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.
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
I would use a scatter plot to display the relationship between runs and at_bats. The relationship looks linear with some outliners. If I knew a team’s at_bats, I would be comfortable using a 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.
## [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.
Answer
There appears to be a positive relationship between runs and at bats. Correlation coefficient( 0.610627) indicates a positive correlation but not very strong. The at bats increases number of runs also increases. There are some outliners present in the data.
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.
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.
# Note that this chunk will only run in interactive mode
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
I ran the function many times and every time I got the same sum of squares: 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).
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.
##
## 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?Answer
##
## Call:
## lm(formula = mlb11$homeruns ~ mlb11$runs)
##
## Coefficients:
## (Intercept) mlb11$runs
## -85.1566 0.3415
Regression line equation: \[\hat{y} = -85.1566 + 0.3415 * homeruns\]
Let’s create a scatterplot with the least squares line laid on top.
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.
Answer
## [1] 728
According to regression line, the manager would predict 728 runs for a team with 5578 at-bats. For actual data, Philadelphia Phillies has a close at-bats of 5579 and runs of 713, which is 15 runs less than the prediction. So this is an overestimation by 15 runs.
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.
Answer
There is no such pattern in the residuals plot. There is probably a linear relationship between runs and at-bats.
Nearly normal residuals: To check this condition, we can look at a histogram
or a normal probability plot of the residuals.
Answer
Based on the histogram and normal probability plot, the residuals appear to be normal.
Constant variability:
Answer
Yes, the constant variability condition appears to be met.
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?Answer
The above scatter plot shows linear relationship between hits and runs.
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
##
## 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
##
## Call:
## lm(formula = mlb11$runs ~ mlb11$hits)
##
## 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 *
## mlb11$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
Comparing R2 values from 2 models i.e m1 and m2, hits seems to predict runs better than at_bats because m2 has higher R2 value than m1.
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).Answer
m3 = lm(runs ~ bat_avg, data = mlb11)
m4 = lm(runs ~ strikeouts, data = mlb11)
m5 = lm(runs ~ homeruns, data = mlb11)
m6 = lm(runs ~ wins, data = mlb11)
m7 = lm(runs ~ stolen_bases, data = mlb11)
summary(m3)##
## 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
##
## Call:
## lm(formula = runs ~ strikeouts, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -132.27 -46.95 -11.92 55.14 169.76
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1054.7342 151.7890 6.949 1.49e-07 ***
## strikeouts -0.3141 0.1315 -2.389 0.0239 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 76.5 on 28 degrees of freedom
## Multiple R-squared: 0.1694, Adjusted R-squared: 0.1397
## F-statistic: 5.709 on 1 and 28 DF, p-value: 0.02386
##
## 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
##
## Call:
## lm(formula = runs ~ wins, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -145.450 -47.506 -7.482 47.346 142.186
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 342.121 89.223 3.834 0.000654 ***
## wins 4.341 1.092 3.977 0.000447 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 67.1 on 28 degrees of freedom
## Multiple R-squared: 0.361, Adjusted R-squared: 0.3381
## F-statistic: 15.82 on 1 and 28 DF, p-value: 0.0004469
##
## Call:
## lm(formula = runs ~ stolen_bases, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -139.94 -62.87 10.01 38.54 182.49
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 677.3074 58.9751 11.485 4.17e-12 ***
## stolen_bases 0.1491 0.5211 0.286 0.777
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 83.82 on 28 degrees of freedom
## Multiple R-squared: 0.002914, Adjusted R-squared: -0.0327
## F-statistic: 0.08183 on 1 and 28 DF, p-value: 0.7769
m3 model (bat_avg vs runs) has best predicting value for the runs because this model has higher \(R^2\) value.
runs? Using the limited (or not so limited) information you know about these baseball statistics, does your result make sense?Answer
m8 = lm(runs ~ new_onbase, data = mlb11)
m9 = lm(runs ~ new_obs, data = mlb11)
m10 = lm(runs ~ new_slug, data = mlb11)
summary(m8)##
## 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
##
## 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
##
## 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
All 3 models are better predictions for runs. m9 (runs ~ new_obs) has higher \(R^2\) than among all. m9 has 93.5% variability accounted for.
Answer
The relationship looks linear. This indicates a linear relationship between the two variables.
The residuals seems mostly normal according to histogram plot.
The points seem to fit the qq plot line well.
The points fall above and below the line constantly. This seems to be constant variability.