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, better predict 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.
Let’s load up the data for the 2011 season.
load("more/mlb11.RData")
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.
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?Answer: Scatterplot can be used to display the relationship between these 2 variables.
plot(mlb11$runs ~ mlb11$at_bats, main = "Runs and atBats", xlab = "At Bats", ylab = "Runs")
Answer (cont.): The relationship looks somewhat linear even though correlation doesnt appear very strong. I would feel 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
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.
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.
Answer: There seams to be a positive correlation between the 2 variables, it is not very strong and there are some outliers.
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.
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?Answer: After running the function 4 times, the smallest sum of squares I got is 125969.5, which is pretty close to what R generated: 123721.9.
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.
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?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
m2 <- lm(runs ~ homeruns, data = mlb11)
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
Answer:
y=415.2389 + 1.8345*x
The slope (1.8345) is positive, so it indicates a positive co-relation. For each additional homerun, we expect an increase of 1.8345 runs.
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.
round(-2789.2429 + 0.6305 * 5578,1)
## [1] 727.7
Answer: Philadelphia Phillies is a team which has 5579 at_bats and their runs are 713.
Our prediction was an overestimate by 14.7. Residual is -14.7.
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
Answer: There is no apparent pattern in the residual plot, which indicates that the relationship can be considered linear and we can use linear model for this data.
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$residuals) # adds diagonal line to the normal prob plot
Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?
Answer: Yes, the data appears nearly normal based on the plots above.
Constant variability:
Answer: Yes, based on the plot, the variability of points around the least squares line remains roughly constant.
mlb11
that you think might be a good predictor of runs
. Produce a scatterplot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?I think that ‘hits’ variable will be a good predictor of ‘runs’ so I will use that variable for my model.
plot(mlb11$runs ~ mlb11$hits, main = "Runs and Hits", xlab = "Hits", ylab = "Runs")
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
At a glance, the relationship appears to be linear.
y = -375.56 + 0.7589*x
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?Answer: R^2 for ‘hits’ is 0.6419 and R^2 for ‘at_bats’ is 0.3729.
Since more of the variability of ‘runs’ can be explained by hits - that variable is a better predictor.
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).Batting Average is the strongest predictor.
plot(mlb11$runs ~ mlb11$bat_avg, main = "Runs and BatAvg", xlab = "BatAvg", ylab = "Runs")
cor(mlb11$runs, mlb11$bat_avg)
## [1] 0.8099859
m4 <- lm(runs ~ bat_avg, data = mlb11)
summary(m4)
##
## Call:
## lm(formula = runs ~ bat_avg, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -94.676 -26.303 -5.496 28.482 131.113
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -642.8 183.1 -3.511 0.00153 **
## bat_avg 5242.2 717.3 7.308 5.88e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 49.23 on 28 degrees of freedom
## Multiple R-squared: 0.6561, Adjusted R-squared: 0.6438
## F-statistic: 53.41 on 1 and 28 DF, p-value: 5.877e-08
runs
? Using the limited (or not so limited) information you know about these baseball statistics, does your result make sense?cor(mlb11$runs, mlb11$new_onbase)
## [1] 0.9214691
cor(mlb11$runs, mlb11$new_slug)
## [1] 0.9470324
cor(mlb11$runs, mlb11$new_obs)
## [1] 0.9669163
m5 <- lm(runs ~ new_onbase, data = mlb11)
summary(m5)
##
## 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
m6 <- lm(runs ~ new_slug, data = mlb11)
summary(m6)
##
## 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
m7 <- lm(runs ~ new_obs, data = mlb11)
summary(m7)
##
## 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
plot(mlb11$runs ~ mlb11$new_onbase, main = "Runs and On Base", xlab = "On Base", ylab = "Runs")
abline(m5)
plot(mlb11$runs ~ mlb11$new_slug, main = "Runs and Slugging", xlab = "Slugging", ylab = "Runs")
abline(m6)
plot(mlb11$runs ~ mlb11$new_obs, main = "Runs and OnBase+Slugging", xlab = "OnBase+Slugging", ylab = "Runs")
abline(m7)
Answer: The new variables are more effective than old variables. The strongest predcitor is “on-base plus slugging” variable - R squared is 0.9349. My result makes sense based on my knowledge of baseball statistics.
Check Linearity:
plot(m7$residuals ~ mlb11$new_obs)
abline(h = 0, lty = 3)
There is no apparent pattern in the residual plot, which indicates that the relationship can be considered linear and we can use linear model for this data.
hist(m7$residuals)
qqnorm(m7$residuals)
qqline(m7$residuals)
The data appears nearly normal based on the plots above.
plot(mlb11$runs ~ mlb11$new_obs)
abline(m7)
Based on the plot, the variability of points around the least squares line remains roughly constant.
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.