Complete the questions below. Compile your answers using Markdown and upload to your RPubs page. Submit a link to your RPubs page (through MOODLE) for grading.

Question 1

Below you will find a list of functions and what they do. Use any number of these functions to perform a descriptive analysis of the mtcars data set.

sum(x) - Sums the elements in row x
prod(x) - multiplies the elements in row x
max(x) - finds the maximum elemnt in x
min(x) - finds the minimum element in x
range(x) - finds the range of elemnts in x
length(x) - finds the number of entries in x
mean (x) - finds the mean of the entries in x
median(x) - finds the middle value in x
var(x) - finds the varience of x
sd(x) - finds the standard deviation of x
cor(x,y) - finds the correlation between x and y

carsmat <- as.matrix(cars)

length(cars)

## [1] 11

mean(cars[, c("wt")])

## [1] 3.21725

min(cars[, c("mpg")])

## [1] 10.4

range(cars[, c("cyl")])

## [1] 4 8

mean(cars[, c("mpg")])

## [1] 20.09062

sd(cars[, c("disp")])

## [1] 123.9387

Question 2

Use the command plot(cars[, c(___)]) to create a bunch of plots. Note that the c(___) indicates a list of the columns you wish to include. Select 4 or 5 of the most interesting values to investigate.

plot(cars[, c("mpg")])

plot(cars[, c("hp")])

plot(cars[, c("mpg","wt","hp","drat")])

Question 3

We can use the command “plot(A,B)” to make a scatterplot of column A versus column B. Do an exploratory analysis on the weight of the car versus the miles per gallon of the car. Form a hypothesis that you could test.

plot(cars[, c("wt","mpg")])

After plotting the two columns, it can be plausible to formulate a hypothesis that states that the weight of cars and their miles per gallon are inversely correlated.

Question 4

Test the hypothesis that you formed in 3. You may need to consult your instructor as to what r commands you may need.

cor(cars[, c("wt","mpg")])

##             wt        mpg
## wt   1.0000000 -0.8676594
## mpg -0.8676594  1.0000000

min(cars[, c("mpg")])

## [1] 10.4

min(cars[, c("wt")])

## [1] 1.513

max(cars[, c("mpg")])

## [1] 33.9

max(cars[, c("wt")])

## [1] 5.424

plot(cars[, c("wt","mpg")])

These were just trials^

x <- (cars[, c("wt")])
y <- (cars[, c("mpg")])
LinearEquation <- lm(y~x)
#Summary statistics
summary(LinearEquation)

## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4.5432 -2.3647 -0.1252  1.4096  6.8727 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  37.2851     1.8776  19.858  < 2e-16 ***
## x            -5.3445     0.5591  -9.559 1.29e-10 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.046 on 30 degrees of freedom
## Multiple R-squared:  0.7528, Adjusted R-squared:  0.7446 
## F-statistic: 91.38 on 1 and 30 DF,  p-value: 1.294e-10

#Generate range of 30 numbers starting from 0 and ending at 5
XSequence <- seq(0,30, length=5)
#Plot
plot(x,y, xlab="x axis", ylab="y axis", main="Regression", ylim=c(0,30), xlim=c(0,5), pch=15, col="blue")
#Add trend lines to the plot
lines(XSequence, predict(LinearEquation, data.frame(x=XSequence)), col="red")

As shown above, the linear regression equation follows a trajectory with a negative slope. This indicates that the two variables are inversely correlated to one another. That would make sense; as the weight of the car increases, the fuel efficiency of the vehicles worsens.