Rescaling a Predictor and Its Confidence Interval

Why Rescaling Changes the Numbers but Not the Fit

A regression slope is expressed in units of response per unit of predictor. If distance is measured in centimetres instead of metres, the numerical value of the slope changes because one centimetre is only one hundredth of a metre. This change of units does not alter the underlying relationship described by the fitted line: the same observations still receive the same fitted responses.

The same principle applies to a slope’s standard error and confidence interval. Their numerical values change along with the slope, but they remain expressed on the new scale. Understanding this connection helps prevent misleading comparisons between regression outputs that use different units.

Model and Rescaling Rule

Begin with a simple linear regression model for a quantitative response \(Y\) and predictor \(x\):

Y_i=\beta_0+\beta_1x_i+\varepsilon_i,\qquad i=1,\ldots,n.

Here \(\beta_0\) is the population intercept, \(\beta_1\) is the population slope, and \(\varepsilon_i\) is the error for observation \(i\). For the usual regression coefficient confidence interval, assume the errors are independent, have mean zero and common variance \(\sigma^2\), and are normally distributed. The model also assumes that the mean response follows a straight-line relationship with \(x\). These conditions support the conventional \(t\)-based interval; rescaling does not make an unsuitable model appropriate.

Definition: A positive multiplicative rescaling replaces each predictor value \(x_i\) by \(x_i^*=c x_i\), where \(c>0\) is a unit-conversion factor. For example, converting metres to centimetres uses \(c=100\). The response and all other predictors, if present, are left unchanged.

In the rescaled predictor’s units, the fitted model uses a new slope. The coefficient estimate, standard error, and confidence interval obey the same conversion rule.

\hat{\beta}_1^*=\frac{\hat{\beta}_1}{c}, \qquad SE(\hat{\beta}_1^*)=\frac{SE(\hat{\beta}_1)}{c}, \qquad \left(\frac{L}{c},\frac{U}{c}\right)

In this expression, \((L,U)\) is the original confidence interval and \(c>0\), so dividing both endpoints preserves their order. The interval’s endpoints, like the estimate, are now in response units per rescaled predictor unit. For a pure multiplication of \(x\), the intercept stays the same.

Why the Fitted Values Stay the Same

The rescaling rule follows by expressing the original model in terms of \(x_i^*\). Since \(x_i=x_i^*/c\), substituting this into the original mean gives:

\beta_0+\beta_1x_i =\beta_0+\beta_1\left(\frac{x_i^*}{c}\right) =\beta_0+\left(\frac{\beta_1}{c}\right)x_i^*.

Thus, the population intercept is unchanged and the population slope on the rescaled predictor is \(\beta_1/c\). The same relationship holds for the least-squares estimates: each possible fitted line on the original scale has an exactly equivalent representation on the new scale. The predictions, and therefore the residuals (observed response minus fitted response), are unchanged.

The standard error changes consistently too. In simple regression, the estimated variance of the slope is the residual variance divided by the predictor’s sum of squared deviations from its mean. Rescaling multiplies that sum by \(c^2\), while leaving the residual variance unchanged:

\sum_{i=1}^n(x_i^*-\bar{x}^*)^2 =c^2\sum_{i=1}^n(x_i-\bar{x})^2, \qquad SE(\hat{\beta}_1^*)^2 =\frac{SE(\hat{\beta}_1)^2}{c^2}.

Taking positive square roots gives \(SE(\hat{\beta}_1^*)=SE(\hat{\beta}_1)/c\). The degrees of freedom and the chosen \(t\) critical value do not change, so the interval’s margin of error is also divided by \(c\). Equivalently, applying the same positive division to the original interval’s two endpoints gives the interval on the new scale.

Worked Conversions

Worked Example: Converting a Distance Predictor from Metres to Centimetres

Hypothetical example. A simple regression relates a measured outcome in points to distance in metres. Suppose the fitted intercept is \(12.0\) points, the slope estimate is \(2.40\) points per metre, and its standard error is \(0.60\) points per metre. There are \(n=20\) observations, so \(df=18\). For a 95% confidence interval, use \(t^*=2.101\). Assume the observations are independent, the straight-line mean relationship is plausible, and residual checks show no major departure from constant variance or approximate normality.

On the metre scale, the margin of error is:

2.101(0.60)=1.2606\text{ points per metre}.

The 95% confidence interval is therefore:

2.40\pm1.2606=(1.1394,3.6606)\text{ points per metre}.

Now set \(x^*=100x\), because there are 100 centimetres in a metre. Divide the slope, standard error, and both interval endpoints by 100:

\hat{\beta}_1^*=\frac{2.40}{100}=0.0240,\qquad SE(\hat{\beta}_1^*)=\frac{0.60}{100}=0.00600,

\left(\frac{1.1394}{100},\frac{3.6606}{100}\right) =(0.011394,0.036606)\text{ points per centimetre}.

Rounded suitably, the interval is \((0.0114,0.0366)\) points per centimetre. For the same distance, the fitted contribution from the predictor is unchanged: at \(1.35\) metres, it is \(2.40(1.35)=3.24\) points; at \(135\) centimetres, it is \(0.0240(135)=3.24\) points. With the intercept of \(12.0\), both scales give a fitted response of \(15.24\) points. The units changed; the fitted line did not.

Worked Example: Rescaling a Predictor in Multiple Regression

Hypothetical example. A multiple regression predicts a response in outcome units from age in years and other predictors. Suppose the estimated age coefficient is \(1.80\) outcome units per year, with standard error \(0.70\), and \(df=24\). Use \(t^*=2.064\) for a 95% interval. Assume the regression’s linearity and error conditions are reasonably supported, including independence, approximately constant error variance, and residuals without a serious departure from normality.

First calculate the age coefficient interval in its original units:

1.80\pm2.064(0.70) =1.80\pm1.4448 =(0.3552,3.2448)\text{ outcome units per year}.

Suppose age is instead recorded in decades, so \(x^*=x/10\). This is the rescaling \(x^*=cx\) with \(c=0.1\). Dividing the coefficient, its standard error, and interval endpoints by \(0.1\) gives:

\hat{\beta}_1^*=\frac{1.80}{0.1}=18.0,\qquad SE(\hat{\beta}_1^*)=\frac{0.70}{0.1}=7.0,

\left(\frac{0.3552}{0.1},\frac{3.2448}{0.1}\right) =(3.552,32.448)\text{ outcome units per decade}.

The interval is about the age slope conditional on the other predictors in the model. Its interpretation changes in units: the estimated difference in mean response associated with a one-year increase is \(1.80\) outcome units, while the equivalent estimated difference for a one-decade increase is \(18.0\) units. For age 47, the age contribution is \(1.80(47)=84.6\) on the year scale and \(18.0(4.7)=84.6\) on the decade scale. Other terms and the fitted response remain unchanged.

Worked Example: Converting an Interval from Hours to Minutes

Hypothetical example. In a simple regression of a response measured in kilograms on a quantitative predictor measured in hours, suppose the slope estimate is \(-0.085\) kilograms per hour and its standard error is \(0.030\) kilograms per hour. With \(n=16\), \(df=14\); use \(t^*=2.145\) for a 95% interval. Assume the observations are independent and that a straight-line mean relationship and the usual residual conditions are reasonable for this hypothetical analysis.

The margin of error and interval in kilograms per hour are:

2.145(0.030)=0.06435,\qquad -0.085\pm0.06435=(-0.14935,-0.02065).

Convert hours to minutes using \(x^*=60x\), so \(c=60\). The coefficient and standard error become:

\hat{\beta}_1^*=\frac{-0.085}{60}=-0.0014167,\qquad SE(\hat{\beta}_1^*)=\frac{0.030}{60}=0.000500.

Dividing the interval endpoints by 60 gives:

\left(\frac{-0.14935}{60},\frac{-0.02065}{60}\right) =(-0.0024892,-0.0003442)\text{ kilograms per minute}.

At \(2.5\) hours, the predictor contribution is \(-0.085(2.5)=-0.2125\) kilograms. The same duration is \(150\) minutes, giving \(-0.0014167(150)\approx-0.2125\) kilograms. The interval’s negative endpoints indicate that the coefficient is estimated on a negative scale; the interval’s meaning is not changed by expressing time in minutes rather than hours.

A Common Mistake: Changing the Units of Only the Slope

A common mistake is to convert the slope but leave its standard error and confidence interval unchanged. For instance, after converting metres to centimetres, someone might report a slope of \(0.024\) points per centimetre alongside the original standard error of \(0.60\) points per metre. Those two numbers use different predictor units, so they cannot be combined to form the correct interval.

The correction is to apply the same scale conversion to the coefficient, its standard error, and both interval endpoints. In the metres-to-centimetres example, all four are divided by 100. This keeps the ratio of the estimate to its standard error, the \(t\) statistic, unchanged, as well as leaving the fitted values unchanged.

A rescaled slope’s numerical size alone does not indicate that the relationship has become stronger or weaker. It reflects the size of one unit on the predictor’s scale. Always report the predictor and response units, and do not interpret a regression association as evidence that changing the predictor causes a change in response.

Key takeaway: If \(x^*=cx\) for \(c>0\), divide the slope estimate, its standard error, and both confidence interval endpoints by \(c\). The coefficient’s units change, but the fitted responses, residuals, and substantive relationship do not.

Check Your Understanding

Work through these questions by tracking the predictor units in every coefficient and interval.

A predictor is converted from kilometres to metres, so \(x^*=1000x\). How does its slope estimate change?
A coefficient is \(4.5\) response units per litre, with standard error \(1.2\). What are both values when volume is recorded in millilitres?
Why do a regression’s fitted values remain unchanged when one predictor is multiplied by a positive constant?
If both endpoints of a slope confidence interval are converted to new predictor units, what happens to its order when the conversion factor is positive?
When rescaling one predictor in a multiple regression, which quantities remain unchanged if all other predictors and the response are left as recorded?