Interpreting two-variable data – Two-Variable Statistics (Senior 6 Basic Mathematics) | Ganzaa

Two-Variable Statistics

Lessons

Senior 6 · Basic Mathematics

Two-Variable Statistics

4 lessons

Lesson 4 of 4

Interpreting two-variable data

25 min

From numbers to decisions

A potato cooperative near Musanze tested how much compost to apply. On eight plots of equal size it applied different masses of compost and weighed the harvest (illustrative data):

Compost, $x$ (kg)	0	10	20	30	40	50	60	70
Yield, $y$ (kg)	62	75	83	98	104	118	121	135

Summary values: $n = 8$ , $\sum x = 280$ , $\sum y = 796$ , $\sum x^2 = 14\,000$ , $\sum y^2 = 83\,568$ , $\sum xy = 32\,120$ .

All the tools of this unit now come together. The task in this lesson is to judge what the numbers do and do not allow us to say.

Worked example 1: a full analysis

Means. $\bar{x} = \frac{280}{8} = 35$ , $\bar{y} = \frac{796}{8} = 99.5$ .

Spread and covariance.

\sigma_x^2 = \frac{14\,000}{8} - 35^2 = 1750 - 1225 = 525

\sigma_y^2 = \frac{83\,568}{8} - 99.5^2 = 10\,446 - 9900.25 = 545.75

\operatorname{cov}(x, y) = \frac{32\,120}{8} - 35 \times 99.5 = 4015 - 3482.5 = 532.5

Correlation. $r = \frac{532.5}{\sqrt{525}\,\sqrt{545.75}} = \frac{532.5}{22.913 \times 23.361} = \frac{532.5}{535.27} \approx 0.995$ : a very strong positive linear correlation.

Regression of $y$ on $x$ . $b_{yx} = \frac{532.5}{525} = 1.014$ ; $y - 99.5 = 1.014 (x - 35)$ , so $y = 1.014 x + 64$ .

Meaning. The gradient says each extra kilogram of compost went with about 1.01 kg more potatoes. The intercept, 64 kg, estimates the yield of a plot with no compost (the real plot gave 62 kg).

Definition

Interpolation is predicting inside the range of the observed $x$ -values. Extrapolation is predicting outside that range.

Interpolation from a strongly correlated data set is usually reliable. Extrapolation assumes the straight-line pattern continues where no data were collected, so it can be badly wrong.

Example: the cooperative's data run from 0 to 70 kg. A prediction at 45 kg is interpolation; at 200 kg it is extrapolation.

Worked example 2: two predictions

At $x = 45$ : $y = 1.014 \times 45 + 64 = 109.6$ kg. This lies inside the data and close to the neighbouring plots (104 kg at 40 and 118 kg at 50), so it is a reasonable estimate.

At $x = 200$ : $y = 1.014 \times 200 + 64 \approx 267$ kg. This is extrapolation far beyond 70 kg. Crops do not respond to compost without limit: past some amount, yield levels off, and other limits (water, light, space) take over. The value should not be trusted.

Outliers

An outlier is a point far from the pattern of the others. Because $r$ and the regression line are built from means, one outlier can change them a lot.

Data used	$r$ (calculator)
the six points $(1, 2), (2, 4), (3, 5), (4, 4), (5, 6), (6, 7)$	$0.916$
the same six points plus $(12, 1)$	$- 0.254$

One point turned a strong positive correlation into a weak negative one. When you find an outlier, check whether it is a recording error. If it is genuine, report the analysis with and without it, and say why it is unusual.

Correlation and cause

A high $∣ r ∣$ shows that two variables move together. It does not prove that one causes the other. There are three other explanations to rule out:

A third (lurking) variable drives both. Across the districts of a country, the number of schools and the number of shops tend to rise together, because both grow with the population.
Reverse direction. The cause may run from $y$ to $x$ .
Coincidence, especially with few data points.

The compost study is stronger evidence of cause than most, because the cooperative chose how much compost each plot received and kept the plots alike in other ways: a controlled experiment, not just observation.

Crash Course Statistics #8: why correlation does not equal causation, with examples of third variables and of chance correlations. Watch for the difference between an experiment and an observational study.

Worked example

Worked example 3: evaluating a claim. A report finds $r = 0.82$ between the number of mobile-money agents in a sector and the average household income there, and concludes: "Opening more agents will make households richer." Evaluate this.

Answer. The correlation is strong and positive, so the two do tend to rise together. But the conclusion claims a cause. Agents set up where people already have more money to move (reverse direction), and towns have both more agents and higher incomes (third variable: urbanisation). The data support "richer sectors have more agents", not "agents make sectors richer". A test would need, for example, agents opened in some comparable sectors but not others.

Warning

Read $r$ with its sample size. With 4 or 5 points, a large $∣ r ∣$ can arise by chance. And $r$ near 0 does not mean "no relationship": look at the scatter diagram for a curve before you conclude.

Tip

An exam answer that interprets data does four things: states the direction and strength from $r$ ; explains the gradient in the units of the problem; says whether a prediction is interpolation or extrapolation; and refuses to claim a cause unless the data come from a controlled experiment.

✦

Practice

Using $y = 1.014 x + 64$ , predict the yield for 25 kg of compost. Is this interpolation or extrapolation?
Interpret the gradient of a regression line $y = - 0.07 x + 64.2$ , where $x$ is the price of tomatoes in Frw per kg and $y$ the mass sold in kg.
For the line in question 2, predict the mass sold at 1000 Frw per kg. Comment on your answer if the data ran from 200 to 600 Frw.
Over 20 years, the number of smartphones in a country and the number of people who own a car are strongly correlated. Give a likely third variable.
Two data sets have $r = 0.30$ and $r = - 0.85$ . Which shows the stronger linear relationship? Explain.
A scatter diagram shows ten points on a clear rising line and one point far below it. Describe two things you should do before reporting $r$ .

Show answersHide answers

$y = 1.014 \times 25 + 64 = 89.35 \approx 89$ kg. Interpolation, since 25 is between 0 and 70.
Each increase of 1 Frw per kg goes with about 0.07 kg less sold; an increase of 100 Frw per kg goes with about 7 kg less.
$y = -0.07 \times 1000 + 64.2 = -5.8$ kg, which is impossible. It is extrapolation far outside 200–600 Frw; the line cannot be used there.
Rising average income (or time and economic growth) drives both.
$r = - 0.85$ : strength depends on $∣ r ∣$ , and $0.85 > 0.30$ . The sign only gives the direction.
Check whether the point is a recording error; if it is real, compute $r$ with and without it and report both, explaining the unusual point.

Important

Interpolate, do not extrapolate. Look for outliers. A strong correlation is evidence of association, and only a controlled experiment is good evidence of cause.

1From numbers to decisions

2Outliers

3Correlation and cause

From numbers to decisions

Outliers

Correlation and cause