Two Numbers, Two Questions
A regression slope and a correlation can both describe a linear relationship, but they answer different questions. The slope describes the change in the response’s conditional mean for a one-unit increase in the predictor. Correlation describes the direction and strength of a linear association on a standardized, unit-free scale. Treating the two as interchangeable can produce an answer with the wrong interpretation, and sometimes the wrong units.
This distinction matters when comparing settings or changing how a variable is measured. A slope of 2 might mean two dollars per additional hour, two points per additional hour, or two dollars per additional minute. Its numerical value depends on the units. Correlation, by contrast, is unchanged in magnitude when either variable is converted using a positive scale factor.
The slope’s interpretation concerns the model’s conditional mean, not a guarantee that every observed response changes by exactly \(\beta_1\). Correlation summarizes linear association; it does not have units such as points per hour. Neither measure alone establishes that changing \(X\) causes a change in \(Y\).
Why the Slope Depends on Scale
Under a linear conditional mean and the usual zero conditional-mean error condition, the population slope can be written as \(\operatorname{Cov}(X,Y)/\operatorname{Var}(X)\). Combining this with the definition of correlation gives a useful relationship. The standard deviations in the relationship carry the variables’ units, which is why the slope does too.
This identity requires a finite, positive variance for \(X\); correlation also requires a finite, positive variance for \(Y\). It does not require normally distributed variables. In particular, \(\beta_1\) equals \(\rho\) only when the standard-deviation ratio \(\sigma_Y/\sigma_X\) happens to be one in the chosen units. Even then, the units still need to be considered before interpreting either number.
For the linear conditional-mean model with zero conditional-mean error, \(\beta_1=\operatorname{Cov}(X,Y)/\operatorname{Var}(X)\).
Since \(\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y\) and \(\operatorname{Var}(X)=\sigma_X^2\), the slope becomes \(\rho\sigma_Y/\sigma_X\).
The ratio \(\sigma_Y/\sigma_X\) has units of \(Y\) per \(X\). Thus a fixed correlation can correspond to different slopes if the variables’ spreads, or their measurement units, differ.
Worked Applications
Worked Example: Converting Correlation to a Slope
A hypothetical study of greenhouse plants uses water supplied per week as \(X\), measured in liters, and weekly height increase as \(Y\), measured in centimeters. Suppose the conditional mean is linear, the zero conditional-mean error condition holds, \(\rho=0.50\), \(\sigma_X=4\) liters, and \(\sigma_Y=6\) centimeters. The means are \(E[X]=8\) liters and \(E[Y]=20\) centimeters. Find the slope and interpret the fitted conditional mean at \(X=10\) liters. No distributional form is assumed.
Method. Use \(\beta_1=\rho\sigma_Y/\sigma_X\), then use the means to determine the intercept.
Step 1: Calculate the slope. The standard-deviation ratio is \(6/4=1.5\) centimeters per liter. Therefore,
Step 2: Calculate the intercept. The conditional-mean line passes through \((E[X],E[Y])\), so \(\beta_0=20-(0.75)(8)=20-6=14\) centimeters. The model’s conditional mean is \(E[Y\mid X]=14+0.75X\).
Step 3: Evaluate at 10 liters. The fitted conditional mean is \(14+(0.75)(10)=14+7.5=21.5\) centimeters.
Check. The slope is in centimeters per liter, while the correlation remains \(0.50\) and has no units. The predicted conditional mean at 10 liters is 21.5 centimeters. The common mistake avoided here is to report \(0.50\) centimeters per liter: \(0.50\) is the correlation, not the slope.
Worked Example: Changing Both Measurement Units
In a hypothetical study of travel, \(X\) is distance in miles and \(Y\) is travel time in minutes. Assume a linear conditional mean and the zero conditional-mean error condition. Suppose \(\rho=-0.60\), \(\sigma_X=5\) miles, and \(\sigma_Y=12\) minutes. Convert the slope to the units hours per kilometer, using 1 mile = 1.6 kilometers and 60 minutes = 1 hour. Find the correlation in the new units as well. No distributional form is assumed.
Method. First calculate the original slope from the standard-deviation ratio. Then convert both standard deviations to the new units and calculate the slope again.
Step 1: Calculate the original slope.
Step 2: Convert the standard deviations. The predictor’s standard deviation becomes \(5(1.6)=8\) kilometers. The response’s standard deviation becomes \(12/60=0.2\) hours.
Step 3: Calculate the slope in the new units.
Step 4: Check the conversion directly. One kilometer is \(1/1.6\) miles, and one minute is \(1/60\) hour. Thus the numerical slope converts to \((-1.44)(1/60)(1/1.6)=-1.44/96=-0.015\) hours per kilometer. Both unit conversions are needed.
Correlation. Both conversions use positive factors, so the correlation remains \(-0.60\), including its sign. The common mistake avoided is dividing by 60 but forgetting the predictor’s conversion from miles to kilometers.
Worked Example: Equal Correlations, Different Slopes
Two hypothetical training programs relate weekly practice hours \(X\) to assessment points \(Y\). In each program, assume a linear conditional mean and the zero conditional-mean error condition. Program A has \(\rho=0.70\), \(\sigma_X=2\) hours, and \(\sigma_Y=10\) points. Program B has \(\rho=0.70\), \(\sigma_X=5\) hours, and \(\sigma_Y=10\) points. Find and compare their slopes. No distributional form is assumed.
Method. Compute each slope as correlation multiplied by the response standard deviation divided by the predictor standard deviation.
Program A: The standard-deviation ratio is \(10/2=5\) points per hour. Therefore, \(\beta_{1,A}=0.70(5)=3.5\) points per hour.
Program B: The standard-deviation ratio is \(10/5=2\) points per hour. Therefore, \(\beta_{1,B}=0.70(2)=1.4\) points per hour.
Interpretation and check. Both programs have the same positive correlation, \(0.70\), but their slopes differ because the predictor’s spread differs. The model in Program A has a greater fitted change in conditional mean per additional practice hour. Correlation alone does not determine that change; the ratio of standard deviations is also needed. The common mistake avoided is concluding that equal correlations imply equal slopes.
Rescaling: The General Relationship
The examples can be summarized algebraically. Let \(X^{*}=a+cX\) and \(Y^{*}=d+eY\), where \(c\) and \(e\) are nonzero constants. Adding constants changes the origins of measurement but not spreads or covariance. Multiplying by a constant changes the corresponding spread, and a negative constant also reverses that variable’s direction. The following calculation shows the consequences for the slope and correlation.
\(\operatorname{Cov}(X^{*},Y^{*})=ce\operatorname{Cov}(X,Y)\), while \(\operatorname{Var}(X^{*})=c^2\operatorname{Var}(X)\).
\(\beta_1^{*}=\operatorname{Cov}(X^{*},Y^{*})/\operatorname{Var}(X^{*})=(e/c)\beta_1\). With ordinary unit conversions, \(c\) and \(e\) are positive, so the slope’s direction is unchanged but its numerical value and units change.
The standard deviations become \(|c|\sigma_X\) and \(|e|\sigma_Y\), giving \(\rho^{*}=\{ce/(|c||e|)\}\rho\). Thus \(|\rho^{*}|=|\rho|\); if \(ce<0\), the sign reverses.
For a positive change of units, such as meters to centimeters, \(c>0\). The correlation is then unchanged, not merely unchanged in magnitude. A negative recoding reverses the direction of one variable, so the sign of correlation reverses too. A zero scale factor is excluded because it would make that variable constant and its correlation undefined.
Common Exam Trap: Treating Correlation as a Slope
Suppose \(\rho=0.40\), \(\sigma_X=3\) hours, and \(\sigma_Y=12\) points. The incorrect calculation is to write “the slope is \(0.40\) points per hour.” This confuses a unit-free standardized measure with a change in the response’s conditional mean.
The correction is to include the standard-deviation ratio:
A useful check is dimensional: the slope must have response units divided by predictor units. Correlation must have no units. If a purported slope has no units, or a purported correlation is reported in points per hour, the quantities have been mixed up.
Check Your Understanding
For each problem, identify the units as well as the numerical result. Assume the stated linear conditional-mean conditions when a population slope is requested; no normal distribution is needed.
- A hypothetical crop study has \(\rho=0.30\), predictor standard deviation 4 kilograms, and response standard deviation 15 dollars. Find the population slope and its units. (Answer: \(1.125\) dollars per kilogram.)
- In a hypothetical staffing study, the predictor is measured in hours, the response in completed claims, \(\rho=-0.75\), \(\sigma_X=3\) hours, and \(\sigma_Y=8\) claims. Find the slope per hour, then express it per minute. (Answer: \(-2\) claims per hour; \(-1/30\), or approximately \(-0.0333\), claims per minute.)
- A hypothetical travel model has slope \(-0.9\) minutes per mile and correlation \(-0.50\). Convert the slope to hours per kilometer using 1 mile = 1.6 kilometers and 60 minutes = 1 hour. What is the correlation after conversion? (Answer: \(-0.009375\) hours per kilometer; correlation \(-0.50\).)
- Two hypothetical programs have the same correlation, \(0.60\), and both measure response in points. Program A has predictor and response standard deviations of 2 hours and 5 points; Program B has standard deviations of 8 hours and 5 points. Find and compare the slopes. (Answer: Program A, \(1.5\) points per hour; Program B, \(0.375\) points per hour.)
- Suppose a predictor and response are recoded as \(X^{*}=-2X\) and \(Y^{*}=3Y\), with original slope \(4\) and correlation \(0.80\). Find the transformed slope and correlation. (Answer: slope \(-6\) in the transformed units; correlation \(-0.80\), with unchanged magnitude.)