Untitled

library(agricolae)
MixTechnique<-c(3129,3000,2865,2890,3200,3300,2975,3150,2800,2900,2985,3050,2600,2700,2600,2765)
dat<-c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
str(dat)

##  num [1:16] 1 1 1 1 2 2 2 2 3 3 ...

dat<-as.factor(dat)
str(dat)

##  Factor w/ 4 levels "1","2","3","4": 1 1 1 1 2 2 2 2 3 3 ...

z<-cbind.data.frame(MixTechnique,dat)
Z<-aov(MixTechnique~dat)
summary(z)

##   MixTechnique  dat  
##  Min.   :2600   1:4  
##  1st Qu.:2791   2:4  
##  Median :2938   3:4  
##  Mean   :2932   4:4  
##  3rd Qu.:3070        
##  Max.   :3300

a<-LSD.test(MixTechnique,dat,12,12826,console = TRUE)

## 
## Study: MixTechnique ~ dat
## 
## LSD t Test for MixTechnique 
## 
## Mean Square Error:  12826 
## 
## dat,  means and individual ( 95 %) CI
## 
##   MixTechnique       std r      LCL      UCL  Min  Max
## 1      2971.00 120.55704 4 2847.623 3094.377 2865 3129
## 2      3156.25 135.97641 4 3032.873 3279.627 2975 3300
## 3      2933.75 108.27242 4 2810.373 3057.127 2800 3050
## 4      2666.25  80.97067 4 2542.873 2789.627 2600 2765
## 
## Alpha: 0.05 ; DF Error: 12
## Critical Value of t: 2.178813 
## 
## least Significant Difference: 174.482 
## 
## Treatments with the same letter are not significantly different.
## 
##   MixTechnique groups
## 2      3156.25      a
## 1      2971.00      b
## 3      2933.75      b
## 4      2666.25      c

plot(a)

### Testing Hypothesis ### Ho=The Tensile strength of the cement are the same for all mixing techniques ### Ha=At least one mean tensile strength differs ### as P is less than 0.05, Hence we can conclude that we reject the null hypothesis ### c)All pairs of means are significantly greater than 3.74 except for pair (1 and 5),(2 and 3) ### d)from the plots we can see that the data is normally distributed

library(agricolae)
cweight<-c(7,7,15,11,9,12,17,12,18,18,14,19,19,18,18,19,25,22,19,23,7,10,11,15,11)
da<-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)
da<-as.factor(da)
db<-cbind.data.frame(cweight,da)
db<-aov(cweight~da)
summary(db)
dt<-LSD.test(cweight,da,20,8.06,console = TRUE)
plot(db)
### As per LSD result we can see that minimu significance difference is 3.745452
### We can see its its normally distributed so we can say model is adequate

Q3

library(pwr)
pwr.anova.test(k=4,n=NULL,f=sqrt((5)^2/25),sig.level=0.05,power=0.9)

## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 4.658119
##               f = 1
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

### The required sample size is 5

### we can see that hypothesis of population means is 0.9 and alpha is o.05 and error variance is 25

Q4

Ho=Means are same

Ha=At least one of the mean differs

pwr.anova.test(k=4, n=NULL,f=sqrt((5)^2/49),sig.level=0.05,power=0.9)

## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 7.998751
##               f = 0.7142857
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

### The required sample size is 8
pwr.anova.test(k=4, n=NULL,f=sqrt((5)^2/36),sig.level=0.05,power=0.9)

## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 6.180857
##               f = 0.8333333
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

### The required sample size is 7