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4.10 Regression line of x on yIB Maths: Analysis and Approaches SL: Revision notes

Section 1

Why are there two regression lines?

The regression line of yy on xx is the least squares line that minimises the squared vertical distances from the points to the line. It is built to predict yy from a given xx.

The regression line of xx on yy, written x=cy+dx=cy+d, minimises the squared horizontal distances. It is built to predict xx from a given yy.

The two lines are different unless the correlation is perfect (r=±1r=\pm1). The weaker the correlation, the further apart they are. So an xx on yy line cannot reliably predict yy from xx, and a yy on xx line cannot reliably predict xx from yy.

Key termsregression line of x on yleast squares
Common mistake

Rearranging y=ax+by=ax+b to make xx the subject and calling it the xx on yy line. It is a different line unless r=±1r=\pm1.

Exam tip

Ask 'which variable am I predicting?' The variable you want goes on the left-hand side of the line you use.

Section 2

Finding the x on y line with technology

On a GDC, enter the data in two lists and run linear regression with the lists swapped: yy as the independent (explanatory) list and xx as the dependent list. The calculator returns an equation in its own letters. Rewrite it as x=cy+dx=cy+d.

For example, for sleep xx hours and reaction time yy ms, the calculator gives x=−0.0624y+24.2x=-0.0624y+24.2, while the yy on xx line is y=−13.3x+369y=-13.3x+369. Give coefficients to 3 significant figures, but keep full accuracy in the calculator for predictions.

Key termsindependent variable
Common mistake

Reporting the equation as 'y=−0.0624x+24.2y=-0.0624x+24.2' because the calculator uses yy for its output. The line must be written as x=…y+…x=\ldots y+\ldots.

Example

Data (2,5),(4,9),(6,10),(8,16),(10,20)(2,5),(4,9),(6,10),(8,16),(10,20): x=0.521y−0.254x=0.521y-0.254 and y=1.85x+0.9y=1.85x+0.9.

Section 3

Both lines pass through the mean point

Every least squares regression line passes through the mean point (xˉ,yˉ)(\bar{x},\bar{y}). So:

  • the yy on xx and xx on yy lines intersect at the mean point, and solving them simultaneously gives xˉ\bar{x} and yˉ\bar{y};
  • if you know one mean and one line, substitute to find the other mean;
  • an unknown data value can be found from a given mean, e.g. 5+9+p+16+205=12\frac{5+9+p+16+20}{5}=12 gives p=10p=10.

With y=2x+10y=2x+10 and x=0.4y−2x=0.4y-2: x=0.4(2x+10)−2x=0.4(2x+10)-2, so 0.2x=20.2x=2, giving x=10x=10 and y=30y=30. The mean point is (10,30)(10,30).

Key termsmean point
Exam tip

To verify that a line passes through the mean point, substitute one mean and show that you get the other. Keep full calculator accuracy so the check comes out exactly.

Section 4

Prediction and reliability

A prediction from a regression line is more reliable when:

  • the correct line is used (xx on yy to predict xx);
  • the correlation is strong (∣r∣|r| close to 1);
  • the given value lies within the range of the data (interpolation).

Predicting outside the data range is extrapolation. The linear pattern may not continue there, so the estimate is unreliable. For example, the reaction times in the data run from 248 ms to 305 ms, so estimating sleep for a reaction time of 340 ms is extrapolation. In an exam comment, refer to both the value of rr and the range of the data.

Key termsinterpolationextrapolation
Common mistake

Checking the range of the wrong variable. For the xx on yy line, check that the given yy lies within the range of the yy-data.

Exam tip

A strong correlation does not rescue an extrapolation. Both conditions matter.

Must know

  • yy on xx predicts yy from xx; xx on yy predicts xx from yy. Never swap them.
  • Find x=cy+dx=cy+d on the GDC with yy as the independent list; do not rearrange the yy on xx line.
  • Both lines pass through (xˉ,yˉ)(\bar{x},\bar{y}), so they intersect there.
  • The lines are the same only when r=±1r=\pm1.
  • Predictions are reliable only with strong correlation and interpolation.

That's the notes covered.

Carry on to the next subtopic.