Homework_1

Author

John Carter

Exercise 1

The four common data types of vectors are as follows:

Logical, integer, double, and character

A matrix can only contain numeric values whereas a data frame can contain other values, such as characters.
It coerces logical values into integer values.
It will create a character matrix.

Exercise 2

a <- c(1, 2, 3, 4, 5)
print(a)

[1] 1 2 3 4 5

typeof(a)

[1] "double"

b <- 1:5
print(b)

[1] 1 2 3 4 5

typeof(b)

[1] "integer"

c <- c(sqrt(2), 4.7e4, 1/0)
print(c)

[1]     1.414214 47000.000000          Inf

typeof(c)

[1] "double"

d <- c(T, T, T, T)
print(d)

[1] TRUE TRUE TRUE TRUE

typeof(d)

[1] "logical"

e <- c("1", 2, 3)
print(e)

[1] "1" "2" "3"

typeof(e)

[1] "character"

f <- c(7L, NA, NA, 5L, 3L)
print(f)

[1]  7 NA NA  5  3

typeof(f)

[1] "integer"

g <- c(7L, "NA", "NA", 5L, 3L)
print(g)

[1] "7"  "NA" "NA" "5"  "3"

typeof(g)

[1] "character"

h <- c()
print(h)

NULL

typeof(h)

[1] "NULL"

Exercise 3

head(airquality)

  Ozone Solar.R Wind Temp Month Day
1    41     190  7.4   67     5   1
2    36     118  8.0   72     5   2
3    12     149 12.6   74     5   3
4    18     313 11.5   62     5   4
5    NA      NA 14.3   56     5   5
6    28      NA 14.9   66     5   6

Ozone1 <- airquality$Ozone

Ozone2 <- is.na(Ozone1)
table(Ozone2)

Ozone2
FALSE  TRUE 
  116    37

There are 37 NA values in the dataset.

Ozone3 <- as.integer(Ozone2)
typeof(Ozone3)

[1] "integer"

summary(Ozone3)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0000  0.0000  0.0000  0.2418  0.0000  1.0000

summary(airquality$Ozone)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
   1.00   18.00   31.50   42.13   63.25  168.00      37

sd1 <- sd(airquality$Ozone)
sd1

[1] NA

sd2 <- sd(airquality$Ozone, na.rm = TRUE)
sd2

[1] 32.98788

Summary of air quality Ozone returns the five number summary for the variable Ozone from the original dataset, without NA values removed but it accounts for the na values and removes them automatically.

sd of airqality Ozone returns the standard deviation for the variable Ozone from the original dataset without NA values removed. Thus an NA is created by an error.

sd(airqualityOzone, na.rm = TRUE) removes the NAs, allowing it to calculate the standard deviation. As shown by the code below

var1 <- sd(airquality$Ozone, na.rm = TRUE)
var2 <- na.omit(airquality$Ozone)
print(var1)

[1] 32.98788

sd(var2)

[1] 32.98788

var3 <- na.omit(airquality)
var4 <- nrow(airquality) - nrow(var3)
var4

[1] 42

There are 42 NA values in the dataset. As show by the code above.

Exercise 4)
a

p1 <- seq(0, 1, (0.2))

fofp1 <- (p1)*(1-p1)
var5 <- plot(p1,fofp1,type='b',ylab='f(p1) = p1*(1-p1)', col="RED", main ="Graph 1")

var5

NULL

p2 <- seq(0, 1, (0.01))

fofp2 <- (p2)*(1-p2)
var6 <- plot(p2,fofp2,type='o',ylab='f(p2) = p2*(1-p2)', col= "BLUE", main = "Graph 2")

var6

NULL

Exercise 5)

movies <- read.csv("https://ericwfox.github.io/data/movies.csv")

hist(movies$runtime, main="Histogram of Runtime", xlab="Runtime", col="PINK")

By the histogram above it appears to be skewed left, with mean roughtly centered around 100 minutes.
There are several potential outliers as shown by the boxplot below, which is also skewed.

boxplot(movies$runtime, main="Boxplot of Runtime", xlab="Runtime", col="LIGHTBLUE")

lmmodel <- lm(movies$critics_score~movies$runtime)
plot(movies$critics_score, movies$runtime)
abline(lmmodel, col="BLUE", lwd=4)

There does not appear to be a linear association between movie critic scores and movie runtimes.

movies$mpaa_rating <- factor(movies$mpaa_rating , levels=c("G", "PG", "PG-13", "R", "NC-17", "Unrated"))


boxplot(movies$imdb_rating ~ movies$mpaa_rating, xlab="MPAA Rating", ylab="IMDB Rating", col='RED')