Introduction 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

Confirm the average by calculating the mean

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

[1] 20

In 2023, Luis Rober Jr. was 6 HR short of that 44 HR mark with a career high of 38 HR

sd(Robert_HRs)

[1] 16.02082

max(Robert_HRs)

[1] 44

min(Robert_HRs)

[1] 11

summary(Robert_HRs)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  11.00   11.75   12.50   20.00   20.75   44.00 

fivenum(Robert_HRs)#Tukey's five number summary

[1] 11.0 11.5 12.5 28.5 44.0

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?

Soto_Walks_Before <- c(79,108,41,145,135)
# Average Number of Home-runs per season wanted
wanted_BB <- 100
# Number of seasons
n_soto_seasons <- 6
# Needed Home-runs on season 4
Soto_Walks_6 <- n_soto_seasons*wanted_BB - sum(Soto_Walks_Before)
# Minimum number of Home-runs needed by Robert
Soto_Walks_6

[1] 92

Soto would need to receive 92 walks in order to average 100 walks per season.

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.

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_3<-7
y_3<-102000
n_4<-9
y_4<-91000
salary_ave_2<-(n_3*y_3+n_4*y_4)/(n_3+n_4)
salary_ave_2

[1] 95812.5

Case-scenario 3

The frequency distribution below lists the number of active players in the Barclays Premier League and the time left in their contract.

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

# Mean 
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)

contracts_n

[1] 499

contracts_sd

[1] 1.69686

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

• The empirical rule assumes the data follows a normal distribution (symmetric, bell-shaped, continuous, and unbounded).

The gap from 0.68 is expected here since contract-year data is bounded, discrete, and right-skewed, so it doesn’t meet that assumption.

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

100% of the data is within 2 standard deviations of the mean.

## Difference from empirical 
contracts_w2sd - 0.95

[1] 0.05

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

[1] 1

As we already knew 100% of the data is within 2 standard deviation of the mean so this calculation was not necessary.

Histogram

# Create histogram
hist(contract_years,xlab = "Years Left in Contract",col = "green",border = "red", xlim = c(0,8), ylim = c(0,250),
   breaks = 3)

# Create histogram
hist(contract_years,xlab = "Years Left in Contract",col = "blue",border = "black", xlim = c(0,6), ylim = c(0,100),
   breaks = 8)

#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 <- read.table("doubles_hit.csv", header = TRUE, sep = ",")
doublesnumber <- doubles$doubles_hit

#mean
doubles_mean <- mean(doublesnumber)
doubles_mean

[1] 23.55

# Median

doubles_median <- median(doublesnumber)
doubles_median

[1] 23.5

# Find number of observations
doubles_n <- length(doublesnumber)
# Find standard deviation
doubles_sd <- sd(doublesnumber)

doubles_n

[1] 100

doubles_sd

[1] 13.37371

doubles_w1sd <- sum((doublesnumber - doubles_mean)/doubles_sd < 1)/ doubles_n
# Percentage of observation within one standard deviation of the mean
doubles_w1sd

[1] 0.79

## Difference from empirical 
doubles_w1sd-0.68

[1] 0.11

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

## Within 2 sd
doubles_w2sd <- sum((doublesnumber - doubles_mean)/doubles_sd < 2) / doubles_n
doubles_w2sd

[1] 1

100% of the data is within 2 standard deviations of the mean.

## Difference from empirical 
doubles_w2sd - 0.95

[1] 0.05

## Within 3 sd
doubles_w3sd <- sum((doublesnumber - doubles_mean)/doubles_sd < 3) / doubles_n
doubles_w3sd

[1] 1

## Difference from empirical 
doubles_w3sd - 0.997

[1] 0.003

# Create histogram
hist(doublesnumber ,xlab = "x",col = "green",border = "red", xlim = c(0,60), ylim = c(0,20),
   breaks = 15)

The distribution being right-skewed (most players hit few doubles, a small group hits many), which is why 100% of data falls within 2 SD but the shape is not symmetric