type1 <- c(65,82,81,67,57,59,66,75,82,70)
type2 <- c(64,56,71,69,83,74,59,82,65,79)
a<-c(65,82,81,67,57,59,66,75,82,70,64,56,71,69,83,74,59,82,65,79)
b<-c("1","1","1","1","1","1","1","1","1","1","2","2","2","2","2","2","2","2","2","2")
library(lawstat)
type<-data.frame(a,b)
colnames(type)<-c("Values","types")
qqnorm(a, main = "normality plot of type1 and type2")
qqline(a)

# plot shows that its normally distributed and the sample size is also small and its normally distributed.

boxplot(type1,type2,col=c("light blue","red"),names=c("type1","type2"))

variance are not equal.

levene.test(type$Values,type$types,location = "mean")
## 
##  Classical Levene's test based on the absolute deviations from the mean
##  ( none not applied because the location is not set to median )
## 
## data:  type$Values
## Test Statistic = 0.0014598, p-value = 0.9699

p value is 0..9699 which is larger than 0.05 so it failes to reject the null hypothesis.

t.test(type1,type2,alternative = c("two.sided","less","greater"))
## 
##  Welch Two Sample t-test
## 
## data:  type1 and type2
## t = 0.048008, df = 17.998, p-value = 0.9622
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -8.552517  8.952517
## sample estimates:
## mean of x mean of y 
##      70.4      70.2

mean of x and y are different and the p value is 0.9622

type1 <- c(65,82,81,67,57,59,66,75,82,70)
type2 <- c(64,56,71,69,83,74,59,82,65,79)
install.packages("lawstat")
library(lawstat)
a<-c(65,82,81,67,57,59,66,75,82,70,64,56,71,69,83,74,59,82,65,79)
b<-c("1","1","1","1","1","1","1","1","1","1","2","2","2","2","2","2","2","2","2","2")
type<-data.frame(a,b)
colnames(type)<-c("Values","types")
qqnorm(a, main = "normality plot of type1 and type2")
qqline(a)
boxplot(type1,type2,col=c("light blue","red"),names=c("type1","type2"))
levene.test(type$Values,type$types,location = "mean")
t.test(dat1,dat2,alternative = c("two.sided","less","greater"))