Batter up

The movie Moneyball focuses on the “quest for the secret of success in baseball”. It follows a low-budget team, the Oakland Athletics, who believed that underused statistics, such as a player’s ability to get on base, betterpredict the ability to score runs than typical statistics like home runs, RBIs (runs batted in), and batting average. Obtaining players who excelled in these underused statistics turned out to be much more affordable for the team.

In this lab we’ll be looking at data from all 30 Major League Baseball teams and examining the linear relationship between runs scored in a season and a number of other player statistics. Our aim will be to summarize these relationships both graphically and numerically in order to find which variable, if any, helps us best predict a team’s runs scored in a season.

The data

Let’s load up the data for the 2011 season.

load("more/mlb11.RData")
head(mlb11)
##                  team runs at_bats hits homeruns bat_avg strikeouts
## 1       Texas Rangers  855    5659 1599      210   0.283        930
## 2      Boston Red Sox  875    5710 1600      203   0.280       1108
## 3      Detroit Tigers  787    5563 1540      169   0.277       1143
## 4  Kansas City Royals  730    5672 1560      129   0.275       1006
## 5 St. Louis Cardinals  762    5532 1513      162   0.273        978
## 6       New York Mets  718    5600 1477      108   0.264       1085
##   stolen_bases wins new_onbase new_slug new_obs
## 1          143   96      0.340    0.460   0.800
## 2          102   90      0.349    0.461   0.810
## 3           49   95      0.340    0.434   0.773
## 4          153   71      0.329    0.415   0.744
## 5           57   90      0.341    0.425   0.766
## 6          130   77      0.335    0.391   0.725

In addition to runs scored, there are seven traditionally used variables in the data set: at-bats, hits, home runs, batting average, strikeouts, stolen bases, and wins. There are also three newer variables: on-base percentage, slugging percentage, and on-base plus slugging. For the first portion of the analysis we’ll consider the seven traditional variables. At the end of the lab, you’ll work with the newer variables on your own.

  1. What type of plot would you use to display the relationship between runs and one of the other numerical variables? Plot this relationship using the variable at_bats as the predictor. Does the relationship look linear? If you knew a team’s at_bats, would you be comfortable using a linear model to predict the number of runs?

I would use scatter plot, as we are looking at two numerical variables, trying to find relationship between them.

library(ggplot2)
theme_set(theme_bw())
ggplot(mlb11, aes(at_bats, runs))+
  geom_point()+
  geom_smooth(method="lm", se=F)

The linearship looks linear and positive. I would look into conditions of linear but so far I think I would be comfortable using linear model to predict the number of runs.

If the relationship looks linear, we can quantify the strength of the relationship with the correlation coefficient.

cor(mlb11$runs, mlb11$at_bats)
## [1] 0.610627

Sum of squared residuals

Think back to the way that we described the distribution of a single variable. Recall that we discussed characteristics such as center, spread, and shape. It’s also useful to be able to describe the relationship of two numerical variables, such as runs and at_bats above.

  1. Looking at your plot from the previous exercise, describe the relationship between these two variables. Make sure to discuss the form, direction, and strength of the relationship as well as any unusual observations.

The relationship between the two variables are positive and linear. The correlation is not very strong (0.61) but also not weak.

Just as we used the mean and standard deviation to summarize a single variable, we can summarize the relationship between these two variables by finding the line that best follows their association. Use the following interactive function to select the line that you think does the best job of going through the cloud of points.

plot_ss(x = mlb11$at_bats, y = mlb11$runs)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##  -2789.2429       0.6305  
## 
## Sum of Squares:  123721.9

After running this command, you’ll be prompted to click two points on the plot to define a line. Once you’ve done that, the line you specified will be shown in black and the residuals in blue. Note that there are 30 residuals, one for each of the 30 observations. Recall that the residuals are the difference between the observed values and the values predicted by the line:

\[ e_i = y_i - \hat{y}_i \]

The most common way to do linear regression is to select the line that minimizes the sum of squared residuals. To visualize the squared residuals, you can rerun the plot command and add the argument showSquares = TRUE.

plot_ss(x = mlb11$at_bats, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##  -2789.2429       0.6305  
## 
## Sum of Squares:  123721.9

Note that the output from the plot_ss function provides you with the slope and intercept of your line as well as the sum of squares.

  1. Using plot_ss, choose a line that does a good job of minimizing the sum of squares. Run the function several times. What was the smallest sum of squares that you got? How does it compare to your neighbors?
plot_ss(x = mlb11$strikeouts, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##   1054.7342      -0.3141  
## 
## Sum of Squares:  163870.1
plot_ss(x = mlb11$stolen_bases, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##    677.3074       0.1491  
## 
## Sum of Squares:  196706.3
plot_ss(x = mlb11$homeruns, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##     415.239        1.835  
## 
## Sum of Squares:  73671.99

The smallest sum of squares I found is 73671.

The linear model

It is rather cumbersome to try to get the correct least squares line, i.e. the line that minimizes the sum of squared residuals, through trial and error. Instead we can use the lm function in R to fit the linear model (a.k.a. regression line).

m1 <- lm(runs ~ at_bats, data = mlb11)

The first argument in the function lm is a formula that takes the form y ~ x. Here it can be read that we want to make a linear model of runs as a function of at_bats. The second argument specifies that R should look in the mlb11 data frame to find the runs and at_bats variables.

The output of lm is an object that contains all of the information we need about the linear model that was just fit. We can access this information using the summary function.

summary(m1)
## 
## Call:
## lm(formula = runs ~ at_bats, data = mlb11)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -125.58  -47.05  -16.59   54.40  176.87 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -2789.2429   853.6957  -3.267 0.002871 ** 
## at_bats         0.6305     0.1545   4.080 0.000339 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 66.47 on 28 degrees of freedom
## Multiple R-squared:  0.3729, Adjusted R-squared:  0.3505 
## F-statistic: 16.65 on 1 and 28 DF,  p-value: 0.0003388

Let’s consider this output piece by piece. First, the formula used to describe the model is shown at the top. After the formula you find the five-number summary of the residuals. The “Coefficients” table shown next is key; its first column displays the linear model’s y-intercept and the coefficient of at_bats. With this table, we can write down the least squares regression line for the linear model:

\[ \hat{y} = -2789.2429 + 0.6305 * atbats \]

One last piece of information we will discuss from the summary output is the Multiple R-squared, or more simply, \(R^2\). The \(R^2\) value represents the proportion of variability in the response variable that is explained by the explanatory variable. For this model, 37.3% of the variability in runs is explained by at-bats.

  1. Fit a new model that uses homeruns to predict runs. Using the estimates from the R output, write the equation of the regression line. What does the slope tell us in the context of the relationship between success of a team and its home runs?
m2 <- lm(runs ~ homeruns, data = mlb11)
m2
## 
## Call:
## lm(formula = runs ~ homeruns, data = mlb11)
## 
## Coefficients:
## (Intercept)     homeruns  
##     415.239        1.835
summary(m2)
## 
## Call:
## lm(formula = runs ~ homeruns, data = mlb11)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -91.615 -33.410   3.231  24.292 104.631 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 415.2389    41.6779   9.963 1.04e-10 ***
## homeruns      1.8345     0.2677   6.854 1.90e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 51.29 on 28 degrees of freedom
## Multiple R-squared:  0.6266, Adjusted R-squared:  0.6132 
## F-statistic: 46.98 on 1 and 28 DF,  p-value: 1.9e-07

predicted runs = intercept + (slope*homeruns)

predicted runs = 415.239 + 1.835*homeruns

For each additional homerun we would expect 1.835 increase in run. The positive strong linear relationship between homerun and run is very strong (i guess as expected)

Prediction and prediction errors

Let’s create a scatterplot with the least squares line laid on top.

plot(mlb11$runs ~ mlb11$at_bats)
abline(m1)

The function abline plots a line based on its slope and intercept. Here, we used a shortcut by providing the model m1, which contains both parameter estimates. This line can be used to predict \(y\) at any value of \(x\). When predictions are made for values of \(x\) that are beyond the range of the observed data, it is referred to as extrapolation and is not usually recommended. However, predictions made within the range of the data are more reliable. They’re also used to compute the residuals.

  1. If a team manager saw the least squares regression line and not the actual data, how many runs would he or she predict for a team with 5,578 at-bats? Is this an overestimate or an underestimate, and by how much? In other words, what is the residual for this prediction?
# find predicted runs
intercept_1 <- -2789.2429
slope_1 <- + 0.6305
predicted_runs <- intercept_1 + slope_1 * 5578
predicted_runs
## [1] 727.6861
# find over or under estimate by observed-predicted
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
observed_1 <- filter(mlb11, at_bats ==5578)
observed_1
##  [1] team         runs         at_bats      hits         homeruns    
##  [6] bat_avg      strikeouts   stolen_bases wins         new_onbase  
## [11] new_slug     new_obs     
## <0 rows> (or 0-length row.names)

In our dataset, we do not have at-bats count of 5,578 that corresponds to runs. We can see if there is something close.

observed_1 <- filter(mlb11, at_bats <=5579 & at_bats >=5577)
observed_1
##                    team runs at_bats hits homeruns bat_avg strikeouts
## 1 Philadelphia Phillies  713    5579 1409      153   0.253       1024
##   stolen_bases wins new_onbase new_slug new_obs
## 1           96  102      0.323    0.395   0.717

For 5579 at-bats, we are overestimating by 728 runs (727.6861) - 713 = 14.68 ~ 15 runs.

Model diagnostics

To assess whether the linear model is reliable, we need to check for (1) linearity, (2) nearly normal residuals, and (3) constant variability.

Linearity: You already checked if the relationship between runs and at-bats is linear using a scatterplot. We should also verify this condition with a plot of the residuals vs. at-bats. Recall that any code following a # is intended to be a comment that helps understand the code but is ignored by R.

plot(m1$residuals ~ mlb11$at_bats)
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

  1. Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between runs and at-bats?

The residual plot shows us that the residuals are distributied pretty normally as there seems to be similar amount of points above and below the line. This meets one of the conditions of linearity of the relationship between runs and at-bats.

Nearly normal residuals: To check this condition, we can look at a histogram

hist(m1$residuals)

or a normal probability plot of the residuals.

qqnorm(m1$residuals)
qqline(m1$residuals)  # adds diagonal line to the normal prob plot

  1. Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?

The histogram of the residuals shows us the distribution of residuals are some what normal. The normal qqnormal and qqline shows us; even though towards the top, bottom and center of the line has some points not within the line, the normal distribution condition is met.

Constant variability:

  1. Based on the plot in (1), does the constant variability condition appear to be met?

The variability of the residuals around the 0 line are rougly constant and noise is approximately the same across all of the values of the variables.

On Your Own

# hits as the predictor variable.

theme_set(theme_bw())
ggplot(mlb11, aes(hits, runs))+
  geom_point()+
  geom_smooth(method="lm", se=F)

At a glance , i see a positive, linear relationship between hits and runs.

plot_ss(x = mlb11$hits, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##   -375.5600       0.7589  
## 
## Sum of Squares:  70638.75

Sum of squares for hits as the predictor variable and run is the response variable is 70638 which is smaller than at_bats as the predictor. (it is even smaller than homeruns as the predictor). So there is a stronger positive linear relationship using hits as the predictor variable compare to using at-bats as the predictor variable.

m3 <- lm(runs ~ hits, data = mlb11)
summary(m3)
## 
## Call:
## lm(formula = runs ~ hits, data = mlb11)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -103.718  -27.179   -5.233   19.322  140.693 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -375.5600   151.1806  -2.484   0.0192 *  
## hits           0.7589     0.1071   7.085 1.04e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 50.23 on 28 degrees of freedom
## Multiple R-squared:  0.6419, Adjusted R-squared:  0.6292 
## F-statistic:  50.2 on 1 and 28 DF,  p-value: 1.043e-07

However, R squared using hits as the predictor variable is 0.6419, R square using at-bats as the predictor variable is 0.3729. In general the higher the R-square, the better the model fits in the data. So i think my predictor variable hits is NOT a better predictor variable compare to at-bats.

# iterate over each variable within mlb11 rsquared and return as a dataframe

df <- data.frame(NULL)
for (i in 2:length(mlb11)){
  m <- summary(lm(runs ~ mlb11[,i], data = mlb11))
  df[i, 1] <- names(mlb11)[i]
  df[i, 2] <- m$r.squared
}
## Warning in summary.lm(lm(runs ~ mlb11[, i], data = mlb11)): essentially
## perfect fit: summary may be unreliable
colnames(df) <- c("variable", "rsquared")
head(df)
##   variable  rsquared
## 1     <NA>        NA
## 2     runs 1.0000000
## 3  at_bats 0.3728654
## 4     hits 0.6419388
## 5 homeruns 0.6265636
## 6  bat_avg 0.6560771
df <- na.omit(df)
df
##        variable    rsquared
## 2          runs 1.000000000
## 3       at_bats 0.372865390
## 4          hits 0.641938767
## 5      homeruns 0.626563570
## 6       bat_avg 0.656077135
## 7    strikeouts 0.169357932
## 8  stolen_bases 0.002913993
## 9          wins 0.360971179
## 10   new_onbase 0.849105251
## 11     new_slug 0.896870368
## 12      new_obs 0.934927126

The highest r squared is new_obs, 0.934927126 which is the best predictor variable for runs.

The newwer variables are new_onbase, new_slug, new_obs.

theme_set(theme_bw())
ggplot(mlb11, aes(new_onbase, runs))+
  geom_point()+
  geom_smooth(method="lm", se=F)

theme_set(theme_bw())
ggplot(mlb11, aes(new_slug, runs))+
  geom_point()+
  geom_smooth(method="lm", se=F)

theme_set(theme_bw())
ggplot(mlb11, aes(new_obs, runs))+
  geom_point()+
  geom_smooth(method="lm", se=F)

mnew_1 <- lm(runs ~ new_onbase, data = mlb11)
mnew_2 <- lm(runs ~ new_slug, data = mlb11)
mnew_3 <- lm(runs ~ new_obs, data = mlb11)
summary(mnew_1)
## 
## Call:
## lm(formula = runs ~ new_onbase, data = mlb11)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -58.270 -18.335   3.249  19.520  69.002 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -1118.4      144.5  -7.741 1.97e-08 ***
## new_onbase    5654.3      450.5  12.552 5.12e-13 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 32.61 on 28 degrees of freedom
## Multiple R-squared:  0.8491, Adjusted R-squared:  0.8437 
## F-statistic: 157.6 on 1 and 28 DF,  p-value: 5.116e-13
summary(mnew_2)
## 
## Call:
## lm(formula = runs ~ new_slug, data = mlb11)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -45.41 -18.66  -0.91  16.29  52.29 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -375.80      68.71   -5.47 7.70e-06 ***
## new_slug     2681.33     171.83   15.61 2.42e-15 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 26.96 on 28 degrees of freedom
## Multiple R-squared:  0.8969, Adjusted R-squared:  0.8932 
## F-statistic: 243.5 on 1 and 28 DF,  p-value: 2.42e-15
summary(mnew_3)
## 
## Call:
## lm(formula = runs ~ new_obs, data = mlb11)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -43.456 -13.690   1.165  13.935  41.156 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -686.61      68.93  -9.962 1.05e-10 ***
## new_obs      1919.36      95.70  20.057  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 21.41 on 28 degrees of freedom
## Multiple R-squared:  0.9349, Adjusted R-squared:  0.9326 
## F-statistic: 402.3 on 1 and 28 DF,  p-value: < 2.2e-16

R-squared values are a lot higher compare to the old variables. new-obs seems to be the best predictor for run response variable.

plot(m3$residuals ~ mlb11$new_obs)
abline(h = 0, lty = 3)

plot_ss(x = mlb11$new_obs, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##      -686.6       1919.4  
## 
## Sum of Squares:  12837.66
qqnorm(m3$residuals)
qqline(m3$residuals)

hist(m3$residuals)

The regression model meets the linearity conditions. There are slight deviations on the upper end of the qq plot but it is not significant. There is a strong positive linear relationship between the two variables.