hw_5
pavithran
10/3/2021
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)
ts<- c(t1,t2,t3,t4)
t <- c(rep(1,4),rep(2,4),rep(3,4),rep(4,4))
d <- cbind(t,ts)
d <- as.data.frame(d)
d$t <- as.factor(d$t)
aov.model<-aov(ts~t,data=d)
plot(aov.model)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## t 3 489740 163247 12.73 0.000489 ***
## Residuals 12 153908 12826
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1library(agricolae)
l <- LSD.test(aov.model,"t",console = TRUE)##
## Study: aov.model ~ "t"
##
## LSD t Test for ts
##
## Mean Square Error: 12825.69
##
## t, means and individual ( 95 %) CI
##
## ts 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.
##
## ts groups
## 2 3156.25 a
## 1 2971.00 b
## 3 2933.75 b
## 4 2666.25 cplot(LSD.test(aov.model,"t"))
library(car)## Loading required package: carDatascatterplot(ts~t,data=d)
#### (c)by using LSD the mean t1 and t2 differes gradter than 174. t2 and t3 is less than 174. #### (d)the probability plot is normally distributed. #### (e)variance are equal.
question 2 (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)
ts <- c(t1,t2,t3,t4,t5)
t <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
d <- cbind(t,ts)
d <- as.data.frame(d)
d$t <- as.factor(d$t)
aov.model<-aov(ts~t,data=d)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## t 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 ' ' 1library(agricolae)
LSD.test(aov.model,"t",p.adj = "none",console=TRUE)##
## Study: aov.model ~ "t"
##
## LSD t Test for ts
##
## Mean Square Error: 8.06
##
## t, means and individual ( 95 %) CI
##
## ts 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.
##
## ts groups
## 4 21.6 a
## 3 17.6 b
## 2 15.4 b
## 5 10.8 c
## 1 9.8 cplot(aov.model)
#### by using lsd the t1 and t2 differs grater then 3.74 whereas t2 & t3 is less than 3.74 #### our data satusfies normality and also have constant varience from graph
question 3 (3.44)
library(pwr)
m1 = 50
m2 = 60
m3 = 50
m4 = 60
varience = 25
m_avg = (m1+m2+m3+m4)/4
pwr.anova.test(k=4,n=NULL,f=5/sqrt(varience),sig.level=0.05,power=0.90)##
## 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####n=4.65 rounding off it is 5 observation needed for each population. ### question (3.45)
m1 = 50
m2 = 60
m3 = 50
m4 =60
varience = 36
m_avg = (m1+m2+m3+m4)/4
pwr.anova.test(k=4,n=NULL,f=5/sqrt(36),sig.level=0.05,power=0.90)##
## 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 groupvarience = 49
pwr.anova.test(k=4,n=NULL,f=5/sqrt(49),sig.level=0.05,power=0.90)##
## 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