question 3.7

library(agricolae)
t1 <- c(3129,3000,2865,2890)
t2 <- c(3200,3300,2975,3150)    
t3 <- c(2800,2900,2985,3050)
t4 <- c(2600,2700,2600,2765)
strength <- c(t1,t2,t3,t4)
technique <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
d1 <- cbind(technique,strength)
d1 <- as.data.frame(d1)
d1$technique <- as.factor(d1$technique)
aov.model<-aov(strength~technique,data=d1)
summary(aov.model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## technique    3 489740  163247   12.73 0.000489 ***
## Residuals   12 153908   12826                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
t <- 2.179 
MSE <- 12826 
n = 4 
LSD = t*sqrt(2*MSE/n)
library(car)
## Loading required package: carData
scatterplot(strength~technique)

plot(aov.model)

#mu1,mu2,mu3,mu4 all the means are equal of null hypothesis,alternative hypothesis = AT least one mean is different #FROM the result, The only pair of means that we would reject is mu1 and mu3, because the difference in means is less than the LSD #The plot shows the variances are close to each other, that has constant variance

problem 3.10

T1 <- c(7,7,15,11,9)
T2 <- c(12,17,12,18,18)
T3 <- c(14,19,19,18,18)
T4 <- c(19,25,22,19,23)
T5 <- c(7,10,11,15,11)
strength <- c(T1,T2,T3,T4,T5)
group <- c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5)
df <- cbind(group,strength)
df <- as.data.frame(df)
df$group <- as.factor(df$group)
aov.model<-aov(strength~group,data=df)
summary(aov.model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## group        4  475.8  118.94   14.76 9.13e-06 ***
## Residuals   20  161.2    8.06                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
library(agricolae)
LSD.test(aov.model,"group",p.adj = "none",console=TRUE)
## 
## Study: aov.model ~ "group"
## 
## LSD t Test for strength 
## 
## Mean Square Error:  8.06 
## 
## group,  means and individual ( 95 %) CI
## 
##   strength      std r       LCL      UCL Min Max
## 1      9.8 3.346640 5  7.151566 12.44843   7  15
## 2     15.4 3.130495 5 12.751566 18.04843  12  18
## 3     17.6 2.073644 5 14.951566 20.24843  14  19
## 4     21.6 2.607681 5 18.951566 24.24843  19  25
## 5     10.8 2.863564 5  8.151566 13.44843   7  15
## 
## Alpha: 0.05 ; DF Error: 20
## Critical Value of t: 2.085963 
## 
## least Significant Difference: 3.745452 
## 
## Treatments with the same letter are not significantly different.
## 
##   strength groups
## 4     21.6      a
## 3     17.6      b
## 2     15.4      b
## 5     10.8      c
## 1      9.8      c
plot(aov.model)

#mu1,mu2,mu3,mu4 all the means are equal of null hypothesis,alternative hypothesis = AT least one mean is different # The residuals have normal shape, so the variance are equal # we got the normality assumptions and variance assumptions are satisfied and graphs plotted.

problem 3.44

library(pwr)
mu1 = 50
mu2 = 60
mu3 = 50  
mu4 = 60
v = 25  
avg_u = (mu1+mu2+mu3+mu4)/4  
pwr.anova.test(k=4,n=NULL,f=sqrt(10^2/v),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

#n = 2.170367, so 3 observations are needed from each population

problem 3.45

mu1 = 50
mu2 = 60
mu3 = 50   
mu4 = 60
v = 36  
avg_u = (mu1+mu2+mu3+mu4)/4  
pwr.anova.test(k=4,n=NULL,f=sqrt(10^2/v),sig.level=0.05,power=0.90)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 2.518782
##               f = 1.666667
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group
v=49
pwr.anova.test(k=4,n=NULL,f=sqrt(10^2/49),sig.level=0.05,power=0.90)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 2.939789
##               f = 1.428571
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

#n = 2.518782, so 3 observations are needed from each population #n = 2.939789, so 3 observations are needed from each population # As the variance incresres the number of samples increases for the samples needed for the possible variance # To estimate standard variance we dont have enough data