Hypothesis Testing in R : Exploring 2017 MLB Data


In this project, I aimed to practice different hypothesis tests in R; while exploring data from the 2017 MLB season. I will briefly walk through the data exploration and cleaning but will focus on the statistical tests. After exploring the data, I chose the following questions to ask, leading to four different statistical tests.

  1. Do the high-paid players win more games? : Z-Test

Organizing and Exploring Data

Lets first load all of the data into the environment. Here we will be working with two different Rdata files. One with data about the players (pls2017), and one about the games (gls2017).

# the following line is just renaming the data so the names match.
pls2017 <- pls

Let’s take a look at it; starting with the players data (pls2017).


Okay, so we have a list with 168 items in it. What are these lists holding?


Okay, so its a list of lists… hmm. let’s explore a bit more.

The image below is a screenshot of how the data is organized. (same as view(pls201))

As you can see the data is organized as a list of lists of dataframes. Where each list contains the same 4 dataframes:
1. leagues
2. playing_positions
3. teams
4. players

note: if you look at the descriptions to the right you can see that each equivalent dataframe have the same number of columns. Although it’s not necessary I aggregate all of the like dataframes into one dataframe. This will just make easier for me to explore the data and to perform my tests.

Starting With the Players Data
We will first create the empty data frames, into which we will organize the data.

pls17.players <-  data.frame()
pls17.playing_positions <- data.frame()
pls17.teams <- data.frame()
pls17.leagues <- data.frame()

Now we will create a for-loop to iterate through each sublist and bind the like dataframes to the empty dataframes initialized above.

for( i in 1:length(pls2017)) {
pls.17.players <- rbind(pls17.players, pls2017[[i]]$players)
pls17.playing_positions <- rbind(pls17.playing_positions, pls2017[[i]]$playing_positions)
pls17.teams <- rbind(pls17.teams, pls2017[[i]]$teams)
pls17.leagues <- rbind(pls17.leagues, pls2017[[i]]$leagues)

Just to be safe, we will remove any duplicate entries.

pls.17.players <- unique(pls.17.players)
pls17.playing_positions <- unique(pls17.playing_positions)
pls17.teams <- unique(pls17.teams)
pls17.leagues <- unique(pls17.leagues)

Okay, great now we have all of the players data organized into the 4 data frames.

now we will repeat the process with the games data
The games data is organized the exact same way however there are 12 different data frames. -> Same steps as above:

gls.17.games          <-  data.frame()
gls.17.home_teams <- data.frame()
gls.17.leagues <- data.frame()
gls.17.away_teams <- data.frame()
gls.17.winning_teams <- data.frame()
gls.17.seasons <- data.frame()
gls.17.venues <- data.frame()
gls.17.officials <- data.frame()
gls.17.players <- data.frame()
gls.17.teams <- data.frame()
gls.17.opponents <- data.frame()
gls.17.game_logs <- data.frame()
# and we again will iterate through the list of lists creating the dataframes
for( i in 1:length(gls2017)) {
gls.17.games <- rbind(gls.17.games, gls2017[[i]]$games)
gls.17.home_teams <- rbind(gls.17.home_teams, gls2017[[i]]$home_teams)
gls.17.leagues <- rbind(gls.17.leagues, gls2017[[i]]$leagues)
gls.17.away_teams <- rbind(gls.17.away_teams, gls2017[[i]]$away_teams)
gls.17.winning_teams <- rbind(gls.17.winning_teams, gls2017[[i]]$winning_teams)
gls.17.seasons <- rbind(gls.17.seasons, gls2017[[i]]$seasons)
gls.17.venues <- rbind(gls.17.venues, gls2017[[i]]$venues)
gls.17.officials <- rbind(gls.17.officials, gls2017[[i]]$officials)
gls.17.players <- rbind(gls.17.players, gls2017[[i]]$players)
gls.17.teams <- rbind(gls.17.teams, gls2017[[i]]$teams)
gls.17.opponents <- rbind(gls.17.opponents, gls2017[[i]]$opponents)
gls.17.game_logs <- rbind(gls.17.game_logs, gls2017[[i]]$game_logs)
# and again we will remove any duplicate rows
gls.17.games <- unique(gls.17.games)
gls.17.home_teams <- unique(gls.17.home_teams)
gls.17.leagues <- unique(gls.17.leagues)
gls.17.away_teams <- unique(gls.17.away_teams)
gls.17.winning_teams <- unique(gls.17.winning_teams)
gls.17.seasons <- unique(gls.17.seasons)
gls.17.venues <- unique(gls.17.venues)
gls.17.officials <- unique(gls.17.officials)
gls.17.players <- unique(gls.17.players)
gls.17.teams <- unique(gls.17.teams)
gls.17.opponents <- unique(gls.17.opponents)
gls.17.game_logs <- unique(gls.17.game_logs)

Now we have all of our data loaded into easy-to-read and navigate data frames. I found that the game_log data (gls.17.game_logs) had the most interesting data, as each row depicted a particular game from an individual player’s perspective. I chose to merge this dataframe with the players dataframe (pls.17.players) to get a single dataframe with the most interesting information. And i conducted my tests from that larger data frame.

Just to show you what we’re actually working with here: here are all of the variables from the players dataframe (pls.17.players).

And here are all the variables in the game_log data (gls.17.game_logs).

Merging the player information and the game log information
Merging these two dataframes was easy to do as each player had its own unique player id.

gls.17.logs <- merge(gls.17.game_logs, gls.17.players, by.x = "player_id", by.y = "id")

Great now we have the majority of the interesting data all in one dataframe.

The majority of this process was exploring this data and looking for interesting things to test. However, I decided to cut that out to focus on the actual hypothesis testing. Let’s get started.

1 ] Salary x Total_Wins : Z-Test

Do the high-paid players on average win more games then the population?

Ho: average wins of players_paid <= average wins of players
H1: average wins of players_paid > average wins of players
Test: One-sample Z-Test with 95% confidence level.

We are going to start out by trying to see if the highest paid players (players_paid) actually win more games. We need to start by making the total wins attribute as it does not already exist.

Adding the Total Wins Attribute
For this step, all we need to do is use the which() function inside of a for-loop to select each game log for that particular player which was a win. Then add up the number of instances.

gls.17.players$wins_total <- NA
for(i in gls.17.logs$player_id){
team_wins <- which(gls.17.logs$player_id == i & gls.17.logs$game_played == TRUE & gls.17.logs$team_outcome == "win")
total_wins <- length(team_wins)
index <- which(gls.17.players$id == i)
gls.17.players$wins_total[index] <- total_wins
# just saving progress
gls.17.players2 <- gls.17.players
# now lets take a quick look at the summary of the wins_total column:

Great, so it looks like we have a range from 0–99 wins.

Now, let’s take a look at the salary data:

# first have to convert the salary into an integer
gls.17.players2$salary <- as.integer(gls.17.players2$salary)
# now I am going to remove any players with a salary equal to zero (which must be inaccurate) as well as remove NA's
noSal <- which(gls.17.players2$salary == 0)
gls.17.players3 <- gls.17.players2[-noSal,]
gls.17.players3 <- gls.17.players3[!is.na(gls.17.players3$salary),]
# lets see how many are left:
# I should also check to take out anything that is not in US currency:
which(gls.17.players3$salary_currency != "USD")
# and lets look at the salaries column now:

So there are only 914 players with salaries not equal to zero or NA out of the original 1357, and all of the salaries are in USD. I decided to just remove these players all together as I am simply looking to compare salary and total wins for practice purposes, therefore there is not a big risk of just removing data. It looks like the max salary was almost 35 mil wow crazy.

Are any NA’s in the game wins column?

gls.17.players3 <- gls.17.players3[!is.na(gls.17.players3$wins_total),]

#saving progress again:
gls.17.players4 <- gls.17.players3

Cool, so there are no players with a missing wins_total.

Okay now lets take a look and see if there is a correlation with salary and total number of wins
There are a couple ways to do this, lets first just look at the plot and see if we can see any correlation ourselves.

ggplot(gls.17.players4, aes(x=salary, y=wins_total)) + geom_point() + scale_fill_brewer() + theme_minimal() + theme(legend.position="none", panel.grid.major = element_blank()) + labs( x ="Salary (USD)", y =  "Total # of Wins") + ggtitle(" Total Wins x Salary") + theme(plot.title = element_text(hjust=0.5))

We can see that there does not appear to be any correlation here. now lets try using the `cor()` function which will actually give us the correlation coefficient.

cor(gls.17.players4$salary, gls.17.players4$wins_total)

Okay, so with 0.17 we can re-enforce the fact that there isnt really a correlation between salary and the number of wins.

I do want to run a statistical test however, so I am going to split the population up into two groups, lets label players that make more than 5,000,000 usd as high-paid players or players_paid.

We can compare these high-paid players to the general population and see if their total wins are statistically greater then the population. note: I chose the number $5,000,000 as it represents the majority of the upper quartile and it a nice round number to discuss.

Let’s subset these high paid players.

paid <- subset(gls.17.players4, gls.17.players4$salary > 5000000)
# okay so how many players fall into this category?
# what percent is that of the population?

Okay, so it looks like we have roughly the top quarter of highest paid players.

Performing the Z-Test

Ho : average wins of playerspaid <= average wins of all players
H1 : average wins of playerspaid > average wins of all players

Lets first pull out the numbers we will need for the test; the population mean and standard deviation.

popmu <- mean(gls.17.players4$wins_total)
popsd <- sd(gls.17.players4$wins_total)
paidmu <- mean(paid$wins_total)
# average population game wins
# average high-paid players game wins

We can see that the average number of game wins is higher for the high paid players, Let’s go ahead and perform the Z-test to see if the means are statistically different.

note: I chose to conduct a one-sample Z-test here because with a sample size of n = 215, is large enough to represent the population over the Students T Distribution.

z.test(paid$wins_total, mu = popmu, sigma.x = popsd, alternative = "greater")

The p-value is clearly less than 0.05 and we can reject the Null hypothesis. Therefore, we can confidently state that the high-paid players do on average win more games than the general population. I would hope this is would be the case otherwise the baseball teams would be doing a bad job allocating their funds.

2] Hometown Advantage : Chi-squared

Is the fact that a game is home or away independent from that team winning or losing the game?

Ho: whether or not any particular game is a home game for a team is independent to a team winning a game
H1: they are not independent
Test: Chi-squared test for independence with 95% confidence level.

Here we are simply going to look at whether or not the hometown advantage really helps a team win. I am going to start by removing any games at a neutral site:

gls.17.games2 <- gls.17.games[is.na(gls.17.games$at_neutral_site),]

So there were actually no games at neutral sites, great.

Making the Contingency Table
In order to run the chi-squared test for independence, all I need to do is create a contingency table with home/away x win/lose. To do this I am first going to create an empty matrix, and then populate it by counting the length of the selected wins/losses.

matrix1 <- matrix(c(NA, NA, NA, NA), nrow = 2)
colnames(matrix1) <- c("Win", "loss")
rownames(matrix1) <- c("Home_Game", "Away_Game")
nhome_win <- length(which(gls.17.games2$home_team_outcome == "win"))
nhome_loss <- length(which(gls.17.games2$home_team_outcome == "loss"))

naway_win <- length(which(gls.17.games2$away_team_outcome == "win"))
naway_loss <- length(which(gls.17.games2$away_team_outcome == "loss"))
matrix1[,1] <- c(nhome_win, naway_win)
matrix1[,2] <- c(nhome_loss, naway_loss)

As you can see, I used each game in the game logs as two data points, one for the home team and one for the away team. This is probably not a good idea since other factors that might have come into play in a certain game would be doubled. However for practice purposes I decided to just use each game from each teams perspective.

Performing the Chi-squared Test for Independence

Ho: Whether or not any particular game is a home game for a team is independent to a team winning a game
H1: They are not independent


We can clearly reject the null hypothesis and conclude that the hometown advantage is real!

3] Home Runs x Strike Outs

Do the players who strike out the most, hit more home runs on average than the population?

Ho : average home runs of players_strikers is equal to or less than the average home runs of all players
H1 : average home runs of players_strikers is greater than the average home runs of all players
Test: One-sample T-test with 95% confidence level.

Let’s see if people who strike out more also hit more home runs. First I am going to add columns to the players table and get all the correct attributes.

#saving progress
players.lm <- gls.17.players
players.lm$slg_avg <- NA
players.lm$hr_tot <- NA
players.lm$strk_tot <- NA
players.lm$rbi_tot <- NA

for(i in players.lm$id){
subset <- subset(gls.17.game_logs, gls.17.game_logs$player_id == i)
avg <- mean(subset$slugging_percentage)
hr_tot <- sum(subset$home_runs)
strk_tot <- sum(subset$strikeouts)
rbi_tot <- sum(subset$runs_batted_in)

players.lm[which(players.lm$id == i),"slg_avg" ] <- avg
players.lm[which(players.lm$id == i),"hr_tot" ] <- hr_tot
players.lm[which(players.lm$id == i),"strk_tot" ] <- strk_tot
players.lm[which(players.lm$id == i),"rbi_tot" ] <- rbi_tot

Now lets remove any NA’s’and look at the summary of the two variables.

players.lm <- subset(players.lm, !is.na(players.lm$strk_tot))
players.lm <- subset(players.lm, !is.na(players.lm$hr_tot))
# lets see how many are left after removing NA's:
# lets take a look at the strikeout distribution
# and lets take a look at the home run distribution

Wow so there are only 710 left after removing NA’s, which is a little over 50% of the total players. Thats a bummer but were just gunna go with it.

Lets make a quick plot to see if we can see any correlation visually.

ggplot(players.lm, aes(x=strk_tot, y=hr_tot)) + geom_point() + scale_fill_brewer() + theme_minimal() + theme(legend.position="none", panel.grid.major = element_blank()) + labs( x = "Total # of Strikeouts", y = "Total # of Home Runs") + ggtitle(" Home Runs x Strikeouts") + theme(plot.title = element_text(hjust=0.5))

Cool, So you can clearly see that players who strikeout more also hit more homeruns, again no surprises here. now lets take a look at the correlation coefficient. *note: In-case you were wondering, Aaron Judge came in at the most strikeouts with a whopping 209 strikeouts for the 2017 season (according to this data-set).*

cor(players.lm$hr_tot, players.lm$strk_tot)

Wow, .87 thats a pretty high correlation coefficient. Now even though we can clearly see that there is a correlation between the two attributes, I still want to run a statistical test…

For this projects purposes; I am going to again subset the players based on how many strike outs they hit. I will simply look to see if players who struck out more than 150 times in the 2017 season also hit more home runs than the population.

most_strk <- subset(players.lm, players.lm$strk_tot >= 150)

Okay so there are 24 players who struck out more than 150 times. perfect for a T-test.

Performing the T-Test

Let’s pull out the numbers we need to run the test.

mu.pop.hr <- mean(players.lm$hr_tot)
mu.most.strk <- mean(most_strk$hr_tot)
# now lets look at the average homeruns hit by the population?
# and the averege for the high-strikeout players?

Well the means are extremely different, lets go ahead and perform the T-test to see if they are significantly different.

Ho : average Home Runs of Players_strikers is equal to or less than the average home runs of all players
H1 : average Home Runs of Players_strikers is greater than the average home runs of all players
Test: One-sample T-Test with 95% confidence level.

t.test(most_strk$hr_tot, alternative = “greater”, mu = mu.pop.hr, var.equal = T)

Great so we can again reject the null hypothesis and state that players who strike out more than 150 times on average hit more home runs than the general population.

4] Years of Experience x Runs Batted In : ANOVA

Do players with different levels of experience have different RBI’s?

Ho: Average RBI is the same across the three experience levels
H1: Average RBI is not the same across the three experience levels
Test: One-Way ANOVA with 95% confidence level. plus a tukey post-hoc.

Now let’s explore how years of experience might affect the rbi of a player. We will again subset all the players based on their years of experience and compare these groups average RBI’s. I will split the level of experience into three groups explained below.

Let’s take a look at the experience variable

# to keep organized i decided to just pull the necessary variables into a new dataframe 
exp_x_rbi <- data.frame(as.numeric( players.lm$rbi_tot), as.numeric(players.lm$years_of_experience))
exp_x_rbi <- na.omit(exp_x_rbi)
colnames(exp_x_rbi) <- c("rbi_total", "years_of_experience")

Okay, so I have 557 data points with a range from 0 to 17 years of experience and a mean of 3.4 years.

I will split the data into three groups
1. Low Experience = less than or equal to 5 years
2. Moderate Experience = greater than 5 and less than or equal to 10 years
3. High Experience = greater than 10 years of experience

I will make a simple for-loop to do so:

exp_x_rbi$exp_scale <- NAfor(i in 1:nrow(exp_x_rbi))
if(exp_x_rbi$years_of_experience[i] <= 5)
exp_x_rbi[i, “exp_scale”] <- “Low Experience”
if(exp_x_rbi$years_of_experience[i] > 5 & exp_x_rbi$years_of_experience[i] <= 10)
exp_x_rbi[i, “exp_scale”] <- “Moderate Experience”
if(exp_x_rbi$years_of_experience[i] > 10)
exp_x_rbi[i, “exp_scale”] <- “High Experience”

Lets go ahead and look at these different groups:

exp.low <- subset(exp_x_rbi, exp_x_rbi$exp_scale == “Low Experience”)
exp.mod <- subset(exp_x_rbi, exp_x_rbi$exp_scale == “Moderate Experience”)
exp.high <- subset(exp_x_rbi, exp_x_rbi$exp_scale == “High Experience”)
mu.rbi.pop <- mean(exp_x_rbi$rbi_total)
mu.rbi.low <- mean(exp.low$rbi_total)
mu.rbi.mod <- mean(exp.mod$rbi_total)
mu.rbi.high <- mean(exp.high$rbi_total)
# lets look at the average rbi of population:
# average rbi of low experience players:
# average rbi of moderate experience players:
# average rbi of high experience players:

So there are clearly differences in the means. Let’s create a simple bar plot to compare the means visually.

class1 <- c(“All Players”,”Low Experience” , “Moderate Experience”,”High Experience”)
class1 <- ordered(class1, levels = c(“All Players”,”Low Experience” , “Moderate Experience”,”High Experience”) )
mean_rbi_tot <- c(33.44, 28.96 , 48.48, 46.64)
df_rbi_tot <- data.frame(class1, mean_rbi_tot)
ggplot(data = df_rbi_tot, aes(x=class1, y = mean_rbi_tot, fill = class1)) + geom_bar(stat = “identity”) + scale_fill_brewer() + theme_minimal() + theme(legend.position=”none”, panel.grid.major = element_blank()) + labs( x = “Experience Level”, y = “Average RBI”) + ggtitle(“Total RBI by Experience Level”) + theme(plot.title = element_text(hjust=0.5))

You can see that players within the moderate experience class have the highest average RBI, i think that this makes sense considering their experience and relative age. now lets go ahead and see if these means are statistically different.

Performing A One-Way ANOVA

Ho: Average RBI is the same across the three experience levels
H1: Average RBI is not the same across the three experience levels
Test: One-Way ANOVA with 95% confidence level. plus a tukey post-hoc.

# first we have to make sure that the groups are in-fact factors
exp_x_rbi$exp_scale <- as.factor(exp_x_rbi$exp_scale)
experience_compare <- aov(exp_x_rbi$rbi_total ~ exp_x_rbi$exp_scale , data = exp_x_rbi)
# the ANOVA Results
# and the Summary

We reject the Null hypothesis; and state that the means of rbi total are not the same for the different levels of experience.

Now let’s do a post-hoc to determine which means were statistically different:

tukeyResults <- TukeyHSD(experience_compare)

Post-Hoc Conclusion
We can state that the means of total rbi’s are different for low Experience — High experience, and Moderate — Low,
but not Moderate — high experience

5] Night Attendance vs. Day Attendance

Do more people attend more Night Games?

Ho : average attendance of night games is equal to the average attendance of all games
H1 : average attendance of night games is NOT equal to the average attendance of all games
Test: One-sample Z-Test with 95% confidence level.

For my last test, I am going to explore whether or not people attend more night games than day games. Another simple question to ask, that I can run a statistical test on.

day.night <- gls.17.games[,c(“attendance”, “daytime”)]
day.night <- na.omit(day.night)

We have 2421 games out of 2431, so we lost 10 games which must have had NA’s.

Now lets subset the games into night and day games:

night <- subset(day.night, day.night$daytime == F)
day <- subset(day.night, day.night$daytime == T)
DN.pop.mu <- mean(day.night$attendance)
DN.pop.sd <- sd(day.night$attendance)
night.mu <- mean(night$attendance)
day.mu <- mean(day$attendance)

Huh, so the average attendance to day games is actually more than at night. Thats weird, lets make a quick barchart to compare these means visually.

D.N <- c(“All”, “Day”, “Night”)
atnd <- c(DN.pop.mu, day.mu, night.mu)
df.DN <- data.frame(D.N, atnd)
df.DN$D.N <- as.factor(D.N)
DN.plot <- ggplot(data = df.DN, aes(x=D.N, y = atnd, fill = D.N)) + geom_bar(stat = “identity”) +
scale_fill_brewer() + theme_minimal() + theme(legend.position=”none”, panel.grid.major = element_blank()) + labs( x = “Game Type”, y = “Average Attendance”) + ggtitle(“Average Attendance of Day/Night Games”) + theme(plot.title = element_text(hjust=0.5))

Performing a Z-test to compare the means

Ho : attendance of night games = the attendance of all games
H1 : attendance of night games != attendance of all games
Test: One-sample Z-Test with 95% confidence level.

z.test(night$attendance, mu = DN.pop.mu, sigma.x = DN.pop.sd, alternative = “two.sided”)

We can again reject the null hypothesis that the means are the same, although the average attendance of night games is actually less than the average attendance of day games, which surprised me.. I guess most people dont work.

M.Sc Computer Science at Northeastern University | Data Analysis: R, MySQL, Tableau | https://www.linkedin.com/in/brianward1428/