Probelm 3.7

library(agricolae)

MixingTech<-c(3129,3000,2865,2890,3200,3300,2975,3150,2800,2900,2985,3050,2600,2700,2600,2765)
a<-c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
str(a)
##  num [1:16] 1 1 1 1 2 2 2 2 3 3 ...
a<-as.factor(a)
str(a)
##  Factor w/ 4 levels "1","2","3","4": 1 1 1 1 2 2 2 2 3 3 ...
b<-cbind.data.frame(MixingTech,a)


b<-aov(MixingTech~a)
summary(b)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## a            3 489740  163247   12.73 0.000489 ***
## Residuals   12 153908   12826                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
residual<-MixingTech-a
## Warning in Ops.factor(MixingTech, a): '-' not meaningful for factors
t<-LSD.test(MixingTech,a,12,12826,console=TRUE)
## 
## Study: MixingTech ~ a
## 
## LSD t Test for MixingTech 
## 
## Mean Square Error:  12826 
## 
## a,  means and individual ( 95 %) CI
## 
##   MixingTech       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.
## 
##   MixingTech groups
## 2    3156.25      a
## 1    2971.00      b
## 3    2933.75      b
## 4    2666.25      c
plot(b)

### Least significant difference is 174.482

c)

Difference between means of technique 2 and 1 is more than 174.482

Difference between means of technique 2 and 4 is more than 174.482

Difference between means of technique 2 and 3 is more than 174.482

Difference between means of technique 3 and 1 is less than 174.482

Difference between means of technique 3 and 4 is more than 174.482

d) Here we can say that the data is appoximately normal.

e) Variences are apporximately normal.

Problem 3.10

library(agricolae)

CottonWeight<-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)
e<-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)
e<-as.factor(e)
f<-cbind.data.frame(CottonWeight,e)

f<-aov(CottonWeight~e)
summary(f)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## e            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
t<-LSD.test(CottonWeight,e,20,8.06,console = TRUE)
## 
## Study: CottonWeight ~ e
## 
## LSD t Test for CottonWeight 
## 
## Mean Square Error:  8.06 
## 
## e,  means and individual ( 95 %) CI
## 
##   CottonWeight      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.
## 
##   CottonWeight groups
## 4         21.6      a
## 3         17.6      b
## 2         15.4      b
## 5         10.8      c
## 1          9.8      c
plot(f)

### Least significant difference is 3.745

b)

Difference between cotton group 4 and 3 is more than 3.745

Difference between cotton group 2 and 4 is more than 3.745

Difference between cotton group 4 and 5 is more than 3.745

Difference between cotton group 4 and 1 is more than 3.745

Difference between cotton group 1 and 5 is less than 3.745

Difference between cotton group 2 and 3 is less than 3.745

c) After analyzing the residuals from this experiment, we can say that the model is adequate.

Problem 3.44

library(pwr)

pwr.anova.test(k=4,n=NULL,f=2,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

Here value of n=2.17, so we can say that we can take approximately 3 observations from each population

Problem 3.45

library(pwr)

pwr.anova.test(k=4,n=NULL,f=1.66,sig.level=0.05,power=0.90)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 2.528053
##               f = 1.66
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group
pwr.anova.test(k=4,n=NULL,f=1.42,sig.level=0.05,power=0.90)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 2.959231
##               f = 1.42
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

For error varience 36 the value of n=2.52, so we can say that we can take approximately 3 observations from each population

For error varience 49 the value of n=2.52, so we can say that we can take approximately 3 observations from each population

```