Regression Line of x on y

The usual line of best fit is built to predict $y$ from $x$. But sometimes you know $y$ and want to estimate $x$: you know a student’s test score and want to estimate how long they revised. For that job there’s a different line, the regression line of $x$ on $y$. This page shows why it’s different, how to find it, and how to pick the right line for each prediction.

Key ideas

Two different “best” lines

The regression line of $y$ on $x$, $y = ax + b$, is found by making the sum of the squared vertical distances from the points to the line as small as possible. That makes it the best line for predicting $y$ when you know $x$.

The regression line of $x$ on $y$ is written with $x$ as the subject:

$$x = cy + d$$

It makes the sum of the squared horizontal distances as small as possible, so it’s the best line for predicting $x$ when you know $y$.

Because they minimize different distances, the two lines are different (unless the points lie exactly on a straight line, $r = \pm 1$, when they coincide). The weaker the correlation, the further apart they are.

Both lines pass through the mean point

Both regression lines pass through the mean point $(\bar{x}, \bar{y})$. So if you’re given both equations, solving them simultaneously gives $\bar{x}$ and $\bar{y}$.

The line of $y$ on $x$ (blue) and the line of $x$ on $y$ (orange) are different lines, but both pass through the mean point $(\bar{x}, \bar{y}) = (6.6, 66)$.

Which line to use

You know	You want	Use
$x$	$y$	the line of $y$ on $x$: $\ y = ax + b$
$y$	$x$	the line of $x$ on $y$: $\ x = cy + d$

You cannot get the line of $x$ on $y$ by rearranging $y = ax + b$ to make $x$ the subject. That gives the same line written differently, which is still the best line for predicting $y$, not $x$. Likewise, you can’t reliably predict $y$ from $x$ using the line of $x$ on $y$.

Finding the line with technology

On a GDC, the line of $x$ on $y$ is found by running the usual linear regression with the lists swapped: put the $y$-values in the list the calculator treats as "$x$" and the $x$-values in the list it treats as "$y$". The calculator reports something like "$y = ax + b$", but remember its "$x$" is really your $y$, so rewrite the result as $x = cy + d$.

The correlation coefficient $r$ is the same either way round.

Making predictions

The same cautions as for $y$ on $x$ apply:

• Interpolation (the known value is inside the range of the data) is reasonably reliable when the correlation is strong.
• Extrapolation (the known value is outside the range of the data) is unreliable: the pattern may not continue, and it can give impossible answers.
• If the correlation is weak, no prediction is very reliable.

For a line of $x$ on $y$, “inside the range” means inside the range of the $y$-values, since $y$ is what you’re given.

Worked examples

Example 1: Choosing the right line

A café records the daily maximum temperature $x$ ($^\circ\text{C}$) and the number of iced drinks sold, $y$, for $30$ summer days. Which regression line should be used to:

• (a) predict the number of iced drinks sold on a day forecast to reach $27\,^\circ\text{C}$;
• (b) estimate the maximum temperature on a day when $140$ iced drinks were sold?

Solution.

(a) The temperature $x$ is known and the sales $y$ are wanted: use the line of $y$ on $x$.

(b) The sales $y$ are known and the temperature $x$ is wanted: use the line of $x$ on $y$.

Example 2: Finding both lines and predicting

The table shows the hours of revision, $x$, and the test score (%), $y$, for $10$ students.

$x$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$12$
$y$	$48$	$60$	$52$	$66$	$58$	$74$	$65$	$80$	$72$	$85$

• (a) Find the regression line of $y$ on $x$ and use it to predict the score of a student who revised for $6.5$ hours.
• (b) Find the regression line of $x$ on $y$ and use it to estimate how long a student who scored $70\%$ revised.

Solution.

(a) Enter $x$ in list 1 and $y$ in list 2 and run linear regression:

$$y = 3.29x + 44.3 \qquad (r = 0.885, \text{ to 3 s.f.})$$

For $x = 6.5$: $y \approx 3.29(6.5) + 44.3 \approx 65.7$. The predicted score is about $65.7\%$.

(b) Run the regression again with the lists swapped ($y$ as the input list, $x$ as the output list):

$$x = 0.238y - 9.10$$

For $y = 70$: $x \approx 0.238(70) - 9.10 \approx 7.55$ hours (using the unrounded coefficients). Both predictions are interpolations ($6.5$ is inside $2$ to $12$, and $70$ is inside $48$ to $85$) and $r = 0.885$ shows strong positive correlation, so both are reasonably reliable.

Check: the mean point is $(\bar{x}, \bar{y}) = (6.6, 66)$. Both lines pass through it: $3.29(6.6) + 44.3 \approx 66.0$ and $0.238(66) - 9.10 \approx 6.61$. ✓

Example 3: Using the mean point

For a set of data, the regression line of $y$ on $x$ is $y = 0.5x + 8$, and the regression line of $x$ on $y$ is $x = 1.5y - 7$.

• (a) Find $\bar{x}$ and $\bar{y}$.
• (b) Estimate $x$ when $y = 22$, and estimate $y$ when $x = 24$.

Solution.

(a) Both lines pass through $(\bar{x}, \bar{y})$, so solve them simultaneously. Substitute the first into the second:

$$\begin{aligned} x &= 1.5(0.5x + 8) - 7 \\ x &= 0.75x + 5 \\ 0.25x &= 5 \\ x &= 20 \end{aligned}$$

Then $y = 0.5(20) + 8 = 18$. So $\bar{x} = 20$ and $\bar{y} = 18$.

(b) Given $y$, use the line of $x$ on $y$: $x = 1.5(22) - 7 = 26$.

Given $x$, use the line of $y$ on $x$: $y = 0.5(24) + 8 = 20$.

Example 4: When predictions go wrong

Using the revision data from Example 2, a student uses $x = 0.238y - 9.10$ to estimate the revision time for scores of (a) $95\%$ and (b) $30\%$. Comment on each answer.

Solution.

(a) $x \approx 0.238(95) - 9.10 \approx 13.5$ hours. The score $95$ is outside the observed scores ($48$ to $85$), so this is extrapolation and unreliable. The pattern might not continue: scores can’t rise forever as revision time increases.

(b) $x \approx 0.238(30) - 9.10 \approx -1.96$ hours. A negative revision time is impossible. Again, $30$ is far outside the observed scores, and the model shouldn’t be used there.

Common mistakes

Rearranging the y-on-x line. Making $x$ the subject of $y = 3.29x + 44.3$ gives $x = \dfrac{y - 44.3}{3.29}$, which predicts $x \approx 7.82$ for $y = 70$. That isn’t the line of $x$ on $y$, and its answer differs from the correct $7.55$. Always run the regression with the lists swapped.

Using the x-on-y line to predict y. The guide is clear that predicting $y$ from $x$ with an $x$-on-$y$ line isn’t reliable. Match the line to what you know.

Reading the GDC output literally. After swapping the lists, the calculator still writes "$y = ax + b$". Rewrite it as $x = ay + b$ so you don’t plug a value into the wrong letter.

Mixing up which equation is which. A line written as $x = \ldots$ is the line of $x$ on $y$. If both equations are given as "$y = \ldots$", read the question carefully to see which is which.

Checking the wrong range for extrapolation. For an $x$-on-$y$ prediction, compare the given $y$-value with the range of the observed $y$-values, not the $x$-values.

Forgetting the mean point. Questions often give both lines and ask for $\bar{x}$ and $\bar{y}$. Solve the equations simultaneously; you don’t need the original data.

Practice

1. (Warm-up) At an outdoor pool, $x$ is the daily maximum temperature ($^\circ\text{C}$) and $y$ is the number of swimmers. Which regression line should be used to:

• (a) predict the number of swimmers when the forecast is $28\,^\circ\text{C}$;
• (b) estimate the temperature on a day when $350$ swimmers came?

Solution

(a) $x$ is known, $y$ is wanted: the line of $y$ on $x$.

(b) $y$ is known, $x$ is wanted: the line of $x$ on $y$.

2. (Warm-up) The regression line of $y$ on $x$ is $y = 2x + 3$, and the regression line of $x$ on $y$ is $x = 0.4y + 1$. Find $\bar{x}$ and $\bar{y}$.

Solution

Both lines pass through $(\bar{x}, \bar{y})$. Substitute:

$$x = 0.4(2x + 3) + 1 = 0.8x + 2.2 \quad\Rightarrow\quad 0.2x = 2.2 \quad\Rightarrow\quad x = 11$$

$y = 2(11) + 3 = 25$. So $\bar{x} = 11$ and $\bar{y} = 25$.

3. (Core) The age $x$ (years) and price $y$ (thousands of dollars) of $7$ used cars of the same model are shown.

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$8$
$y$	$28$	$25$	$23$	$19$	$18$	$14$	$10$

Use your GDC to find the regression line of $x$ on $y$. Use it to estimate the age of a car of this model priced at $21 000.

Solution

Run linear regression with the prices as the input list and the ages as the output list:

$$x = -0.381y + 11.6 \qquad (r = -0.995, \text{ to 3 s.f.})$$

For a price of $21 000, $y = 21$:

$$x \approx -0.381(21) + 11.6 \approx 3.60 \text{ years}$$

(using the unrounded coefficients). This is interpolation ($21$ is between $10$ and $28$) and the correlation is very strong, so it’s reliable.

4. (Core) For the cars in question 3, a buyer wants to predict the price of a $7$-year-old car. Explain why the line from question 3 shouldn’t be used, find the correct line, and make the prediction.

Solution

The age $x$ is known and the price $y$ is wanted, so the line of $y$ on $x$ is needed. The line of $x$ on $y$ isn’t designed to predict $y$ from $x$.

GDC (ages as input, prices as output):

$$y = -2.60x + 30.3$$

For $x = 7$: $y \approx -2.60(7) + 30.3 \approx 12.1$, so about $12 100 (using unrounded coefficients). Since $7$ is inside the ages $1$ to $8$, this is interpolation.

5. (Core) Using the line from question 3, estimate the age of a car priced at $5000, and comment on the reliability of your answer.

Solution

$y = 5$: $x \approx -0.381(5) + 11.6 \approx 9.69$ years.

The prices in the data run from $10$ to $28$ thousand dollars, so $5$ is outside the range: this is extrapolation. Even though $r$ is very close to $-1$, the straight-line pattern may not continue (car prices usually level off rather than keep dropping at the same rate), so the estimate is unreliable.

6. (Core) For a data set, $\bar{x} = 20$. The regression line of $x$ on $y$ is $x = 0.8y - 4$, and the regression line of $y$ on $x$ has gradient $1.05$.

• (a) Find $\bar{y}$.
• (b) Find the equation of the regression line of $y$ on $x$.

Solution

(a) The line of $x$ on $y$ passes through $(\bar{x}, \bar{y})$: $20 = 0.8\bar{y} - 4$, so $0.8\bar{y} = 24$ and $\bar{y} = 30$.

(b) The line of $y$ on $x$ also passes through $(20, 30)$: $30 = 1.05(20) + b$, so $b = 9$.

$$y = 1.05x + 9$$

7. (Core) For the revision data in Example 2, a student wants to estimate the revision time for a score of $70\%$. She rearranges $y = 3.29x + 44.3$ to get $x = \dfrac{y - 44.3}{3.29}$.

• (a) Find her estimate.
• (b) Explain why it differs from the estimate in Example 2, and which estimate is better.

Solution

(a) $x = \dfrac{70 - 44.3}{3.29} \approx 7.81$ hours. (With the unrounded GDC coefficients, $7.82$ hours.)

(b) Rearranging doesn’t change the line: it’s still the line of $y$ on $x$, which minimizes vertical distances and is designed for predicting $y$. The line of $x$ on $y$ minimizes horizontal distances, so it’s the best line for predicting $x$. The estimate of $7.55$ hours from Example 2 is the better one.

8. (Challenge) The gradients of the two regression lines can be written as $a = r\dfrac{s_y}{s_x}$ (for $y$ on $x$) and $c = r\dfrac{s_x}{s_y}$ (for $x$ on $y$), where $s_x$, $s_y$ are the standard deviations of $x$ and $y$.

• (a) Show that $ac = r^2$.
• (b) For Example 3, the lines are $y = 0.5x + 8$ and $x = 1.5y - 7$. Find $r$.
• (c) Use (a) to explain why the two lines are the same line when $r = \pm 1$.

Solution

(a)

$$ac = r\frac{s_y}{s_x} \times r\frac{s_x}{s_y} = r^2$$

(b) $r^2 = 0.5 \times 1.5 = 0.75$. Both gradients are positive, so $r$ is positive: $r = \sqrt{0.75} \approx 0.866$ (3 s.f.).

(c) If $r = \pm 1$, then $ac = 1$, so $c = \dfrac{1}{a}$. The line of $x$ on $y$ is $x = \dfrac{1}{a}y + d$, which rearranges to $y = ax - ad$: a line with the same gradient as $y = ax + b$. Both lines also pass through $(\bar{x}, \bar{y})$, and two lines with the same gradient through the same point are the same line.

9. (Challenge) The two regression lines for a data set are $y = -0.5x + 20$ and $x = -1.6y + 34$.

• (a) State which line is the line of $x$ on $y$.
• (b) Find the mean point.
• (c) Using the result of question 8, find $r$.
• (d) Estimate $y$ when $x = 12$.

Solution

(a) $x = -1.6y + 34$ has $x$ as the subject, so it’s the line of $x$ on $y$.

(b) Substitute $y = -0.5x + 20$ into the other line:

$$x = -1.6(-0.5x + 20) + 34 = 0.8x + 2 \quad\Rightarrow\quad 0.2x = 2 \quad\Rightarrow\quad x = 10$$

$y = -0.5(10) + 20 = 15$. The mean point is $(10, 15)$.

(c) $r^2 = (-0.5)(-1.6) = 0.8$. Both gradients are negative, so $r = -\sqrt{0.8} \approx -0.894$ (3 s.f.).

(d) $x$ is known, so use the line of $y$ on $x$: $y = -0.5(12) + 20 = 14$.