3  Bivariate Descriptive Statistics

Status: ported 2026-05-19. Reviewed by editor: pending.

Learning outcomes

By the end of this chapter the reader should be able to:

• Organise paired observations \((x_i, y_i)\) into a two-way frequency table and read off joint, marginal, and conditional distributions.
• Distinguish statistical independence from dependence using the descriptive criterion \(f_{ij} = f_{i\cdot}\cdot f_{\cdot j}\).
• Read a scatter plot and describe the direction, form, and strength of an association.
• Compute the sample covariance \(S_{XY}\) and Pearson’s correlation \(r_{XY}\) from a small dataset and interpret their sign and magnitude.
• Fit the OLS line \(\hat{y} = a + b x\) by hand using \(b = S_{XY}/S_X^2\) and \(a = \bar{y} - b\bar{x}\), and interpret slope and intercept descriptively.
• Compute \(R^2\) and interpret it as the proportion of the variance of \(Y\) explained by its linear relationship with \(X\).
• Recognise common mistakes: \(r=0\) does not mean no association; high \(r\) does not imply causation; extrapolation is dangerous.
• Reproduce the calculations in R with cov(), cor(), lm(), and summary().

Motivating empirical question

Does spending more on advertising actually go hand in hand with higher sales — and how strongly?

A Granada tapas restaurant has 15 months of paired data on advertising spend and revenue. Looking at the two columns side by side does not answer the question: we need tools that summarise the joint behaviour of two variables. This chapter builds those tools — contingency tables, scatter plots, covariance, correlation, and the descriptive OLS line — and applies them to the restaurant dataset in Section 2.7. Every method here is purely descriptive: it summarises the sample. Whether a relationship is causal or transferable to the population is a separate question and the subject of later courses (Wooldridge 2020).

3.1 2.1 From one variable to two

In Chapter 1 we described a single variable: its frequency distribution, centre, spread, and shape. In economics, however, the interesting questions are almost always about how variables move together. Does education affect income? Does advertising spending track sales? Does GDP growth coincide with falling unemployment?

To answer such questions we need bivariate descriptive statistics: tools that summarise the joint behaviour of two variables observed on the same set of individuals.

Throughout this chapter we consider two statistical variables, \(X\) and \(Y\), observed on a sample of \(n\) individuals. Each individual \(i\) contributes a pair \((x_i, y_i)\), so the data take the form

\[ (x_1, y_1),\; (x_2, y_2),\; \ldots,\; (x_n, y_n). \]

WarningPair the values, never mix them

The order inside each pair matters: \((x_i, y_i)\) means both values come from the same individual \(i\). Re-sorting the \(x\)-column independently of the \(y\)-column destroys the bivariate information and turns the dataset into two unrelated univariate samples.

A small bivariate dataset to keep in mind: a labour economist surveys \(n=8\) recent graduates and records years of post-secondary education \(X\) and monthly starting salary \(Y\) (in euros):

\(i\)	1	2	3	4	5	6	7	8
\(x_i\) (yrs)	3	4	4	5	5	5	6	6
\(y_i\) (EUR)	1200	1400	1350	1600	1500	1550	1800	1900

The pair \((x_3, y_3) = (4, 1350)\) tells us that the third graduate studied for 4 years and earns 1350 EUR per month.

3.2 2.2 Contingency tables and marginal distributions

When \(X\) and \(Y\) take only a moderate number of distinct values (categorical variables, or quantitative variables that have been grouped), the bivariate data fit neatly into a contingency table, also called a two-way frequency table.

Let \(X\) take values \(x_1, \ldots, x_k\) and \(Y\) take values \(y_1, \ldots, y_p\).

NoteDefinition: joint, marginal, and conditional frequencies

• \(n_{ij}\) is the joint absolute frequency: the number of individuals with \(X = x_i\) and \(Y = y_j\).
• \(f_{ij} = n_{ij}/n\) is the corresponding joint relative frequency.
• \(n_{i\cdot} = \sum_{j=1}^{p} n_{ij}\) is the row marginal frequency (total for \(X = x_i\)).
• \(n_{\cdot j} = \sum_{i=1}^{k} n_{ij}\) is the column marginal frequency (total for \(Y = y_j\)).
• \(f_{i\cdot} = n_{i\cdot}/n\) and \(f_{\cdot j} = n_{\cdot j}/n\) are the marginal relative frequencies.
• The marginal distribution of \(X\) is the collection \(\{(x_i, f_{i\cdot})\}_{i=1}^k\); the marginal distribution of \(Y\) is \(\{(y_j, f_{\cdot j})\}_{j=1}^p\).

By construction, \(\sum_i n_{i\cdot} = \sum_j n_{\cdot j} = n\) and \(\sum_i f_{i\cdot} = \sum_j f_{\cdot j} = 1\).

The general contingency table looks like

	\(y_1\)	\(y_2\)	\(\cdots\)	\(y_p\)	\(n_{i\cdot}\)
\(x_1\)	\(n_{11}\)	\(n_{12}\)	\(\cdots\)	\(n_{1p}\)	\(n_{1\cdot}\)
\(x_2\)	\(n_{21}\)	\(n_{22}\)	\(\cdots\)	\(n_{2p}\)	\(n_{2\cdot}\)
\(\vdots\)	\(\vdots\)	\(\vdots\)	\(\ddots\)	\(\vdots\)	\(\vdots\)
\(x_k\)	\(n_{k1}\)	\(n_{k2}\)	\(\cdots\)	\(n_{kp}\)	\(n_{k\cdot}\)
\(n_{\cdot j}\)	\(n_{\cdot 1}\)	\(n_{\cdot 2}\)	\(\cdots\)	\(n_{\cdot p}\)	\(n\)

The right-most column collects the row marginals, the bottom row collects the column marginals, and the bottom-right cell holds the total sample size.

NoteExample: study mode by faculty

A university surveys \(n = 200\) students about their faculty (\(X\)) and their preferred study mode (\(Y\): in-person IP, hybrid H, online O):

	IP	H	O	\(n_{i\cdot}\)
Economics	40	25	15	80
Law	30	20	10	60
Sciences	20	25	15	60
\(n_{\cdot j}\)	90	70	40	200

The cell \(n_{11} = 40\) says 40 Economics students prefer in-person mode. The row marginal \(n_{1\cdot} = 80\) says there are 80 Economics students overall. The joint relative frequency of “Economics and Online” is \(f_{13} = 15/200 = 7.5\%\).

Reading off the marginal distribution of faculty: 40% Economics, 30% Law, 30% Sciences. Reading off the marginal distribution of study mode: 45% IP, 35% H, 20% O. These are exactly the univariate frequency distributions of Chapter 1 — the joint table contains both.

WarningMarginals do not pin down the joint distribution

Two different bivariate distributions can have the same marginal distributions but completely different joint distributions. The marginals on their own say nothing about the relationship between \(X\) and \(Y\) — for that we need the full table.

3.3 2.3 Conditional distributions

A conditional distribution answers questions of the form “Among students in the Economics faculty, what fraction prefers each study mode?” We fix one variable and look at how the other distributes within that subgroup.

NoteDefinition: conditional relative frequency

The conditional distribution of \(Y\) given \(X = x_i\) consists of the frequencies

\[ f_{j|i} = \frac{n_{ij}}{n_{i\cdot}}, \qquad j = 1, 2, \ldots, p, \]

which satisfy \(\sum_{j=1}^p f_{j|i} = 1\) for each fixed \(i\). Symmetrically, the conditional distribution of \(X\) given \(Y = y_j\) uses \(f_{i|j} = n_{ij}/n_{\cdot j}\).

For the faculty/mode table, the conditional distribution of mode given Faculty = Economics (\(n_{1\cdot} = 80\)) is

\[ f_{1|1} = \tfrac{40}{80} = 0.50, \quad f_{2|1} = \tfrac{25}{80} = 0.3125, \quad f_{3|1} = \tfrac{15}{80} = 0.1875. \]

So 50% of Economics students prefer IP, 31.25% prefer Hybrid, 18.75% prefer Online. Repeating the calculation for Sciences students (\(n_{3\cdot}=60\)) gives \(0.333\), \(0.417\), \(0.250\) — a noticeably different profile. Whenever conditional distributions differ across the values of \(X\), there is a relationship between \(X\) and \(Y\).

When \(Y\) is quantitative, we can also compute the conditional mean of \(Y\) given \(X = x_i\):

\[ \bar{y}_{|X=x_i} = \frac{\sum_{j} n_{ij}\, y_j}{n_{i\cdot}}. \]

If \(\bar{y}_{|X=x_i}\) changes systematically with \(i\), that is again evidence of association.

3.4 2.4 Statistical independence (descriptive)

Two variables are statistically independent when knowing one gives no information about the other.

NoteDefinition: independence (descriptive)

Variables \(X\) and \(Y\) are statistically independent if and only if, for every \(i\) and \(j\),

\[ f_{ij} = f_{i\cdot}\cdot f_{\cdot j}. \]

Equivalently, in absolute frequencies, \(n_{ij} = n_{i\cdot}\cdot n_{\cdot j}/n\) for all \((i,j)\). An equivalent condition is that every conditional distribution equals the corresponding marginal: \(f_{j|i} = f_{\cdot j}\) for all \(i\), \(j\).

If even one cell fails the equality, the variables are dependent.

WarningDescriptive vs. inferential

In this chapter we check independence by literally comparing \(f_{ij}\) with \(f_{i\cdot}\cdot f_{\cdot j}\) in the sample. In practice, observed and expected frequencies will rarely match exactly because of random sampling variation. The formal \(\chi^2\) test of independence, which decides whether the gap is “too large to be chance,” requires inferential machinery developed in TC2.

Worked check. A company records salary level (\(Y\): Low, Medium, High) and gender (\(X\): Male, Female) for \(n=100\) employees:

	Low	Medium	High	\(n_{i\cdot}\)
Male	15	25	20	60
Female	10	20	10	40
\(n_{\cdot j}\)	25	45	30	100

Marginals: \(f_{1\cdot} = 0.60\), \(f_{2\cdot} = 0.40\), \(f_{\cdot 1} = 0.25\), \(f_{\cdot 2} = 0.45\), \(f_{\cdot 3} = 0.30\).

Cell \((1,1)\): \(f_{11} = 15/100 = 0.15\). Under independence: \(0.60 \times 0.25 = 0.15\). Match.

Cell \((1,2)\): \(f_{12} = 25/100 = 0.25\). Under independence: \(0.60 \times 0.45 = 0.27\). No match.

Because the equality fails for at least one cell, \(X\) and \(Y\) are dependent in this sample.

3.5 2.5 Scatter plots: direction, form, strength

When both \(X\) and \(Y\) are quantitative, the natural graphical tool is the scatter plot: each pair \((x_i, y_i)\) is a point in the Cartesian plane. By convention the explanatory variable goes on the horizontal axis, the response variable on the vertical axis.

A scatter plot is read along three dimensions:

1. Direction — positive (upward trend), negative (downward trend), or no apparent trend.
2. Form — approximately linear, curved, or none.
3. Strength — tight cluster around a simple pattern (strong) vs. wide scatter (weak).

These three features are exactly what the numerical measures in the next section try to capture. The scatter plot should always come first, because:

WarningAlways plot before computing

Summary statistics (covariance, correlation, regression) compress a cloud of points into a few numbers. Different clouds can produce identical numbers — see Section 2.6 on Anscombe’s quartet. A 30-second look at the scatter plot prevents disasters that no formula can warn you about.

3.6 2.6 Covariance, correlation, and Anscombe’s quartet

The scatter plot gives a visual impression. To quantify the direction and strength of a linear relationship, we use two numerical measures: the covariance and Pearson’s correlation.

3.6.1 2.6.1 Covariance

NoteDefinition: sample covariance

The sample covariance of \(X\) and \(Y\) (using the divisor \(n\), consistent with our convention in Chapter 1) is

\[ S_{XY} = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y}) = \overline{xy} - \bar{x}\,\bar{y}, \]

where \(\overline{xy} = \tfrac{1}{n}\sum_i x_i y_i\) is the mean of the products.

Sign interpretation via quadrants. Place the point \((\bar{x}, \bar{y})\) at the centre of the scatter plot and split the plane into four quadrants. Points in quadrants I and III (both deviations have the same sign) contribute positive terms \((x_i - \bar{x})(y_i - \bar{y}) > 0\); points in quadrants II and IV contribute negative terms. Whichever side wins on average determines the sign of \(S_{XY}\):

• Positive contributions dominate \(\Rightarrow S_{XY} > 0\) (positive association).
• Negative contributions dominate \(\Rightarrow S_{XY} < 0\) (negative association).
• They roughly cancel \(\Rightarrow S_{XY} \approx 0\) (no linear association).

WarningTwo limitations of the covariance

1. Unit-dependent. \(S_{XY}\) is in the product of the units of \(X\) and \(Y\). Re-expressing salary in thousands of euros multiplies the covariance by \(1/1000\). Covariances cannot be compared across datasets that use different units.
2. No natural scale. A covariance of 500 is “big” or “small” only relative to the spread of \(X\) and \(Y\). The covariance alone cannot tell you how strong the linear association is.

Both limitations disappear once we normalise — which is exactly what the correlation coefficient does.

3.6.2 2.6.2 Pearson’s correlation coefficient

NoteDefinition: Pearson’s linear correlation coefficient

The Pearson correlation of \(X\) and \(Y\) is

\[ r_{XY} = \operatorname{Corr}(X,Y) = \frac{S_{XY}}{S_X\, S_Y} = \frac{\overline{xy} - \bar{x}\,\bar{y}}{\sqrt{\overline{x^2}-\bar{x}^2}\,\sqrt{\overline{y^2}-\bar{y}^2}}. \]

Its properties (provable from the definition; see Exercise 2.7) are:

1. \(-1 \leq r_{XY} \leq 1\).
2. \(r_{XY} = 1\) iff all points lie on a line with positive slope (perfect positive linear association).
3. \(r_{XY} = -1\) iff all points lie on a line with negative slope.
4. \(r_{XY} = 0\) means no linear association — a nonlinear relationship may still be present.
5. \(r_{XY}\) is dimensionless and symmetric: \(r_{XY} = r_{YX}\).
6. \(r_{XY}\) is invariant under positive linear rescalings: if \(X' = a + bX\) and \(Y' = c + dY\) with \(b,d > 0\), then \(r_{X'Y'} = r_{XY}\). Changing currency, °C to °F, etc., leaves the correlation unchanged.

Warning\(r = 0\) does not mean “no relationship”

It means no linear relationship. A perfect inverted-U pattern can have \(r \approx 0\) yet \(Y\) is fully determined by \(X\). The correlation coefficient is blind to curvature. Always plot the data.

3.6.3 2.6.3 Worked example: study hours and exam scores

Six students report weekly study hours \(X\) and exam score \(Y\):

\(i\)	1	2	3	4	5	6
\(x_i\)	2	5	8	10	12	15
\(y_i\)	45	55	62	70	78	85

Step 1: means.

\[ \bar{x} = \tfrac{52}{6} \approx 8.667, \qquad \bar{y} = \tfrac{395}{6} \approx 65.833. \]

Step 2: mean of products.

\[ \sum_i x_i y_i = 90 + 275 + 496 + 700 + 936 + 1275 = 3772, \qquad \overline{xy} = \tfrac{3772}{6} \approx 628.667. \]

Step 3: covariance.

\[ S_{XY} = \overline{xy} - \bar{x}\,\bar{y} = 628.667 - 8.667 \times 65.833 \approx 628.667 - 570.556 = 58.111. \]

Step 4: standard deviations.

\[ S_X^2 = \overline{x^2} - \bar{x}^2 = \tfrac{562}{6} - 8.667^2 \approx 93.667 - 75.111 = 18.556, \qquad S_X \approx 4.306, \]

\[ S_Y^2 = \tfrac{27103}{6} - 65.833^2 \approx 4517.167 - 4334.028 = 183.139, \qquad S_Y \approx 13.533. \]

Step 5: correlation.

\[ r_{XY} = \frac{58.111}{4.306 \times 13.533} = \frac{58.111}{58.279} \approx 0.997. \]

An extremely strong positive linear association: in this sample, study hours and exam scores essentially line up.

3.6.4 2.6.4 Anscombe’s quartet — a cautionary tale

In 1973, Francis Anscombe constructed four datasets that share nearly identical summary statistics — means, variances, correlation \(r \approx 0.816\), and OLS line \(\hat y \approx 3.0 + 0.5x\) — yet look completely different on a scatter plot:

1. Dataset I: a clean linear pattern. The OLS line is a sensible summary.
2. Dataset II: a clear quadratic curve. A straight line is the wrong model even though \(r\) is high.
3. Dataset III: an almost-perfect line plus one outlier that distorts the slope.
4. Dataset IV: all observations share the same \(x\)-value except one — a single high-leverage point determines the entire regression.

WarningThe moral

Identical \(r\), identical \(a\) and \(b\), identical \(R^2\) — four completely different stories. Never trust summary statistics alone. Always look at the data. We replicate Anscombe’s quartet in R in Section 2.7.

3.7 2.7 The OLS regression line (descriptive)

The correlation summarises how strongly two variables move together. To go further and model the relationship — to predict \(Y\) from \(X\) — we fit a straight line.

3.7.1 2.7.1 The regression problem

Given the \(n\) pairs \((x_1,y_1), \ldots, (x_n, y_n)\), we look for the line

\[ \hat{y} = a + b x \]

that best fits the data. For observation \(i\), the fitted value is \(\hat{y}_i = a + b x_i\) and the residual (vertical gap from the line to the point) is

\[ e_i = y_i - \hat{y}_i = y_i - (a + b x_i). \]

3.7.2 2.7.2 The least-squares criterion

NoteDefinition: Ordinary Least Squares (OLS)

The OLS line minimises the sum of squared residuals:

\[ \mathrm{SSR}(a, b) = \sum_{i=1}^n e_i^2 = \sum_{i=1}^n (y_i - a - b x_i)^2. \]

Setting \(\partial\mathrm{SSR}/\partial a = 0\) and \(\partial\mathrm{SSR}/\partial b = 0\) and solving the resulting two linear equations gives the closed-form OLS coefficients

\[ b = \frac{S_{XY}}{S_X^2}, \qquad a = \bar{y} - b\,\bar{x}. \]

The lowercase \(a\) and \(b\) are deliberate: in this descriptive setting they are features of the sample, not estimators of unknown population parameters. The inferential framing (\(\hat\beta_0\), \(\hat\beta_1\), sampling distribution, standard errors) waits for a course on inference.

A few immediate observations:

• The slope \(b\) has the same sign as \(S_{XY}\) — and therefore the same sign as \(r_{XY}\).
• The OLS line always passes through the point of means \((\bar{x}, \bar{y})\): plug \(x = \bar{x}\) into \(\hat y = a + b\bar x = (\bar y - b\bar x) + b\bar x = \bar y\).
• Slope \(b\): in this sample, individuals who differ by one unit in \(X\) differ on average by \(b\) units in the predicted \(Y\).
• Intercept \(a\): the predicted \(Y\) when \(X = 0\). Only meaningful if \(X = 0\) is inside (or at least near) the observed range of \(X\).

WarningTwo regression lines, two slopes

The line “\(Y\) on \(X\)” minimises vertical squared distances and uses slope \(b_{Y|X} = S_{XY}/S_X^2\). The line “\(X\) on \(Y\)” minimises horizontal squared distances and uses slope \(b_{X|Y} = S_{XY}/S_Y^2\). The two lines are different (unless \(r = \pm 1\)); they only coincide when all points lie on a single line. Pick the regression that matches your prediction direction.

3.7.3 2.7.3 Worked example: GDP growth and unemployment

GDP growth rate (%, \(X\)) and change in unemployment (percentage points, \(Y\)) for 11 European countries:

\(i\)	1	2	3	4	5	6	7	8	9	10	11
\(x_i\)	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	1.2	2.8
\(y_i\)	1.2	0.8	0.5	0.1	-0.3	-0.5	-0.8	-1.2	-1.5	0.6	-0.2

Step 1: means. \(\sum x_i = 26.5\), \(\sum y_i = -1.3\), so \(\bar x \approx 2.409\) and \(\bar y \approx -0.118\).

Step 2: sums of squares and products.

\[ \sum x_i^2 = 80.53,\qquad \overline{x^2} = 7.321, \qquad S_X^2 = 7.321 - 2.409^2 = 1.518, \]

\[ \sum x_i y_i = -14.09,\qquad \overline{xy} = -1.281, \qquad S_{XY} = -1.281 - 2.409 \times (-0.118) = -0.997. \]

Step 3: slope and intercept.

\[ b = \frac{S_{XY}}{S_X^2} = \frac{-0.997}{1.518} \approx -0.657,\qquad a = \bar{y} - b\,\bar{x} = -0.118 - (-0.657)(2.409) \approx 1.465. \]

The fitted line is \(\hat{y} = 1.465 - 0.657\,x\). Descriptively: countries that differ by one percentage point of GDP growth differ on average by \(-0.657\) percentage points in unemployment change — the cross-country shape of Okun’s law.

3.8 2.8 Goodness of fit: \(R^2\)

The OLS line is the best linear summary — but how good is the best? The answer is the coefficient of determination.

NoteDefinition: coefficient of determination

The coefficient of determination is

\[ R^2 = r_{XY}^2 = \left(\frac{S_{XY}}{S_X\, S_Y}\right)^2. \]

It measures the proportion of the variance of \(Y\) that is captured by the linear fit on \(X\).

The interpretation comes from a variance decomposition. Define the variance of the fitted values \(\hat y_i\) and the variance of the residuals \(e_i\):

\[ S_{\hat Y}^2 = b^2 \cdot S_X^2 \quad \text{(explained variance)}, \qquad S_e^2 = \frac{1}{n}\sum_i e_i^2 \quad \text{(residual variance)}. \]

Then

\[ S_Y^2 = \underbrace{b^2 \cdot S_X^2}_{\text{explained}} + \underbrace{S_e^2}_{\text{unexplained}}, \]

so

\[ R^2 = \frac{\text{explained variance}}{\text{total variance}} = 1 - \frac{S_e^2}{S_Y^2}. \]

A reading guide:

• \(R^2 = 0\): the line explains none of the variation. The OLS “line” is flat at \(\bar y\) and has slope \(b = 0\) (which happens when \(S_{XY} = 0\)).
• \(R^2 = 1\): the line explains all the variation. Every point lies exactly on the line.
• \(R^2 = 0.85\): 85% of the variance of \(Y\) in this sample is matched by the linear fit on \(X\). The remaining 15% is residual — noise plus whatever non-linear or omitted structure the line does not capture.

Warning\(R^2\) does not measure causation

A high \(R^2\) says the line fits the sample well. It says nothing about why \(X\) and \(Y\) move together, nor about whether the relationship will hold outside this sample. We come back to this in Section 2.9.

Worked \(R^2\). For the GDP/unemployment example above we need \(S_Y^2\):

\[ \sum y_i^2 = 7.41,\qquad \overline{y^2} = 0.6736, \qquad S_Y^2 = 0.6736 - (-0.118)^2 = 0.6597, \qquad S_Y \approx 0.8123. \]

Then \(r_{XY} = -0.997 / (1.232 \times 0.8123) \approx -0.996\) and \(R^2 \approx 0.992\). About 99% of the cross-country variability in unemployment changes is captured by the linear fit on GDP growth in this sample.

3.8.1 2.8.1 Prediction and the risk of extrapolation

The natural use of the regression line is prediction: given a new \(x_0\), predict \(\hat y_0 = a + b x_0\). Predictions are most trustworthy when \(x_0\) lies near the centre of the observed \(X\) range and when \(R^2\) is high.

WarningExtrapolation is dangerous

Using \(\hat y = a + b x\) for an \(x_0\) outside the observed range of \(X\) is called extrapolation. It is risky because:

1. The linear pattern observed within the data may not continue outside the range.
2. Small wobbles in \(b\) get amplified over long horizontal distances.
3. There is no data to check the prediction against.

A vivid case: in Exercise 2.4 below, fitting a line to hotel prices of 40–140 EUR and pushing the prediction to 250 EUR yields a negative number of overnight stays — physically impossible. The model said it because the user asked it to, not because it is true.

Within the observed \(X\) range (interpolation) predictions are far more reliable.

3.9 2.9 Correlation does not imply causation

This is perhaps the single most important slogan in applied statistics:

A high correlation between \(X\) and \(Y\) says they move together. It does not say why.

There are at least four reasons two variables can be strongly correlated:

1. \(X\) causes \(Y\). More advertising literally drives more sales.
2. \(Y\) causes \(X\). Companies with high sales reinvest in advertising. (Reverse causality.)
3. A confounder \(Z\) drives both. Ice-cream sales and drowning deaths are positively correlated, not because ice cream is dangerous but because both rise with summer temperature.
4. Pure coincidence. With enough variables, some pairs will be correlated by chance — “spurious” correlations.

WarningFamous example: chocolate and Nobel prizes

A well-known study found \(r \approx 0.79\) between a country’s per-capita chocolate consumption and the number of Nobel Prize laureates per 10 million population. Eating chocolate clearly does not earn one a Nobel Prize. The likely confounders are national wealth, education spending, and research infrastructure — all of which correlate with chocolate consumption.

The lesson: no matter how tempting the causal story, ask whether a confounder could explain the association before claiming an effect.

Establishing causation requires either a controlled experiment (random assignment of \(X\)) or more sophisticated econometric methods (instrumental variables, panel data, natural experiments) developed in later courses (Wooldridge 2020). Descriptive statistics like covariance, correlation, and OLS regression measure an association — nothing more.

3.10 2.10 R Lab — restaurant sales and ad spend

We now reproduce the entire chapter on a small simulated dataset: 15 months of advertising spend (hundreds of euros) and revenue (thousands of euros) for a Granada tapas restaurant. As in Chapter 1, data are simulated inline with a fixed seed for reproducibility.

library(dplyr)

3.10.1 2.10.1 The dataset

set.seed(2026)
ad_spend <- round(runif(15, min = 2, max = 12), 1)
sales    <- round(1.8 + 0.9 * ad_spend + rnorm(15, mean = 0, sd = 1.2), 1)

restaurant <- data.frame(
  Month    = month.abb[1:15],
  Ad_Spend = ad_spend,
  Sales    = sales
)
restaurant

   Month Ad_Spend Sales
1    Jan      9.0   9.5
2    Feb      7.6   5.0
3    Mar      3.4   2.3
4    Apr      4.9   5.7
5    May      7.6   6.8
6    Jun      2.3   2.8
7    Jul      6.7   6.5
8    Aug     10.6  11.6
9    Sep      4.5   4.7
10   Oct      7.8   6.3
11   Nov      2.1   3.5
12   Dec      8.9   9.7
13  <NA>      4.3   5.5
14  <NA>     10.5  12.6
15  <NA>      3.5   4.8

Quick univariate summary of each column — the bivariate analysis will use what we already know from Chapter 1:

summary(restaurant[, c("Ad_Spend", "Sales")])

    Ad_Spend          Sales       
 Min.   : 2.100   Min.   : 2.300  
 1st Qu.: 3.900   1st Qu.: 4.750  
 Median : 6.700   Median : 5.700  
 Mean   : 6.247   Mean   : 6.487  
 3rd Qu.: 8.350   3rd Qu.: 8.150  
 Max.   :10.600   Max.   :12.600  

3.10.2 2.10.2 Scatter plot

plot(ad_spend, sales,
     pch = 19, col = "steelblue", cex = 1.3,
     xlab = "Advertising spend (hundreds of EUR)",
     ylab = "Sales (thousands of EUR)",
     main = "Tapas restaurant -- 15 months",
     las  = 1)

The cloud rises from lower-left to upper-right with no obvious curvature and no extreme outliers — positive, approximately linear, fairly strong.

3.10.3 2.10.3 Covariance and correlation by hand and by R

# By hand using the divisor n (chapter convention)
n     <- length(ad_spend)
x_bar <- mean(ad_spend)
y_bar <- mean(sales)

S_XY  <- sum((ad_spend - x_bar) * (sales - y_bar)) / n
S_X   <- sqrt(sum((ad_spend - x_bar)^2) / n)
S_Y   <- sqrt(sum((sales    - y_bar)^2) / n)
r_xy  <- S_XY / (S_X * S_Y)

cat("S_XY (by hand, /n)    :", round(S_XY, 3), "\n")

S_XY (by hand, /n)    : 7.52 

cat("Correlation (by hand) :", round(r_xy, 3), "\n")

Correlation (by hand) : 0.912 

R’s built-in cov() and cor() use the divisor \(n-1\), so the covariance differs by the factor \(n/(n-1) \approx 1.07\) for \(n=15\); the correlation is unaffected because \(r\) is dimensionless and the ratio cancels:

cat("cov() (divisor n-1)   :", round(cov(ad_spend, sales), 3), "\n")

cov() (divisor n-1)   : 8.057 

cat("cor()                 :", round(cor(ad_spend, sales), 3), "\n")

cor()                 : 0.912 

A correlation around 0.9 indicates a strong positive linear association: in this sample, months with higher ad spending also have higher sales.

3.10.4 2.10.4 The OLS line

By hand from \(b = S_{XY}/S_X^2\) and \(a = \bar y - b\bar x\), then with lm():

b <- S_XY / S_X^2
a <- y_bar - b * x_bar
cat("Slope (b)     :", round(b, 4), "\n")

Slope (b)     : 0.9807 

cat("Intercept (a) :", round(a, 4), "\n")

Intercept (a) : 0.3605 

fit <- lm(sales ~ ad_spend)
round(coef(fit), 4)

(Intercept)    ad_spend 
     0.3605      0.9807 

The two outputs agree (up to rounding): OLS solves the same first-order conditions either way. Reading \(b\) descriptively: months that differ by 100 EUR more in advertising have on average \(b \cdot 1000\) EUR more in sales.

plot(ad_spend, sales,
     pch = 19, col = "steelblue", cex = 1.3,
     xlab = "Ad spend (hundreds of EUR)",
     ylab = "Sales (thousands of EUR)",
     main = "OLS regression line",
     las  = 1)
abline(fit, col = "red", lwd = 2)
legend("topleft",
       legend = paste0("y = ", round(a, 2), " + ", round(b, 2), " * x"),
       col = "red", lwd = 2, bty = "n")

3.10.5 2.10.5 \(R^2\) two ways

R2_a <- summary(fit)$r.squared
R2_b <- cor(ad_spend, sales)^2
cat("R^2 from summary() :", round(R2_a, 4), "\n")

R^2 from summary() : 0.8322 

cat("r_xy^2             :", round(R2_b, 4), "\n")

r_xy^2             : 0.8322 

The two values agree: in simple regression, \(R^2 = r_{XY}^2\). Roughly four-fifths of the in-sample variation in monthly sales is captured by the linear fit on ad spend; the rest is residual.

3.10.6 2.10.6 Residual plot (sanity check)

A scatter of residuals against fitted values is the simplest diagnostic. If the line is the right shape and the spread of \(Y\) does not change with \(X\), the residuals should look like a featureless cloud around zero:

plot(fitted(fit), resid(fit),
     pch = 19, col = "steelblue",
     xlab = "Fitted values (predicted sales)",
     ylab = "Residuals",
     main = "Residual plot",
     las  = 1)
abline(h = 0, col = "red", lty = 2, lwd = 2)

The residuals scatter around zero with no obvious pattern: the linear model looks adequate for this sample.

3.10.7 2.10.7 Anscombe’s quartet in four lines

R ships with the anscombe dataset. Four scatter plots, four “identical” regression lines:

par(mfrow = c(2, 2), mar = c(4, 4, 2, 1))
for (i in 1:4) {
  x <- anscombe[[paste0("x", i)]]
  y <- anscombe[[paste0("y", i)]]
  plot(x, y, pch = 19, col = "steelblue",
       main = paste("Dataset", i),
       xlab = paste0("x", i), ylab = paste0("y", i), las = 1)
  abline(lm(y ~ x), col = "red", lwd = 2)
}

par(mfrow = c(1, 1))

for (i in 1:4) {
  x <- anscombe[[paste0("x", i)]]
  y <- anscombe[[paste0("y", i)]]
  cat(sprintf("Dataset %d -- cor: %.3f  R2: %.3f\n",
              i, cor(x, y), summary(lm(y ~ x))$r.squared))
}

Dataset 1 -- cor: 0.816  R2: 0.667
Dataset 2 -- cor: 0.816  R2: 0.666
Dataset 3 -- cor: 0.816  R2: 0.666
Dataset 4 -- cor: 0.817  R2: 0.667

Identical correlations, identical \(R^2\), four very different stories. Plot first.

Self-check

TipQ1. Sign of the covariance

If most points of a scatter plot lie in the upper-right and lower-left of the cloud (using \((\bar x, \bar y)\) as the origin), then \(S_{XY}\) is:

• A. Positive, because most terms \((x_i - \bar x)(y_i - \bar y)\) are positive.
• B. Negative, because deviations cancel.
• C. Zero, because the cloud has a mean.
• D. Undefined, because covariance requires categorical data.

Answer: A. Upper-right (\(x_i > \bar x\), \(y_i > \bar y\)) and lower-left (\(x_i < \bar x\), \(y_i < \bar y\)) points both contribute positive terms to the sum, so \(S_{XY} > 0\).

TipQ2. Units of correlation

If \(X\) is measured in euros and \(Y\) in years, the Pearson correlation \(r_{XY}\) has units of:

• A. EUR \(\cdot\) year.
• B. EUR / year.
• C. Year / EUR.
• D. No units at all.

Answer: D. \(r_{XY}\) is dimensionless: dividing \(S_{XY}\) (units EUR\(\cdot\)year) by \(S_X \cdot S_Y\) (also units EUR\(\cdot\)year) cancels the units. This is precisely what makes \(r\) comparable across datasets.

TipQ3. Interpretation of \(r=0\)

If \(r_{XY} = 0\) in a sample, the strongest correct conclusion is:

• A. \(X\) and \(Y\) are statistically independent.
• B. There is no linear association between \(X\) and \(Y\); a nonlinear relationship may still exist.
• C. The OLS regression line is undefined.
• D. \(X\) and \(Y\) have the same mean.

Answer: B. A correlation of zero rules out a linear pattern only. A perfect \(Y = X^2\) relationship can have \(r \approx 0\). Independence is a stronger condition; the OLS line is well-defined (it is just flat).

TipQ4. OLS slope formula

In the descriptive simple regression \(\hat y = a + b x\) fit by OLS, the slope \(b\) equals:

• A. \(S_{XY}/(S_X\, S_Y) = r_{XY}\).
• B. \(\bar y / \bar x\).
• C. \(S_{XY}/S_X^2\).
• D. \(S_X / S_Y\).

Answer: C. This follows directly from the first-order conditions: \(b = S_{XY}/S_X^2\). Note: it is not the correlation, except by accident when \(S_X = S_Y\).

TipQ5. The line through the means

The OLS regression line \(\hat y = a + b x\) always passes through:

• A. The origin \((0, 0)\).
• B. The point of means \((\bar x, \bar y)\).
• C. The point \((0, \bar y)\).
• D. The point \((\bar x, 0)\).

Answer: B. Substituting \(x = \bar x\): \(\hat y = a + b\bar x = (\bar y - b\bar x) + b\bar x = \bar y\). The intercept formula \(a = \bar y - b\bar x\) is exactly the condition that makes this true.

TipQ6. \(R^2\) in simple regression

In simple regression, the coefficient of determination \(R^2\) equals:

• A. The Pearson correlation \(r_{XY}\) itself.
• B. The square of the slope \(b^2\).
• C. The square of the Pearson correlation \(r_{XY}^2\).
• D. The variance of the residuals divided by the variance of \(Y\).

Answer: C. \(R^2 = r_{XY}^2\) holds in simple regression. The variance ratio in option D equals \(1 - R^2\), not \(R^2\).

TipQ7. Anscombe’s lesson

The main lesson of Anscombe’s quartet is that:

• A. The correlation coefficient is wrong when the relationship is nonlinear.
• B. Summary statistics can hide curvature, outliers, and high-leverage points — always look at the scatter plot.
• C. Four datasets always give four different OLS lines.
• D. The OLS slope changes when you transform \(Y\) to \(\log Y\).

Answer: B. Anscombe’s four datasets have nearly identical means, variances, correlation, and OLS line, yet only one of them is genuinely well-summarised by a straight line.

TipQ8. Correlation vs. causation

A regression on the 15-month restaurant sample gives \(\hat{\text{Sales}} \approx 1.8 + 0.9\cdot\text{AdSpend}\). The most defensible descriptive reading of the slope is:

• A. Spending 100 EUR more on ads causes sales to rise by 900 EUR.
• B. Each extra 100 EUR of advertising will raise next month’s sales by exactly 900 EUR.
• C. In this sample, months with 100 EUR more in advertising have on average 900 EUR more in sales.
• D. 90% of sales is explained by ad spend.

Answer: C. The slope is a descriptive average comparison within the sample. Causal language requires more than a single regression on observational data; “next month” requires the relationship to be transferable; “90% of sales” confuses the slope with \(R^2\).

Exercises

3.10.8 Exercise 2.1 ★ — Contingency table and marginals

A survey of 200 workers classifies them by department and level of job satisfaction:

	Low	Medium	High
Production	30	25	15
Logistics	10	20	25
Sales	15	30	30

1. Add the row and column marginals and the grand total.
2. What proportion of all workers report high satisfaction?
3. What proportion of all workers belong to the Sales department?

TipSolution

1. Row totals: Production 70, Logistics 55, Sales 75. Column totals: Low 55, Medium 75, High 70. Grand total 200.
2. \(70/200 = 0.35 = 35\%\).
3. \(75/200 = 0.375 = 37.5\%\).

3.10.9 Exercise 2.2 ★ — Conditional means

A company records years of experience \(X\) and monthly salary (hundreds of EUR) \(Y\) for 10 employees: \(X = 1, \ldots, 10\) and \(Y = 14, 15, 16, 17, 18, 20, 21, 23, 24, 26\). Group employees as “Junior” (\(X \le 5\)) and “Senior” (\(X > 5\)).

1. Compute the conditional mean \(\bar y_{|\text{Junior}}\).
2. Compute the conditional mean \(\bar y_{|\text{Senior}}\).
3. What does the comparison of the two conditional means suggest about the association between experience and salary?

TipSolution

1. \(\bar y_{|\text{Junior}} = (14+15+16+17+18)/5 = 16.0\) (hundreds EUR).
2. \(\bar y_{|\text{Senior}} = (20+21+23+24+26)/5 = 22.8\) (hundreds EUR).
3. Senior employees earn on average 680 EUR more per month than Junior employees. The conditional means differ across the two experience groups, so \(X\) and \(Y\) are descriptively related: more experience goes with higher salary in this sample.

3.10.10 Exercise 2.3 ★★ — Independence check

A market researcher records preferred payment method and age group for 300 consumers:

	Cash	Card	Mobile
18–35	20	50	30
36–55	40	60	20
56+	50	20	10

1. Compute the marginal distributions of age and payment method.
2. Under independence, compute the expected frequency \(n_{ij}^{\text{exp}} = n_{i\cdot} n_{\cdot j}/n\) for at least three cells.
3. Comment, in plain English, on whether the variables look independent. (This chapter only asks for a descriptive comparison; the formal \(\chi^2\) test comes later.)

3.10.11 Exercise 2.4 ★★ — Full bivariate analysis

A tourism board has data on average hotel price per night (\(X\), in euros) and overnight stays (\(Y\), in thousands) for 6 price levels:

\(X\) (price)	40	60	80	100	120	140
\(Y\) (stays)	52	45	38	30	24	18

1. Compute \(\bar x\), \(\bar y\), \(S_X^2\), \(S_{XY}\), and the OLS line \(\hat y = a + b x\).
2. Predict the number of overnight stays when the price is 90 EUR. Is this interpolation or extrapolation?
3. Predict the number of overnight stays when the price is 250 EUR. Comment on the result and what it teaches you about extrapolation.

3.10.12 Exercise 2.5 ★★ — Slope is not correlation

For a dataset of \(n = 50\) observations a student reports \(S_X = 4\), \(S_Y = 2\), and \(S_{XY} = 6\).

1. Compute \(r_{XY}\), \(b\) (slope of \(Y\) on \(X\)), and \(b'\) (slope of \(X\) on \(Y\)).
2. Verify the relation \(b \cdot b' = r_{XY}^2 = R^2\) and interpret it geometrically.

3.10.13 Exercise 2.6 ★★★ — Why the OLS line passes through \((\bar x, \bar y)\)

Starting from the SSR

\[ \mathrm{SSR}(a, b) = \sum_{i=1}^n (y_i - a - b x_i)^2, \]

1. Differentiate with respect to \(a\) and set the result to zero. Show that this single first-order condition is equivalent to \(\bar y = a + b\bar x\).
2. Explain in one sentence why this guarantees that the OLS line passes through the point of means for any dataset.

3.10.14 Exercise 2.7 ★★★ — \(-1 \le r_{XY} \le 1\) via Cauchy–Schwarz

Use the Cauchy–Schwarz inequality

\[ \left(\sum_{i=1}^n u_i v_i\right)^2 \le \left(\sum_{i=1}^n u_i^2\right)\left(\sum_{i=1}^n v_i^2\right) \]

applied to \(u_i = x_i - \bar x\) and \(v_i = y_i - \bar y\) to prove that \(|S_{XY}| \le S_X\, S_Y\), and hence \(-1 \le r_{XY} \le 1\). Identify the equality case and explain why it corresponds to a perfect linear relationship.

3.10.15 Exercise 2.8 ★★ — Correlation vs. causation

For each scenario, state whether the correlation is most plausibly (i) causal, (ii) reverse causal, (iii) driven by a confounder, or (iv) likely spurious. Give one sentence of justification.

1. Ice cream sales and drownings are positively correlated across Spanish months.
2. In a randomised clinical trial, the dose of a painkiller and self-reported pain are negatively correlated.
3. Across countries, per-capita chocolate consumption and Nobel laureates per capita are positively correlated.
4. In a company without random assignment, training hours and productivity are positively correlated.

TipSolution (Exercise 2.8)

1. Confounder: summer temperature drives both.
2. Causal: randomisation breaks the link with any confounder, so the negative association is genuinely from the dose.
3. Confounder: national wealth and R&D spending drive both.
4. Confounder or selection: more motivated employees may choose more training and also be more productive; without random assignment we cannot tell.

Wooldridge, Jeffrey M. 2020. Introductory Econometrics: A Modern Approach. 7th ed. Cengage Learning.