Worked Unit Conversion and Standardization Problems

Why Work Through Scale Changes?

Regression coefficients depend on how variables are measured. A slope expressed as dollars per year is numerically different from the slope expressed as thousands of dollars per year, even though both describe the same fitted relationship. Similarly, changing a predictor from kilometers to meters changes the numerical slope without changing the predictions.

This tutorial works through unit conversions and standardization in additive multiple regression. The central check is simple: if a variable is transformed consistently, the model's fitted values must still represent the same outcomes. Coefficients change to match the new units; predictions do not change when converted back to the original outcome scale.

A General Rule for Transforming Coefficients

Begin with an additive fitted regression model for \(n\) observations and \(p\) predictors. Write the fitted value for observation \(i\) as \(\hat{y}_i\), and let \(b_0\) and \(b_j\) denote the fitted intercept and slope for predictor \(j\).

\hat{y}_i=b_0+\sum_{j=1}^{p} b_j x_{ij}

Suppose each predictor is expressed in new units according to \(x_{ij}^{*}=(x_{ij}-a_j)/c_j\), and the outcome is expressed as \(y_i^{*}=(y_i-d)/r\), where each \(c_j\) and \(r\) is positive. Thus \(x_{ij}=a_j+c_jx_{ij}^{*}\) and \(y_i=d+ry_i^{*}\). Substituting these relationships into the fitted equation gives the transformed model.

\hat{y}_i^{*} = \frac{\hat{y}_i-d}{r} = \frac{b_0+\sum_{j=1}^{p}b_j a_j-d}{r} +\sum_{j=1}^{p}\frac{b_j c_j}{r}x_{ij}^{*}

Coefficient transformation: Under these affine changes of units, the new intercept is \((b_0+\sum_{j=1}^{p}b_j a_j-d)/r\), and the new slope for predictor \(j\) is \(b_jc_j/r\). These formulas transform a fitted model; they do not change its fitted relationship when the conversions are applied consistently.

The general rule includes ordinary unit conversion and standardization. For a pure predictor conversion with no shift, \(a_j=0\). If the outcome stays in its original units, \(d=0\) and \(r=1\). If the outcome is rescaled, the slopes and intercept must be adjusted for that change too.

Worked Example: Converting Predictor Units

Suppose a hypothetical fitted model predicts a clinic wait time, in minutes, from distance to the clinic in kilometers and age in decades:

\widehat{\text{wait}}=18+4.5(\text{distance in km})+2.0(\text{age in decades})

Find the equivalent model when distance is recorded in meters and age in years, then check the fitted value for a person living 3.2 km away who is 40 years old.

Let distance in meters be \(D\) and age in years be \(A\). Since 1 km equals 1,000 m, distance in kilometers is \(D/1000\). Since one decade is 10 years, age in decades is \(A/10\). Substitute these expressions into the original equation:

\widehat{\text{wait}} =18+4.5\left(\frac{D}{1000}\right)+2.0\left(\frac{A}{10}\right) =18+0.0045D+0.20A

The new distance slope is \(0.0045\) minutes per meter, and the new age slope is \(0.20\) minutes per year. To check the prediction, first use the original units:

18+4.5(3.2)+2.0(4) =18+14.4+8 =40.4\text{ minutes}

Now use \(D=3{,}200\) meters and \(A=40\) years:

18+0.0045(3200)+0.20(40) =18+14.4+8 =40.4\text{ minutes}

Both versions give the same fitted wait time. The slopes have different numerical values because their units differ: one is per meter rather than per kilometer, and the other is per year rather than per decade.

Rescaling the Outcome

Changing the outcome scale changes the units of the intercept and every slope. For instance, if the outcome is divided by 1,000 to express dollars in thousands of dollars, then the intercept and all slopes are divided by 1,000. Predictor values do not change in this conversion.

Worked Example: Expressing an Outcome in Thousands

Consider a hypothetical model for annual cost in dollars, using age in years and an indicator \(R\) that equals 1 for a remote service plan and 0 otherwise:

\widehat{\text{cost}}=24000+1250(\text{age})-800R

Rewrite the model with annual cost measured in thousands of dollars. For an individual aged 40 with \(R=1\), compare the fitted value on both scales.

The outcome in thousands of dollars is the original outcome divided by 1,000. Divide the whole fitted equation by 1,000:

\widehat{\text{cost}}_{\text{thousands}} =\frac{24000}{1000}+\frac{1250}{1000}(\text{age})-\frac{800}{1000}R =24+1.25(\text{age})-0.80R

For age 40 and \(R=1\), the original-scale prediction is:

24000+1250(40)-800 =24000+50000-800 =73200\text{ dollars}

On the thousands-of-dollars scale, the prediction is:

24+1.25(40)-0.80 =24+50-0.80 =73.2\text{ thousand dollars}

Because \(73.2\) thousand dollars equals \(73{,}200\) dollars, the transformed equation gives the same fitted cost. Its slopes now have units of thousands of dollars per year and thousands of dollars for the remote-plan difference.

Standardizing Predictors and Outcome

Standardization expresses a variable in standard deviation units. For a sample with mean \(\bar{x}\) and sample standard deviation \(s_x\), the standardized value of an observation \(x_i\) is \(z_{x,i}=(x_i-\bar{x})/s_x\). It tells how many sample standard deviations the observation is above or below the sample mean. The same calculation applies to the outcome, using its mean and standard deviation.

For a fitted model with an intercept, standardize each predictor and the outcome using the same observations that were used to fit the model. The general transformation rule gives a standardized slope equal to the original slope multiplied by the predictor's standard deviation and divided by the outcome's standard deviation. If the fitted model is an ordinary least-squares regression with an intercept, the standardized intercept is zero: the fitted outcome has sample mean equal to the observed outcome mean.

b_j^{*}=b_j\frac{s_{x_j}}{s_y}, \qquad b_0^{*}=0

The intercept result relies on fitting an ordinary least-squares model with an intercept to the data being standardized. In that setting, the residuals sum to zero, so the mean fitted outcome is \(\bar{y}\). The transformed intercept is therefore zero when all predictors and the outcome are standardized using their sample means. Standard deviations must be positive; a variable with no variation cannot be standardized this way.

Worked Example: Standardizing a Two-Predictor Model

Suppose a hypothetical fitted model predicts an outcome \(y\) from weekly exercise hours \(x_1\) and age in decades \(x_2\):

\hat{y}=122+2.4x_1-1.5x_2

Assume the sample means and standard deviations are \(\bar{x}_1=5\), \(s_{x_1}=2\), \(\bar{x}_2=4\), \(s_{x_2}=0.5\), \(\bar{y}=128\), and \(s_y=10\). These values are consistent with the fitted intercept because \(122+2.4(5)-1.5(4)=128\). Find the standardized model and verify its fitted value when \(x_1=7\) and \(x_2=3.5\).

The standardized predictor values at this observation are:

z_{x_1}=\frac{7-5}{2}=1, \qquad z_{x_2}=\frac{3.5-4}{0.5}=-1

The standardized slopes are the original slopes multiplied by the predictor standard deviation and divided by the outcome standard deviation:

b_1^{*}=2.4\left(\frac{2}{10}\right)=0.48, \qquad b_2^{*}=-1.5\left(\frac{0.5}{10}\right)=-0.075

With an intercept and all three variables standardized, the standardized intercept is zero, so the transformed model is:

\hat{z}_y=0.48z_{x_1}-0.075z_{x_2}

At \(z_{x_1}=1\) and \(z_{x_2}=-1\), this model gives:

\hat{z}_y=0.48(1)-0.075(-1)=0.555

To check against the original fitted value, calculate using the unstandardized model:

\hat{y}=122+2.4(7)-1.5(3.5) =122+16.8-5.25 =133.55

Convert this prediction to standard deviation units:

\frac{133.55-128}{10}=0.555

The standardized and original-scale calculations agree. Here the fitted outcome is 0.555 sample standard deviations above its sample mean.

Why the Predictions Agree: A Short Derivation

The agreement in the examples is not a coincidence. Under an affine transformation, each original predictor can be recovered exactly from its transformed value, and the original outcome can be recovered from the transformed outcome. Substituting those relationships into the original fitted equation produces the transformed coefficients while preserving the prediction.

x_{ij}=a_j+c_jx_{ij}^{*}, \qquad y_i=d+ry_i^{*}

Starting with the original fitted value, substitute the predictor conversions and then solve for its equivalent on the new outcome scale:

\begin{aligned} \hat{y}_i &=b_0+\sum_{j=1}^{p}b_j(a_j+c_jx_{ij}^{*})\\ &=b_0+\sum_{j=1}^{p}b_ja_j+\sum_{j=1}^{p}b_jc_jx_{ij}^{*},\\ \frac{\hat{y}_i-d}{r} &=\frac{b_0+\sum_{j=1}^{p}b_ja_j-d}{r} +\sum_{j=1}^{p}\frac{b_jc_j}{r}x_{ij}^{*}. \end{aligned}

The left side is the original fitted outcome expressed in the new outcome units. The right side has exactly the transformed intercept and slopes from the general rule. Multiplying the transformed prediction by \(r\) and adding \(d\) returns the original fitted prediction. This argument is algebraic: it does not require a new regression fit.

Common Mistakes and a Practical Check

A frequent error is to change a predictor's units but leave its slope unchanged. If a predictor is converted from kilometers to meters, the numerical value of the predictor becomes 1,000 times larger, so its slope must become 1,000 times smaller to keep the product of slope and predictor unchanged. For an outcome conversion, the direction is different: dividing the outcome by 1,000 divides the intercept and slopes by 1,000 as well.

Another mistake is to treat a standardized slope as if it had the original units. A standardized slope describes the fitted change in the outcome, in outcome standard deviations, for a one-standard-deviation increase in that predictor, holding the other predictors fixed in the additive model. It is not a change in the original outcome units. Standardization changes the numerical scale of coefficients, not the fitted relationship or the evidence from the same fitted model.

Key check: After any conversion, choose a concrete predictor combination, calculate the fitted value with both equations, and express both predictions in the same outcome units. If they disagree, check whether each slope was adjusted for both predictor and outcome conversions.

These calculations describe fitted associations, not necessarily causal effects. A unit conversion or standardization cannot by itself establish causation, improve a model's assumptions, or make coefficients directly comparable without regard to which predictors are included and how they are measured.

Check Your Understanding

Use the transformation rules to solve these questions. Include units when interpreting a coefficient.

A fitted slope is 6 minutes per kilometer. What is the corresponding slope per meter, and why must its numerical value change?
An outcome is converted from dollars to hundreds of dollars by dividing by 100. How do the intercept and slopes change?
A predictor has sample mean 12 and sample standard deviation 3. What standardized value corresponds to an observed value of 18?
In a model where both a predictor and an outcome are standardized, what does a slope of \(-0.4\) describe?
Why should a fitted value calculated from a consistently transformed model agree with the original fitted value after converting back to the original outcome units?