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.
Let’s load up the data for the 2011 season.
library(ggplot2)
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?Ans: I will use a scatter plot to look the relation. Based on below scatter plot, it looks to be linear relation. Yes, we are comfortable with go head with linear model.
ggplot(mlb11, aes(x=mlb11$at_bats, y=mlb11$runs)) + geom_point()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.
Ans: From the scatterplot, there is positive upward correlation. The correlation is moderate, but not strong. There are few outliers too which might be affecting the overall corelation.
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?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.
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?
Ans: Equation.
\[ \hat{y} = 415.2389 + 1.8345 * homeruns \]
Slope is positive which means, every 1% increase in homeruns would contribute 1.8345% of Y
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
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.
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?
Use below formulae for preidiction.
\[ \hat{y} = -2789.2429 + 0.6305 * atbats \]
The approximate prediction is 727.6861.
atbats <- 5578
y_pred <- -2789.2429 + 0.6305 * atbats
y_pred## [1] 727.6861
He would arrive at an estimate of 728 runs. We see that the Phillies have an actual data point very close to this number of at-bats, with 5579 at-bats and only 713 runs. So we have an overestimate and a negative residual: -14.
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 = 0Ans: There is no pattern for residuals plot that indicates a linear relationship.
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$residuals) # adds diagonal line to the normal prob plotAns: This appears to be rougly normally distributed although there is some slight skewing pattern from histogram and the qq plot.
Constant variability:
Ans : Yes, the constant variability condition appears to be met.
Choose another traditional variable from mlb11 that you think might be a good predictor of runs. Produce a scatterplot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?
Ans:: I use another variable called homeruns. which has high R^2 value(0.6266) and high correlation (0.7915577)
m2 <- lm(runs ~ homeruns, data = mlb11)
summary(m2)##
## Call:
## lm(formula = runs ~ homeruns, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -91.615 -33.410 3.231 24.292 104.631
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 415.2389 41.6779 9.963 1.04e-10 ***
## homeruns 1.8345 0.2677 6.854 1.90e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 51.29 on 28 degrees of freedom
## Multiple R-squared: 0.6266, Adjusted R-squared: 0.6132
## F-statistic: 46.98 on 1 and 28 DF, p-value: 1.9e-07
cor(mlb11$runs, mlb11$homeruns)## [1] 0.7915577How 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?
The correlation for new variable homeruns is strong compared to the relationship between runs and at_bats Also the R^2 value is higher that the R^2 value of relationship between runs and at_bats
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 : Out of 7 traditional variables, bat_avg is having high R^2 and string correlation of .80. Below shown is the scatter plot of relationship and the best fit regression line of the relatonship.
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
cor(mlb11$runs, mlb11$bat_avg)## [1] 0.8099859
plot(mlb11$runs ~ mlb11$bat_avg)
abline(m4)runs? Using the limited (or not so limited) information you know about these baseball statistics, does your result make sense?Ans: There are also three newer variables: on-base percentage, slugging percentage, and on-base plus slugging. Out of the new_obs is having high R^2 and strong correlation(0.96)
n3 <- lm(runs ~ new_obs, data = mlb11)
summary(n3)##
## 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
cor(mlb11$runs, mlb11$new_obs)## [1] 0.9669163
plot(mlb11$runs ~ mlb11$new_obs)
abline(n3)Check the model diagnostics for the regression model with the variable you decided was the best predictor for runs.
plot(n3$residuals ~ mlb11$new_obs)
abline(h = 0, lty = 3) Below is the histogram that shows it is symettrical.
hist(n3$residuals)Now, lets look at the Q-Q plot, that proves the normality of disribution of residuals and clear linear relation ship of two variables.
qqnorm(n3$residuals)
qqline(n3$residuals)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.