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")
library(plotly)
## Warning: package 'plotly' was built under R version 3.3.3
## Loading required package: ggplot2
## Warning: package 'ggplot2' was built under R version 3.3.3
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout

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?

Ans: I use scatter plot to display the relationship between at_bat and runs. It looks like linear and shows positive relationship between these two variables.Yes, it would be useful for linear model even though the data do not all fall exactly on the line. The corelation coefficient 0.61 for 30 observations is not strong, since there are some outliners far from the mean of the linear regression, to enlarge the variance.

plot(mlb11$at_bat, mlb11$runs, xlab = "at_bat", ylab = "runs",main="at_bat average vs runs", frame.plot=TRUE,col="blue")
abline(lm(mlb11$runs ~ mlb11$at_bat))

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.

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, 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

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?

Ans:The samllese sum of the squares is close to 0. Comparing its neighors, all distance between the points and mean are further from the line.

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?
home.runs<- lm(runs ~ homeruns, data = mlb11)
home.runs
## 
## Call:
## lm(formula = runs ~ homeruns, data = mlb11)
## 
## Coefficients:
## (Intercept)     homeruns  
##     415.239        1.835

\[ \hat{y} = 415.2389 + 1.835 * homeruns \]

Ans:With zero homeruns,the average run is 415.24. The slop tells for every homerus there is 1.835 runs needed.

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?
pred = -2789.2429+0.63058*(5578)
pred
## [1] 728.1323
r=0.63058*(5578) #residual
r
## [1] 3517.375

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?

Ans:The residuals are plotted at their original horizontal locations but with the vertical coordinate as the residual.

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?

Ans: Generally the residuals must be nearly normal. When this condition is found to be unreasonable, it is usually because of outliers or concerns about influential points. In this case, it is nearly normal.

Constant variability:

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

Ans:Yes, the constant variability condition appear to be met sicne the variability of points around the least squares line remains roughly constant.

On Your Own

Ans: I use at_bat to predict bat_avg. The scatterplot shows these two variables having linear relationship.

str(mlb11)
## 'data.frame':    30 obs. of  12 variables:
##  $ team        : Factor w/ 30 levels "Arizona Diamondbacks",..: 28 4 10 13 26 18 19 16 9 12 ...
##  $ runs        : int  855 875 787 730 762 718 867 721 735 615 ...
##  $ at_bats     : int  5659 5710 5563 5672 5532 5600 5518 5447 5544 5598 ...
##  $ hits        : int  1599 1600 1540 1560 1513 1477 1452 1422 1429 1442 ...
##  $ homeruns    : int  210 203 169 129 162 108 222 185 163 95 ...
##  $ bat_avg     : num  0.283 0.28 0.277 0.275 0.273 0.264 0.263 0.261 0.258 0.258 ...
##  $ strikeouts  : int  930 1108 1143 1006 978 1085 1138 1083 1201 1164 ...
##  $ stolen_bases: int  143 102 49 153 57 130 147 94 118 118 ...
##  $ wins        : int  96 90 95 71 90 77 97 96 73 56 ...
##  $ new_onbase  : num  0.34 0.349 0.34 0.329 0.341 0.335 0.343 0.325 0.329 0.311 ...
##  $ new_slug    : num  0.46 0.461 0.434 0.415 0.425 0.391 0.444 0.425 0.41 0.374 ...
##  $ new_obs     : num  0.8 0.81 0.773 0.744 0.766 0.725 0.788 0.75 0.739 0.684 ...
plot(mlb11$hits, mlb11$runs, xlab = "hits", ylab = "runs",main="hits average vs runs", frame.plot=TRUE,col="blue")
abline(lm(mlb11$runs ~ mlb11$hits))

  • How does this relationship compare to the relationship between runs and at_bats? Use the R\(^2\) values from the two model summaries to compare. Does your variable seem to predict runs better than at_bats? How can you tell?

Ans: Yes, ‘hits’ does better prediction for ‘runs’, since there are more points on lm line, and lesser outliners. For ‘at_bat’ and ‘runs’, its multiple R-squared is 0.3729, which means only 37.29% ‘runs’ is explained by the linear model. For ‘hits’ and ‘runs’, its multiple R-squared is 0.6419, which means only 64.19% ‘runs’ is explained by the linear model.Therefore, ‘hits’ is better for predicting ‘runs’.

m2 <- lm(runs ~ hits, data = mlb11)
summary(m2)
## 
## 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
  • Now that you can summarize the linear relationship between two variables, investigate the relationships between runs and each of the other five traditional variables. Which variable best predicts runs? Support your conclusion using the graphical and numerical methods we’ve discussed (for the sake of conciseness, only include output for the best variable, not all five).

Ans: The following five variables ‘hits’,‘homeruns’,‘bat_avg’,‘new_onbase’, ‘new_obs’ are better to predite ‘runs’. Among these five variables, ‘new_obs’ has the highest \(R^2\) which is the best variable, inaddition, the scatto plot also shows all points are on or close to the linear regression line.

par(mfrow = c(2, 3))

plot(mlb11$hits, mlb11$runs, xlab = "hits", ylab = "runs", frame.plot=TRUE,col="blue")


plot(mlb11$bat_avg, mlb11$runs, xlab = "bat_avg", ylab = "runs", frame.plot=TRUE,col="deeppink")

plot(mlb11$new_slug, mlb11$runs, xlab = "new_slug", ylab = "runs", frame.plot=TRUE,col="orange")

plot(mlb11$new_onbase, mlb11$runs, xlab = "new_onbase", ylab = "runs", frame.plot=TRUE,col="brown")

plot(mlb11$new_obs, mlb11$runs, xlab = "new_obs", ylab = "runs", frame.plot=TRUE,col="darkgreen")

l<-data.frame(mlb11$runs,mlb11$hits,mlb11$bat_avg,mlb11$new_slug,mlb11$new_onbase,mlb11$new_obs)

for (i in 1:5){
    lms<-lm(l[,i] ~ l[,i+1], data = l)
    print(summary(lms))
}
## 
## Call:
## lm(formula = l[, i] ~ l[, i + 1], data = l)
## 
## 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 *  
## l[, i + 1]     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
## 
## 
## Call:
## lm(formula = l[, i] ~ l[, i + 1], data = l)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -27.855  -8.840   1.141  10.086  21.899 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   -312.1       51.0   -6.12 1.32e-06 ***
## l[, i + 1]    6750.9      199.8   33.79  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 13.71 on 28 degrees of freedom
## Multiple R-squared:  0.9761, Adjusted R-squared:  0.9752 
## F-statistic:  1142 on 1 and 28 DF,  p-value: < 2.2e-16
## 
## 
## Call:
## lm(formula = l[, i] ~ l[, i + 1], data = l)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -0.0120811 -0.0038072 -0.0007623  0.0050569  0.0142072 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.11038    0.01851   5.962 2.02e-06 ***
## l[, i + 1]   0.36244    0.04630   7.828 1.58e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.007263 on 28 degrees of freedom
## Multiple R-squared:  0.6864, Adjusted R-squared:  0.6752 
## F-statistic: 61.28 on 1 and 28 DF,  p-value: 1.582e-08
## 
## 
## Call:
## lm(formula = l[, i] ~ l[, i + 1], data = l)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.035295 -0.008481  0.000156  0.010515  0.024257 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.20671    0.06434  -3.213   0.0033 ** 
## l[, i + 1]   1.88957    0.20059   9.420 3.54e-10 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.01452 on 28 degrees of freedom
## Multiple R-squared:  0.7601, Adjusted R-squared:  0.7516 
## F-statistic: 88.74 on 1 and 28 DF,  p-value: 3.538e-10
## 
## 
## Call:
## lm(formula = l[, i] ~ l[, i + 1], data = l)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -0.0074684 -0.0042599  0.0009995  0.0024718  0.0127444 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.10243    0.01535   6.674 3.05e-07 ***
## l[, i + 1]   0.30321    0.02131  14.229 2.42e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.004768 on 28 degrees of freedom
## Multiple R-squared:  0.8785, Adjusted R-squared:  0.8742 
## F-statistic: 202.5 on 1 and 28 DF,  p-value: 2.425e-14
  • Now examine the three newer variables. These are the statistics used by the author of Moneyball to predict a teams success. In general, are they more or less effective at predicting runs that the old variables? Explain using appropriate graphical and numerical evidence. Of all ten variables we’ve analyzed, which seems to be the best predictor of runs? Using the limited (or not so limited) information you know about these baseball statistics, does your result make sense?

Ans: Three new variables ‘new_slug’,‘new_onbase’,and ‘new_obs’ all are predicte better on ‘runs’ than old variables. From above statistical result, ‘new_obs’ is the best which can explain 87.85% of ‘runs’. ‘new_onbase’ is second best which can explain 77.85% of ‘runs’. The third best is ‘new_slug’ which can explian 68.84%.

  • Check the model diagnostics for the regression model with the variable you decided was the best predictor for runs.
plot(mlb11$new_obs, mlb11$runs, xlab = "new_obs", ylab = "runs",main="new_obs average vs runs", frame.plot=TRUE,col="blue")
abline(lm(mlb11$runs ~ mlb11$new_obs))

This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was adapted for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel from a lab written by the faculty and TAs of UCLA Statistics.