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TASK 3d
ddd.df<-read.csv(paste("Data - Deans Dilemma.csv",sep=""))
View(ddd.df)TASK 3d_1
placed.df<-ddd.df[which(ddd.df$Placement=="Placed"),]
aggregate(placed.df$Salary, by=list(Gender=placed.df$Gender), mean)## Gender x
## 1 F 253068.0
## 2 M 284241.9mean salary of males and females, who were placed are 284241.9 and 253068.0 respectively
TASK 3d_2
placedM.df<-ddd.df[which(ddd.df$Placement_B=="1"),]
aggregate(placed.df$Salary, by=list(Gender=placed.df$Gender), mean)## Gender x
## 1 F 253068.0
## 2 M 284241.9average salary of male MBAs who were placed = 284241.9
average salary of female MBAs who were placed = 253068.0
TASK 3d_4
t.test(placed.df$Salary~placed.df$Gender,data = placed.df)##
## Welch Two Sample t-test
##
## data: placed.df$Salary by placed.df$Gender
## t = -3.0757, df = 243.03, p-value = 0.00234
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -51138.42 -11209.22
## sample estimates:
## mean in group F mean in group M
## 253068.0 284241.9p-value = 0.00234
Since p-value < 0.05, we would reject our null hypothesis. Thus, there’s significant difference between the means of our sample population i.e. it is true that the average salary of the male MBAs is higher than the average salary of female MBAs.