titanic day5

setting up working directory to the folder of file

setwd(“C:/Users/Taiyyab Ali/Desktop/R language”)

task 2b

#reading data
Titanic <-read.csv("Titanic Data.csv")
View(Titanic)

task 3a

Total no. of passengers in titanic

length(Titanic$Survived)

## [1] 889

Total no. of passengers = 889

task 3b

no.of passengers survived equal value under 1

table(Titanic$Survived)

## 
##   0   1 
## 549 340

no. of passenger survived= 340

task 3c

prop.test(340,889)

## 
##  1-sample proportions test with continuity correction
## 
## data:  340 out of 889, null probability 0.5
## X-squared = 48.666, df = 1, p-value = 3.035e-12
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.3505253 0.4154084
## sample estimates:
##         p 
## 0.3824522

percentage of passengers survived = 38.24%

task 3d

aggregate(Survived~Pclass,data=Titanic,sum)

##   Pclass Survived
## 1      1      134
## 2      2       87
## 3      3      119

no.of 1st class passenger survived= 134 # sum of not survived is “0” for all class #task 3e

prop.test(134,889)

## 
##  1-sample proportions test with continuity correction
## 
## data:  134 out of 889, null probability 0.5
## X-squared = 432.4, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.1281932 0.1763462
## sample estimates:
##         p 
## 0.1507312

persentage of 1st class passenger survived = 15.07%

task 3f

ftable(Titanic$Survived,Titanic$Pclass,Titanic$Sex)

##      female male
##                 
## 0 1       3   77
##   2       6   91
##   3      72  300
## 1 1      89   45
##   2      70   17
##   3      72   47

no. female passenger from 1st class survived=89

task 3g

table(Titanic$Survived,Titanic$Sex)

##    
##     female male
##   0     81  468
##   1    231  109

prop.test(231,340)

## 
##  1-sample proportions test with continuity correction
## 
## data:  231 out of 340, null probability 0.5
## X-squared = 43.062, df = 1, p-value = 5.304e-11
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.6265239 0.7281727
## sample estimates:
##         p 
## 0.6794118

persentage of female- survived passenger=67.94% :P quite high #task 3h

prop.test(231,312)

## 
##  1-sample proportions test with continuity correction
## 
## data:  231 out of 312, null probability 0.5
## X-squared = 71.157, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.6873432 0.7873876
## sample estimates:
##         p 
## 0.7403846

persentage of female passenger that survived = 74.04%

task3i

Pearson’s Chi-squared test to test the following hypothesis:

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.

library(gmodels) # for funtion crossTable
CrossTable(Titanic$Survived,Titanic$Sex,chisq = TRUE)

## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## | Chi-square contribution |
## |           N / Row Total |
## |           N / Col Total |
## |         N / Table Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  889 
## 
##  
##                  | Titanic$Sex 
## Titanic$Survived |    female |      male | Row Total | 
## -----------------|-----------|-----------|-----------|
##                0 |        81 |       468 |       549 | 
##                  |    64.727 |    35.000 |           | 
##                  |     0.148 |     0.852 |     0.618 | 
##                  |     0.260 |     0.811 |           | 
##                  |     0.091 |     0.526 |           | 
## -----------------|-----------|-----------|-----------|
##                1 |       231 |       109 |       340 | 
##                  |   104.515 |    56.514 |           | 
##                  |     0.679 |     0.321 |     0.382 | 
##                  |     0.740 |     0.189 |           | 
##                  |     0.260 |     0.123 |           | 
## -----------------|-----------|-----------|-----------|
##     Column Total |       312 |       577 |       889 | 
##                  |     0.351 |     0.649 |           | 
## -----------------|-----------|-----------|-----------|
## 
##  
## Statistics for All Table Factors
## 
## 
## Pearson's Chi-squared test 
## ------------------------------------------------------------
## Chi^2 =  260.7563     d.f. =  1     p =  1.173958e-58 
## 
## Pearson's Chi-squared test with Yates' continuity correction 
## ------------------------------------------------------------
## Chi^2 =  258.4266     d.f. =  1     p =  3.77991e-58 
## 
## 

interpretation of result

Based on p-value we can neglect the null hypothesis that female and male survived almost equally.