Intro to R Part 2

Case-scenario 1

This is the fourth season of outfielder Luis Robert with the Chicago White Socks. If during the first three seasons he hit 11, 13, and 12 home runs, how many does he need on this season for his overall average to be at least 20?

# Home-runs so far
HR_before <- c(11, 13, 12)
# Average Number of Home-runs per season wanted
wanted_HR <- 20
# Number of seasons
n_seasons <- 4
# Needed Home-runs on season 4
x_4 <- n_seasons*wanted_HR - sum(HR_before)
# Minimum number of Home-runs needed by Robert
x_4

[1] 44

According to the calculations above, Robert must hit 44 home-runs or better on this season to get an average number of home-runs per season of at least 20.

We could confirm this, by using the function mean() in R

# Robert's performance
Robert_HRs <- c(11, 13, 12,44)
# Find mean
mean(Robert_HRs)

[1] 20

# Robert's actual performance 
Robert_HR <- c(11, 13, 12,38)
mean(Robert_HR)

[1] 18.5

# Find standard deviation
sd(Robert_HR)

[1] 13.02562

# Find the maximum number of home-runs during the four seasons period
max(Robert_HR)

[1] 38

# Find the minimum number of home-runs during the four seasons period
min(Robert_HR)

[1] 11

We can also use the summary() function to find basic statistics, including the median!

summary(Robert_HR)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  11.00   11.75   12.50   18.50   19.25   38.00 

Question 1

Now, you must complete the problem below which represents a similar case scenario. You may use the steps that we executed in Case-scenario 1 as a template for your solution.

This is the sixth season of outfielder Juan Soto in the majors. If during the first five seasons he received 79, 108,41,145, and 135 walks, how many does he need on this season for his overall number of walks per season to be at least 100?

# Overall # of walks
walk_per_season <- c(79,108,41,145,135)

wanted_walk <- 100

n_seasons <- 6

w_6 <- n_seasons*wanted_walk - sum(walk_per_season)

w_6

[1] 92

Case-scenario 2

The average salary of 10 baseball players is 72,000 dollars a week and the average salary of 4 soccer players is 84,000. Find the mean salary of all 14 professional players.

Solution

n_1 <- 10
n_2 <- 4
y_1 <- 72000
y_2 <- 84000
# Mean salary overall
salary_ave <-  (n_1*y_1 + n_2*y_2)/(n_1+n_2)
salary_ave

[1] 75428.57

Question 2 The average salary of 7 basketball players is 102,000 dollars a week and the average salary of 9 NFL players is 91,000. Find the mean salary of all 16 professional players.

n_1 <- 7
n_2 <- 9
y_1 <- 102000
y_2 <- 91000
# Mean salary overall
salary_ave <-  (n_1*y_1 + n_2*y_2)/(n_1+n_2)
salary_ave

[1] 95812.5

getwd()

[1] "/cloud/project"

contract_length <- read.table("allcontracts.csv", header = TRUE, sep = ",")
contract_years <- contract_length$years

1. To find the mean and the standard deviation

contracts_mean  <- mean(contract_years)
contracts_mean

[1] 3.458918

# Median
contracts_median <- median(contract_years)
contracts_median

[1] 3

# Find number of observations
contracts_n <- length(contract_years)
# Find standard deviation
contracts_sd <- sd(contract_years)

2. What percentage of the data lies within one standard deviation of the mean?

contracts_w1sd <- sum((contract_years - contracts_mean)/contracts_sd < 1)/ contracts_n
# Percentage of observation within one standard deviation of the mean
contracts_w1sd

[1] 0.8416834

## Difference from empirical 
contracts_w1sd - 0.68

[1] 0.1616834

3. What percentage of the data lies within two standard deviations of the mean?

## Within 2 sd
contracts_w2sd <- sum((contract_years - contracts_mean)/ contracts_sd < 2)/contracts_n
contracts_w2sd

[1] 1

## Difference from empirical 
contracts_w2sd - 0.95

[1] 0.05

contracts_w3sd - 0.9973

[1] 0.0027

4. What percent of the data lies within three standard deviations of the mean?

## Within 3 sd 
contracts_w3sd <- sum((contract_years - contracts_mean)/ contracts_sd < 3)/contracts_n
contracts_w3sd

[1] 1

## Difference from empirical 
contracts_w3sd - 0.9973

[1] 0.0027

5. Draw a histogram

# Create histogram
hist(contract_years,xlab = "Years Left in Contract",col = "green",border = "red", xlim = c(0,6), ylim = c(0,210),
   breaks = 3)

Question 3

Use the skills learned in case scenario number 3 on one the following data sets. You may choose only one dataset. They are both available in Canvas.

doubles_hit.csv and triples_hit.csv

triples <- read.csv("triples_hit.csv", header = TRUE, sep = ",")
triplesnumber <- triples$triples_hit

# Mean 
triplesnumber_mean  <- mean(triplesnumber)
triplesnumber_mean

[1] 4.96

# Median
triplesnumber_median <- median(triplesnumber)
triplesnumber_median

[1] 5

# Find number of observations
triplesnumber_n <- length(triplesnumber)
# Find standard deviation
triplesnumber_sd <- sd(triplesnumber)

2. Percentage of the data lies within one standard deviation of the mean

triplesnumber_w1sd <- sum((triplesnumber - triplesnumber_mean)/triplesnumber_sd < 1)/ triplesnumber_n
# Percentage of observation within one standard deviation of the mean
triplesnumber_w1sd

[1] 0.88

## Difference from empirical 
triplesnumber_w1sd - 0.68

[1] 0.2

3. Percentage of the data lies within two standard deviation of the mean

## Within 2 sd
triplesnumber_w2sd <- sum((triplesnumber - triplesnumber_mean)/ triplesnumber_sd < 2)/triplesnumber_n
triplesnumber_w2sd

[1] 0.93

## Difference from empirical 
triplesnumber_w2sd - 0.95

[1] -0.02

4. Percentage of the data lies within three standard deviation of the mean

## Within 3 sd 
triplesnumber_w3sd <- sum((triplesnumber - triplesnumber_mean)/ triplesnumber_sd < 3)/triplesnumber_n
triplesnumber_w3sd

[1] 0.98

## Difference from empirical 
triplesnumber_w3sd - 0.9973

[1] -0.0173

5. Histogram

# Create histogram
hist(triplesnumber,xlab = "Number of Triple",col = "green",border = "red", xlim = c(0,15), ylim = c(0,80),
   breaks = 5)