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.

- 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?

A scatter plot is the best visualization for runs vs. at bats.

`plot(mlb11$at_bats,mlb11$runs, xlab = "At Bats", ylab = "Runs", col = 'darkblue')`

The relationship does look linear, and I would be comfortable using at bats in a linear model to predict number of runs since a hit requires an at bat to happen, and hits result in more runs than walks and hit-by-pitches.

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.

The relationships appears to be linear; you can draw a straight line through the data with equal numbers of data points above an below the line and no trend or pattern to the deviations from the line. The direction is positive, as at bats increases so does runs for the most part. The relationship is moderately strong there is scatter, but there is a definite slope to the trend line.

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.

- 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?

`plot_ss(mlb11$at_bats,mlb11$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
```

- Sum of Squares: 148506.2 - by sight
- Sum of Squares: 124275.9 - by lining up on the corners of the plot frame.
- Sum of Squares: 126185 - small adjustment to the above line.
- Sum of Squares: 156007 - tried to match the rmarkdown output.
- Sum of Squares: 127722 - adjusted opposite to 3rd try.

124275.9 was my best attempt further adjustment to the left or right only increased the sum of squares.

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?

```
m2 <- lm(mlb11$runs ~ mlb11$homeruns)
summary(m2)
```

```
##
## Call:
## lm(formula = mlb11$runs ~ mlb11$homeruns)
##
## 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 ***
## mlb11$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
```

\[ \hat{y} = 415.2389 + 1.8345*homeruns \]

The slope tells us that the more home runs a team has the more runs they have. Runs are the point system in baseball so the more runs the more successful the team.

Let’s create a scatter plot 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?

\[ \hat{y} = -2789.2429 + 0.6305 * atbats \\ \hat{y} = -2789.2429 + 0.6305 * 5578 \\ \hat{y} = 727.6861 \]

```
runs_df <- data.frame(mlb11$runs,mlb11$at_bats)
runs_df
```

```
## mlb11.runs mlb11.at_bats
## 1 855 5659
## 2 875 5710
## 3 787 5563
## 4 730 5672
## 5 762 5532
## 6 718 5600
## 7 867 5518
## 8 721 5447
## 9 735 5544
## 10 615 5598
## 11 708 5585
## 12 644 5436
## 13 654 5549
## 14 735 5612
## 15 667 5513
## 16 713 5579
## 17 654 5502
## 18 704 5509
## 19 731 5421
## 20 743 5559
## 21 619 5487
## 22 625 5508
## 23 610 5421
## 24 645 5452
## 25 707 5436
## 26 641 5528
## 27 624 5441
## 28 570 5486
## 29 593 5417
## 30 556 5421
```

From the data we see that 5579 at bats resulted in 713 runs. We would expect 0.6305 fewer runs for 5578 for 712.3695. That means our model over estimated with a residual of

\[ \hat{e} = y_i - \hat{y} \\ \hat{e} = 712.3695 - 727.6861 \\ \hat{e} =-15.3166 \]

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 scatter plot. 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
```

- Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between runs and at-bats?

No, I do not see a pattern in the residuals. The data seems fairly evenly distributed above and below the trend line at 12 above 18 below. This can come about by chance:

`dbinom(18,30,0.5)`

`## [1] 0.08055309`

This does not seem to significantly indicate that something other than chance is responsible for the split, and that the data is linear.

*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?

There does seem to be a slight right skew to the data, but not enough to invalidate the model by itself. The QQ Normal plot seems to follow the theoretical values well enough. I’d say that the normal residual condition have been met.

*Constant variability*:

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

When the residuals are plotted they randomly vary form -100 to 100 with the exception of 1 point at 150. This seems like fairly constant variance.

- Choose another traditional variable from
`mlb11`

that you think might be a good predictor of`runs`

. Produce a scatter plot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?

Stolen Bases. Putting runners in a better position may increase runs scored.

`plot(mlb11$stolen_bases,mlb11$runs, xlab = 'Stolen Bases', ylab = 'Runs' ,col = 'steelblue3')`

There does not seem to be a linear relationship.

- 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?

```
m3 <- lm(mlb11$runs ~ mlb11$stolen_bases)
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
```

`summary(m3)`

```
##
## Call:
## lm(formula = mlb11$runs ~ mlb11$stolen_bases)
##
## 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 ***
## mlb11$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
```

The R\(^2\) for at bats is 0.3729. Meaning approximately 37% of the trend is caused by the predicting variable. The R\(^2\) for stolen_bases is 0.002914. Meaning approximately 0.3% of the trend is caused by the predicting variable. At bats is a better predictor of runs than stolen bases.

- 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).

```
m4 <- lm(mlb11$runs ~ mlb11$bat_avg )
summary(m4)
```

```
##
## Call:
## lm(formula = mlb11$runs ~ mlb11$bat_avg)
##
## 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 **
## mlb11$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
```

```
plot(mlb11$bat_avg, mlb11$runs, xlab = 'Batting Ave', ylab= 'Runs', col = 'darkblue')
abline(m4, col = 'red')
```

`hist(resid(m4))`

```
plot(mlb11$bat_avg, resid(m4))
abline(h = 0, lty = 3)
```

```
qqnorm(resid(m4))
qqline(resid(m4))
```

Batting average, the number of hits / at bats is the best predictor with R\(^2\) of 0.6561. The conditions for validity also seem to be met.

- 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?

```
m5 <- lm(mlb11$runs ~ mlb11$new_obs)
summary(m5)
```

```
##
## Call:
## lm(formula = mlb11$runs ~ mlb11$new_obs)
##
## 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 ***
## mlb11$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$new_obs, mlb11$runs, xlab = 'On Base + Slugging', ylab= 'Runs', col = 'darkblue')
abline(m5, col = 'red')
```

On Base + Slugging, a combination of on base percentage and slugging average is the best predictor of Runs with an impressive R\(^2\) of 0.9349. Note that slugging average had R\(^2\) of 0.8969 and on base percentage had R\(^2\) of 0.8491.

All three new stats are better than the traditional stats which had R\(^2\) ~ 0.6.

- Check the model diagnostics for the regression model with the variable you decided was the best predictor for runs.

`hist(resid(m5))`

```
plot(mlb11$new_obs, resid(m5))
abline(h = 0, lty = 3)
```

```
qqnorm(resid(m5))
qqline(resid(m5))
```

The residuals are randomly scattered about 0, suggesting constant variance and linearity and normally distributed. On base % + slugging % is a valid predictor of Runs.

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.