Rewrite fitted linear models after variable changes

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Difficulty: Advanced
Estimated time: 28 minutes
Domains: Problem Solving and Data Analysis
Techniques: Fitted-model-rewritingVariable-redefinitionUnit-rescalingOrigin-shiftsAffine-changes

What you’ll learn

Tell when a fitted line should be rewritten, not fitted again.
Rescale the input or output of a fitted linear model.
Handle a new zero point, or origin, in either variable.
Change the input and output together in one reliable order.

Prerequisites

Why this matters on the SAT

Same line, new variables

Sometimes the SAT gives you a line of best fit, then changes how one or both variables are measured. Hours become minutes, or thousands of dollars become dollars. The data haven’t changed, so the relationship hasn’t either. Your job is to write the same line in the new variables.

Solution to the example

The fitted equation is written in $x$ , the hours. So before you can use it, you need $x$ in terms of the new variable $t$ . Solve the definition for $x$ :

t=60x \quad\Longrightarrow\quad x=\frac{t}{60}.

Now substitute into the model:

y=14\left(\frac{t}{60}\right)+80 =\frac{7}{30}t+80.

Choice B is correct. You never needed the original data. The question hands you the fitted line, so your job is to rewrite it, not rebuild it.

SAT example

A line of best fit predicts water use $y$ , in gallons, from the number of hours $x$ that a system operates:

y=14x+80.

The operating time is instead recorded as $t$ minutes, where $t=60x$ . Which equation gives the same line of best fit in terms of $t$ and $y$ ?

A
$y=840t+80$
B
$y=\frac{7}{30}t+80$
C
$y=14t+80$
D
$y=\frac{7}{30}t+\frac{4}{3}$

Rewrite in the same order every time

These questions often give you three equations:

the old fitted model;
a definition of the new input; and
a definition of the new output.

With answer choices, try a point first. Pick an easy old input, like $x=0$ , and find its old output. Convert both coordinates to the new variables, then test that new point in the choices. If exactly one choice works, you’re done.

Otherwise, rewrite the whole equation. If the question asks you to write the equation yourself, or the point doesn’t pick out one choice, use the universal rewriting method. It works every time. Give each equation its own job, and go in this order:

Write the old input in terms of the new input.
Substitute that expression into the fitted model.
Rewrite or solve for the new output.
Simplify, and check the variable names.

Why start with the old input? The fitted equation only understands the old variables. It has an $x$ in it, so you have to translate the new input back into $x$ before you can use it. Think of it as old input in, new output out.

Here’s the whole order on a general model. Say the old model is

y=mx+b

and the new variables are

X=ax+c \qquad\text{and}\qquad Y=dy+e.

First, get the old input by itself:

x=\frac{X-c}{a}.

Then substitute:

y=m\left(\frac{X-c}{a}\right)+b.

Finally, use $Y=dy+e$ to rewrite the output:

Y=d\left[m\left(\frac{X-c}{a}\right)+b\right]+e.

You don’t need to memorize that last formula. The order is the part worth keeping, because it works whatever the variables are called and whatever the context is.

Check your understanding:

A fitted model is $y=9x+4$ , and the new input is $t=3x$ . Rewrite the model in terms of $t$ and $y$ .

Common mistake:

It’s easy to read $t=60x$ backward and replace $x$ with $60t$ . That gives $y=840t+80$ , choice A in the opening example. The model contains the old input, so solve for it first: $x=\frac{t}{60}$ .

Rescale the input or output

An input rescale changes the size of one input unit. In the opening example, one hour became $60$ minutes. One minute is only $\frac{1}{60}$ of an hour, so the water use grows only $\frac{1}{60}$ as much per minute. The new slope is the old slope divided by $60$ .

An output rescale changes every predicted output. Suppose

r=-0.08x+4.6

models revenue $r$ in thousands of dollars, and $R$ is revenue in dollars. Since

R=1000r,

multiply the whole output by $1000$ :

\begin{aligned} R&=1000(-0.08x+4.6)\\[1.4em] &=-80x+4600. \end{aligned}

Both the slope and the intercept get multiplied. If you converted only the $4.6$ , the intercept would be in dollars while the slope was still in thousands of dollars, and the equation would mix units.

Check your understanding:

A fitted model is $q=2t+7$ , where $q$ is measured in thousands. If $Q=1000q$ , what is the model for $Q$ in terms of $t$ ?

Handle origin shifts

An origin shift moves the zero: it changes what the value $0$ means. It doesn’t have to change the unit.

Say $t$ counts years since 2015, and $u$ counts years since 2020. When $u=0$ , it’s 2020, and five years have already passed since 2015. So

t=u+5.

If the fitted model is

p=3.2t+18,

then

\begin{aligned} p&=3.2(u+5)+18\\[1.4em] &=3.2u+34. \end{aligned}

The slope stays $3.2$ because both variables still count years. The intercept changes because the two variables put their zero on different dates.

Output origins work the same way. If $v=y-7$ , then $v$ measures how far $y$ is above $7$ . You can use $v=y-7$ as it is, or rewrite it as $y=v+7$ when you need the old output.

Check your understanding:

A fitted model is $y=4x+10$ . New variables are defined by $u=x-3$ and $v=y-7$ . Rewrite the model in terms of $u$ and $v$ .

Common mistake:

2020 comes after 2015, so it can feel like you should subtract and write $t=u-5$ . Don’t guess. Check the zero point instead: $u=0$ is 2020, which is $t=5$ years after 2015. So $t=u+5$ .

Example: Change the input and the output together

Worked example

A fitted linear model relates the total volume of water $W$ , in liters, pumped after $T$ minutes:

\mathbf{W=18T+120}.

For a new report:

$x=60(T-4)$ is the number of seconds after the pump has run for

4

minutes.

$y=1000(W-150)$ is the number of milliliters pumped beyond a baseline of

150

liters.

Which equation gives the same fitted model in terms of $x$ and $y$ ?

A
$y=300x+42{,}000$
B
$y=300x+192{,}000$
C
$y=0.3x+42$
D
$y=1{,}080{,}000x+42{,}000$

Step 1

Write the old input using the new input

The fitted model uses $T$ , so solve the new-input definition for $T$ :

\begin{aligned} x&=60(T-4)\\[1.4em] \frac{x}{60}&=T-4\\[1.4em] T&=\frac{x}{60}+4. \end{aligned}

Check the zero point: $x=0$ should be the moment the pump reaches $4$ minutes, and the formula gives $T=4$ .

Step 2

Substitute into the fitted model

Replace $T$ in the old model:

\begin{aligned} W&=18\left(\frac{x}{60}+4\right)+120\\[1.4em] &=\frac{3}{10}x+72+120\\[1.4em] &=\frac{3}{10}x+192. \end{aligned}

You’re halfway there. The input is new, but the output is still the old $W$ .

Step 3

Apply the new output definition

Now use $y=1000(W-150)$ , with the whole expression in place of $W$ :

\begin{aligned} y&=1000\left(\frac{3}{10}x+192-150\right)\\[1.4em] &=1000\left(\frac{3}{10}x+42\right)\\[1.4em] &=300x+42{,}000. \end{aligned}

Choice A is correct. If you forget the $-150$ , you land on choice B.

You can check the intercept too. At $x=0$ , the old model predicts $W=192$ liters. That’s $42$ liters, or $42{,}000$ milliliters, beyond the new baseline, just as the equation says.

Check your understanding:

Why is the new slope $300$ milliliters per second?

Practice problems

Work these by hand. Once your setup is written down, you can use the calculator to check any arithmetic that isn’t quick in your head.

Rescale the input

Practice problem

A fitted model predicts distance $d$ , in miles, after $h$ hours:

d=42h+15.

Time is instead recorded as $t$ minutes, where $t=60h$ . Which equation gives the same model in terms of $t$ and $d$ ?

Calculator loads as you approach

By hand: solve for

h=t/60

, then substitute. No calculator needed.

Rescale the output

Practice problem

A line of best fit predicts revenue $r$ , in thousands of dollars, from an input $x$ :

r=-0.08x+4.6.

Revenue is instead recorded as $R$ dollars, where $R=1000r$ . Which equation gives the same line of best fit in terms of $x$ and $R$ ?

Calculator loads as you approach

By hand: multiply the whole fitted output by

1000

. No calculator needed.

Move the time origin

Practice problem

A fitted model predicts a quantity $p$ from $t$ , the number of years since the beginning of 2018:

p=3.2t+18.

Let $u$ be the number of years since the beginning of 2023. Which equation gives the same fitted model in terms of $u$ and $p$ ?

Calculator loads as you approach

By hand: check the zero point to find

t=u+5

. No calculator needed.

Change both variables

Practice problem

A line of best fit is

y=-3x+28.

New variables are defined by

X=2x+10 \qquad\text{and}\qquad Y=4y-6.

Which equation gives the same line of best fit in terms of $X$ and $Y$ ?

Calculator loads as you approach

Try the point

x=0

in the choices first. Rewrite the whole equation only if the point doesn’t pick out one choice. No calculator needed.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

When the question gives you a fitted equation, keep it. Don’t refit the data.
Write the old input in terms of the new input before you substitute.
After substituting, rewrite or solve for the new output.
An input rescale changes how much input one new unit stands for.
An output rescale applies to the whole predicted output, slope and intercept alike.
An origin shift changes what $0$ means. Check its sign by matching the two variables at a known zero point.
With answer choices, test an easy old point in the new variables, and stop if exactly one choice fits it.
When you have to write the equation, or the point doesn’t pick out one choice, go in order: old input, substitute, new output, simplify.
Work by hand, and save the calculator for arithmetic that isn’t quick once the setup is done.

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Use counts and observed frequencies to calculate and interpret probabilities.

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