library("ggplot2")

The data

Let’s load up the data for the 2011 season.

download.file("http://www.openintro.org/stat/data/mlb11.RData", destfile = "mlb11.RData")
load("mlb11.RData")

Exercise 1

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?

  • I would use a scatterplot, because it’s one method that shows the relationship between two numerical variables. When plotted, the relationship looks linear. I would be somewhat comfortable using a linear model for this data. It’s clear that a relationship exists between the two variables but it’s unclear whether the relationship is strong.
ggplot(data=mlb11,aes(at_bats,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

Sum of squared residuals

Exercise 2

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.

  • form: linear direction: positive, as x increases, y increases strength: it appears weak to good

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

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

Exercise 3

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(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
  • It’s not prompting me to choose a line. The given sum of squares is 123721.9.

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, 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)
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

we want to make a linear model of runs as a function of at_bats. Least squares regression line for the linear model: ŷ =−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, R2.The R2 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.

Exercise 4

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? equation of the regression lines: ŷ= 415.2389+1.8345*homeruns

  • slope:the slope is 1.8345. For every one unit increase in homeruns the average change in runs is +1.8345.
m2<-lm(formula=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

Prediction and prediction errors

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.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.

Exercise 5

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?

  • From looking at just the least squares regression line the team manager would predict about 700 runs for 5,578 at_bats. The regression line is below the actual data meaning that this is an underestimate.

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.

plot(m1$residuals ~ mlb11$at_bats)
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

## Exercise 6 Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between runs and at-bats?

-The residuals look scattered. There does not appear to be a pattern or form. The residuals cluster moreso towards the middle of the plot and towards the lower digits. They look evenly distributed. In terms of a residual plot, these are good characteristics- it means that the model is good/ has good linearity.

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

## Exercise 7 Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met? - Yes. In terms of residuals there are four assumptions of linear regression: linearity, independence, normal distribution, equal variance. The histogram looks nearly normal.The normal probability plot meets the linearity assumption. the plot follows a linear pattern.

Exercise 8

Based on the plot in (1), does the constant variability condition appear to be met? - Yes the constant variabilty condition seems to be met because the residuals do not fan out in a “triangular” fashion.

On Your Own

1.

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?

  • At a glance, yes there appears to be a linear relationship
ggplot(mlb11,aes(x=homeruns,y=runs))+geom_point()

2.

How does this relationship compare to the relationship between runs and at_bats? Use the R2 values from the two model summaries to compare. Does your variable seem to predict runs better than at_bats? How can you tell?

  • The relationship between runs and at_bats seems similar to runs and homeruns. The shape/form is linear, the direction is positive. The strength looks mid-strong.

  • mdlhomeruns->Multiple R-squared: 0.6266, Adjusted R-squared: 0.6132

  • mdlat_bats->Multiple R-squared: 0.3729, Adjusted R-squared: 0.3505

  • The R^2 for my variable is higher than the R^2 for the at_bats variable. This implies that my variable better predicts runs than at_bats.R^2 is generally interpreted as how well the regression model explains the observed data. A higher R^2 indicates that more variability is explained by the model.

mdlhomeruns<-lm(runs~homeruns,mlb11)
summary(mdlhomeruns)
## 
## 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

3.

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).

  • Out of the other variables, the bat_avg best predicts runs. numerical proof: Multiple R-squared: 0.6561, Adjusted R-squared: 0.6438; it has the highest R^2 of the other variables graphical proof:plotted and fitted with a line, the residuals of bat_avg are clustered closer to the line.
mdlhits<-lm(runs~hits,mlb11)
summary(mdlhits)
## 
## 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
mdlbat_avg<-lm(runs~bat_avg,mlb11)
summary(mdlbat_avg)
## 
## 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
mdlstrikeouts<-lm(runs~strikeouts,mlb11)
summary(mdlstrikeouts)
## 
## 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
mdlstolen_bases<-lm(runs~stolen_bases,mlb11)
summary(mdlstolen_bases)
## 
## 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
mdlwins<-lm(runs~wins,mlb11)
summary(mdlwins)
## 
## 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
plot(mlb11$runs ~ mlb11$hits)
abline(mdlhits)

plot(mlb11$runs ~ mlb11$bat_avg)
abline(mdlbat_avg)

plot(mlb11$runs ~ mlb11$strikeouts)
abline(mdlstrikeouts)

plot(mlb11$runs ~ mlb11$stolen_bases)
abline(mdlstolen_bases)

plot(mlb11$runs ~ mlb11$wins)
abline(mdlwins)

4.

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 new variables are more effective in predicting a team’s success. Their R^2 are higher and the residuals are tighter to the fitted lines. The on-base plus slugging variable appears to be the best overall with an R^2 of more than 93% and a graph with the tightest residuals. I do not know much about baseball so I resorted to Google. Wikipedia lists the top ten MLB players in lifetime On-base plus slugging. The 2020 list consisted of players such as Babe Ruth, Ted Williams, Lou Gehrig, and Jimmie Fox. Great players! It seems like my results make sense.
mdlnew_onbase<-lm(runs~new_onbase,mlb11)
summary(mdlnew_onbase)
## 
## 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
mdlnew_slug<-lm(runs~new_slug,mlb11)
summary(mdlnew_slug)
## 
## 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
mdlnew_obs<-lm(runs~new_obs,mlb11)
summary(mdlnew_obs)
## 
## 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)
abline(mdlnew_onbase)

plot(mlb11$runs ~ mlb11$new_slug)
abline(mdlnew_slug)

plot(mlb11$runs ~ mlb11$new_obs)
abline(mdlnew_obs)

5.

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

summary(mdlnew_obs)
## 
## 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