IB Applications and Interpretation SL · Regression and linear correlation · Section 1 of 4

Bivariate distributions and scatter diagram

3 worked examples · By Álvaro López

Key idea

When you measure two things for each individual (hours of study and mark, temperature and sales), you get data in pairs $(x, y)$ . The first step is to plot them on a scatter diagram and summarise them with a few numbers: the mean and standard deviation of each variable, and the mean point $(x ˉ, y ˉ)$ . They are the basis of the whole topic: correlation and the regression line are built on them. In the IB exam almost everything comes from the graphic display calculator (GDC); here you learn what to ask it for and what its output means.

Paired data and scatter diagram

A bivariate statistical variable is a set of pairs $(x_{1}, y_{1}), (x_{2}, y_{2}), \dots, (x_{n}, y_{n})$ : each pair is the two measurements of the same individual. The variable placed on the horizontal axis, $x$ , is called the independent variable (or explanatory variable), and the one on the vertical axis, $y$ , is the dependent variable (or response variable). For example, if you study how hours of study $x$ affect the mark $y$ , the mark is the one that depends.

What the cloud shows is its trend: whether the points rise from left to right, fall, cluster around a line or are scattered with no order. That is studied in the next section; here it is enough to know how to draw it and calculate its parameters.

Scatter diagram of five points (2, 4), (3, 5), (5, 6), (6, 9) and (9, 11) with an increasing trend; the mean point (x̄, ȳ) = (5, 7) is marked in orange

When there are many individuals with the same values, the data are given in a frequency table with three columns: $x_{i}$ , $y_{i}$ and the frequency $f_{i}$ of the pair $(x_{i}, y_{i})$ , with $\sum f_{i} = N$ the total number of individuals. If there are no repeats, $f_{i} = 1$ in every row and $N = n$ .

Mean and standard deviation of each variable

Each variable is studied separately with the measures you already know from descriptive statistics.

In the IB the standard deviation used is that of the whole population, which the GDC shows as $σ_{x}$ . Do not confuse it with the sample one, $s_{x}$ , which divides by $N - 1$ instead of by $N$ .

Covariance

Means and standard deviations measure each variable separately. What is missing is a number that tells you how the two move together.

Its sign indicates the trend:

$s_{x y} > 0$ : when $x$ is above its mean, $y$ tends to be above its mean too (the cloud rises).
$s_{x y} < 0$ : when $x$ is above its mean, $y$ tends to be below its mean (the cloud falls).
$s_{x y} \approx 0$ : there is no clear linear trend.

Covariance depends on the units, so its size does not say whether the relationship is strong or weak. For that, the next section uses the correlation coefficient, which is obtained by dividing the covariance by $σ_{x} σ_{y}$ .

The mean point

This point is very important: the regression line always passes through it. So, if you are given the mean of one variable and one value is missing, you can use the mean point to recover it.

Worked examples

Example 1 Basic

Advertisements and sales

A shop records, over five weeks, the number of advertisements published ( $x$ ) and the sales of the week in hundreds of euros ( $y$ ):

Advertisements $x$	2	3	5	6	9
Sales $y$ (hundreds of €)	4	5	6	9	11

a) Find the mean point $(x ˉ, y ˉ)$ .

b) Find the standard deviation $σ_{x}$ of the advertisements.

c) Find the standard deviation $σ_{y}$ of the sales.

d) Calculate the covariance $s_{x y}$ .

Solution

a) There are $N = 5$ pairs. The sums are $\sum x_{i} = 25$ and $\sum y_{i} = 35$ , so $x ˉ = \frac{25}{5} = 5 y ˉ = \frac{35}{5} = 7$ The mean point is $(5, 7)$ .

b) With $\sum x_{i}^{2} = 155$ : $σ_{x} = \frac{155}{5} - 5^{2} = 31 - 25 = 6 = 2.45$

c) With $\sum y_{i}^{2} = 279$ : $σ_{y} = \frac{279}{5} - 7^{2} = 55.8 - 49 = 6.8 = 2.61$

d) With $\sum x_{i} y_{i} = 206$ : $s_{x y} = \frac{206}{5} - 5 \cdot 7 = 41.2 - 35 = 6.2$ It is positive: the more advertisements, the more sales.

Example 2 Medium

Hours of study and mark, with frequencies

10 students were asked how many hours they studied for a test ( $x$ ), and their mark out of 10 ( $y$ ) was recorded. The results are summarised as follows:

Hours $x$	Mark $y$	Students $f$
1	2	2
2	3	3
3	5	3
4	6	2

a) Calculate $x ˉ$ and $y ˉ$ .

b) Calculate $σ_{x}$ and $σ_{y}$ .

c) Calculate the covariance and interpret its sign.

Solution

The data are organised with $N = \sum f = 10$ . The weighted sums are:

$\sum x_{i} f_{i} = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 3 + 4 \cdot 2 = 25$
$\sum x_{i}^{2} f_{i} = 1 \cdot 2 + 4 \cdot 3 + 9 \cdot 3 + 16 \cdot 2 = 73$
$\sum y_{i} f_{i} = 2 \cdot 2 + 3 \cdot 3 + 5 \cdot 3 + 6 \cdot 2 = 40$
$\sum y_{i}^{2} f_{i} = 4 \cdot 2 + 9 \cdot 3 + 25 \cdot 3 + 36 \cdot 2 = 182$
$\sum x_{i} y_{i} f_{i} = 1 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 + 3 \cdot 5 \cdot 3 + 4 \cdot 6 \cdot 2 = 115$

a) $x ˉ = \frac{25}{10} = 2.5$ hours and $y ˉ = \frac{40}{10} = 4$ points.

b) $σ_{x} = \frac{73}{10} - 2. 5^{2} = 1.05 = 1.02 σ_{y} = \frac{182}{10} - 4^{2} = 2.2 = 1.48$

c) $s_{x y} = \frac{115}{10} - 2.5 \cdot 4 = 11.5 - 10 = 1.5$ The covariance is positive: students who study more hours than the mean tend to get a higher mark than the mean.

Example 3 Hard

A missing value and a new point

A café records the maximum temperature of the day $x$ (in °C) and the number of hot chocolates sold $y$ (in hundreds) over five days:

$x$ (°C)	1	2	4	5	8
$y$ (hundreds)	6	5	4	2	$k$

It is known that the mean point of the data is $(4; 3.6)$ .

a) Find the value of $k$ .

b) Calculate the covariance $s_{x y}$ of the five days and interpret its sign.

c) A sixth day coincides exactly with the mean point, that is, $(4; 3.6)$ . State the new mean point and calculate the new covariance.

d) Explain why the covariance has changed even though the mean point is the same.

Solution

a) The mean point tells us that $y ˉ = 3.6$ . With $N = 5$ , the sum of the sales is $5 \cdot 3.6 = 18$ : $6 + 5 + 4 + 2 + k = 18 \Rightarrow 17 + k = 18 \Rightarrow k = 1$ (The mean of $x$ confirms the data: $\frac{1 + 2 + 4 + 5 + 8}{5} = 4$ .)

b) With $k = 1$ the products are $1 \cdot 6 = 6$ , $2 \cdot 5 = 10$ , $4 \cdot 4 = 16$ , $5 \cdot 2 = 10$ and $8 \cdot 1 = 8$ , which sum to $50$ : $s_{x y} = \frac{50}{5} - 4 \cdot 3.6 = 10 - 14.4 = - 4.4$ The covariance is negative: the hotter it is, the fewer hot chocolates are sold. This is what we would expect.

c) When a value equal to the mean is added, the mean does not change: the mean point is still $(4; 3.6)$ . Now $N = 6$ and the sum of the products is $50 + 4 \cdot 3.6 = 64.4$ : $s_{x y} = \frac{64.4}{6} - 4 \cdot 3.6 = \frac{64.4}{6} - 14.4 = - \frac{11}{3} = - 3.67$

d) The sum $\sum (x_{i} - x ˉ) (y_{i} - y ˉ) = - 22$ does not change, because the new day contributes a zero deviation in both variables ( $0 \cdot 0 = 0$ ). But now that sum is shared among 6 days instead of 5, so the mean of the products, $- \frac{22}{6}$ , is smaller in absolute value than $- \frac{22}{5}$ .

In the exam: when the mean or the mean point is given, use it to find the unknown value by setting up the equation for the mean; without that set-up, a correct value with no justification may lose marks. Write the covariance formula with the numbers substituted, and in interpretations link the sign to the context (heat and sales), not only say "it is negative".

Common mistakes

Summary

A bivariate distribution is data in pairs $(x_{i}, y_{i})$ , with frequencies $f_{i}$ and $N = \sum f_{i}$ . It is plotted on a scatter diagram, with $x$ (independent) on the horizontal axis and $y$ (dependent) on the vertical axis.
$x ˉ = \frac{\sum x _{i} f _{i}}{N}$ and $y ˉ = \frac{\sum y _{i} f _{i}}{N}$ ; $σ_{x} = \frac{\sum x _{i}^{2} f _{i}}{N} - x ˉ^{2}$ , and similarly $σ_{y}$ . In the IB the population standard deviation $σ$ is used.
Covariance: $s_{x y} = \frac{\sum x _{i} y _{i} f _{i}}{N} - x ˉ y ˉ$ . Its sign indicates the trend (positive: the cloud rises; negative: it falls); its size depends on the units.
The mean point $(x ˉ, y ˉ)$ is the centre of the cloud, and the regression line passes through it.
With the GDC: two lists ordered in pairs (plus one of frequencies if needed) and two-variable statistics; check $N$ and do not mix $σ$ with $s$ .

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