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View(tit.df)
Number of passengers on board =889
dim(tit.df)
## [1] 889 8
mytable <- with(tit.df, table(Survived))
mytable
## Survived
## 0 1
## 549 340
The number of passengers who survived the sinking of the Titanic =340
prop.table(mytable)
## Survived
## 0 1
## 0.6175478 0.3824522
The percentage of passengers who survived the sinking of the Titanic = .3824522
mytable1<-xtabs(~Survived+Pclass, data = tit.df)
mytable1
## Pclass
## Survived 1 2 3
## 0 80 97 372
## 1 134 87 119
The number of first-class passengers who survived the sinking of the Titanic = 134
prop.table(mytable1, 2)
## Pclass
## Survived 1 2 3
## 0 0.3738318 0.5271739 0.7576375
## 1 0.6261682 0.4728261 0.2423625
the percentage of first-class passengers who survived the sinking of the Titanic = .06261682
mytable2 <-xtabs(~Survived+Pclass+Sex, data = tit.df)
mytable2
## , , 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
The number of females from First-Class who survived the sinking of the Titanic =89
mytable3<-margin.table(mytable2, c(1,3))
mytable3
## Sex
## Survived female male
## 0 81 468
## 1 231 109
prop.table(mytable3, 1)*100
## Sex
## Survived female male
## 0 14.75410 85.24590
## 1 67.94118 32.05882
The percentage of survivors who were female= 67.94118 %
prop.table(mytable3, 2)*100
## Sex
## Survived female male
## 0 25.96154 81.10919
## 1 74.03846 18.89081
The percentage of females on board the Titanic who survived= 74.03%
chisq.test(mytable3)
##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: mytable3
## X-squared = 258.43, df = 1, p-value < 2.2e-16
The p-values are the probability of obtaining the sampled results, assuming independence of the row and colum variables in the population. Since the probability is small (p <.05), we reject the Null hypothesis that sex and survival are independent.