Coaching Centre
Interpolation, extrapolation and investigations
Making predictions from a least-squares line, the difference between interpolation and extrapolation, judging reliability, and planning a bivariate statistical investigation.
The whole point of a line of best fit is prediction. But a line doesn't know where the data stops. It will cheerfully predict that a student who studies $20$ hours a week scores $102$ out of $100$. Knowing when to trust a prediction is worth as many marks as making it.
Interpolation and extrapolation
Recall the study data from the last section: hours of study from $2$ to $13$, and the line $\text{mark} = 3.09 \times \text{hours} + 40.72$ with $r \approx 0.99$.
- Interpolation is predicting inside the range of the $x$-data. Here that means $2 \le x \le 13$.
- Extrapolation is predicting outside that range: $x \lt 2$ or $x \gt 13$.
Use $\text{mark} = 3.09 \times \text{hours} + 40.72$ to predict the mark for (a) $9$ hours and (b) $20$ hours. Comment on each prediction.
The more precise calculator line ($m = 3.0862$, $c = 40.718$) gives $68.49$ and $102.44$. Small differences like that come from rounding $m$ and $c$, and a marker accepts either.
How reliable is a prediction?
Check three things, and mention them in your answer:
- Interpolation or extrapolation? Is the $x$-value inside the data range?
- Strength of $r$. A strong $r$ means the points sit close to the line, so predictions are closer to reality.
- Does the answer make sense? Negative heights, marks over $100$ or a price below zero are a red flag.
Drag the vertical marker along the $x$-axis to choose a number of study hours. The prediction appears on the line. The shaded bands show where you are extrapolating.
Find the smallest number of hours where the line predicts more than $100$ marks. Then switch on Show what really happens. What does real data tend to do beyond the range?
Running a statistical investigation
The syllabus expects you to understand the steps of a full bivariate investigation, which is the same cycle you met in Year 11:
| Step | What you do | Example |
|---|---|---|
| 1. Pose a question | A question about two variables | "Do students who sleep more get faster reaction times?" |
| 2. Collect data | Decide the variables, the sample and how to measure | Survey $30$ students: average sleep and an online reaction test |
| 3. Analyse | Scatterplot, $r$, least-squares line | $r = -0.71$; $\text{time} = -18 \times \text{sleep} + 420$ |
| 4. Interpret and report | Answer the question in context, with limitations | Moderate negative association; small sample; can't prove cause |
A good report always includes limitations: sample size, how the sample was chosen, possible lurking variables, and how far the predictions can be trusted.
Your turn
A line of best fit for temperature ($x$, from $18$°C to $34$°C) against ice-creams sold per day ($y$) is $y = 12.5x - 140$, with $r = 0.92$. Predict the sales for $26$°C and for $5$°C, and comment on each.
Show the answer
At $26$°C: $12.5 \times 26 - 140 = 325 - 140 = 185$ ice-creams. This is interpolation with a strong association, so it is reliable.
At $5$°C: $12.5 \times 5 - 140 = 62.5 - 140 = -77.5$. This is extrapolation, and a negative number of sales is impossible. The prediction is not reliable.
The $x$-values in a dataset run from $5$ to $40$. Predicting $y$ when $x = 22$ is:
$22$ is between $5$ and $40$, inside the data, so this is interpolation.
Using the same data ($x$ from $5$ to $40$), which value of $x$ gives an extrapolation?
$52$ is greater than $40$, outside the data range. The others are inside it (including the end point $5$).
Using $y = 12.5x - 140$, predict $y$ when $x = 26$.
$12.5 \times 26 - 140 = 325 - 140 = 185$.
Which prediction is likely to be the most reliable?
The best case is both: inside the data range and a strong association, so the points sit close to the line where we predict.
A line predicts a negative price for a car $25$ years old. The data only covered cars up to $12$ years old. The best explanation is:
$25$ is well outside $0$ to $12$, so this is extrapolation. Real car values level off rather than falling below zero.
Using $\text{mark} = 3.09 \times \text{hours} + 40.72$, predict the mark for $9$ hours of study, to one decimal place.
$3.09 \times 9 + 40.72 = 27.81 + 40.72 = 68.53 \approx 68.5$.
In a statistical investigation, which belongs in the "interpret and report" step?
Reporting includes answering the question in context and stating limitations. Sampling and surveys belong to data collection, and $r$ belongs to analysis.
A survey of $8$ students finds $r = 0.91$ between phone use and screen-time anxiety. One limitation is:
With only $8$ students, a strong $r$ could happen by chance or reflect who was chosen. A larger sample would give more confidence.
Exam style: The line $\text{mark} = 3.09 \times \text{hours} + 40.72$ was found from data with hours between $2$ and $13$. Find the smallest whole number of hours for which the line predicts a mark over $100$.
Solve $3.09h + 40.72 \gt 100$: $3.09h \gt 59.28$, so $h \gt 19.18\ldots$. The smallest whole number is $20$ hours, which is well outside the data range, so this shows the danger of extrapolation.
Exam style: A line $\text{height} = 6.1 \times \text{age} + 77$ was fitted to children aged $2$ to $10$, with $r = 0.97$. A student predicts the height of a $35$-year-old. Which comment is best?
$6.1 \times 35 + 77 = 290.5$ cm, which is impossible. A strong $r$ only supports predictions inside the data range. This is extrapolation far beyond it.