0.1 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.

0.2 The data

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.

0.2.1 Exercise 1

  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?

A scatter plot is good starting point to visualize relationship between runs and at bats.

paste("We would use a scatter plot to display the relationship in between the variables. The linear model predicts the number of runs.")
## [1] "We would use a scatter plot to display the relationship in between the variables. The linear model predicts the number of runs."
plot(x = mlb11$at_bats, y = mlb11$runs, main = "MLB 11", xlab = "At bats", ylab = "Runs")

reg <- lm(mlb11$runs~mlb11$at_bats)
plot(mlb11$at_bats,mlb11$runs)
abline(reg)

ggplot(mlb11, aes(x=at_bats,y=runs, color = team)) + 
  geom_point() + 
  geom_smooth(method="lm") +
  xlab('Number of Runs') + 
  ylab('Number of at Bats') +
  ggtitle('Runs vs At Bats')

mlb11 %>% summarise(cor(runs, at_bats))
##   cor(runs, at_bats)
## 1           0.610627
paste("The plot shows the relationship is linear and positive. The correlation coeffiecient(R) is 0.61. So if we know the team's at_bats, we can approximatly predict the number of runs.")
## [1] "The plot shows the relationship is linear and positive. The correlation coeffiecient(R) is 0.61. So if we know the team's at_bats, we can approximatly predict the number of runs."

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

0.3 Sum 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.

0.3.1 Exercise 2

  1. 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.
par(mfrow=c(2,2))

hist(mlb11$runs, main = "Runs")
hist(mlb11$at_bats, main = "At Bats")

boxplot(mlb11$runs, main = "Runs")
boxplot(mlb11$at_bats, main = "At Bats")

paste("The relationships are positive and linear; we can draw a straight line through the data with almost equal numbers of data points on the both side of the line. The relationship is strong there is scatter, but there is a definite slope to the trend line. Most points lend to a positive linear relationship such that an increased in bats have effect in increased in runs")
## [1] "The relationships are positive and linear; we can draw a straight line through the data with almost equal numbers of data points on the both side of the line. The relationship is strong there is scatter, but there is a definite slope to the trend line. Most points lend to a positive linear relationship such that an increased in bats have effect in increased in runs"
paste("The relationship appears to be positively correlated with somewhat a mild to moderate positive (up and to the right) trend.")
## [1] "The relationship appears to be positively correlated with somewhat a mild to moderate positive (up and to the right) trend."
paste("There is a large correlation between the number of at bats and runs the correlation is .6 it is important to note that the range of both the x and y variables is between a small subset of possible values, because runs are all between 550 and 900 and at bats are between 5300 and 5800")
## [1] "There is a large correlation between the number of at bats and runs the correlation is .6 it is important to note that the range of both the x and y variables is between a small subset of possible values, because runs are all between 550 and 900 and at bats are between 5300 and 5800"

For runs, - the histogram looks like single peak - the distribuion is almost symmetrical. - There are also an outlier.

For at_bats, - the histogram looks like dual peak. - The distribution is little bit right skewed. - As the population size is around 30, we have to make some reasonable assumptions on distributions and consider it as normal distribution - The direction between two variables is positive. But the relationship is moderately weak. The linear regression line did not fall or fit into any actual points. There are some influential points in the plot which alters the regression 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.

0.3.2 Exercise 3

  1. 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?
par(mfrow=c(1,2))
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
plot_ss(x = mlb11$homeruns, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##     415.239        1.835  
## 
## Sum of Squares:  73671.99
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

The lowest sum of squares I came across using this method was 73671.99.

0.4 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).

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.

0.4.1 Exercise 4

  1. 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?
ggplot(mlb11,aes(x=homeruns,y=runs)) + geom_point() + geom_smooth(method="lm")

plot_ss(x = mlb11$homeruns, y = mlb11$runs, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##     415.239        1.835  
## 
## Sum of Squares:  73671.99
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
paste("The linear regression equation is: runs = 415.2389 + 1.8345 * homeruns. The slope of 1.8345 means that the more homeruns the team makes, the runs they make (which makes sense, since a homerun creates more runs).")
## [1] "The linear regression equation is: runs = 415.2389 + 1.8345 * homeruns. The slope of 1.8345 means that the more homeruns the team makes, the runs they make (which makes sense, since a homerun creates more runs)."

0.5 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. 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.

0.5.1 Exercise 5

  1. 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?
mlb11 %>% 
  select(team:at_bats) %>% 
  mutate(atbats_resid = abs(at_bats -5578),
         runs_resid = abs (runs - ceiling(-2789.2429 + .6305 * 5578))) %>% 
  arrange(atbats_resid) %>% 
  filter(atbats_resid == min(atbats_resid))
##                    team runs at_bats atbats_resid runs_resid
## 1 Philadelphia Phillies  713    5579            1         15
predictedRuns  <- -2789.2429 + 0.6305 * 5578
paste("Predicted runs:", predictedRuns)
## [1] "Predicted runs: 727.686099999999"
observed_df = filter(mlb11, at_bats == 5579)
ObservedRuns = observed_df$runs

residual = ObservedRuns - predictedRuns

qplot( mlb11$at_ba, m1$residuals) + geom_hline(yintercept = 20)

paste0(" The residuals look evenly distributed. We can predict that a team with 5,578 at bats would score ", predictedRuns, " runs. If we look at mlb11 dataset we notice that Philadephia Phillies at_bats score is 5,579 (~5,578) with 713 runs. So, the observed runs would be close to 713, but the residual is ",residual,", which overestimates the observation.")
## [1] " The residuals look evenly distributed. We can predict that a team with 5,578 at bats would score 727.686099999999 runs. If we look at mlb11 dataset we notice that Philadephia Phillies at_bats score is 5,579 (~5,578) with 713 runs. So, the observed runs would be close to 713, but the residual is -14.6860999999994, which overestimates the observation."

0.6 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.

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

0.6.1 Exercise 6

  1. Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between runs and at-bats?
par(mfrow=c(1,2))

ggplot(data = m1, aes(x = .resid)) +
  geom_histogram(binwidth = 25) +
  xlab("Residuals")

ggplot(data = m1, aes(sample = .resid)) +
  stat_qq()

paste("There does not appear to be any apparent pattern in the residuals plot which suggest there is linearity between runs and at-bats. The variability appears fairly constant with each other (with the exception ~5525, which appears to be an outlier) and there appears to be nearly normal residuals. Therefore, the linear regression model is likely reliable. We can notice that in the residual plot, there are more points on the left side than the right side. There are more or less similar number of points above and below the line. Therefore, we can say that there is no strong non-linearity in the residual plot.")
## [1] "There does not appear to be any apparent pattern in the residuals plot which suggest there is linearity between runs and at-bats. The variability appears fairly constant with each other (with the exception ~5525, which appears to be an outlier) and there appears to be nearly normal residuals. Therefore, the linear regression model is likely reliable. We can notice that in the residual plot, there are more points on the left side than the right side. There are more or less similar number of points above and below the line. Therefore, we can say that there is no strong non-linearity in the residual plot."

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

0.6.2 Exercise 7

  1. Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?
paste("The distribution looks to be fairly normally distributed. It seems reasonable to conclude that nearly normal residuals condition is met though there are some deviations in the line.")
## [1] "The distribution looks to be fairly normally distributed. It seems reasonable to conclude that nearly normal residuals condition is met though there are some deviations in the line."

Constant variability:

0.6.3 Exercise 8

  1. Based on the plot in (1), does the constant variability condition appear to be met?
par(mfrow=c(1,3))
ggplot(data = m2, aes(x = .fitted, y = .resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  xlab("Fitted values") +
  ylab("Residuals")

ggplot(data = m2, aes(x = .resid)) +
  geom_histogram(binwidth = 25) +
  xlab("Residuals")

ggplot(data = m2, aes(sample = .resid)) +
  stat_qq()

paste("It appears that the condition of constant variability appears to have been met. Condition for constant variability says that the variability of points around the least squares line remains roughly constant. It appears that the constant variability condition is met here.")
## [1] "It appears that the condition of constant variability appears to have been met. Condition for constant variability says that the variability of points around the least squares line remains roughly constant. It appears that the constant variability condition is met here."

0.7 On Your Own

  • 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?

    m3 = lm(runs ~ wins, data = mlb11)
summary(m3) #wins
    ## 
## 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$runs ~ mlb11$wins)
abline(m3)

    plot(mlb11$hits, mlb11$runs, main = "Hits vs. Runs")

    plot_ss(x = mlb11$hits, y = mlb11$runs)

    ## Click two points to make a line.

## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##   -375.5600       0.7589  
## 
## Sum of Squares:  70638.75
    paste("There seems to be a decisive linear relationship between these two hits and runs. The more hits a team makes, the more chances they have for a run. After creating a scatterplot and fitting a linear model, it appeared, at a glance, that there appeared to be a linear relationship between hits and runs.")
    ## [1] "There seems to be a decisive linear relationship between these two hits and runs. The more hits a team makes, the more chances they have for a run. After creating a scatterplot and fitting a linear model, it appeared, at a glance, that there appeared to be a linear relationship between hits and runs."
    ls = lm(runs ~ new_onbase, data = mlb11)
ggplot(data = mlb11, aes(x = new_onbase, y = runs))+
  geom_point() + 
  xlab('On Base %') + 
  ylab('Runs') + 
  geom_smooth(method = lm, se = FALSE)

    paste("We can't completely rule out that there is no relationship between the bases and runs. At best we can say that there is a very week relationship.")
    ## [1] "We can't completely rule out that there is no relationship between the bases and runs. At best we can say that there is a very week 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?

    summary(m1) #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
    summary(ls)$r.squared
    ## [1] 0.8491053
    m = lm(runs ~ at_bats, data=mlb11)
summary(m)$r.squared
    ## [1] 0.3728654
    paste0("The hits variable appears to predict runs better than at_bats since the r-squared value is higher and therefore the variability that can be explained by the model is higher.The R2 value for runs ~ at_bats is 0.37 and 0.64 for runs ~ hits. This means that the linear model for runs ~ at_bats only covers 37% of the variance of the data where runs ~ hits covers 64% of the variance.")
    ## [1] "The hits variable appears to predict runs better than at_bats since the r-squared value is higher and therefore the variability that can be explained by the model is higher.The R2 value for runs ~ at_bats is 0.37 and 0.64 for runs ~ hits. This means that the linear model for runs ~ at_bats only covers 37% of the variance of the data where runs ~ hits covers 64% of the variance."

  • 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).

     ggplot(data=mlb11, aes(x=bat_avg, y=runs)) + 
  geom_point()

    m = lm(runs ~ bat_avg, data=mlb11)
plot(mlb11$runs ~ mlb11$bat_avg)
abline(m)

    paste0("Bat_avg has the highest R^2 value at ", summary(m)$r.squared, ", therefore, best predicts runs. The data is linear, the inner-quartile range for the residuals is a consistent distance from zero, and the R2 is a high 0.65 which means the linear model covers a lot of the variance of the data. It appeared that the batting average was the variable which best predicted runs, even though the r-squared value (at approximately 65.61%) was not significantly higher than the runs' variable (at approximately 64.19%).")
    ## [1] "Bat_avg has the highest R^2 value at 0.656077134646863, therefore, best predicts runs. The data is linear, the inner-quartile range for the residuals is a consistent distance from zero, and the R2 is a high 0.65 which means the linear model covers a lot of the variance of the data. It appeared that the batting average was the variable which best predicted runs, even though the r-squared value (at approximately 65.61%) was not significantly higher than the runs' variable (at approximately 64.19%)."

  • 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?

    p1 <- ggplot(mlb11, aes(x = new_onbase, y = runs)) +
  geom_point() +
  xlab("On-Base Percentage") +
  ylab("Runs") +
  geom_smooth(method = "lm", se = F)

p2 <- ggplot(mlb11, aes(x = new_slug, y = runs)) +
      geom_point() +
      xlab("Slug Percentage") +
      ylab("Runs") +
      geom_smooth(method = "lm", se = F)

p3 <- ggplot(mlb11, aes(x = new_obs, y = runs)) +
      geom_point() +
      xlab("On-Base + Slugging") +
      ylab("Runs") +
      geom_smooth(method = "lm", se = F)

grid.arrange(p1, p2, p3, ncol = 2)

    paste("The three new variable are more effective at predicting runs with r-squared values ranging from 84.91% to 93.49%, a significant increase from our previously closest predicting variable that had the highest r-squared value of 65.61%. Overall, the new_obs variable was the best variable in predicting runs.")
    ## [1] "The three new variable are more effective at predicting runs with r-squared values ranging from 84.91% to 93.49%, a significant increase from our previously closest predicting variable that had the highest r-squared value of 65.61%. Overall, the new_obs variable was the best variable in predicting runs."

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

    ls = lm(runs ~ new_obs, data = mlb11)
plot(mlb11$runs ~ mlb11$new_obs, main = "New_Obs Scatterplot and Regression Line")
abline(ls)

    hist(ls$residuals, main = "New_Obs Residuals")

    paste("The histogram of new_obs is nearly symmetrical and unimodal.")
    ## [1] "The histogram of new_obs is nearly symmetrical and unimodal."
    plot(ls$residuals ~ mlb11$new_obs, main = "Variability")
abline(h = 0, lty = 3)

    paste("There seems to be constant variability as the points do not seem to follow any patterns.")
    ## [1] "There seems to be constant variability as the points do not seem to follow any patterns."
    qqnorm(ls$residuals)
qqline(ls$residuals)

    paste("There are some deviations in the qqplot but that seems normal, linear relationship and the points are still close to the line. The relationship looks linear based on the residual plot (residuals are constant across the distribution).")
    ## [1] "There are some deviations in the qqplot but that seems normal, linear relationship and the points are still close to the line. The relationship looks linear based on the residual plot (residuals are constant across the distribution)."