Normal & regression 1 · The regression line of x on y

Module 29 of 46Lesson 1 of 5 · Normal & regression 1 · The regression line of x on y

Regression x on y, standardisation & the normal distribution · Essential understanding 1

The regression line of x on y

IB AA SL 4.10with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Two variables give you two regression lines. The line of y on x predicts y from x. The line of x on y predicts x from y. They are different lines, they cross at the mean point (x̄, ȳ), and picking the wrong one gives a wrong answer.

By the end of this lesson you can:

explain what x on y means and when to use it instead of y on x
find the regression line of x on y with the GDC (swap the lists)
find both lines by hand from x̄, ȳ, sx, sy and r or sxy
use the fact that both lines pass through (x̄, ȳ)
use two given lines to find the mean point and the correlation r
predict, and comment on whether the prediction is reliable

🧑🏾 Imagine this…

Thato measures how many hours students study (x) and the mark they get (y). The usual line answers: “I study 6 hours. What mark should I expect?” But Thato may ask the other way round: “I want 70%. How many hours must I study?” Now the mark is the fixed number and the hours are the thing we are predicting. For that question we need a line built to predict hours: the line of x on y.

Regression line of

y

x

the best-fit line for predicting y from a given x. Written

y - \bar{y} = \dfrac{s_{xy}}{s_x^{\,2}}(x - \bar{x})

. It makes the vertical gaps as small as possible.

Regression line of

x

y

the best-fit line for predicting x from a given y. Written

x - \bar{x} = \dfrac{s_{xy}}{s_y^{\,2}}(y - \bar{y})

. It makes the horizontal gaps as small as possible.

Covariance

s_{xy}

a number that measures how x and y move together. It is linked to r by

r = \dfrac{s_{xy}}{s_x\, s_y}

, so

s_{xy} = r\, s_x\, s_y

Mean point

the point (x̄, ȳ). Both regression lines pass through it.

★ Two lines, one rule

Predicting y from x: use the line of y on x. Predicting x from y: use the line of x on y.
Both lines pass through the mean point (x̄, ȳ). So where they cross, you are at (x̄, ȳ).
By hand: $y - \bar{y} = \dfrac{s_{xy}}{s_x^{\,2}}(x - \bar{x})$ and $x - \bar{x} = \dfrac{s_{xy}}{s_y^{\,2}}(y - \bar{y})$ . (the y-on-x form is in the IB formula booklet; the x-on-y line is the same with x and y swapped)
On the GDC for x on y: run Linear Regression (mx+b) with X List = the y data and Y List = the x data. The answer is then $x = my + b$ .
If the gradients are written as $y = m_1 x + c_1$ and $x = m_2 y + c_2$ , then $r^2 = m_1 \times m_2$ . The sign of r is the sign of the gradients.
A prediction is reliable only if r is strong and the value you put in is inside the range of the data (interpolation).

⚠️ Watch out

Do not rearrange the y-on-x line to find x. It will give a different (worse) answer. If x is the thing you are asked to find, you need the x-on-y line.
On the GDC the swap is easy to forget. If your x-on-y line looks like it has the same gradient as y on x, you did not swap the lists.
An x-on-y line is x = my + c. The gradient m is “change in x per 1 unit of y”, not the slope you would draw on a graph with x across.
Your answer is only trusted if it is inside the data range. Say “interpolation” or “extrapolation” whenever you predict.

💡 Which line? Ask “what is given and what is wanted?”

Underline the number you are given and the quantity you must find. Given x, find y: y on x. Given y, find x: x on y. This one step decides the whole question.

🧪 Try it yourselfTwo lines open and close like scissors

Take x̄ = 10, ȳ = 50, sx = 2 and sy = 6. The slider r is the correlation. The two lines always cross at (10, 50). When r is small the lines are far apart. When r gets close to 1 the two lines almost become one line. Try r = 0.5, then 0.9, then 1.

Worked examples

Example 1.1Both lines on the GDC

A farmer in Palapye tries different amounts of fertiliser x (kg per hectare) on 8 plots and records the maize yield y (tonnes per hectare).
x: 20, 30, 40, 50, 60, 70, 80, 90
y: 2.1, 2.9, 2.4, 3.6, 3.1, 4.4, 3.8, 4.6

(a) Find the regression line of y on x.
(b) Find the regression line of x on y.
(c) The farmer wants a yield of 4 tonnes per hectare. Estimate the fertiliser needed.

🧠 Think it through

Part (a) is the usual regression: X List = x, Y List = y.
Part (b): swap the lists. X List = y, Y List = x. The equation is then x = my + b.
Part (c): the yield (4) is given and the fertiliser is wanted, so use x on y.

✍️ By hand

(a)

y = 0.0327x + 1.56

(b)

x = 24.3y - 26.7

(c)

x = 24.3(4) - 26.7 = 70.5

🔢 On your TI-Nspire CX II

On Lists & Spreadsheet name column A fert and column B yld. Type the data.
Go to menu💻 click menu on keypad▸6: Statistics▸1: Stat Calculations▸3: Linear Regression (mx+b). For (a) set X List = fert, Y List = yld.
X List = fert, Y List = yld
y on x: y = 0.0327x + 1.56
For (b) run it again and swap: X List = yld, Y List = fert.
X List = yld, Y List = fert
x on y: x = 24.3y − 26.7
For (c) type the value into the saved equation: f1(4).

✅ Answer (a) y = 0.0327x + 1.56. (b) x = 24.3y − 26.7. (c) About 70.5 kg per hectare.

r is the same (0.892) for both lines. Using the wrong line, y = 0.0327x + 1.56 solved for y = 4, would give 74.5 kg per hectare: a different answer. 70.5 comes from the x-on-y line.

Example 1.2Which line should I use?

For 40 students, x is the hours of study per week and y the test mark (%). The two regression lines are
y on x: y = 4.2x + 31 x on y: x = 0.19y − 4.8

(a) Estimate the mark of a student who studies 6 hours.
(b) Estimate the hours a student should study to get 70%.
(c) A learner rearranges y = 4.2x + 31 to answer (b). What does he get, and why is it wrong?

🧠 Think it through

(a) Hours (6) is given, the mark is wanted: y on x.
(b) The mark (70) is given, the hours are wanted: x on y.
(c) Rearranging y on x gives $x = \dfrac{y - 31}{4.2}$ . That is a different line from x on y.

✍️ By hand

(a)

y = 4.2(6) + 31 = 56.2

(b)

x = 0.19(70) - 4.8 = 8.5

(c)

x = \dfrac{70 - 31}{4.2} = 9.29

🔢 On your TI-Nspire CX II

Check the three values.

✅ Answer (a) 56.2%. (b) 8.5 hours. (c) He gets 9.29 hours. It is wrong because y on x is built to predict y; rearranging it does not give the best line for predicting x.

The two answers are not the same (8.5 and 9.29), and the x-on-y answer is the one the IB mark scheme wants whenever the question asks you to estimate x.

Example 1.3Both lines by hand from summary statistics

A sample of students has x̄ = 18.5, ȳ = 62, sx = 4, sy = 12 and correlation r = 0.75.

(a) Find the covariance sxy.
(b) Find the equation of the regression line of y on x.
(c) Find the equation of the regression line of x on y.
(d) Use the correct line to estimate x when y = 80.

🧠 Think it through

Covariance: $s_{xy} = r\, s_x\, s_y$ .
y on x: gradient $\dfrac{s_{xy}}{s_x^{\,2}}$ . x on y: gradient $\dfrac{s_{xy}}{s_y^{\,2}}$ . Divide by the variance of the variable on the right of “on”.
Put the mean point into each line. Simplify to the form asked for.
y = 80 is given and x is wanted: x on y.

✍️ By hand
(a) 
(b) 
(c) 
(d) 

🔢 On your TI-Nspire CX II

Check by typing the numbers.

✅ Answer (a) 36. (b) y = 2.25x + 20.4. (c) x = 0.25y + 3. (d) x = 23.

Check that (18.5, 62) satisfies both: 2.25(18.5) + 20.375 = 62 and 0.25(62) + 3 = 18.5. If a mean point does not fit, you used the wrong variance.

Example 1.4Two lines given: find the mean point and r

The regression line of y on x is y = 0.8x + 12. The regression line of x on y is x = 0.45y + 1.

(a) Find the mean of x and the mean of y.
(b) Find the value of r.
(c) Find the value of $\dfrac{s_y}{s_x}$ .

🧠 Think it through

Both lines pass through (x̄, ȳ), so (x̄, ȳ) is where the two lines cross. Solve the two equations together.
Gradients: $m_1 = 0.8$ (y on x) and $m_2 = 0.45$ (x on y). Then $r^2 = m_1 m_2$ and r has the same sign as the gradients.
The gradient of y on x is also $r\dfrac{s_y}{s_x}$ . Use r from part (b).

✍️ By hand
(a) 
(b) 
 (gradients are positive)
(c) 

🔢 On your TI-Nspire CX II

Check part (a) by solving both equations at once.

✅ Answer (a) x̄ = 10, ȳ = 20. (b) r = 0.6. (c)

\dfrac{s_y}{s_x} = \dfrac{4}{3}

Exam question tip: if you are told which equation is which, identify them from the form: y = … is “y on x”, x = … is “x on y”. If the two lines are given in a mixed form such as 2y − 3x = 8, rearrange both so that the variable that is alone is on the left.

Example 1.5Predicting and judging reliability

A dealer in Francistown records the age x (years) and the value y (thousand pula) of 7 used cars.
x: 1, 2, 3, 4, 5, 6, 7
y: 140, 112, 121, 90, 95, 70, 76

(a) Find r and describe the correlation.
(b) Find the regression line of x on y.
(c) A buyer has P85 thousand. Estimate the age of car she should expect, and comment on reliability.
(d) Comment on using the line for a car worth P30 thousand.

🧠 Think it through

Linear Regression with X List = x and Y List = y gives r. The x-on-y line needs the lists swapped.
For (c): y = 85 is given. Is 85 inside the y-values 70 to 140?
For (d): 30 is far outside 70 to 140.

🔢 On your TI-Nspire CX II

Enter lists age and val. Run Linear Regression with X List = val and Y List = age.
X List = val, Y List = age
x on y: x = −0.0798y + 12.0
Type f1(85) and f1(30).

✅ Answer (a) r = −0.928: a strong negative linear correlation (older cars are worth less). (b) x = −0.0798y + 12.0. (c) About 5.24 years. This is interpolation (85 lies between 70 and 140) and r is strong, so it is reliable. (d) 9.63 years is extrapolation (30 is outside 70 to 140; the data only cover cars up to 7 years old), so the estimate is not reliable.

r does not change when you swap x and y. It is the same −0.928 either way. Only the gradient and intercept change.

Challenge questions

⭐ Challenge 1.6Summary statistics, no calculatorPaper 1 style

For a set of bivariate data, x̄ = 14, ȳ = 52, sx = 2.5, sy = 10 and r = −0.8.

(a) Find sxy.
(b) Find the equation of the regression line of x on y, giving your answer in the form x = ay + b.
(c) Estimate x when y = 60.
(d) Explain why the line in (b) should not be used to estimate y.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

$s_{xy} = r\, s_x\, s_y$ .
x on y: gradient $\dfrac{s_{xy}}{s_y^{\,2}}$ , through (14, 52).
Substitute y = 60.
Think: which variable is the line built to predict?

✍️ By hand
(a) 
(b) 
(c) 

✅ Answer (a) −20. (b) x = −0.2y + 24.4. (c) 12.4. (d) The x-on-y line is designed to estimate x from y. To estimate y from x you must use the y-on-x line.

The y-on-x gradient is

\dfrac{-20}{6.25} = -3.2

. Multiply the two gradients: (−0.2)(−3.2) = 0.64 = r². Good check.

⭐ Challenge 1.7Rainfall and river flowPaper 2 style

Rainfall x (mm) in a catchment area and the flow y (m³ per second) in a river were recorded in 8 months.
x: 40, 55, 62, 75, 88, 95, 110, 125
y: 12, 15, 19, 20, 27, 26, 33, 36

(a) Find r and describe the correlation.
(b) Write down the equation of the regression line of y on x and of x on y.
(c) Find the mean point and show that both lines pass through it.
(d) Estimate the rainfall needed for a flow of 30 m³ per second.
(e) Estimate the flow for 100 mm of rain.
(f) Explain why you should not use these lines to estimate the rain for a flow of 5 m³ per second.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Run Linear Regression twice, swapping the lists.
Mean point: Two-Variable Statistics, or average the data.
(d) flow given, rain wanted: x on y. (e) rain given, flow wanted: y on x.
(f) 5 is outside the y-range 12 to 36.

🔢 On your TI-Nspire CX II

Lists rain and flow. First run with X List = rain, Y List = flow, then swap.
y on x: y = 0.292x − 0.191
x on y: x = 3.35y + 2.42
Mean point: type mean(rain) and mean(flow), then test the lines.
Predictions from the x-on-y line and the y-on-x line.

✅ Answer (a) r = 0.989: very strong positive linear correlation. (b) y = 0.292x − 0.191 and x = 3.35y + 2.42. (c) (x̄, ȳ) = (81.25, 23.5); both lines give 23.5 and 81.25 at the mean point. (d) About 103 mm. (e) About 29.0 m³ per second. (f) 5 is outside the data range (12 to 36), so this is extrapolation. The line gives 19.2 mm, but the real relationship may bend or stop near zero flow.

Because r is so close to 1, the two lines are almost the same, and either would give nearly the same answer here. When r is weaker, as in Example 1 (r = 0.89), the choice of line matters much more.

Next: Lesson 2 · Standardisation & z-values

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Normal & regression 1 · The regression line of x on y

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

The regression line of x on y

Worked examples

Challenge questions