The mean point

Every regression line passes through (x̄, ȳ)

In 6 of 20 recent exams
Paper 1 or 2
1 to 3 marks

How it is set in the exam

Paper 2 (calculator) mostly (8 of 9 logged parts), 1–3 marks a part; once on Paper 1 (no calculator) with both lines given. Common (6 of 20 exam versions). It is often a small part inside a longer regression question, and twice it led into a missing data value.

Spot it in the exam

Exam phrases to spot (5)

“find the mean of $x$ and the mean of $y$ ”
“show that the mean of $y$ is …”
“write down the mean point”
“draw the regression line on the scatter diagram”
“the two regression lines intersect at …”

In disguise (3)

"the lines $L_1$ and $L_2$ pass through the same point $(p, q)$ ; find $p$ and $q$ ": that point is $(\bar{x}, \bar{y})$
a line of best fit "by eye": it must go through $(\bar{x}, \bar{y})$
"a line through the origin and the mean point": its gradient is $\frac{\bar{y}}{\bar{x}}$

Variants (2)

given: both lines; one line and one mean (the mean itself found from the table first); the data
asked: both means, one mean as a "show that", the line drawn on a scatter diagram, a line through the origin and the mean point

Looks like it, but isn't (2)

both lines are given and an estimate is asked for: Predict x from y
the question asks for a value missing from the table: Work backwards from the mean or median. With the regression line given, this type is only its first step, the mean from the line (Example 3.2)

Worked example

Two lines given: where they cross

Paper 1 (no calculator)

Question

For a group of trees, $x$ is the height in metres and $y$ is the trunk diameter in centimetres. The regression line of $y$ on $x$ is $y = 0.5x + 20$ , and the regression line of $x$ on $y$ is $x = 1.2y - 16$ .
(a) Find the mean height and the mean trunk diameter of the trees. [3]
(b) The heights of the trees add up to $1000$ m. Find the number of trees in the group. [2]

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Which method?

Given	First move
both regression lines	solve them simultaneously: they cross at $(\bar{x}, \bar{y})$
one line and $\bar{x}$ (or $\bar{y}$ )	substitute the known mean into the line and solve for the other mean (exact, because $(\bar{x}, \bar{y})$ lies on every regression line: not the rearrangement that Predict x from y forbids)
the raw data	the means from the GDC (one-variable statistics on each list, or the regression output)

Steps

Say it:
every regression line passes through the mean point $(\bar{x}, \bar{y})$ . Writing this (or sketching the two lines crossing) is the recognition mark.
Both lines:
substitute one into the other and solve (by hand on ; simultaneous linear equations on ). Label the answers $\bar{x} = \ldots$ and $\bar{y} = \ldots$ , in context.
One line and one mean:
substitute the mean you know. If it isn't given, find it from the complete list first.
A missing value:
total $= n \times$ mean, then subtract the known values (Work backwards). Never put the missing point's $x$ into the line: the line need not pass through any data point, only through $(\bar{x}, \bar{y})$ .
Draw the line:
plot $(\bar{x}, \bar{y})$ and a second point of the line far from it (the $y$ -intercept, or the value at the far end of the axis), and rule a straight line through both across the data.
A line through the origin and the mean point:
its gradient is $\frac{\bar{y}}{\bar{x}}$ (rise over run from $(0, 0)$ ), from the unrounded means. Write $y = \ldots x$ , with no constant.

Common mistakes

Substituting the missing point's $x$ into the line
to find a missing $y$ : the line gives an estimate, not the recorded value. It scores nothing (Example 3.2 would give $10.4$ instead of $9.7$ ).
Taking $\bar{y}$ from the incomplete table
(dividing the known values by $7$ , or by $6$ ).
Rounding a mean before using it
carry $\bar{x}$ and $\bar{y}$ unrounded into the next part.

More mistakes (4)

Unlabelled means
in (a) of a two-lines question: say which is $\bar{x}$ and which is $\bar{y}$ , in context.
Drawing freehand,
through the wrong point, or with two points too close together to fix the line's direction.
A line through the origin with the gradient upside down
it is $\frac{\bar{y}}{\bar{x}}$ , rise over run.
Reading $x = cy + d$ as $y = cx + d$
when solving the two lines (or typing it into a graph screen as $y = cx + d$ ): keep each line's subject and substitute one into the other. In Example 3.1 the misread lines give $0.5x + 20 = 1.2x - 16$ , $x = 51.4$ instead of $20$ .

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