Question 3.7 c

let u1= mean of tensile strength using mixing technique 1

u2= mean of tensile strength using mixing technique 2

u3= mean of tensile strength using mixing technique 3

u4= mean of tensile strength using mixing technique 4

Null hypothesis is that u1=u2=u3=u4

Alternative hypothesis (Ha) is that at least of the u’s differs.

MTechnique<- c("1","1","1","1","2","2","2","2","3","3","3","3","4","4","4","4")
Tdata<- c(3129,3000,2865,2890,3200,3300,2975,3150,2800,2900,2985,3050,2600,2700,2600,2765)
dat<- cbind(MTechnique,Tdata)
dat<- as.data.frame(dat)
dat$Tdata <- as.numeric(dat$Tdata)
dat$MTechnique<- as.factor(dat$MTechnique)
model <- aov(Tdata~MTechnique,data=dat)
summary(model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## MTechnique   3 489740  163247   12.73 0.000489 ***
## Residuals   12 153908   12826                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
library(agricolae)
LSD.test(model,"MTechnique",console = TRUE)
## 
## Study: model ~ "MTechnique"
## 
## LSD t Test for Tdata 
## 
## Mean Square Error:  12825.69 
## 
## MTechnique,  means and individual ( 95 %) CI
## 
##     Tdata       std r      LCL      UCL  Min  Max
## 1 2971.00 120.55704 4 2847.624 3094.376 2865 3129
## 2 3156.25 135.97641 4 3032.874 3279.626 2975 3300
## 3 2933.75 108.27242 4 2810.374 3057.126 2800 3050
## 4 2666.25  80.97067 4 2542.874 2789.626 2600 2765
## 
## Alpha: 0.05 ; DF Error: 12
## Critical Value of t: 2.178813 
## 
## least Significant Difference: 174.4798 
## 
## Treatments with the same letter are not significantly different.
## 
##     Tdata groups
## 2 3156.25      a
## 1 2971.00      b
## 3 2933.75      b
## 4 2666.25      c

Using the LSD test

μ1 = μ3 - we fail to reject Null Hypothesis (Ho).

Also

μ2 differs from μ1,μ3, μ4

μ4 differs from μ1,μ2 ,μ3

QUESTION 3.7 d and e

plot(model)

Answer Q3.7d)

Since the normal probability plot of the residual looks fairly normally distributed, we can now draw a conclusion that the validity of normality is true.

Answer Q3.7)e) from the graph of the standandized residuals vs predicted tensile strength, we can see that all the residual points follows a sequential straight line point which indicates constant variance.

plot(Tdata, ylab = "Tensile Strength")

Answer Q3.7)f)

From the scatter plot, we can make accurate conclusions that the data is normally distributed because the data follows no specific pattern.

Question 3.10 b

cotton<- c("15","15","15","15","15","20","20","20","20","20","25","25","25","25","25","30","30","30","30","30","35","35","35","35","35")
Obs<- 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)
data1<- cbind(cotton,Obs)
data1 <- as.data.frame(data1)
data1$cotton <- as.factor(data1$cotton)
data1$Obs <- as.numeric(data1$Obs)
model1<- aov(Obs~cotton,data=data1)
summary(model1)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## cotton       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(model1,"cotton",console=TRUE)
## 
## Study: model1 ~ "cotton"
## 
## LSD t Test for Obs 
## 
## Mean Square Error:  8.06 
## 
## cotton,  means and individual ( 95 %) CI
## 
##     Obs      std r       LCL      UCL Min Max
## 15  9.8 3.346640 5  7.151566 12.44843   7  15
## 20 15.4 3.130495 5 12.751566 18.04843  12  18
## 25 17.6 2.073644 5 14.951566 20.24843  14  19
## 30 21.6 2.607681 5 18.951566 24.24843  19  25
## 35 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.
## 
##     Obs groups
## 30 21.6      a
## 25 17.6      b
## 20 15.4      b
## 35 10.8      c
## 15  9.8      c

Question 3.10 b

let μ1 = mean value of 15% cotton by Weight.

μ2 = mean value of 20% cotton by Weight.

μ3 = mean value of 25% cotton by Weight.

μ4 = mean value of 30% cotton by Weight.

μ5 = mean value of 35% cotton by Weight.

From the LSD test we can conclude that

μ4 is different from μ1,μ2,μ3,μ5

μ1 is similar to μ5 but different from μ2,μ3,μ4

μ2 is similar to μ3 but different from μ1,μ4,μ5

μ3 is similar to μ2 but different from μ1,μ4,μ5

μ5 is similar to μ1 but different from μ2,μ3,μ4

Question 3.10 c

plot(model1)

Since the normal probability plot of the residual looks fairly normally distributed, we can now draw a conclusion that the validity of normality is true.

Also,from the graph of the standardized residuals we can see also see that all the residual points follows a sequential straight line point which indicates constant variance.

Question 3.44

u1=u3=50 u2=u4=60

power=0.9 , alpha=0.05

variance=25

library(pwr)
## Warning: package 'pwr' was built under R version 4.1.3
pwr.anova.test(k = 4, n = NULL, f = sqrt(((60-50)^2)/25), sig.level = 0.05, power = 0.9)
## 
##      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

The number of samples n=2.17 which would be needed is rounded up to 3 samples in each group

Question 3.45a

pwr.anova.test(k = 4, n = NULL, f = sqrt(((60-50)^2)/36), sig.level = 0.05, power = 0.9)
## 
##      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

An increase in sample size was noted with n=2.5, which would be rounded up to 3 samples in each group.

Question 3.45 b

pwr.anova.test(k = 4, n = NULL, f = sqrt(((60-50)^2)/49), sig.level = 0.05, power = 0.9)
## 
##      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

An increase in sample size was noted with n=2.93, which would be rounded up to 3 samples in each group.

Question 3.44 c

we can see that based on the analysis obtained by increasing the figures of σ, the number of sample size required simultaneously increases to attain the same power of the test.

Question 3.45 d

The best way to chose n value is to take in account of the number of groups, effect size, and power.

We can also see based on our previous example that increasing the σ value increases the number of samples and decreases the effect size.

And this should be taken into account when choosing n value in practice.