Skip to main content
\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
International Baccalaureate · DP 2: Analysis and Approaches SL

Bivariate Statistics

⌚ ~130 min ▢ 14 blocks ✓ 58 exercises ➣ Prerequisites : Straight Lines, Descriptive Statistics
I Association between Two Variables
I.1 Bivariate Data
Recappreviously studied — click to reread
In univariate statistics, we analyse a single variable at a time. Bivariate statistics extends this analysis to explore the relationship between two variables. By examining pairs of data, we can investigate patterns, determine the nature and strength of their relationship, and use this relationship to make predictions. Bivariate statistics studies the relationship between two quantitative variables.
Report an error
Definition — Bivariate Data
Bivariate data consists of pairs of values for two quantitative variables, recorded for each individual in a dataset. When one variable is studied as a possible explanation of the other, the pairs are written \((x, y)\), where:
  • \(x\) is the independent (or explanatory) variable.
  • \(y\) is the dependent (or response) variable.
Example
A teacher records the hours each student studied (\(x\)) and their final exam score (\(y\)).
Hours Studied (\(x\)) 5 10 8 15
Exam Score (\(y\)) 50 85 75 95
Each pair of values, such as \((5, 50)\), is a single bivariate data point.
Skills to practice
  • Identifying Explanatory and Response Variables
I.2 Scatter Diagrams
Recappreviously studied — click to reread
Definition — Scatter Plot
A scatter plot is a graph that displays bivariate data as a collection of points in the Cartesian plane. The independent (explanatory) variable is plotted on the horizontal axis (\(x\)-axis), and the dependent (response) variable is plotted on the vertical axis (\(y\)-axis).
A scatter plot is the primary tool for visually identifying a potential relationship, or correlation, between two quantitative variables.
Report an error
Method — Constructing a Scatter Plot
  1. Identify Variables: Determine which variable is independent (\(x\)) and which is dependent (\(y\)).
  2. Set Up the Axes: Draw and label the horizontal axis for the \(x\)-variable and the vertical axis for the \(y\)-variable. Choose appropriate scales for both axes that cover the range of the data.
  3. Plot the Points: For each pair of (\(x, y\)) values in the dataset, plot a single point on the graph at the corresponding coordinates.
Example
A teacher recorded the number of hours students studied and their corresponding exam scores. The data is shown below:
Hours Studied (\(x\)) 5 10 8 15
Exam Score (\(y\)) 50 85 75 95
Construct a scatter plot to visualise this data.

  1. Variables: “Hours Studied” is the independent variable (\(x\)) and “Exam Score” is the dependent variable (\(y\)).
  2. Axes: The \(x\)-axis is labelled “Hours Studied” and the \(y\)-axis “Exam Score”. The scales cover the ranges of the data.
  3. Plot Points: We plot the four coordinate pairs: (5, 50), (10, 85), (8, 75), and (15, 95).
The resulting scatter plot is shown below:
Figure

Skills to practice
  • Reading Scatter Diagrams
I.3 Describing a Correlation
Recappreviously studied — click to reread
Definition — Correlation
Correlation describes the nature of the relationship between two quantitative variables.
Definition — Direction: Positive or Negative
The direction describes the overall trend of the data.
  • Positive: As the independent variable (\(x\)) increases, the dependent variable (\(y\)) tends to increase. The points trend upward.
  • Negative: As the independent variable (\(x\)) increases, the dependent variable (\(y\)) tends to decrease. The points trend downward.
Figure  Figure
Definition — Form: Linear or Non-linear
The form of the relationship is linear if the data points appear to follow a straight-line pattern. If they follow a curve other than a straight line, the form is non-linear.
FigureFigure
Definition — Strength
The strength of a correlation describes how closely the data points adhere to the identified form.
FigureFigureFigure
Definition — Outliers
An outlier is a data point that deviates significantly from the main pattern of the data.
Figure
Method — Describing a Correlation
To describe the relationship shown in a scatter plot, we always comment on all four features in a concise statement.
  1. Direction: Is the trend upward (positive), downward (negative), or neither?
  2. Form: Is it linear, non-linear, or without any clear form?
  3. Strength: Is it strong, moderate, or weak?
  4. Outliers: Are there any notable outliers?
Example
Describe the correlation between hours studied and exam scores shown in this scatter plot.
Figure

There is a strong, positive correlation between hours studied and exam scores: as the number of hours studied increases, the exam score tends to increase. Its form is not quite linear: the scores rise quickly up to about \(10\) hours, then level off, so the cloud bends. There are no obvious outliers.

Skills to practice
  • Identifying the Strength of a Correlation
  • Plotting and Describing Scatter Diagrams
I.4 Correlation and Causation
Recappreviously studied — click to reread
Correlation Does Not Imply Causation
Observing a statistical relationship (correlation) between two variables, \(x\) and \(y\), is not sufficient evidence to conclude that a change in \(x\) causes a change in \(y\).
Report an error
Definition — Causation
Causation exists when a change in the independent variable produces — directly or through intermediate steps — a change in the dependent variable. The strongest evidence for causation comes from a carefully designed controlled experiment; observational data alone is rarely enough.
Definition — Confounding Variable
Often, a correlation between two variables (\(x\) and \(y\)) is created or distorted by a third factor — observed or not — known as a confounding variable. This variable influences both \(x\) and \(y\), so that part or all of their correlation comes from it rather than from a direct link between \(x\) and \(y\).
Example
Data shows a strong positive correlation between ice cream sales and the number of people who get sunburned.
Does this mean eating ice cream causes sunburn? If not, identify the relationships and the likely confounding variable.

No, eating ice cream does not cause sunburn.
  • The relationship between ice cream sales and sunburn cases is a correlation, not causation.
  • The likely confounding variable is Sunny Weather. Hot and sunny days cause an increase in ice cream sales and also cause an increase in people getting sunburned.
Figure

Skills to practice
  • Analysing Correlation and Causation
II Measuring Correlation
II.1 Pearson's Correlation Coefficient
Recappreviously studied — click to reread
While scatter plots allow us to visually describe a correlation, this assessment is subjective. To provide a precise and objective measure of the strength and direction of a linear relationship, we use numerical coefficients.
Report an error
Definition — Pearson's Correlation Coefficient (\(r\))
The Pearson's correlation coefficient (\(r\)) is a value in the range \([-1, 1]\) that quantifies the direction and strength of a linear relationship between two quantitative variables.
  • The sign of \(r\) indicates the direction (positive or negative).
  • The magnitude (absolute value) of \(r\) indicates the strength. An \(|r|\) value close to 1 implies a strong linear correlation, while a value close to 0 implies a weak or no linear correlation.
Value of \(|r|\) Strength of Correlation
\(|r| = 1\) Perfect
\(0.9 \le |r| < 1\) Very Strong
\(0.7 \le |r| < 0.9\) Strong
\(0.5 \le |r| < 0.7\) Moderate
\(0.3 \le |r| < 0.5\) Weak
\(0 \le |r| < 0.3\) Very Weak or None
Definition — Coefficient of Determination (\(r^2\))
The coefficient of determination (\(r^2\)) is the square of the correlation coefficient. It is a value in the range \([0, 1]\) and is typically expressed as a percentage.
The value of \(r^2\) represents the proportion of the variance in the dependent variable (\(y\)) that is explained by the linear relationship with the independent variable (\(x\)). In simple terms, it tells us how well the linear model fits the data.
Example
A study of hours spent studying and exam scores finds a correlation coefficient of \(r = 0.9\).
Interpret both \(r\) and \(r^2\).

  • Interpretation of \(r\): Since \(r=0.9\), there is a very strong, positive, linear correlation between the hours spent studying and the exam scores.
  • Interpretation of \(r^2\): We calculate \(r^2 = (0.9)^2 = 0.81\). This means that 81\(\pourcent\) of the variation in the exam scores can be explained by the linear relationship with the number of hours spent studying. The remaining 19\(\pourcent\) is unexplained by the linear model (other factors, measurement error, natural variability, etc.).

Method — Obtaining \(r\) and \(r^2\) with Technology
In this method, the correlation coefficient \(r\) is not computed by hand: a calculator or a spreadsheet gives it from the data, together with \(r^2\).
  1. Enter the data: the values of \(x\) in one list (or one column), the values of \(y\) in another, in the same order, so that each pair stays on the same row.
  2. Ask for the two-variable statistics: on a calculator, the two-variable statistics menu gives \(r\) and \(r^2\); on many models they appear on the linear regression screen \(y = ax + b\), beside the coefficients \(a\) and \(b\) of the least squares line (on some models, the display of \(r\) and \(r^2\) must first be switched on in the settings, and some models show only \(r\)); in a spreadsheet, with \(x\) in A2:A9 and \(y\) in B2:B9, the formula =CORREL(A2:A9,B2:B9) gives \(r\) and =RSQ(B2:B9,A2:A9) gives \(r^2\).
  3. Round only at the end: keep the full value on the screen and give \(r\) and \(r^2\) rounded, usually to three decimal places; \(r^2\) is the square of the unrounded \(r\), so when only \(r\) is displayed, square the full value of \(r\).
  4. Check with the scatter plot: the sign of \(r\) should agree with the general direction of the cloud, and \(|r|\) with how closely the points follow a line. When they disagree, check that each pair was entered on the same row, and look for an outlier, which can change \(r\) a lot.
Example
The coach of a running club records, for eight runners, the distance \(x\) run in training each week, in kilometres, and the time \(y\) of each runner in a \(5\) km race, in minutes.
\(x\) (km) \(10\) \(15\) \(20\) \(25\) \(30\) \(35\) \(40\) \(50\)
\(y\) (min) \(31.5\) \(30.2\) \(29.8\) \(27.9\) \(28.3\) \(26.4\) \(25.9\) \(24.8\)
  1. Use technology to find \(r\) and \(r^2\), rounded to three decimal places.
  2. Describe the correlation between \(x\) and \(y\).
  3. Interpret \(r^2\) in the context.

  1. With the distances in L1 and the times in L2, the linear regression gives the screen below.
    Figure
    So \(r \approx -0.980\) and \(r^2 \approx 0.960\). In a spreadsheet, =CORREL(A2:A9,B2:B9) gives the same \(r\).
  2. \(r\) is negative, so the correlation is negative: the runners who train more tend to have shorter race times. \(|r| \approx 0.980\) lies between \(0.9\) and \(1\), so the linear correlation is very strong. The sign agrees with the data: the times go down as the distances go up.
  3. \(r^2 \approx 0.960\): about \(96\pourcent\) of the variation in the race times is explained by the linear relationship with the training distance; the remaining \(4\pourcent\) is not explained by this linear relationship (other factors, measurement errors, natural variation). This is a correlation, not a proof that training alone makes the runners faster.

A correlation coefficient \(r\) is almost always computed from a sample, not from a whole population. Like any sample statistic, it is subject to sampling fluctuation: a different sample would give a slightly different \(r\). A strong \(r\) from a small sample is therefore weaker evidence of a real relationship than the same \(r\) from a large one — and deciding whether a whole population is genuinely correlated is itself a hypothesis test.
Report an error
Skills to practice
  • Estimating the Correlation Coefficient
  • Computing the Correlation Coefficient with Technology
III Lines of Best Fit
III.1 Line of Best Fit by Eye
Recappreviously studied — click to reread
Definition — Mean Point
The mean point of a scatter plot with points \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\) is the point \(G(\bar{x}, \bar{y})\) whose coordinates are the means of the two variables: $$\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} \qquad \text{and} \qquad \bar{y} = \frac{y_1 + y_2 + \cdots + y_n}{n}.$$ The mean point is the “centre of gravity” of the cloud: it is usually not one of the plotted points.
Method — Fitting a Line by Eye
When the points of a scatter plot lie roughly along a straight line, one can fit a line to the cloud, that is, draw a line that follows the trend of the points:
  1. Plot the mean point \(G\) and make the line pass through it.
  2. Follow the cloud: turn the line around \(G\) so that it follows the direction of the points, with all of them as close to it as possible — as a rough check, the points should be spread on both sides of the line.
  3. Read its equation \(y = ax + b\) on the graph (two points of the line give \(a\), then \(b\)).
Two people fitting by eye get two slightly different lines: the fit is a judgement, not a computation.
Skills to practice
  • Estimating the Line of Best Fit by Eye
III.2 The Least Squares Regression Line
Recappreviously studied — click to reread
When a scatter plot indicates a linear correlation between two variables, we can model this relationship using a straight line. This line, known as the regression line, can be used to make predictions. The reliability of this model is often assessed using the coefficient of determination (\(r^2\)). A high \(r^2\) value indicates that a large proportion of the variance in the dependent variable is explained by the independent variable, suggesting the linear model is a good fit for the data.
Report an error
Definition — Least Squares Regression Line
The least squares regression line, written as \(y = ax + b\), is the unique line of best fit (provided the \(x\)-values are not all equal) that models the linear relationship between \(x\) and \(y\). It is calculated by minimising the sum of the squares of the residuals.
A residual is the signed vertical difference (positive above the line, negative below) between an observed data point \((x_i, y_i)\) and the predicted point on the regression line \((x_i, \hat{y}_i)\).$$ \text{Residual} = \text{observed } y - \text{predicted } y = y_i - \hat{y}_i $$ A key property is that the least squares regression line always passes through the mean point, \((\bar{x}, \bar{y})\).
Figure
Method — Obtaining the Least Squares Line with a Calculator or a Spreadsheet
The coefficients \(a\) and \(b\) of the least squares line \(y = ax + b\) are not computed by hand: a calculator or a spreadsheet gives them.
  1. Enter the data: the values of \(x\) in one list (or one column), the values of \(y\) in another, in the same order.
  2. Ask for the line: on a calculator, the two-variable statistics menu, linear regression \(y = ax + b\); in a spreadsheet, with \(x\) in A2:A6 and \(y\) in B2:B6, the formulas =SLOPE(B2:B6,A2:A6) and =INTERCEPT(B2:B6,A2:A6), or a linear trend line added to the scatter plot, with its equation displayed.
  3. Round sensibly: keep about as many significant figures as the measurements have, and write the equation with the letters of the situation.
  4. Check with the mean point: the least squares line passes through \(G(\bar{x}, \bar{y})\); replacing \(x\) by \(\bar{x}\) in the equation must give \(\bar{y}\), up to rounding.
Example
A laboratory prepares five solutions of known concentration \(x\), in milligrams per litre, and measures the absorbance \(y\) of each one.
\(x\) \(2\) \(4\) \(6\) \(8\) \(10\)
\(y\) \(0.11\) \(0.19\) \(0.31\) \(0.40\) \(0.49\)
  1. Give, with a calculator, the equation of the least squares line, with \(a\) rounded to four decimal places and \(b\) to three.
  2. Check that this line passes through the mean point.
  3. Predict the absorbance of a solution of concentration \(7\) mg/L.
  4. A sample of unknown concentration has an absorbance of \(0.35\). Here the measured value is \(y\): use the same equation the other way round to estimate its concentration.

  1. The calculator gives \(a = 0.0485\) and \(b = 0.009\): the calibration line is \(y = 0.0485x + 0.009\) (the gradient is kept with three significant figures).
  2. \(\bar{x} = 6\) and \(\bar{y} = 0.30\). For \(x = 6\) the equation gives \(0.0485 \times 6 + 0.009 = 0.300\): the line passes through \(G(6, 0.30)\).
  3. \(0.0485 \times 7 + 0.009 = 0.3485\): the expected absorbance is about \(0.35\).
  4. Solve \(0.0485x + 0.009 = 0.35\): \(x = \dfrac{0.35 - 0.009}{0.0485} \approx 7.0\). The concentration is about \(7.0\) mg/L. This value lies between \(2\) and \(10\) mg/L, the smallest and the largest concentrations measured: the line is used inside the range where it was fitted.

Definition — Interpolation and Extrapolation
The line of best fit can be used to make predictions:
  • Interpolation is the process of predicting a \(y\)-value for an \(x\)-value that is within the range of the original data. If the correlation is strong, interpolation is generally considered reliable (provided the linear model fits the data well).
  • Extrapolation is the process of predicting a \(y\)-value for an \(x\)-value that is outside the range of the original data. Extrapolation is generally considered unreliable, as we cannot assume the linear trend continues indefinitely.
Example
For a survey of hours studied and exam scores (out of \(100\)), a calculator gives the line of best fit \(y = 3.5x + 40\). In this dataset, the hours studied ranged from 2 to 15.
  1. Predict the exam score for a student who studied for 11 hours.
  2. Predict the exam score for a student who studied for 25 hours.
  3. Comment on the reliability of the prediction in question 2.

  1. Since \(x=11\) is within the data range, this is interpolation. $$ y = 3.5(11) + 40 = 38.5 + 40 = 78.5 $$ The predicted score is 78.5.
  2. Since \(x=25\) is outside the data range, this is extrapolation. $$ y = 3.5(25) + 40 = 87.5 + 40 = 127.5 $$ The predicted score is 127.5.
  3. The prediction in question 2 is unreliable. It is an extrapolation, and we cannot assume the linear relationship holds for 25 hours of study. Furthermore, the model predicts a score over 100, which is impossible in this context, highlighting the danger of extrapolation.

Example
The average weekly study time and the final mathematics test score for a selection of students are shown below:
Study Time (\(x\) hours) 2 5 1 7 4 9 6 3
Test Score (\(y\) marks) 45 65 30 85 60 95 70 55
  1. Construct a scatter diagram to illustrate the data.
  2. Find the equation of the regression line \(y\) on \(x\). State and interpret its gradient in the context of the problem.
  3. Estimate the test score for a student who studies for 8 hours per week.
  4. Explain why the regression line of \(y\) on \(x\) should not be used to estimate the weekly study time of a student who scores 40 marks.
  5. Comment on the reliability of your estimate in 3, and say whether the score of 40 marks in 4 lies within the range of the data.

  1. Scatter Diagram:
    Figure
  2. Regression Line:
    Using a calculator, the equation is approximately: $$y = 7.71x + 27.5$$ Interpretation:
    The gradient is \(7.71\). This means that for every additional hour spent studying per week, the test score increases by approximately 7.7 marks on average.
  3. Estimation (Score for 8 hours):
    Substitute \(x = 8\) into the equation: $$y = 7.71(8) + 27.5 = 89.2$$ The estimated score is roughly 89 marks.
  4. Predicting in the Other Direction (40 marks):
    Here the study time \(x\) would be estimated from a given score \(y = 40\). The line of \(y\) on \(x\) is built to estimate \(y\) from \(x\): it makes the sum of the squares of the vertical distances between the points and the line as small as possible. Estimating \(x\) from \(y\) would require the horizontal distances to be small, and solving \(y = 7.71x + 27.5\) for \(x\) does not give the line that achieves this. A regression line of \(y\) on \(x\) does not reliably predict \(x\) from \(y\).
  5. Reliability:
    • The estimate in 3 (\(x=8\)) is reliable because 8 hours is within the range of the given data (\(1 \le x \le 9\)). This is interpolation.
    • The score \(y = 40\) in 4 lies within the observed scores (\(30 \le y \le 95\)): an estimate of the study time for this score would not be an extrapolation. What rules out the line of \(y\) on \(x\) in 4 is the direction of the prediction, not the range of the data.

Skills to practice
  • Comparing Lines with the Sum of the Squares
  • Fitting a Line in Context with a Calculator
III.3 The Regression Line of \(x\) on \(y\)
Predicting in the Other Direction
The least squares line of \(y\) on \(x\) makes the sum of the squares of the vertical distances between the points and the line as small as possible: it is built to estimate \(y\) from a given value of \(x\). Estimating \(x\) from a given value of \(y\) is another question. The distances that matter are then the horizontal ones, and solving the equation of the line of \(y\) on \(x\) for \(x\) does not give the line that makes the sum of their squares smallest.
Report an error
Definition — Regression Line of \(x\) on \(y\)
The regression line of \(x\) on \(y\) is the line \(x = cy + d\) (provided the values of \(y\) are not all equal) that minimises the sum of the squares of the horizontal residuals \(x_i - \hat{x}_i\), where \(\hat{x}_i = cy_i + d\) is the value of \(x\) that the line gives for \(y_i\). It is the least squares line of the data with the roles of the two variables exchanged: \(y\) becomes the explanatory variable, and \(x\) the response variable.
Exchanging the roles of \(x\) and \(y\) does not move the mean point: like the least squares line of \(y\) on \(x\), the line of \(x\) on \(y\) passes through the mean point \(G(\bar{x}, \bar{y})\), that is \(\bar{x} = c\bar{y} + d\).
Figure
Method — Predicting with the Right Regression Line
A regression line is used in the direction for which it was built.
  1. Name the variable to estimate. To estimate \(y\) from a given value of \(x\), use the line of \(y\) on \(x\); to estimate \(x\) from a given value of \(y\), use the line of \(x\) on \(y\). Solving \(y = ax + b\) for \(x\) (when \(a \neq 0\)) gives another line, which makes the vertical distances small, not the horizontal ones.
  2. Obtain the line of \(x\) on \(y\) with technology. Enter the data as for the line of \(y\) on \(x\), then, in the linear regression, give the list of the \(y\)-values as the first list (Xlist) and the list of the \(x\)-values as the second (Ylist). The screen still displays \(y = ax + b\), with the calculator's own letters: its \(a\) is \(c\) and its \(b\) is \(d\), and the line reads \(x = cy + d\). In a spreadsheet, with \(x\) in A2:A11 and \(y\) in B2:B11, the formulas =SLOPE(A2:A11,B2:B11) and =INTERCEPT(A2:A11,B2:B11) give \(c\) and \(d\).
  3. Check with the mean point: replacing \(y\) by \(\bar{y}\) in the stored equation \(x = cy + d\) must give \(\bar{x}\); a rounded equation only gives it approximately.
  4. Substitute and judge: substitute the given value in the stored equation, round only at the end, and judge the estimate: it is generally reliable when the correlation is strong and the given value lies within the range of the data, and to be treated with caution outside it (extrapolation) — an estimate, never a certainty for one individual.
The two lines cross at the mean point. They coincide only when the points are exactly aligned (\(r = 1\) or \(r = -1\)), and the weaker the correlation, the further apart they are: near \(G\) their estimates almost agree, far from \(G\) they differ. When the values of \(x\) are chosen in advance by the experimenter, such as the concentrations of the solutions used to calibrate an instrument, \(x\) is not a random variable: only \(y\) varies from one measurement to the next, the model is the line of \(y\) on \(x\), and the usual practice is to solve it for \(x\). The line of \(x\) on \(y\) is meant for data in which both variables vary from one individual to another.
The mock-up fits the line of \(x\) on \(y\) to the marks of ten students, with the values of \(x\) in L1 and those of \(y\) in L2.
Figure
The line of \(x\) on \(y\) is \(x \approx 1.03y - 0.344\).
Example
Ten students sat the two papers of an examination. Their marks out of \(100\) are \(x\) on Paper 1 and \(y\) on Paper 2.
\(x\) \(45\) \(52\) \(58\) \(60\) \(63\) \(68\) \(72\) \(75\) \(81\) \(88\)
\(y\) \(50\) \(48\) \(62\) \(55\) \(66\) \(61\) \(70\) \(78\) \(74\) \(85\)
  1. Use technology to find the equation of the regression line of \(y\) on \(x\) and the equation of the regression line of \(x\) on \(y\).
  2. Check that both lines pass through the mean point.
  3. A student was ill on the day of Paper 1 and scored \(80\) on Paper 2. Assuming that the relationship between the two papers applies to this student, estimate the mark this student would have scored on Paper 1.
  4. Another student missed Paper 2 and scored \(70\) on Paper 1. Estimate this student's mark on Paper 2.
  5. Explain why the answer to question 3 is not found by solving the equation of the line of \(y\) on \(x\) for \(x\).

  1. With \(x\) in L1 and \(y\) in L2, the linear regression gives \(a = 0.8556740\ldots\) and \(b = 8.2543793\ldots\): the line of \(y\) on \(x\) is \(y \approx 0.856x + 8.25\). With the lists exchanged (Xlist L2, Ylist L1), it gives \(c = 1.0253251\ldots\) and \(d = -0.3436003\ldots\): the line of \(x\) on \(y\) is \(x \approx 1.03y - 0.344\). Both regressions display \(r \approx 0.937\): the correlation is strong and positive.
  2. The means are \(\bar{x} = 66.2\) and \(\bar{y} = 64.9\). With the stored coefficients, $$a\bar{x} + b = 0.8556740\ldots \times 66.2 + 8.2543793\ldots = 64.9 = \bar{y}$$ and $$c\bar{y} + d = 1.0253251\ldots \times 64.9 - 0.3436003\ldots = 66.2 = \bar{x},$$ so both lines pass through \(G(66.2, 64.9)\), where they cross.
    Figure
  3. The mark to estimate is \(x\), from \(y = 80\): the line of \(x\) on \(y\) is used. $$\begin{aligned}[t] x &= cy + d && \text{(line of } x \text{ on } y)\\ &= 1.0253251\ldots \times 80 - 0.3436003\ldots\\ &\approx 81.7. \end{aligned}$$ The student would have scored about \(82\) on Paper 1. The value \(80\) lies within the range of the Paper 2 marks, from \(48\) to \(85\), and the correlation is strong: the estimate is reasonable, though an individual student may of course differ from it.
  4. The mark to estimate is \(y\), from \(x = 70\): the line of \(y\) on \(x\) is used. $$\begin{aligned}[t] y &= ax + b && \text{(line of } y \text{ on } x)\\ &= 0.8556740\ldots \times 70 + 8.2543793\ldots\\ &\approx 68.2. \end{aligned}$$ The student would have scored about \(68\) on Paper 2; \(70\) lies within the range of the Paper 1 marks.
  5. Solving \(0.8556740\ldots x + 8.2543793\ldots = 80\) gives \(x \approx 83.8\), not \(81.7\). The line of \(y\) on \(x\) makes the vertical distances between the points and the line small, the right criterion to estimate \(y\); to estimate \(x\), the horizontal distances must be small, and this is what the line of \(x\) on \(y\) does. The two lines cross at \(G\) and move apart away from it: at \(y = 80\), far above \(\bar{y} = 64.9\), the two answers differ by about \(2\) marks.

Skills to practice
  • Estimating a Value with the Right Regression Line
IV Going further
Skills to practice
  • Revisiting Correlation and Regression
  • Analysing Bivariate Data with Technology
  • Reasoning about Correlation and Regression
  • Quiz