When Adjustment Reveals a Stronger Association

When Adjustment Makes an Association Stronger

A predictor can have a modest overall association with an outcome even when its association at a fixed value of another predictor is stronger. This can happen when the predictors are related to each other and the second predictor also helps account for differences in the outcome. In that situation, the overall comparison partly mixes together differences in both predictors.

The distinction is between a marginal slope, which summarizes the linear association of an outcome with one predictor by itself, and an adjusted slope, which summarizes that predictor’s association while the other predictor is held fixed in an additive regression model. Adjustment does not always increase a slope; here we focus on the case where the adjusted slope is larger in magnitude.

The Model and the Meaning of Adjustment

Definition: In a two-predictor linear regression model, the conditional mean of an outcome $Y$, given predictors $X$ and $Z$, is represented as $$ E(Y\mid X,Z)=\beta_0+\beta_1X+\beta_2Z. $$ The coefficient $\beta_1$ is the adjusted slope for $X$: the model’s expected change in $Y$ for a one-unit increase in $X$, comparing observations at the same value of $Z$. The marginal slope describes the linear association between $Y$ and $X$ without including $Z$ in the model.

The observed-data estimates of these slopes are often written $b_1$ for the adjusted slope and $b_{\mathrm{marginal}}$ for the marginal slope. They are estimates, not necessarily the corresponding population coefficients. Both slopes have units of outcome units per $X$-unit, so their magnitudes can be compared directly.

For the regression calculations below, assume the conditional mean is adequately represented by the stated additive linear model, observations are independent, and the predictors vary enough that their separate coefficients can be estimated. For the usual interpretation of regression coefficients as estimates of the conditional mean, the errors must have mean zero given the predictors. Constant variance and normally distributed errors are relevant to some inference procedures, but no confidence intervals or tests are calculated here. The examples describe associations; these assumptions alone do not establish that changing a predictor causes an outcome to change.

A helpful way to anticipate the direction of the difference is to consider how $X$ and $Z$ move together. If $X$ tends to be lower when $Z$ is higher, and higher $Z$ is associated with higher $Y$, then the $Z$-related part of the overall $X$-to-$Y$ pattern can offset some of the positive association of $X$ at a fixed $Z$. Including $Z$ separates these two patterns in the additive model.

Worked Example: A Marginal Slope of 1 and an Adjusted Slope of 3

Suppose a hypothetical class exercise records $X$, each student’s centered weekly practice hours; $Z$, the student’s centered prior-assessment score; and $Y$, the student’s centered final-assessment score. Centering means subtracting the sample mean, so negative values indicate measurements below the sample mean. The values are invented for illustration.

Student	$X$ (centered hours)	$Z$ (centered points)	$Y$ (centered points)
1	-2	3	1
2	-1	-1	-5
3	0	1	1
4	0	1	1
5	1	-3	-3
6	2	-1	5

First fit the marginal regression of $Y$ on $X$ alone. The sample means of $X$ and $Y$ are both zero, so the least-squares slope is the cross-product sum divided by the sum of squared $X$ values:

b_{\mathrm{marginal}} =\frac{\sum X_iY_i}{\sum X_i^2} =\frac{(-2)(1)+(-1)(-5)+(0)(1)+(0)(1)+(1)(-3)+(2)(5)} {(-2)^2+(-1)^2+0^2+0^2+1^2+2^2} =\frac{10}{10}=1.

In this data set, the marginal fitted association is 1 final-assessment point per centered practice hour. Now fit the additive model with both predictors. The needed cross-product sums are $\sum X_i^2=10$, $\sum X_iZ_i=-10$, $\sum Z_i^2=22$, $\sum X_iY_i=10$, and $\sum Z_iY_i=14$. Because the variables are centered, the least-squares equations for the two slopes are

10b_1-10b_2=10,\qquad -10b_1+22b_2=14.

The values $b_1=3$ and $b_2=2$ satisfy both equations: $10(3)-10(2)=10$ and $-10(3)+22(2)=14$. The predictor cross-product matrix is nonsingular because $10(22)-(-10)^2=120>0$, so these coefficients are uniquely determined. The adjusted slope for practice hours is therefore 3 final-assessment points per centered practice hour, holding prior-assessment score fixed. Its magnitude is greater than the marginal slope’s magnitude: 3 rather than 1.

This comparison is about fitted slopes in the same hypothetical data. It does not say that the effect of adding one hour of practice is 3 points. It says that the additive model estimates a stronger association between practice and the outcome when comparing at the same prior-assessment score.

Why the Marginal and Adjusted Slopes Can Differ

The difference can be shown directly for centered variables. Suppose the fitted additive relationship is $Y=b_1X+b_2Z+e$, where $e$ is the regression residual. In least squares with an intercept, the residuals have zero cross-product with $X$, so $\sum X_ie_i=0$. Multiplying the fitted relationship by $X_i$, summing over observations, and dividing by $\sum X_i^2$ gives

b_{\mathrm{marginal}} =\frac{\sum X_iY_i}{\sum X_i^2} =b_1+b_2\frac{\sum X_iZ_i}{\sum X_i^2}.

The ratio $\sum X_iZ_i/\sum X_i^2$ describes how $Z$ varies with $X$ in the sample, in the units of $Z$ per unit of $X$. The product of this ratio and $b_2$ is the part of the marginal slope associated with how $Z$ changes alongside $X$, according to the fitted model. If that product is negative while $b_1$ is positive, it subtracts from the adjusted slope to give a smaller marginal slope.

For the first example, the ratio is $-10/10=-1$. Thus $b_{\mathrm{marginal}}=3+2(-1)=1$. The negative predictor relationship and the positive adjusted association of $Z$ with $Y$ combine to make the marginal slope smaller than the adjusted slope. This calculation is a description of how the fitted slopes relate; it does not determine which predictors should be included for a particular scientific question.

Worked Example: Checking the Adjusted Slope by Residualizing

Use the same hypothetical data to verify the adjusted slope for $X$ by removing the linear association with $Z$ from both $X$ and $Y$. This is a calculation for checking the adjusted slope, not a separate model interpretation.

The least-squares slope from regressing $X$ on $Z$ is $\sum X_iZ_i/\sum Z_i^2=-10/22=-5/11$. The corresponding residual for each $X_i$ is $X_i^*=X_i-(-5/11)Z_i=X_i+(5/11)Z_i$. The slope from regressing $Y$ on $Z$ is $\sum Y_iZ_i/\sum Z_i^2=14/22=7/11$, so its residual is $Y_i^*=Y_i-(7/11)Z_i$.

Regressing these two sets of residuals through the origin gives their cross-product sum divided by the squared-residual sum. The cross-product and squared-residual sums can be calculated from the original sums:

\sum X_i^*Y_i^* =\sum X_iY_i-\frac{(\sum X_iZ_i)(\sum Y_iZ_i)}{\sum Z_i^2} =10-\frac{(-10)(14)}{22} =\frac{180}{11},

\sum (X_i^*)^2 =\sum X_i^2-\frac{(\sum X_iZ_i)^2}{\sum Z_i^2} =10-\frac{100}{22} =\frac{60}{11}.

Therefore the residual-on-residual slope is $(180/11)/(60/11)=3$. It agrees with the adjusted coefficient from the two-predictor regression. The check shows why the adjusted slope concerns variation in $X$ not linearly accounted for by $Z$, paired with the corresponding remaining variation in $Y$.

Worked Example: Comparing Slopes From Regression Summaries

Suppose a second hypothetical analysis studies centered practice hours $X$, centered baseline-score points $Z$, and centered final-score points $Y$. A two-predictor least-squares fit gives $b_1=2.4$ final-score points per practice hour and $b_2=1.5$ final-score points per baseline-score point. In the same sample, suppose $\sum X_i^2=25$ and $\sum X_iZ_i=-20$. These are hypothetical summaries for illustrating the slope relationship.

The predictor ratio is $(-20)/25=-0.8$ baseline-score points per practice hour. Applying the slope identity gives

b_{\mathrm{marginal}} =b_1+b_2\frac{\sum X_iZ_i}{\sum X_i^2} =2.4+(1.5)(-0.8) =1.2.

The marginal association is 1.2 final-score points per practice hour, whereas the adjusted association is 2.4 points per hour at a fixed baseline score. The adjusted estimate is twice the marginal estimate in magnitude. The calculation is consistent with a negative relationship between practice hours and baseline score in this sample, combined with a positive adjusted association between baseline score and final score.

Interpretation and Common Pitfall

An increase in magnitude after adjustment is a change in the comparison being summarized. The marginal slope compares observations across their observed values of both $X$ and $Z$. The adjusted slope compares fitted outcomes across values of $X$ while holding $Z$ fixed in the model. If those two ways of comparing observations differ, the two slopes can differ substantially.

A common mistake is to say, “The adjusted coefficient is larger, so the marginal calculation was wrong.” That reasoning confuses two distinct quantities. The marginal slope is not a faulty version of the adjusted slope; it answers a different descriptive question. Another mistake is to conclude, “Adjustment proves that increasing $X$ will increase $Y$ by the adjusted slope.” A fitted adjusted association alone does not justify a causal interpretation. Any causal claim requires additional design and assumptions beyond the regression calculation.

Key takeaway: Adjustment can reveal a stronger association when the part of the marginal $X$-to-$Y$ pattern associated with $Z$ offsets the $X$ association at a fixed $Z$. Compare slopes in context, keep their units clear, and do not treat a larger adjusted coefficient as automatic evidence of causation.

Check Your Understanding

Use the model and calculations in this tutorial to answer the following.

In the first example, what does the marginal slope of 1 mean in context, and what does the adjusted slope of 3 mean?
For the first example, calculate $b_2\sum X_iZ_i/\sum X_i^2$. How does its sign help explain the difference between the marginal and adjusted slopes?
In the second example, why are the adjusted and marginal slopes different quantities rather than competing calculations of the same slope?
If the marginal slope is 1.2 and the adjusted slope is 2.4 final-score points per practice hour, which comparison holds baseline score fixed?
Why does a stronger adjusted association, by itself, not establish that changing practice hours causes the outcome to change?

When Adjustment Reveals a Stronger Association

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