# Read CSV file
mydata <- read.csv('./train.csv')
str(mydata)
## 'data.frame': 891 obs. of 12 variables:
## $ PassengerId: int 1 2 3 4 5 6 7 8 9 10 ...
## $ Survived : int 0 1 1 1 0 0 0 0 1 1 ...
## $ Pclass : int 3 1 3 1 3 3 1 3 3 2 ...
## $ Name : chr "Braund, Mr. Owen Harris" "Cumings, Mrs. John Bradley (Florence Briggs Thayer)" "Heikkinen, Miss. Laina" "Futrelle, Mrs. Jacques Heath (Lily May Peel)" ...
## $ Sex : chr "male" "female" "female" "female" ...
## $ Age : num 22 38 26 35 35 NA 54 2 27 14 ...
## $ SibSp : int 1 1 0 1 0 0 0 3 0 1 ...
## $ Parch : int 0 0 0 0 0 0 0 1 2 0 ...
## $ Ticket : chr "A/5 21171" "PC 17599" "STON/O2. 3101282" "113803" ...
## $ Fare : num 7.25 71.28 7.92 53.1 8.05 ...
## $ Cabin : chr "" "C85" "" "C123" ...
## $ Embarked : chr "S" "C" "S" "S" ...
head(mydata)
## PassengerId Survived Pclass
## 1 1 0 3
## 2 2 1 1
## 3 3 1 3
## 4 4 1 1
## 5 5 0 3
## 6 6 0 3
## Name Sex Age SibSp Parch
## 1 Braund, Mr. Owen Harris male 22 1 0
## 2 Cumings, Mrs. John Bradley (Florence Briggs Thayer) female 38 1 0
## 3 Heikkinen, Miss. Laina female 26 0 0
## 4 Futrelle, Mrs. Jacques Heath (Lily May Peel) female 35 1 0
## 5 Allen, Mr. William Henry male 35 0 0
## 6 Moran, Mr. James male NA 0 0
## Ticket Fare Cabin Embarked
## 1 A/5 21171 7.2500 S
## 2 PC 17599 71.2833 C85 C
## 3 STON/O2. 3101282 7.9250 S
## 4 113803 53.1000 C123 S
## 5 373450 8.0500 S
## 6 330877 8.4583 Q
summary(mydata)
## PassengerId Survived Pclass Name
## Min. : 1.0 Min. :0.0000 Min. :1.000 Length:891
## 1st Qu.:223.5 1st Qu.:0.0000 1st Qu.:2.000 Class :character
## Median :446.0 Median :0.0000 Median :3.000 Mode :character
## Mean :446.0 Mean :0.3838 Mean :2.309
## 3rd Qu.:668.5 3rd Qu.:1.0000 3rd Qu.:3.000
## Max. :891.0 Max. :1.0000 Max. :3.000
##
## Sex Age SibSp Parch
## Length:891 Min. : 0.42 Min. :0.000 Min. :0.0000
## Class :character 1st Qu.:20.12 1st Qu.:0.000 1st Qu.:0.0000
## Mode :character Median :28.00 Median :0.000 Median :0.0000
## Mean :29.70 Mean :0.523 Mean :0.3816
## 3rd Qu.:38.00 3rd Qu.:1.000 3rd Qu.:0.0000
## Max. :80.00 Max. :8.000 Max. :6.0000
## NA's :177
## Ticket Fare Cabin Embarked
## Length:891 Min. : 0.00 Length:891 Length:891
## Class :character 1st Qu.: 7.91 Class :character Class :character
## Mode :character Median : 14.45 Mode :character Mode :character
## Mean : 32.20
## 3rd Qu.: 31.00
## Max. :512.33
##
#Q1 a. What are the types of variable (quantitative / qualitative) and levels of measurementLinks to an external site. (nominal / ordinal / interval / ratio) for PassengerId and Age?
str(mydata$PassengerId)
## int [1:891] 1 2 3 4 5 6 7 8 9 10 ...
str(mydata$Age)
## num [1:891] 22 38 26 35 35 NA 54 2 27 14 ...
# PassengerId is qualitative variable and Age is quantitative variable.
#The levels of measurement for PassengerId is nominal and Age is Interval.
#Q1b. Which variable has the most missing observations? You could have Googled for "Count NA values in R for all columns in a dataframeLinks to an external site." or something like that.
mydata[mydata == ""
|mydata == " "] <- NA
colSums(is.na(mydata))
## PassengerId Survived Pclass Name Sex Age
## 0 0 0 0 0 177
## SibSp Parch Ticket Fare Cabin Embarked
## 0 0 0 0 687 2
# In the data, there are certian blank column which contain in the orginal data. The Cabin section have the most missing observations.
#Q2. Impute missing observations for Age, SibSp, and Parch with the column median (ordinal, interval, or ratio), or the column mode. To do so, use something like this: mydata$Age[is.na(mydata$Age)] <- median(mydata$Age, na.rm=TRUE) . You can read up on indexing in dataframeLinks to an external site. (i.e. "dataframe_name$variable_name").
summary(mydata$Age)
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 0.42 20.12 28.00 29.70 38.00 80.00 177
summary(mydata$SibSp)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 0.000 0.000 0.523 1.000 8.000
mydata$Age[is.na(mydata$Age)] <- median(mydata$Age, na.rm=TRUE)
mydata$SibSp[is.na(mydata$SibSp)] <- median(mydata$SibSp, na.rm=TRUE)
mydata$Parch[is.na(mydata$Parch)] <- median(mydata$Parch, na.rm=TRUE)
summary(mydata$Age)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.42 22.00 28.00 29.36 35.00 80.00
summary(mydata$SibSp)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 0.000 0.000 0.523 1.000 8.000
#Q3. Install the psych package in R: install.packages('pscyh'). Invoke the package using library(psych). Then provide descriptive statistics for Age, SibSp, and Parch (e.g., describe(mydata$Age) .
library(psych)
describe(mydata$Age)
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 891 29.36 13.02 28 28.83 8.9 0.42 80 79.58 0.51 0.97 0.44
describe(mydata$SibSp)
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 891 0.52 1.1 0 0.27 0 0 8 8 3.68 17.73 0.04
describe(mydata$Parch)
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 891 0.38 0.81 0 0.18 0 0 6 6 2.74 9.69 0.03
#Q4. Provide a cross-tabulation of Survived and Sex (e.g., table(mydata$Survived, mydata$Sex). What do you notice?
cross_tabulation <- table (mydata$Survived, mydata$Sex)
print (cross_tabulation)
##
## female male
## 0 81 468
## 1 233 109
# What I noticed with the code is that the female passengers had higher survived rate than the male passengers
#Q5. Provide notched boxplots for Survived and Age (e.g., boxplot(mydata$Age~mydata$Survived, notch=TRUE, horizontal=T). What do you notice?
boxplot(mydata$Age~mydata$Survived, notch=TRUE, horizontal=T,
xlab = "Age", ylab = "Survived", main = "Age vs Survived boxplot graph")
# boxplot help me testify my perspective in question 4 is correct which is that the female passengers had higher survived rate than the male passengers