3 Introduction to linear models
Linear regression is a very powerful statistical technique. Many people have some familiarity with regression just from reading the news, where straight lines are overlaid on scatterplots. Linear models can be used for prediction or to evaluate whether there is a linear relationship between two numerical variables.
3.1 Fitting a line, residuals, and correlation
It’s helpful to think deeply about the line fitting process. In this section, we define the form of a linear model, explore criteria for what makes a good fit, and introduce a new statistic called correlation.
3.1.1 Fitting a line to data
Figure 3.1 shows two variables whose relationship can be modeled perfectly with a straight line. The equation for the line is \(y = 5 + 64.96 x.\) Consider what a perfect linear relationship means: we know the exact value of \(y\) just by knowing the value of \(x.\) A perfect linear relationship unrealistic in almost any natural process. For example, if we took family income (\(x\)), this value would provide some useful information about how much financial support a college may offer a prospective student (\(y\)). However, the prediction would be far from perfect, since other factors play a role in financial support beyond a family’s finances.
Linear regression is the statistical method for fitting a line to data where the relationship between two variables, \(x\) and \(y,\) can be modeled by a straight line with some error:
\[ y = b_0 + b_1 \ x + e\]
The values \(b_0\) and \(b_1\) represent the model’s intercept and slope, respectively, and the error is represented by \(e\). These values are calculated based on the data, i.e. they are sample statistics. If the observed data is a random sample from a target population that we are interested in making inferences about, these values are considered to be point estimates for the population parameters \(\beta_0\) and \(\beta_1\). We will discuss how to make inferences about parameters of a linear model based on sample statistics in Chapter 8.
The Greek letter \(\beta\) is pronounced beta, listen to the pronunciation here.
When we use \(x\) to predict \(y,\) we usually call \(x\) the predictor variable and we call \(y\) the outcome. We also often drop the \(e\) term when writing down the model since our main focus is often on the prediction of the average outcome.
It is rare for all of the data to fall perfectly on a straight line. Instead, it’s more common for data to appear as a cloud of points, such as those examples shown in Figure 3.2. In each case, the data fall around a straight line, even if none of the observations fall exactly on the line. The first plot shows a relatively strong downward linear trend, where the remaining variability in the data around the line is minor relative to the strength of the relationship between \(x\) and \(y.\) The second plot shows an upward trend that, while evident, is not as strong as the first. The last plot shows a very weak downward trend in the data, so slight we can hardly notice it. In each of these examples, we will have some uncertainty regarding our estimates of the model parameters, \(\beta_0\) and \(\beta_1.\) For instance, we might wonder, should we move the line up or down a little, or should we tilt it more or less? As we move forward in this chapter, we will learn about criteria for linefitting, and we will also learn about the uncertainty associated with estimates of model parameters.
There are also cases where fitting a straight line to the data, even if there is a clear relationship between the variables, is not helpful. One such case is shown in Figure 3.3 where there is a very clear relationship between the variables even though the trend is not linear. We discuss nonlinear trends in this chapter and the next, but details of fitting nonlinear models are saved for a later course.
3.1.2 Using linear regression to predict possum head lengths
Brushtail possums are a marsupial that lives in Australia, and a photo of one is shown in Figure 3.4. Researchers captured 104 of these animals and took body measurements before releasing the animals back into the wild. We consider two of these measurements: the total length of each possum, from head to tail, and the length of each possum’s head.
Figure 3.5 shows a scatterplot for the head length (mm) and total length (cm) of the possums. Each point represents a single possum from the data. The head and total length variables are associated: possums with an above average total length also tend to have above average head lengths. While the relationship is not perfectly linear, it could be helpful to partially explain the connection between these variables with a straight line.
We want to describe the relationship between the head length and total length variables in the possum data set using a line. In this example, we will use the total length as the predictor variable, \(x,\) to predict a possum’s head length, \(y.\) We could fit the linear relationship by eye, as in Figure 3.6.
The equation for this line is
\[\hat{y} = 41 + 0.59x\]
A “hat” on \(y\) is used to signify that this is an estimate. We can use this line to discuss properties of possums. For instance, the equation predicts a possum with a total length of 80 cm will have a head length of
\[\hat{y} = 41 + 0.59 \times 80 = 88.2\]
The estimate may be viewed as an average: the equation predicts that possums with a total length of 80 cm will have an average head length of 88.2 mm. Absent further information about an 80 cm possum, the prediction for head length that uses the average is a reasonable estimate.
There may be other variables that could help us predict the head length of a possum besides its length. Perhaps the relationship would be a little different for male possums than female possums, or perhaps it would differ for possums from one region of Australia versus another region. Plot A in Figure 3.7 shows the relationship between total length and head length of brushtail possums, taking into consideration their sex. Male possums (represented by blue triangles) seem to be larger in terms of total length and head length than female possums (represented by red circles). Plot B in Figure 3.7 shows the same relationship, taking into consideration their age. It’s harder to tell if age changes the relationship between total length and head length for these possums.
In Chapter 4, we’ll learn about how we can include more than one predictor in our model. Before we get there, we first need to better understand how to best build a simple linear model with one predictor.
3.1.3 Residuals
Residuals are the leftover variation in the data after accounting for the model fit:
\[\text{Data} = \text{Fit} + \text{Residual}\]
Each observation will have a residual, and three of the residuals for the linear model we fit for the data are shown in Figure 3.8. If an observation is above the regression line, then its residual, the vertical distance from the observation to the line, is positive. Observations below the line have negative residuals. One goal in picking the right linear model is for these residuals to be as small as possible.
Figure 3.8 is almost a replica of Figure 3.6, with three points from the data highlighted. The observation marked by a red circle has a small, negative residual of about 1; the observation marked by a green diamond has a large positive residual of about +7; and the observation marked by a yellow triangle has a moderate negative residual of about 4. The size of a residual is usually discussed in terms of its absolute value. For example, the residual for the observation marked by a yellow triangle is larger than that of the observation marked by a red circle because \(4\) is larger than \(1.\)
Residual: Difference between observed and expected. The residual of the \(i^{th}\) observation \((x_i, y_i)\) is the difference of the observed outcome (\(y_i\)) and the outcome we would predict based on the model fit (\(\hat{y}_i\)):
\[e_i = y_i  \hat{y}_i\]
We typically identify \(\hat{y}_i\) by plugging \(x_i\) into the model.
The linear fit shown in Figure 3.8 is given as \(\hat{y} = 41 + 0.59x.\) Based on this line, formally compute the residual of the observation \((76.0, 85.1).\) This observation is marked by a red circle in Figure 3.8. Check it against the earlier visual estimate, 1.
We first compute the predicted value of the observation marked by a red circle based on the model:
\[\hat{y} = 41+0.59x = 41+0.59\times 76.0 = 85.84\]
Next we compute the difference of the actual head length and the predicted head length:
\[e = y  \hat{y} = 85.1  85.84 = 0.74\]
The model’s error is \(e = 0.74\) mm, which is very close to the visual estimate of 1 mm. The negative residual indicates that the linear model overpredicted head length for this particular possum.
If a model underestimates an observation, will the residual be positive or negative? What about if it overestimates the observation?^{68}
Compute the residuals for the observation marked by a green diamond, \((85.0, 98.6),\) and the observation marked by a yellow triangle, \((95.5, 94.0),\) in the figure using the linear relationship \(\hat{y} = 41 + 0.59x.\)^{69}
Residuals are helpful in evaluating how well a linear model fits a data set. We often display them in a scatterplot such as the one shown in Figure 3.9 for the regression line in Figure 3.8. The residuals are plotted with their predicted outcome variable value as the horizontal coordinate, and the vertical coordinate as the residual. For instance, the point \((85.0, 98.6)\) (marked by the green diamond) had a predicted value of 91.4 mm and had a residual of 7.45 mm, so in the residual plot it is placed at \((91.4, 7.45).\) Creating a residual plot is sort of like tipping the scatterplot over so the regression line is horizontal.
One purpose of residual plots is to identify characteristics or patterns still apparent in data after fitting a model. Figure 3.10 shows three scatterplots with linear models in the first row and residual plots in the second row. Can you identify any patterns remaining in the residuals?
In the first data set (first column), the residuals show no obvious patterns. The residuals appear to be scattered randomly around the dashed line that represents 0.
The second data set shows a pattern in the residuals. There is some curvature in the scatterplot, which is more obvious in the residual plot. We should not use a straight line to model these data. Instead, a more advanced technique should be used to model the curved relationship, such as the variable transformations discussed in Section 2.1.7.
The last plot shows very little upwards trend, and the residuals also show no obvious patterns. It is reasonable to try to fit a linear model to the data. However, it is unclear whether there is evidence that the slope parameter is different from zero. The point estimate of the slope parameter, labeled \(b_1,\) is not zero, but we might wonder if this could just be due to chance. We will address this sort of scenario in Chapter 8.
3.1.4 Describing linear relationships with correlation
We’ve seen plots with strong linear relationships and others with very weak linear relationships. It would be useful if we could quantify the strength of these linear relationships with a statistic.
Correlation: strength of a linear relationship. Correlation which always takes values between 1 and 1, describes the strength of the linear relationship between two variables. We denote the correlation by \(r.\)
The correlation value has no units and will not affected by a linear change in the units (e.g., going from inches to centimeters).
We can compute the correlation using a formula, just as we did with the sample mean and standard deviation. This formula is rather complex^{70}, and like with other statistics, we generally perform the calculations on a computer or calculator. One important aspect of the correlation can be seen in the formula, however. The value of the correlation is unitless. That is, unlike a measurement of the slope of a line (see the next section) which provides an increase in the ycoordinate for a one unit increase in the xcoordinates (in units of the x and y variable), there are no units associated with the correlation. Consider a plot with inches on the xaxis and pounds on the yaxis. If the points are transformed to be cm versus kg, the spread of the points will not change. The only visual change to the plot itself will be the axis labelling of the points.
Figure 3.11 shows eight plots and their corresponding correlations. Only when the relationship is perfectly linear is the correlation either 1 or 1. If the relationship is strong and positive, the correlation will be near +1. If it is strong and negative, it will be near 1. If there is no apparent linear relationship between the variables, then the correlation will be near zero.
The correlation is intended to quantify the strength of a linear trend. Nonlinear trends, even when strong, sometimes produce correlations that do not reflect the strength of the relationship; see three such examples in Figure 3.12.
No straight line is a good fit for any of the data sets represented in Figure 3.12. Try drawing nonlinear curves on each plot. Once you create a curve for each, describe what is important in your fit.^{71}
Consider a handful of crop yields taken from Our World in Data. In each plot a point represents a different country. The x and y variables represent the proportion of total yield in the last 50 years which is due to that crop type.
Order the six scatterplots from strongest negative to strongest positive linear relationship.
As seen in the chart below, the order of most negative correlation to most positive correlation is:
\[A \rightarrow D \rightarrow B \rightarrow C \rightarrow E \rightarrow F\]
Graph  xvariable  yvariable  correlation 

A  potatoes  bananas  0.54 
B  soybeans  cassava  0.16 
C  maize  cassava  0.46 
D  cocoa  bananas  0.44 
E  peas  barley  0.69 
F  wheat  barley  0.85 
3.1.5 Exercises

Visualize the residuals. The scatterplots shown below each have a superimposed regression line. If we were to construct a residual plot (residuals versus \(x\)) for each, describe what those plots would look like.

Trends in the residuals. Shown below are two plots of residuals remaining after fitting a linear model to two different sets of data. Describe important features and determine if a linear model would be appropriate for these data. Explain your reasoning.

Identify relationships, Part I. For each of the six plots, identify the strength of the relationship (e.g. weak, moderate, or strong) in the data and whether fitting a linear model would be reasonable.

Identify relationships, Part II. For each of the six plots, identify the strength of the relationship (e.g. weak, moderate, or strong) in the data and whether fitting a linear model would be reasonable.

Midterms and final. The two scatterplots below show the relationship between the overall course average and two midterm exams (Exam 1 and Exam 2) recorded for 233 students during several years for a statistics course at a university.^{72}
Based on these graphs, which of the two exams has the strongest correlation with the course grade? Explain.
Can you think of a reason why the correlation between the exam you chose in part (a) and the course grade is higher?

Partners’ ages and heights. The Great Britain Office of Population Census and Surveys collected data on a random sample of 170 married couples in Britain, recording the age (in years) and heights (converted here to inches) of the partners. The scatterplot on the left shows the heights of the partners plotted against each other and the plot on the right shows the ages of the partners plotted against each other.^{73}
Describe the relationship between partners’ ages.
Describe the relationship between partners’ heights.
Which plot shows a stronger correlation? Explain your reasoning.
Data on heights were originally collected in centimeters, and then converted to inches. Does this conversion affect the correlation between partners’ heights?

Match the correlation, Part I. Match each correlation to the corresponding scatterplot.^{74}
\(r = 0.7\)
\(r = 0.45\)
\(r = 0.06\)
\(r = 0.92\)

Match the correlation, Part II. Match each correlation to the corresponding scatterplot.^{75}
\(r = 0.49\)
\(r = 0.48\)
\(r = 0.03\)
\(r = 0.85\)

Body measurements, correlation. Researchers studying anthropometry collected body and skeletal diameter measurements, as well as age, weight, height and sex for 507 physically active individuals. The scatterplot below shows the relationship between height and shoulder girth (circumference of shoulders measured over deltoid muscles), both measured in centimeters.^{76}
Describe the relationship between shoulder girth and height.
How would the relationship change if shoulder girth was measured in inches while the units of height remained in centimeters?
Guess the correlation. Eduardo and Rosie are both collecting data on number of rainy days in a year and the total rainfall for the year. Eduardo records rainfall in inches and Rosie in centimeters. How will their correlation coefficients compare?

The Coast Starlight, correlation. The Coast Starlight Amtrak train runs from Seattle to Los Angeles. The scatterplot below displays the distance between each stop (in miles) and the amount of time it takes to travel from one stop to another (in minutes).^{77}
Describe the relationship between distance and travel time.
How would the relationship change if travel time was instead measured in hours, and distance was instead measured in kilometers?
Correlation between travel time (in miles) and distance (in minutes) is \(r = 0.636\). What is the correlation between travel time (in kilometers) and distance (in hours)?

Crawling babies, correlation. A study conducted at the University of Denver investigated whether babies take longer to learn to crawl in cold months, when they are often bundled in clothes that restrict their movement, than in warmer months. Infants born during the study year were split into twelve groups, one for each birth month. We consider the average crawling age of babies in each group against the average temperature when the babies are six months old (that’s when babies often begin trying to crawl). Temperature is measured in degrees Fahrenheit (F) and age is measured in weeks.^{78}
Describe the relationship between temperature and crawling age.
How would the relationship change if temperature was measured in degrees Celsius (C) and age was measured in months?
The correlation between temperature in F and age in weeks was \(r=0.70\). If we converted the temperature to C and age to months, what would the correlation be?

Partners’ ages. What would be the correlation between the ages of partners if people always dated others who are
3 years younger than themselves?
2 years older than themselves?
half as old as themselves?

Graduate degrees and salaries. What would be the correlation between the annual salaries of people with and without a graduate degree at a company if for a certain type of position someone with a graduate degree always made
$5,000 more than those without a graduate degree?
25% more than those without a graduate degree?
15% less than those without a graduate degree?
3.2 Least squares regression
Fitting linear models by eye is open to criticism since it is based on an individual’s preference. In this section, we use least squares regression as a more rigorous approach to fitting a line to a scatterplot.
3.2.1 Gift aid for freshman at Elmhurst College
This section considers a dataset on family income and gift aid data from a random sample of fifty students in the freshman class of Elmhurst College in Illinois. Gift aid is financial aid that does not need to be paid back, as opposed to a loan. A scatterplot of the data is shown in Figure 3.13 along with a linear fit. The lines follows a negative trend in the data; students who have higher family incomes tended to have lower gift aid from the university.
3.2.2 An objective measure for finding the best line
We begin by thinking about what we mean by the “best” line. Mathematically, we want a line that has small residuals. But beyond the mathematical reasons, hopefully it also makes sense intuitively that whatever line we fit, the residuals should be small (i.e., the points should be close to the line). The first option that may come to mind is to minimize the sum of the residual magnitudes:
\[e_1 + e_2 + \dots + e_n\]
which we could accomplish with a computer program. The resulting dashed line shown in Figure 3.14 demonstrates this fit can be quite reasonable.
However, a more common practice is to choose the line that minimizes the sum of the squared residuals:
\[e_{1}^2 + e_{2}^2 + \dots + e_{n}^2\]
The line that minimizes this least squares criterion is represented as the solid line in Figure 3.14 and is commonly called the least squares line. The following are three possible reasons to choose the least squares option instead of trying to minimize the sum of residual magnitudes without any squaring:
 It is the most commonly used method.
 Computing the least squares line is widely supported in statistical software.
 In many applications, a residual twice as large as another residual is more than twice as bad. For example, being off by 4 is usually more than twice as bad as being off by 2. Squaring the residuals accounts for this discrepancy.
 The analyses which link the model to inference about a population are most straightforward when the line is fit through least squares.
The first two reasons are largely for tradition and convenience; the third and fourth reasons explain why the least squares criterion is typically most helpful when working with real data.^{80}
3.2.3 Finding and interpreting the least squares line
For the Elmhurst data, we could write the equation of the least squares regression line as \[\widehat{\texttt{aid}} = \beta_0 + \beta_{1}\times \texttt{family_income}\]
Here the equation is set up to predict gift aid based on a student’s family income, which would be useful to students considering Elmhurst. These two values, \(\beta_0\) and \(\beta_1,\) are the parameters of the regression line.
The parameters are estimated using the observed data. In practice, this estimation is done using a computer in the same way that other estimates, like a sample mean, can be estimated using a computer or calculator.
The dataset where these data are stored is called elmhurst
.
The first 5 rows of this dataset are given in Table 3.2.
family_income  gift_aid  price_paid 

92.92  21.7  14.28 
0.25  27.5  8.53 
53.09  27.8  14.25 
50.20  27.2  8.78 
137.61  18.0  24.00 
We can see that family income is recorded in a variable called family_income
and gift aid from university is recorded in a variable called gift_aid
.
For now, we won’t worry about the price_paid
variable.
We should also note that these data are from the 20112012 academic year, and all monetary amounts are given in $1,000s, i.e., the family income of the first student in the data shown in Table 3.2 is $92,920 and they received a gift aid of $21,700.
(The data source states that all numbers have been rounded to the nearest whole dollar.)
Statistical software is usually used to compute the least squares line and the typical output generated as a result of fitting regression models looks like the one shown in Table 3.3. For now we will focus on the first column of the output, which lists \({b}_0\) and \({b}_1.\) In Chapter 8 we will dive deeper into the remaining columns which give us information on how accurate and precise these values of intercept and slope that are calculated from a sample of 50 students are in estimating the population parameters of intercept and slope for all students.
term  estimate  std.error  statistic  p.value 

(Intercept)  24.319  1.291  18.83  0 
family_income  0.043  0.011  3.98  0 
The model output tells us that the intercept is approximately 24.319 and the slope on family_income
is approximately 0.043.
But what do these values mean? Interpreting parameters in a regression model is often one of the most important steps in the analysis.
The intercept and slope estimates for the Elmhurst data are \(b_0\) = 24.319 and \(b_1\) = 0.043. What do these numbers really mean?
Interpreting the slope parameter is helpful in almost any application. For each additional $1,000 of family income, we would expect a student to receive a net difference of 1,000 \(\times\) (0.0431) = $43.10 in aid on average, i.e., $43.10 less. Note that a higher family income corresponds to less aid because the coefficient of family income is negative in the model. We must be cautious in this interpretation: while there is a real association, we cannot interpret a causal connection between the variables because these data are observational. That is, increasing a particular student’s family income may not cause the student’s aid to drop. (Although it would be reasonable to contact the college and ask if the relationship is causal, i.e., if Elmhurst College’s aid decisions are partially based on students’ family income.)
The estimated intercept \(b_0\) = 24.319 describes the average aid if a student’s family had no income, $2.432^{4}. The meaning of the intercept is relevant to this application since the family income for some students at Elmhurst is $0. In other applications, the intercept may have little or no practical value if there are no observations where \(x\) is near zero.
Interpreting parameters estimated by least squares. The slope describes the estimated difference in the predicted \(y\) variable if the predictor variable \(x\) for a case happened to be one unit larger. The intercept describes the average outcome of \(y\) if \(x=0\) and the linear model is valid all the way to \(x=0\) (values of \(x=0\) are not observed or relevant in many applications).
If you would like to learn more about using R to fit linear models, see Section 3.4.3 for the interactive R tutorials. An alternative way of calculating the values of intercept and slope of a least squares line is manual calculations using formulas. While manual calculations are not commonly used by practising statisticians and data scientists, it is useful to work through the first time you’re learning about the least squares line and modeling in general. Calculating the values by hand leverages two properties of the least squares line:
 The slope of the least squares line can be estimated by
\[b_1 = \frac{s_y}{s_x} r \]
where \(r\) is the correlation between the two variables, and \(s_x\) and \(s_y\) are the sample standard deviations of the predictor and outcome, respectively.
 If \(\bar{x}\) is the sample mean of the predictor variable and \(\bar{y}\) is the sample mean of the outcome variable, then the point \((\bar{x}, \bar{y})\) falls on the least squares line.
Table 3.4 shows the sample means for the family income and gift aid as $101,780 and $19,940, respectively. We could plot the point \((102, 19.9)\) on Figure 3.13 to verify it falls on the least squares line (the solid line).
mean  sd  mean  sd  r 

102  63.2  19.9  5.46  0.499 
Next, we formally find the point estimates \(b_0\) and \(b_1\) of the parameters \(\beta_0\) and \(\beta_1.\)
Using the summary statistics in Table 3.4, compute the slope for the regression line of gift aid against family income.
Compute the slope using the summary statistics from Table 3.4:
\[b_1 = \frac{s_y}{s_x} r = \frac{5.46}{63.2}(0.499) = 0.0431\]
You might recall the form of a line from math class, which we can use to find the model fit, including the estimate of \(b_0.\) Given the slope of a line and a point on the line, \((x_0, y_0),\) the equation for the line can be written as
\[y  y_0 = slope\times (x  x_0)\]
Identifying the least squares line from summary statistics.
To identify the least squares line from summary statistics:
 Estimate the slope parameter, \(b_1 = (s_y / s_x) r.\)
 Note that the point \((\bar{x}, \bar{y})\) is on the least squares line, use \(x_0 = \bar{x}\) and \(y_0 = \bar{y}\) with the pointslope equation: \(y  \bar{y} = b_1 (x  \bar{x}).\)
 Simplify the equation, we get \(y = \bar{y}  b_1 \bar{x} + b_1 x,\) which reveals that \(b_0 = \bar{y}  b_1 \bar{x}.\)
Using the point (102, 19.9) from the sample means and the slope estimate \(b_1 = 0.0431,\) find the leastsquares line for predicting aid based on family income.
Apply the pointslope equation using \((102, 19.9)\) and the slope \(b_1 = 0.0431\):
\[ \begin{align*} y  y_0 &= b_1 (x  x_0) \\ y  19.9 &= 0.0431(x  102) \end{align*} \]
Expanding the right side and then adding 19.9 to each side, the equation simplifies:
\[ \widehat{\texttt{aid}} = 24.3  0.0431 \times \texttt{family_income} \]
Here we have replaced \(y\) with \(\widehat{\texttt{aid}}\) and \(x\) with \(\texttt{family_income}\) to put the equation in context. The final least squares equation should always include a “hat” on the variable being predicted, whether it is a generic \(``y"\) or a named variable like \(``aid"\).
Suppose a high school senior is considering Elmhurst College. Can they simply use the linear equation that we have estimated to calculate her financial aid from the university?
She may use it as an estimate, though some qualifiers on this approach are important. First, the data all come from one freshman class, and the way aid is determined by the university may change from year to year. Second, the equation will provide an imperfect estimate. While the linear equation is good at capturing the trend in the data, no individual student’s aid will be perfectly predicted (as can be seen from the individual data points in the cloud around the line).
3.2.4 Extrapolation is treacherous
When those blizzards hit the East Coast this winter, it proved to my satisfaction that global warming was a fraud. That snow was freezing cold. But in an alarming trend, temperatures this spring have risen. Consider this: On February \(6^{th}\) it was 10 degrees. Today it hit almost 80. At this rate, by August it will be 220 degrees. So clearly folks the climate debate rages on.^{81}
Stephen Colbert
April 6th, 2010
Linear models can be used to approximate the relationship between two variables. However, like any model, they have real limitations. Linear regression is simply a modeling framework. The truth is almost always much more complex than a simple line. For example, we do not know how the data outside of our limited window will behave.
Use the model \(\widehat{\texttt{aid}} = 24.3  0.0431 \times \texttt{family_income}\) to estimate the aid of another freshman student whose family had income of $1 million.
We want to calculate the aid for a family with $1 million income. Note that in our model this will be represented as 1,000 since the data are in $1,000s.
\[24.3  0.0431 \times 1000 = 18.8 \]
The model predicts this student will have $18,800 in aid (!). However, Elmhurst College does not offer negative aid where they select some students to pay extra on top of tuition to attend.
Applying a model estimate to values outside of the realm of the original data is called extrapolation. Generally, a linear model is only an approximation of the real relationship between two variables. If we extrapolate, we are making an unreliable bet that the approximate linear relationship will be valid in places where it has not been analyzed.
3.2.5 Describing the strength of a fit
We evaluated the strength of the linear relationship between two variables earlier using the correlation, \(r.\) However, it is more common to explain the strength of a linear fit using \(R^2,\) called Rsquared. If provided with a linear model, we might like to describe how closely the data cluster around the linear fit.
The \(R^2\) of a linear model describes the amount of variation in the outcome variable that is explained by the least squares line. For example, consider the Elmhurst data, shown in Figure 3.13. The variance of the outcome variable, aid received, is about \(s_{aid}^2 \approx 29.8\) million (calculated from the data, some of which is shown in Table 3.2). However, if we apply our least squares line, then this model reduces our uncertainty in predicting aid using a student’s family income. The variability in the residuals describes how much variation remains after using the model: \(s_{_{RES}}^2 \approx 22.4\) million. In short, there was a reduction of \[\frac{s_{aid}^2  s_{_{RES}}^2}{s_{aid}^2} = \frac{29800  22400}{29800} = \frac{7500}{29800} \approx 0.25,\] or about 25%, of the outcome variable’s variation by using information about family income for predicting aid using a linear model. It turns out that \(R^2\) corresponds exactly to the squared value of the correlation:
\[r = 0.499 \rightarrow R^2 = 0.25\]
If a linear model has a very strong negative relationship with a correlation of 0.97, how much of the variation in the outcome is explained by the predictor?^{82}
\(R^2\) is also called the coefficient of determination.
Coefficient of determination: proportion of variability in the outcome variable explained by the model.
Since \(r\) is always between 1 and 1, \(R^2\) will always be between 0 and 1. This statistic is called the coefficient of determination, and it measures the proportion of variation in the outcome variable, \(y,\) that can be explained by the linear model with predictor \(x.\)
More generally, \(R^2\) can be calculated as a ratio of a measure of variability around the line divided by a measure of total variability.
Sums of squares to measure variability in \(y.\)
We can measure the variability in the \(y\) values by how far they tend to fall from their mean, \(\bar{y}.\) We define this value as the total sum of squares, calculated using the formula below, where \(y_i\) represents each \(y\) value in the sample, and \(\bar{y}\) represents the mean of the \(y\) values in the sample.
\[ SST = (y_1  \bar{y})^2 + (y_2  \bar{y})^2 + \cdots + (y_n  \bar{y})^2. \]
Leftover variability in the \(y\) values if we know \(x\) can be measured by the sum of squared errors, or sum of squared residuals, calculated using the formula below, where \(\hat{y}_i\) represents the predicted value of \(y_i\) based on the least squares regression.^{83},
\[\begin{align} SSE &= (y_1  \hat{y}_1)^2 + (y_2  \hat{y}_2)^2 + \cdots + (y_n  \hat{y}_n)^2\\ &= e_{1}^2 + e_{2}^2 + \dots + e_{n}^2 \end{align}\]The coefficient of determination can then be calculated as
\[ R^2 = \frac{SST  SSE}{SST} = 1  \frac{SSE}{SST} \]
Among 50 students in the elmhurst
dataset, the total variability in gift aid is \(SST = 1461\)^{84} The sum of squared residuals is \(SSE = 1098.\) Find \(R^2.\)
Since we know \(SSE\) and \(SST,\) we can calculate \(R^2\) as \[R^2 = 1  \frac{SSE}{SST} = 1  \frac{1098}{1461} = 0.25,\] the same value we found when we squared the correlation: \(R^2 = (0.499)^2 = 0.25.\)
3.2.6 Categorical predictors with two levels
Categorical variables are also useful in predicting outcomes.
Here we consider a categorical predictor with two levels (recall that a level is the same as a category).
We’ll consider Ebay auctions for a video game, Mario Kart for the Nintendo Wii, where both the total price of the auction and the condition of the game were recorded.
Here we want to predict total price based on game condition, which takes values used
and new
.
A plot of the auction data is shown in Figure 3.15. Note that the original dataset contains some Mario Kart games being sold at prices above $100 but for this analysis we have limited our focus to the 141 Mario Kart games that were sold below $100.
To incorporate the game condition variable into a regression equation, we must convert the categories into a numerical form.
We will do so using an indicator variable called condnew
, which takes value 1 when the game is new and 0 when the game is used.
Using this indicator variable, the linear model may be written as
\[\widehat{\texttt{price}} = b_0 + b_1 \times \texttt{condnew}\]
The parameter estimates are given in Table 3.5.
term  estimate  std.error  statistic  p.value 

(Intercept)  42.9  0.814  52.67  0 
condnew  10.9  1.258  8.66  0 
Using values from Table 3.5, the model equation can be summarized as
\[\widehat{\texttt{price}} = 42.871 + 10.90 \times \texttt{condnew}\]
Interpret the two parameters estimated in the model for the price of Mario Kart in eBay auctions.
The intercept is the estimated price when condnew
has a value 0, i.e., when the game is in used condition.
That is, the average selling price of a used version of the game is $42.87.
The slope indicates that, on average, new games sell for about $10.90 more than used games.
Interpreting model estimates for categorical predictors.
The estimated intercept is the value of the response variable for the first category (i.e., the category corresponding to an indicator value of 0). The estimated slope is the average change in the outcome variable between the two categories.
Note that, fundamentally, the intercept and slope interpretations don’t change when modeling categorical variables with two levels. However, when the predictor variable is binary, the coefficient estimates (\(b_0\) and \(b_1\)) are directly interpretable with respect to the dataset at hand.
We’ll elaborate further on modeling categorical predictors in Chapter 4, where we examine the influence of many predictor variables simultaneously using multiple regression.
3.2.7 Exercises
Units of regression. Consider a regression predicting the number of calories (cal) from width (cm) for a sample of square shaped chocolate brownies. What are the units of the correlation coefficient, the intercept, and the slope?
Which is higher? Determine if (I) or (II) is higher or if they are equal: “For a regression line, the uncertainty associated with the slope estimate, \(b_1\), is higher when (I) there is a lot of scatter around the regression line or (II) there is very little scatter around the regression line.” Explain your reasoning.
Overunder, Part I. Suppose we fit a regression line to predict the shelf life of an apple based on its weight. For a particular apple, we predict the shelf life to be 4.6 days. The apple’s residual is 0.6 days. Did we over or under estimate the shelflife of the apple? Explain your reasoning.
Overunder, Part II. Suppose we fit a regression line to predict the number of incidents of skin cancer per 1,000 people from the number of sunny days in a year. For a particular year, we predict the incidence of skin cancer to be 1.5 per 1,000 people, and the residual for this year is 0.5. Did we over or under estimate the incidence of skin cancer? Explain your reasoning.

Tourism spending. The Association of Turkish Travel Agencies reports the number of foreign tourists visiting Turkey and tourist spending by year. Three plots are provided: scatterplot showing the relationship between these two variables along with the least squares fit, residuals plot, and histogram of residuals.^{85}
Describe the relationship between number of tourists and spending.
What are the explanatory and response variables?
Why might we want to fit a regression line to these data?
Do the data meet the conditions required for fitting a least squares line? In addition to the scatterplot, use the residual plot and histogram to answer this question.

Starbucks, calories and carbs. The scatterplot below shows the relationship between the number of calories and amount of carbohydrates (in grams) Starbucks food menu items contain. Since Starbucks only lists the number of calories on the display items, we are interested in predicting the amount of carbs a menu item has based on its calorie content.^{86}
Describe the relationship between number of calories and amount of carbohydrates (in grams) that Starbucks food menu items contain.
In this scenario, what are the explanatory and response variables?
Why might we want to fit a regression line to these data?
Do these data meet the conditions required for fitting a least squares line?

The Coast Starlight, regression. The Coast Starlight Amtrak train runs from Seattle to Los Angeles. The scatterplot below displays the distance between each stop (in miles) and the amount of time it takes to travel from one stop to another (in minutes). The mean travel time from one stop to the next on the Coast Starlight is 129 mins, with a standard deviation of 113 minutes. The mean distance traveled from one stop to the next is 108 miles with a standard deviation of 99 miles. The correlation between travel time and distance is 0.636.^{87}
Write the equation of the regression line for predicting travel time.
Interpret the slope and the intercept in this context.
Calculate \(R^2\) of the regression line for predicting travel time from distance traveled for the Coast Starlight, and interpret \(R^2\) in the context of the application.
The distance between Santa Barbara and Los Angeles is 103 miles. Use the model to estimate the time it takes for the Starlight to travel between these two cities.
It actually takes the Coast Starlight about 168 mins to travel from Santa Barbara to Los Angeles. Calculate the residual and explain the meaning of this residual value.
Suppose Amtrak is considering adding a stop to the Coast Starlight 500 miles away from Los Angeles. Would it be appropriate to use this linear model to predict the travel time from Los Angeles to this point?

Body measurements, regression. Researchers studying anthropometry collected body and skeletal diameter measurements, as well as age, weight, height and sex for 507 physically active individuals. The scatterplot below shows the relationship between height and shoulder girth (circumference of shoulders measured over deltoid muscles), both measured in centimeters. The mean shoulder girth is 107.20 cm with a standard deviation of 10.37 cm. The mean height is 171.14 cm with a standard deviation of 9.41 cm. The correlation between height and shoulder girth is 0.67.^{88}
Write the equation of the regression line for predicting height.
Interpret the slope and the intercept in this context.
Calculate \(R^2\) of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.
A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of this student using the model.
The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.
A one year old has a shoulder girth of 56 cm. Would it be appropriate to use this linear model to predict the height of this child?

Murders and poverty, regression. The following regression output is for predicting annual murders per million (
annual_murders_per_mil
) from percentage living in poverty (perc_pov
) in a random sample of 20 metropolitan areas.^{89}term
estimate
std.error
statistic
p.value
(Intercept)
29.90
7.79
3.84
0.001
perc_pov
2.56
0.39
6.56
0.000
s
Rsquared
Adjusted Rsquared
5.51
70.52%
68.89%
Write out the linear model.
Interpret the intercept.
Interpret the slope.
Interpret \(R^2\).
Calculate the correlation coefficient.

Cats, regression. The following regression output is for predicting the heart weight (
Hwt
, in g) of cats from their body weight (Bwt
, in kg). The coefficients are estimated using a dataset of 144 domestic cats.^{90}term
estimate
std.error
statistic
p.value
(Intercept)
0.357
0.692
0.515
0.607
Bwt
4.034
0.250
16.119
0.000
s
Rsquared
Adjusted Rsquared
1.45
64.66%
64.41%
Write out the linear model.
Interpret the intercept.
Interpret the slope.
Interpret \(R^2\).
Calculate the correlation coefficient.
3.3 Outliers in linear regression
In this section, we identify criteria for determining which outliers are important and influential. Outliers in regression are observations that fall far from the cloud of points. These points are especially important because they can have a strong influence on the least squares line.
3.3.1 Types of outliers
There are three plots shown in Figure 3.16 along with the corresponding least squares line and residual plots. For each scatterplot and residual plot pair, identify the outliers and note how they influence the least squares line. Recall that an outlier is any point that doesn’t appear to belong with the vast majority of the other points.
A: There is one outlier far from the other points, though it only appears to slightly influence the line.
B: There is one outlier on the right, though it is quite close to the least squares line, which suggests it wasn’t very influential.
C: There is one point far away from the cloud, and this outlier appears to pull the least squares line up on the right; examine how the line around the primary cloud doesn’t appear to fit very well.
There are three plots shown in Figure 3.17 along with the least squares line and residual plots. As you did in previous exercise, for each scatterplot and residual plot pair, identify the outliers and note how they influence the least squares line. Recall that an outlier is any point that doesn’t appear to belong with the vast majority of the other points.
D: There is a primary cloud and then a small secondary cloud of four outliers. The secondary cloud appears to be influencing the line somewhat strongly, making the least square line fit poorly almost everywhere. There might be an interesting explanation for the dual clouds, which is something that could be investigated.
E: There is no obvious trend in the main cloud of points and the outlier on the right appears to largely (and problematically) control the slope of the least squares line.
F: There is one outlier far from the cloud. However, it falls quite close to the least squares line and does not appear to be very influential.
Examine the residual plots in Figures 3.16 and 3.17. You will probably find that there is some trend in the main clouds of Plots C, D, and E. In these cases, the outliers influenced the slope of the least squares lines. In Plot E, data with no clear trend were assigned a line with a large trend simply due to one outlier (!).
Leverage. Points that fall horizontally away from the center of the cloud tend to pull harder on the line, so we call them points with high leverage.
Points that fall horizontally far from the line are points of high leverage; these points can strongly influence the slope of the least squares line. If one of these high leverage points does appear to actually invoke its influence on the slope of the line – as in Plots C, D, and E of Figures 3.16 and 3.17 – then we call it an influential point. Usually we can say a point is influential if, had we fitted the line without it, the influential point would have been unusually far from the least squares line.
It is tempting to remove outliers. Don’t do this without a very good reason. Models that ignore exceptional (and interesting) cases often perform poorly. For instance, if a financial firm ignored the largest market swings – the “outliers” – they would soon go bankrupt by making poorly thoughtout investments.
3.3.2 Exercises

Outliers, Part I. Identify the outliers in the scatterplots shown below, and determine what type of outliers they are. Explain your reasoning.

Outliers, Part II. Identify the outliers in the scatterplots shown below and determine what type of outliers they are. Explain your reasoning.

Urban homeowners, outliers. The scatterplot below shows the percent of families who own their home vs. the percent of the population living in urban areas. There are 52 observations, each corresponding to a state in the US. Puerto Rico and District of Columbia are also included.^{91}
Describe the relationship between the percent of families who own their home and the percent of the population living in urban areas.
The outlier at the bottom right corner is District of Columbia, where 100% of the population is considered urban. What type of an outlier is this observation?

Crawling babies, outliers. A study conducted at the University of Denver investigated whether babies take longer to learn to crawl in cold months, when they are often bundled in clothes that restrict their movement, than in warmer months. The plot below shows the relationship between average crawling age of babies born in each month and the average temperature in the month when the babies are six months old. The plot reveals a potential outlying month when the average temperature is about 53F and average crawling age is about 28.5 weeks. Does this point have high leverage? Is it an influential point?^{92}
3.4 Chapter review
3.4.1 Terms
We introduced the following terms in the chapter. If you’re not sure what some of these terms mean, we recommend you go back in the text and review their definitions. We are purposefully presenting them in alphabetical order, instead of in order of appearance, so they will be a little more challenging to locate. However you should be able to easily spot them as bolded text.
coefficient of determination  indicator variable  outlier  sum of squared error 
correlation  influential point  predictor  total sum of squares 
extrapolation  least squares line  Rsquared  
high leverage  outcome  residuals 
3.4.2 Chapter exercises

True / False. Determine if the following statements are true or false. If false, explain why.
A correlation coefficient of 0.90 indicates a stronger linear relationship than a correlation of 0.5.
Correlation is a measure of the association between any two variables.

Trees. The scatterplots below show the relationship between height, diameter, and volume of timber in 31 felled black cherry trees. The diameter of the tree is measured 4.5 feet above the ground.^{93}
Describe the relationship between volume and height of these trees.
Describe the relationship between volume and diameter of these trees.
Suppose you have height and diameter measurements for another black cherry tree. Which of these variables would be preferable to use to predict the volume of timber in this tree using a simple linear regression model? Explain your reasoning.
3.4.3 Interactive R tutorials
Navigate the concepts you’ve learned in this chapter in R using the following selfpaced tutorials. All you need is your browser to get started!
You can also access the full list of tutorials supporting this book here.