Introduction to linear regression
Amit Kapoor
11/4/2019
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.
## 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
## 7 New York Yankees 867 5518 1452 222 0.263 1138
## 8 Milwaukee Brewers 721 5447 1422 185 0.261 1083
## 9 Colorado Rockies 735 5544 1429 163 0.258 1201
## 10 Houston Astros 615 5598 1442 95 0.258 1164
## 11 Baltimore Orioles 708 5585 1434 191 0.257 1120
## 12 Los Angeles Dodgers 644 5436 1395 117 0.257 1087
## 13 Chicago Cubs 654 5549 1423 148 0.256 1202
## 14 Cincinnati Reds 735 5612 1438 183 0.256 1250
## 15 Los Angeles Angels 667 5513 1394 155 0.253 1086
## 16 Philadelphia Phillies 713 5579 1409 153 0.253 1024
## 17 Chicago White Sox 654 5502 1387 154 0.252 989
## 18 Cleveland Indians 704 5509 1380 154 0.250 1269
## 19 Arizona Diamondbacks 731 5421 1357 172 0.250 1249
## 20 Toronto Blue Jays 743 5559 1384 186 0.249 1184
## 21 Minnesota Twins 619 5487 1357 103 0.247 1048
## 22 Florida Marlins 625 5508 1358 149 0.247 1244
## 23 Pittsburgh Pirates 610 5421 1325 107 0.244 1308
## 24 Oakland Athletics 645 5452 1330 114 0.244 1094
## 25 Tampa Bay Rays 707 5436 1324 172 0.244 1193
## 26 Atlanta Braves 641 5528 1345 173 0.243 1260
## 27 Washington Nationals 624 5441 1319 154 0.242 1323
## 28 San Francisco Giants 570 5486 1327 121 0.242 1122
## 29 San Diego Padres 593 5417 1284 91 0.237 1320
## 30 Seattle Mariners 556 5421 1263 109 0.233 1280
## 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
## 7 147 97 0.343 0.444 0.788
## 8 94 96 0.325 0.425 0.750
## 9 118 73 0.329 0.410 0.739
## 10 118 56 0.311 0.374 0.684
## 11 81 69 0.316 0.413 0.729
## 12 126 82 0.322 0.375 0.697
## 13 69 71 0.314 0.401 0.715
## 14 97 79 0.326 0.408 0.734
## 15 135 86 0.313 0.402 0.714
## 16 96 102 0.323 0.395 0.717
## 17 81 79 0.319 0.388 0.706
## 18 89 80 0.317 0.396 0.714
## 19 133 94 0.322 0.413 0.736
## 20 131 81 0.317 0.413 0.730
## 21 92 63 0.306 0.360 0.666
## 22 95 72 0.318 0.388 0.706
## 23 108 72 0.309 0.368 0.676
## 24 117 74 0.311 0.369 0.680
## 25 155 91 0.322 0.402 0.724
## 26 77 89 0.308 0.387 0.695
## 27 106 80 0.309 0.383 0.691
## 28 85 86 0.303 0.368 0.671
## 29 170 71 0.305 0.349 0.653
## 30 125 67 0.292 0.348 0.640In 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.
- What type of plot would you use to display the relationship between
runsand one of the other numerical variables? Plot this relationship using the variableat_batsas the predictor. Does the relationship look linear? If you knew a team’sat_bats, would you be comfortable using a linear model to predict the number of runs?
A scatter plot would be good to diplay the relationship between runs and any other numerical variable. Below is the plot for runs and at_bats. The relationship seems positive and linear but not strong enough to predict runs from at_bats.
If the relationship looks linear, we can quantify the strength of the relationship with the correlation coefficient.
## [1] 0.610627Sum 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.
- 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 runs and at_bats seems linear and positive though not strong enough. There are few 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.
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## -2789.2429 0.6305
##
## Sum of Squares: 123721.9After 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.
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## -2789.2429 0.6305
##
## Sum of Squares: 123721.9Note 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.
- 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?
Here are the sum of squares values I tried Sum of Squares: 137958.3 Sum of Squares: 169908.7 Sum of Squares: 126575.7 Sum of Squares: 160804.4 Sum of Squares: 142495.2
I got the smallest sum of squares as 126575.7.
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).
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: 0.0003388Let’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
homerunsto predictruns. 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?
\[ \hat{y} = 415.2389 + 1.8345 * homeruns \]
##
## 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-07Here is the slope is 1.8345 which means runs increase by 1.8345 for every increase in homerun.
Prediction and prediction errors
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.
- 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?
\[ \hat{y} = -2789.2429 + 0.6305 * atbats \]
## [1] 727.6861Based on regression line, the prediction is 727. Given that we dont have data for at_bats as 5578, if we could approximate it as 5579, the runs of the team will be 713 which says model overestimated the run by about 14.
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.
- Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between runs and at-bats?
It doesnt seem any apparent pattern in the residuals plot. The data seems fairly distributed so relationship is linear.
Nearly normal residuals: To check this condition, we can look at a histogram
or a normal probability plot of the residuals.
- Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?
Seeing histogram, it seems slightly right skewed and the QQ plot seems to follow theoretical values so I can say the nearly normal residuals condition appears to be met.
Constant variability:
- Based on the plot in (1), does the constant variability condition appear to be met?
Yes the constant variability conditions appears to be met though there are a few outliers.
On Your Own
- Choose another traditional variable from
mlb11that you think might be a good predictor ofruns. Produce a scatterplot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?
Lets consider ‘hits’ as a predictor here. Below scatter plot shows a positive, linear relationship.
- How does this relationship compare to the relationship between
runsandat_bats? Use the R\(^2\) values from the two model summaries to compare. Does your variable seem to predictrunsbetter thanat_bats? How can you tell?
R\(^2\) is 0.6419 for runs~hits and 0.3729 for runs~at_bats. So ‘hits’ predicts ‘runs’ better as compared to ‘at_bats’.
##
## 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 = 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
runsand each of the other five traditional variables. Which variable best predictsruns? 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).
It seems ‘bat_avg’ best predicts ‘runs’ as its R\(^2\) is the highest among others.
##
## 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
- 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?
The newer variables seem to predict runs better than older traditional variables. The R\(^2\) value for newer variables is higher than the older variables. Of all ten variables, new_obs variable has the maximum R\(^2\) so new_obs seems best predictor.
n1 <- lm(runs ~ new_onbase, data = mlb11)
n2 <- lm(runs ~ new_slug, data = mlb11)
n3 <- lm(runs ~ new_obs, data = mlb11)
par(mfrow=c(1,3))
plot(mlb11$runs ~ mlb11$new_onbase)
abline(n1)
plot(mlb11$runs ~ mlb11$new_slug)
abline(n2)
plot(mlb11$runs ~ mlb11$new_obs)
abline(n3)
##
## 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_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##
## 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- Check the model diagnostics for the regression model with the variable you decided was the best predictor for runs.
The below histogram of residuals shows nearly normal distribution. I see constant variability below as well.
