# 2a and 2b
titanic <- read.csv(paste("Titanic Data.csv"),sep = ",")
View(titanic)
# 3a 

dim(titanic)
## [1] 889   8

total number of enteries(rows) is equal to total passengers=889

#3b
table(titanic$Survived)
## 
##   0   1 
## 549 340

true is 1 and hence 340 survived

#3c
prop.table(table(titanic$Survived))*100
## 
##        0        1 
## 61.75478 38.24522

people who survived after the sink are 38.25%

#3d
xtabs(~Pclass + Survived, data  = titanic)
##       Survived
## Pclass   0   1
##      1  80 134
##      2  97  87
##      3 372 119

total number of passengers who survived after the sink of titanic is 134

#3e

prop.table(xtabs(~Pclass + Survived, data = titanic))*100
##       Survived
## Pclass         0         1
##      1  8.998875 15.073116
##      2 10.911136  9.786277
##      3 41.844769 13.385827

first passengers survived after the sink of titanic is 15.07%

#3f
xtabs(~Sex+Survived+Pclass, data = titanic)
## , , Pclass = 1
## 
##         Survived
## Sex        0   1
##   female   3  89
##   male    77  45
## 
## , , Pclass = 2
## 
##         Survived
## Sex        0   1
##   female   6  70
##   male    91  17
## 
## , , Pclass = 3
## 
##         Survived
## Sex        0   1
##   female  72  72
##   male   300  47

total first class survived females = 89

#3g
prop.table(xtabs(~Sex+Survived, data = titanic),2)*100
##         Survived
## Sex             0        1
##   female 14.75410 67.94118
##   male   85.24590 32.05882

total female survivors are 67.94%

#3h
prop.table(xtabs(~Sex+Survived, data = titanic),1)*100
##         Survived
## Sex             0        1
##   female 25.96154 74.03846
##   male   81.10919 18.89081

total 74.03 females in the board survived in titanic

#3i 
hold <- xtabs(~Sex+Survived, data = titanic)
addmargins(hold)
##         Survived
## Sex        0   1 Sum
##   female  81 231 312
##   male   468 109 577
##   Sum    549 340 889
chisq.test(hold)
## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  hold
## X-squared = 258.43, df = 1, p-value < 2.2e-16

since p < 0.05 it conclude to be null hypothesis