90% family-wise confidence level
## 
## Fit: aov(formula = Life ~ FT, data = dat)
## 
## $FT
##               diff        lwr       upr     p adj
## FT2-FT1 -0.7000000 -3.2670196 1.8670196 0.9080815
## FT3-FT1  2.3000000 -0.2670196 4.8670196 0.1593262
## FT4-FT1  0.1666667 -2.4003529 2.7336862 0.9985213
## FT3-FT2  3.0000000  0.4329804 5.5670196 0.0440578
## FT4-FT2  0.8666667 -1.7003529 3.4336862 0.8413288
## FT4-FT3 -2.1333333 -4.7003529 0.4336862 0.2090635

plot(TukeyHSD(model,conf.level=.9))

Fluids 2 and 3 significantly differ in mean lifetime.