This is an R Markdown document which describes the analysis of the Titanic case study
setwd("F:/FOR DA INTERN/TASKS/WEEK 1/D5")
titanic.df <- read.csv(paste("Titanic Data.csv", sep=""))
length(titanic.df$Survived)
## [1] 889
Thus there were 899 passengers on board the Titanic
table(titanic.df$Survived)
##
## 0 1
## 549 340
Thus 340 passengers survived the sinking of the Titanic
table1 <- with(titanic.df, table(Survived))
table1
## Survived
## 0 1
## 549 340
prop.table(table1)*100
## Survived
## 0 1
## 61.75478 38.24522
Thus 38.24522% of the passengers survived the sinking of the Titanic
table1 <- xtabs(~ Survived+Pclass, data=titanic.df)
table1
## Pclass
## Survived 1 2 3
## 0 80 97 372
## 1 134 87 119
Thus 134 first-class passengers survived the sinking of the Titanic
prop.table(table1,2)*100
## Pclass
## Survived 1 2 3
## 0 37.38318 52.71739 75.76375
## 1 62.61682 47.28261 24.23625
Thus 62.61682% of first-class passengers survived the sinking of the Titanic
table1 <- xtabs(~ Survived+Pclass+Sex, data=titanic.df)
table1
## , , Sex = female
##
## Pclass
## Survived 1 2 3
## 0 3 6 72
## 1 89 70 72
##
## , , Sex = male
##
## Pclass
## Survived 1 2 3
## 0 77 91 300
## 1 45 17 47
Thus the number of females from First-Class who survived the sinking of the Titanic was 89
table1 <- xtabs(~ Survived+Sex, data=titanic.df)
table1
## Sex
## Survived female male
## 0 81 468
## 1 231 109
prop.table(table1,1)*100
## Sex
## Survived female male
## 0 14.75410 85.24590
## 1 67.94118 32.05882
Thus 67.94118% of the survivors were females
table1 <- xtabs(~ Survived+Sex, data=titanic.df)
prop.table(table1,2)*100
## Sex
## Survived female male
## 0 25.96154 81.10919
## 1 74.03846 18.89081
Thus the percentage of females on board the Titanic who survived was 74.03846%
Hypothesis: The proportion of females onboard who survived the sinking of the Titanic was higher than the proportion of males onboard who survived the sinking of the Titanic.
table2 <- xtabs(~Sex+Survived, data=titanic.df)
table2
## Survived
## Sex 0 1
## female 81 231
## male 468 109
chisq.test(table2)
##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: table2
## X-squared = 258.43, df = 1, p-value < 2.2e-16
Thus from the Chi-squared test, we see that the value of p is lesser than 0.01 which suggests that we have to reject the null hypothesis