This project will demonstrate your ability to do exploratory data analysis on single variables of data in R and RStudio. The entire project will use the NSCC Student Dataset, which you will download and load into R in question 1.
Download the “nscc_student_data.csv” file from MyOpenMath and use the read.csv() function to store it into an object called “nscc_student_data”. Print the first few lines of each data set using the head() function. Print a summary of each data set using the str() function.
# Load dataset in and store as object "nscc_student_data"
nscc_student_data <- read.csv("~/Desktop/honorsStats/nscc_student_data.csv")
# Preview of dataframe
head(nscc_student_data)
## Gender PulseRate CoinFlip1 CoinFlip2 Height ShoeLength Age Siblings
## 1 Female 64 5 5 62 11.00 19 4
## 2 Female 75 4 6 62 11.00 21 3
## 3 Female 74 6 1 60 10.00 25 2
## 4 Female 65 4 4 62 10.75 19 1
## 5 Female NA NA NA 66 NA 26 6
## 6 Female 72 6 5 67 9.75 21 1
## RandomNum HoursWorking Credits Birthday ProfsAge Coffee VoterReg
## 1 797 35 13 July 5 31 No Yes
## 2 749 25 12 December 27 30 Yes Yes
## 3 13 30 6 January 31 29 Yes No
## 4 613 18 9 6-13 31 Yes Yes
## 5 53 24 15 02-15 32 No Yes
## 6 836 15 9 april 14 32 No Yes
# Summary of dataframe with str()
str(nscc_student_data)
## 'data.frame': 40 obs. of 15 variables:
## $ Gender : Factor w/ 2 levels "Female","Male": 1 1 1 1 1 1 2 2 1 2 ...
## $ PulseRate : int 64 75 74 65 NA 72 72 60 66 60 ...
## $ CoinFlip1 : int 5 4 6 4 NA 6 6 3 7 6 ...
## $ CoinFlip2 : int 5 6 1 4 NA 5 6 5 8 5 ...
## $ Height : num 62 62 60 62 66 ...
## $ ShoeLength : num 11 11 10 10.8 NA ...
## $ Age : int 19 21 25 19 26 21 19 24 24 20 ...
## $ Siblings : int 4 3 2 1 6 1 2 2 3 1 ...
## $ RandomNum : int 797 749 13 613 53 836 423 16 12 543 ...
## $ HoursWorking: int 35 25 30 18 24 15 20 0 40 30 ...
## $ Credits : int 13 12 6 9 15 9 15 15 13 16 ...
## $ Birthday : Factor w/ 40 levels "02-15","03.14.1984",..: 32 25 30 18 1 21 19 27 35 31 ...
## $ ProfsAge : int 31 30 29 31 32 32 28 28 31 28 ...
## $ Coffee : Factor w/ 2 levels "No","Yes": 1 2 2 2 1 1 2 2 2 1 ...
## $ VoterReg : Factor w/ 2 levels "No","Yes": 2 2 1 2 2 2 2 2 2 1 ...
a.) What are the dimensions of the nscc_student_data dataframe?
# Find the dimensions of the nscc_student_data dataframe.
dim(nscc_student_data)
## [1] 40 15
The dimensions of the nscc_student_data dataframe are 40 by 15.
b.) The chunk of code below will tell you how many values in the PulseRate variable exist (FALSE) and how many are NA (TRUE). How many values are in the variable are missing?
# How many values in PulseRate variable are missing
table(is.na(nscc_student_data$PulseRate))
##
## FALSE TRUE
## 38 2
Two values in the PulseRate variable are missing.
What is the mean of the pulse rate variable? What is the median of the pulse rate variable? Do they differ by much? If yes, explain why and which would be a better choice as the “center” or “average” of this variable.
# Find the mean of the pulse rate variable, ignoring the NA variables.
mean(nscc_student_data$PulseRate, na.rm = TRUE)
## [1] 73.47368
# Find the median of the pulse rate variable, ignoring the NA variables.
median(nscc_student_data$PulseRate, na.rm = TRUE)
## [1] 70.5
These values do not differ very much. The mean is 73.47 and the median is 70.50 and therefore either value would be a valid “center” or “average” for this variable.
What is the sample standard deviation of the pulse rate variable?
# Find the sample standard deviation of the pulse rate variable.
sd(nscc_student_data$PulseRate, na.rm = TRUE)
## [1] 12.51105
The sample standard deviation of the pulse rate variable is 12.51.
What is the Five Number Summary and IQR of the pulse rate variable? Based on the definition of outliers being more than 1.5 IQRs above Q3 or below Q1, are there any values in the pulse rate variable that are considered outliers?
# Five Number Summary of the PulseRate variable
summary(nscc_student_data$PulseRate)
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 50.00 64.25 70.50 73.47 83.75 98.00 2
The IQR of the PulseRate variable is the Q3-Q1, so 83.75-64.25, so the answer is 19.5.
Any outliers that are below 64.25 or above 83.75 would be considered to be outliers. Based on that, there are outliers and those outliers are 50, 56, 60, 61, 62, 64, 87, 88, 89, 85, 98.
Five number summary:
| Value | Number |
|---|---|
| Min | 50.00 |
| 1st Qu. | 64.25 |
| Median | 70.50 |
| 3rd Qu. | 83.75 |
| Max | 98.00 |
The Gender variable gives whether students identified as male or female. Create a table and a barplot of that variable.
# Table of results of Gender variable
table(nscc_student_data$Gender)
##
## Female Male
## 27 13
# Barplot of results
barplot(table(nscc_student_data$Gender))
Split the dataframe into two subsets – one that has all the males and another that has all the females. Store them into objects called “NSCC_males” and “NSCC_females”. The first one has been done for you as a template.
# Create males subset
NSCC_males <- subset(nscc_student_data, nscc_student_data$Gender == "Male")
# Create females subset
NSCC_females <- subset(nscc_student_data, nscc_student_data$Gender == "Female")
What is the Five Number Summary for the pulse rate variable for each of the male and female subsets.
# Five Number Summary of males subset
summary(NSCC_males$PulseRate)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 50.00 60.00 71.00 70.85 80.00 96.00
# Five Number Summary of females subset
summary(NSCC_females$PulseRate)
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 56.00 65.00 70.00 74.84 88.00 98.00 2
The Five Number Summary of each dataset is:
Males:
| Value | Number |
|---|---|
| Min | 50.00 |
| 1st Qu. | 60.00 |
| Median | 71.00 |
| 3rd Qu. | 80.00 |
| Max | 96.00 |
Females:
| Value | Number |
|---|---|
| Min | 56.00 |
| 1st Qu. | 65.00 |
| Median | 70.00 |
| 3rd Qu. | 88.00 |
| Max | 98.00 |
Create side-by-side boxplots for the pulse rate variable each of the male and female subsets. Is there any noticeable difference between the spread of the variables?
# Create side-by-side boxplots for each subset
boxplot(NSCC_females$PulseRate, NSCC_males$PulseRate)
Yes, there is a noticable difference between the spread of the variables. The medians of the datasets are very close and so are the maximum values of each dataset. However, the minimum number in the male dataset is at least five less than the minimum number of the female dataset, and both the first and the third quartiles differ greatly in each dataset.
Create a frequency distribution for how many males and females answered “Yes” or “No” to the variable “Coffee” by using the table() function. What percent of this sample of NSCC students drink coffee? Is there any noticeable difference in coffee drinking based on gender?
# Male Coffee Drinkers
table(NSCC_males$Coffee)
##
## No Yes
## 3 10
# Females Coffee Drinkers
table(NSCC_females$Coffee)
##
## No Yes
## 7 20
# Percent of Males that Drink Coffee
10/(10+3)
## [1] 0.7692308
# Percent of Females that Drink Coffee
20/(20+7)
## [1] 0.7407407
77% of the male students in this dataset drink coffee and 74% of female students in the dataset drink coffee. Therefore there is not much of a difference, and definitely not a noticable difference in coffee drinking based on gender.