The Population Slope as an Inferential Target

Why the Population Slope Is the Target

A regression line calculated from observed data summarizes a relationship in that particular sample. But when the purpose is to learn about a wider population, the sample line is not usually the final object of interest. The inferential target is a population slope: a parameter describing the linear association between a predictor \(X\) and a response \(Y\) in the population being studied.

The slope has a useful scale. If \(X\) is measured in hours and \(Y\) in milligrams per litre, the slope is measured in milligrams per litre per hour. It describes how the response tends to vary with the predictor in the population’s linear relationship. It does not, on its own, say that changing \(X\) would cause a change in \(Y\).

Definition: For a population in which \(X\) and \(Y\) have finite variances and \(\operatorname{Var}(X)>0\), the population regression slope is

\beta_1=\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}.

It is the slope of the population’s best-fitting linear prediction of \(Y\) from \(X\), using squared prediction errors as the criterion. Its units are response units per predictor unit.

The definition applies to a population described either as a probability distribution or as a finite collection of units. For a finite population with \(N\) units, the covariance and variance can be calculated as averages over those \(N\) units. Their common divisor cancels in the ratio, so the slope is also the sum of the products of deviations divided by the sum of squared predictor deviations.

If the population conditional mean is exactly linear, so that \(E(Y\mid X=x)=\beta_0+\beta_1x\), then \(\beta_1\) is also the change in the population mean response associated with a one-unit increase in \(X\). If the conditional mean is not linear, the slope still gives a best-fitting linear summary, but it does not mean that the mean response changes by the same amount at every value of \(X\).

A Population Slope and a Sample Slope

The population slope \(\beta_1\) is a parameter: it describes the target population and is usually unknown. A sample provides data with which to estimate it. The slope calculated from the sample is commonly written \(b_1\):

b_1=\frac{\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})} {\sum_{i=1}^{n}(x_i-\bar{x})^2}.

Here, \(x_i\) and \(y_i\) are the observed predictor and response for sampled unit \(i\), and \(\bar{x}\) and \(\bar{y}\) are their sample means. The calculation resembles the population definition, but it uses sample data. A different sample from the same population will generally give a different \(b_1\). Thus \(b_1\) is an estimate of \(\beta_1\), not the population parameter itself.

Worked Example: Calculating a Slope for a Finite Population

Suppose a hypothetical population consists of four clinics. For each clinic, \(X\) is its weekly number of specialist sessions and \(Y\) is the number of completed appointments, in hundreds.

Clinic	\(X\), sessions	\(Y\), hundreds of appointments
A	1	5
B	2	7
C	3	8
D	4	12

Because these four clinics are the entire hypothetical population of interest, we can calculate its population slope directly. The means are \(\mu_X=(1+2+3+4)/4=2.5\) sessions and \(\mu_Y=(5+7+8+12)/4=8\) hundreds of appointments. The sum of cross-products of deviations is

\sum (x_i-\mu_X)(y_i-\mu_Y) =(-1.5)(-3)+(-0.5)(-1)+(0.5)(0)+(1.5)(4)=11.

The sum of squared predictor deviations is

\sum (x_i-\mu_X)^2=(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2=5.

The population slope is therefore

\beta_1=\frac{11}{5}=2.2 \quad\text{hundreds of appointments per weekly session.}

For these clinics, the best-fitting linear summary associates one additional weekly session with 2.2 hundreds, or 220, more completed appointments. This summarizes a population-level linear trend; it does not claim that every clinic has exactly 220 more appointments for each added session.

Worked Example: A Sample Slope Estimates a Different Quantity

Now suppose a hypothetical sample of four clinics has the following data, and the larger population’s slope is not known.

Clinic	\(x_i\), sessions	\(y_i\), hundreds of appointments
A	1	4
B	2	5
C	3	7
D	4	8

The sample means are \(\bar{x}=2.5\) sessions and \(\bar{y}=6\) hundreds of appointments. The numerator and denominator of the sample slope are

\sum (x_i-\bar{x})(y_i-\bar{y}) =(-1.5)(-2)+(-0.5)(-1)+(0.5)(1)+(1.5)(2)=7,

\sum (x_i-\bar{x})^2=5.

Thus \(b_1=7/5=1.4\) hundreds of appointments per weekly session. This is the slope calculated from these four sampled clinics. It is not automatically the population slope: another sample could produce a different value. In an inferential analysis, the sample slope is used to learn about \(\beta_1\), while keeping the distinction between estimate and target clear.

Why the Covariance Ratio Is the Linear Slope

The population slope can be justified by asking which straight line predicts \(Y\) with the smallest average squared error. Consider a line with intercept \(a\) and slope \(b\), predicting \(Y\) as \(a+bX\). Assume \(X\) and \(Y\) have finite second moments and \(\operatorname{Var}(X)>0\). The average squared prediction error is \(E[(Y-a-bX)^2]\).

For any fixed \(b\), the error is minimized when the average residual is zero. Choosing \(a=\mu_Y-b\mu_X\) centers the residuals, so the criterion becomes

E\left[\left\{(Y-\mu_Y)-b(X-\mu_X)\right\}^2\right] =\operatorname{Var}(Y)-2b\operatorname{Cov}(X,Y)+b^2\operatorname{Var}(X).

Complete the square in \(b\):

\operatorname{Var}(X)\left(b-\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}\right)^2 +\operatorname{Var}(Y)-\frac{\operatorname{Cov}(X,Y)^2}{\operatorname{Var}(X)}.

The last term does not depend on \(b\). Since \(\operatorname{Var}(X)>0\), the squared term is smallest, and equals zero, when \(b=\operatorname{Cov}(X,Y)/\operatorname{Var}(X)\). This is the unique slope minimizing the population mean squared prediction error. The corresponding intercept is \(a=\mu_Y-b\mu_X\). This derivation explains why the covariance-to-variance ratio defines the population least-squares slope.

Worked Example: Interpreting a Slope When the Mean Relationship Is Linear

Suppose a hypothetical population of patients has a predictor \(X\), the number of weeks in a treatment program, and a response \(Y\), a symptom score. Assume the population conditional mean is \(E(Y\mid X=x)=30-1.6x\). This is a model for the average score among patients with a given program duration, not a statement about every patient’s score.

For \(x=5\) weeks, the model gives a mean response of \(30-1.6(5)=22\) score points. For \(x=6\) weeks, it gives \(30-1.6(6)=20.4\) points. The difference in these conditional means is

E(Y\mid X=6)-E(Y\mid X=5)=20.4-22=-1.6 \quad\text{score points.}

The population slope is \(-1.6\) score points per week: within this assumed linear mean relationship, the average score is 1.6 points lower at a predictor value one week higher. This interpretation concerns the population conditional means. It does not imply that each patient’s score falls by 1.6 points, and the model alone does not establish that adding a week of treatment causes the decrease.

Association Is Not Automatically a Causal Effect

A regression slope describes how \(Y\) and \(X\) vary together in a population. A causal effect asks a different question: how would the response change if the same units, or comparable units, were exposed to different values of \(X\)? An observed slope alone cannot answer that question. Other characteristics may be related to both \(X\) and \(Y\), creating or changing their association.

Worked Example: A Positive Slope Can Reflect Group Differences

Consider a hypothetical population of four workers. \(X\) is weekly training hours and \(Y\) is a performance score. Two workers have lower prior experience and two have higher prior experience.

Worker	Prior experience group	\(X\), training hours	\(Y\), performance points
A	Lower	0	60
B	Lower	1	62
C	Higher	2	80
D	Higher	3	82

Across all four workers, \(\mu_X=1.5\) hours and \(\mu_Y=71\) points. The sum of cross-products is

$$ (-1.5)(-11)+(-0.5)(-9)+(0.5)(9)+(1.5)(11)=42, $$

and the sum of squared \(X\) deviations is \(5\). The population slope is \(42/5=8.4\) performance points per training hour. Yet within each experience group, the score difference across the two workers is only 2 points for a one-hour difference in training. The overall slope also reflects the large difference in prior experience and performance between the groups.

These data do not show that training has no causal effect. They show why the pooled association of 8.4 points per hour cannot, by itself, be read as a causal effect of one additional training hour. A causal interpretation needs appropriate study design and justification beyond calculating a regression slope.

Common Mistakes

Calling the sample slope the population truth. The sample slope \(b_1\) is calculated from observed data; \(\beta_1\) describes the population. A sample slope is an estimate and may vary across samples.
Giving the slope without its units. A slope of \(-1.6\) is incomplete without saying, for example, “score points per week.” Its units are the response units divided by the predictor units.
Assuming a constant change when the mean relationship is curved. The population slope always describes the best-fitting linear summary under the stated definition. It represents a constant change in conditional means only when the conditional mean is linear.
Treating association as cause. A positive or negative slope describes a population relationship. It does not establish what would happen under an intervention on \(X\).

Key takeaway: The population slope \(\beta_1\) is the slope of the population’s best linear prediction of \(Y\) from \(X\), with units of response per predictor unit. The sample slope \(b_1\) estimates this target, and neither slope alone proves a causal effect.

Check Your Understanding

Try these questions without looking back.

In a study of daily salt intake in grams and blood pressure in millimetres of mercury, what are the units of the population slope?
Explain the difference between the population slope \(\beta_1\) and the sample slope \(b_1\).
If the population conditional mean is \(E(Y\mid X=x)=12+2.5x\), what is the difference between the conditional mean responses at \(x=4\) and \(x=5\)? State the units if \(Y\) is measured in points.
Why does a slope of 8.4 performance points per training hour not, by itself, show that an additional hour of training causes an 8.4-point increase?
What mathematical condition on \(X\) is needed for the covariance-to-variance definition of the population slope to be well-defined?