This note covers how to choose a suitable model for a situation or a data set, how to fit it with your GDC, and how to judge whether the result is believable. A model that fits the data well can still give nonsense outside the range of the data, so judging a model matters as much as building it.
Key ideas
Choosing a model. Start with the context and the shape of the data. Look at a scatter plot, then ask what the situation would do in real life.
- Linear, \(y = mx + c\): equal steps in \(x\) give equal changes in \(y\).
- Quadratic, \(y = ax^2 + bx + c\): the data rise then fall (or fall then rise) once, and the second differences are constant.
- Cubic, \(y = ax^3 + bx^2 + cx + d\): up to two turning points.
- Exponential, \(y = ka^x\): equal steps in \(x\) give equal ratios in \(y\). Growth is slow at first and then very fast, or decay levels off.
- Power, \(y = ax^b\): includes direct variation (\(y \propto x^n\)) and inverse variation (\(y \propto x^{-n}\)). Doubling \(x\) multiplies \(y\) by \(2^b\).
- Sinusoidal, \(y = a\sin(b(x - c)) + d\): the data repeat regularly.
To test between linear and exponential from a table with equally spaced \(x\)-values, compare differences (constant for linear) and ratios (constant for exponential). For a quadratic, the first differences change by a constant amount, so the second differences are constant.
Fitting with the GDC. Enter the data in two lists, choose the regression type (linear, quadratic, cubic, exponential, power or sine), and read off the parameters. Then draw the curve over the scatter plot. A good model passes close to the points with no obvious pattern in where it is above or below them. Keep full accuracy in the stored model, and round only your final answer.
Interpreting parameters. Always say what each parameter means in context, with units. For \(y = 4(1.5)^x\) with \(x\) in weeks, 4 is the starting value and 1.5 means a 50% increase each week. For a linear model, the gradient is the rate of change per unit of \(x\).
Using the model for predictions.
- Interpolation means predicting inside the range of the data. It is usually reliable.
- Extrapolation means predicting outside that range. It is much less reliable, because the pattern may change.
Limitations and domain. Each family has built-in limits. A quadratic with a negative leading coefficient eventually gives negative values, which is impossible for a count or a height. An exponential growth model grows without bound, but real populations run out of food or space. A sinusoidal model repeats forever, which only suits genuinely periodic situations. When you evaluate a model, state the domain and range in which it is sensible, compare model values with real values, and comment on whether the answers are reasonable.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Constant differences | Equal changes in \(y\) for equal steps in \(x\): linear model. |
| Constant ratios | Equal multiplier for equal steps in \(x\): exponential model \(y = ka^x\), with \(a\) equal to the ratio. |
| Constant second differences | The first differences themselves change by a constant: quadratic model. |
| \(y = ax^b\) | Power model. Find \(b\) from \(\dfrac{y_2}{y_1} = \left(\dfrac{x_2}{x_1}\right)^b\). |
| Regression | GDC method for finding the parameters of the chosen type of curve that best fits the data. |
| Interpolation | Prediction within the range of the data values. |
| Extrapolation | Prediction outside the range of the data values. Treat with caution. |
| Domain of validity | Inputs for which the model gives sensible outputs. |
Worked examples
Example 1 The number of followers (in hundreds) of a new channel is recorded each week:
Week \(x\): 0, 1, 2, 3, 4. Followers \(y\): 4, 6, 9, 13.5, 20.25.
(a) Decide which type of model suits the data. (b) Find the model. (c) Predict the followers in week 7 and comment.
(a) The differences are 2, 3, 4.5, 6.75, which are not constant. The ratios are \(\dfrac{6}{4} = 1.5\), \(\dfrac{9}{6} = 1.5\), \(\dfrac{13.5}{9} = 1.5\), \(\dfrac{20.25}{13.5} = 1.5\), which are constant. This points to an exponential model.
(b) Since the week-0 value is 4 and the ratio is 1.5, the model is \(y = 4(1.5)^x\). Exponential regression on the GDC gives the same result: \(a = 4.00\), \(b = 1.50\).
(c) \(y(7) = 4 \times 1.5^7 = 4 \times 17.0859 = 68.3\), so about 6830 followers.
Week 7 is outside the data (weeks 0 to 4), so this is extrapolation. Growth of 50% every week cannot continue for long, because the audience is limited, so the prediction may be too high.
Answer: \(y = 4(1.5)^x\); about 6830 followers in week 7, but this is unreliable.
Example 2 Daylight hours in a northern city, measured in the middle of selected months \(m\) (January is \(m = 1\)): \(m = 1\): 9.6 h; \(m = 6\): 15.1 h; \(m = 12\): 9.3 h. A model is \(L(m) = 2.9\sin\left(\dfrac{\pi}{6}(m - 3)\right) + 12.1\). (a) Compare the model with the data. (b) Explain why a sinusoidal model is suitable here, but a quadratic model would not be.
(a) With the GDC in radian mode:
\(L(1) = 2.9\sin\left(-\dfrac{\pi}{3}\right) + 12.1 = 2.9(-0.8660) + 12.1 = 9.59\) h, compared with 9.6 h.
\(L(6) = 2.9\sin\left(\dfrac{\pi}{2}\right) + 12.1 = 15.0\) h, compared with 15.1 h.
\(L(12) = 2.9\sin\left(\dfrac{3\pi}{2}\right) + 12.1 = 9.2\) h, compared with 9.3 h.
All three values are within 0.1 h of the data, so the model fits well.
(b) Daylight repeats every year, with period \(\dfrac{2\pi}{\pi/6} = 12\) months. A sinusoidal model repeats in the same way, so it can be used for later years. A quadratic model has only one turning point and would eventually go very high or very low, which is impossible for hours of daylight (it must stay between 0 and 24).
Answer: the model is within 0.1 h at each point. Sinusoidal suits periodic data, and a quadratic would give impossible values outside the data.
Example 3 A shop's monthly sales (in thousands of euros) in its first five months were fitted by \(S(t) = -1.2t^2 + 14t + 20\), where \(t\) is the number of months since opening. (a) Find \(S(3)\). (b) Find \(S(15)\) and comment. (c) Find the largest value of \(t\) for which the model gives non-negative sales.
(a) \(S(3) = -1.2(9) + 42 + 20 = 51.2\). Sales are about €51 200. Since \(t = 3\) lies within the data range, this is interpolation and is fairly reliable.
(b) \(S(15) = -1.2(225) + 210 + 20 = -40\). Negative sales make no sense, so the model has failed here. Also \(t = 15\) is extrapolation.
(c) Solve \(-1.2t^2 + 14t + 20 = 0\) on the GDC. The roots are \(t = -1.29\) and \(t = 12.95\ldots\) The negative root is outside the domain.
So the model gives non-negative sales only for \(0 \leq t \leq 12.9\), to 3 s.f. The model also predicts that sales peak at \(t = 5.83\) months (about €60 800) and then fall. This is just beyond the data, so it needs checking against real figures.
Answer: (a) €51 200; (b) \(-40\), impossible; (c) \(t \leq 12.9\) months.
Common errors
- Choosing a model only because it has a high-looking fit, without asking whether it makes sense in the context.
- Calling a prediction inside the data range extrapolation, or the other way round.
- Describing parameters without units or context, such as "\(b = 1.5\)" with no explanation of what it means for the situation.
- Rounding the model parameters before using them in later calculations, which makes the final answer inaccurate.
- Saying a model is "correct" or "wrong". A model is better or worse, and it is valid in a certain domain only.
Exam tip: When asked to comment on a prediction, give two things: whether it is interpolation or extrapolation, and one reason from the context why the pattern may or may not continue.
Check yourself
- A table gives \(x\): 1, 2, 3, 4 and \(y\): 6, 12, 24, 48. Which model is suitable, and what is it?
Answer
The differences are 6, 12, 24 (not constant), but the ratios are all 2. So the model is exponential, \(y = ka^x\) with \(a = 2\). At \(x = 1\), \(2k = 6\), so \(k = 3\). The model is \(y = 3(2)^x\). - The values of \(y\) for \(x = 0, 1, 2, 3, 4\) are 3, 6, 11, 18, 27. Show that a quadratic model is suitable and find it.
Answer
First differences: 3, 5, 7, 9. Second differences: 2, 2, 2, which are constant, so a quadratic is suitable. From \(x = 0\), \(c = 3\). Then \(a + b + 3 = 6\) and \(4a + 2b + 3 = 11\), so \(a + b = 3\) and \(2a + b = 4\). Hence \(a = 1\), \(b = 2\). The model is \(y = x^2 + 2x + 3\). - A model is fitted to data collected for \(2 \leq x \leq 10\). Is a prediction at \(x = 6\) interpolation or extrapolation? What about \(x = 25\)?
Answer
\(x = 6\) is inside the data range, so it is interpolation. \(x = 25\) is outside it, so it is extrapolation and is less reliable. - A power model \(y = ax^b\) passes through \((2, 12)\) and \((4, 48)\). Find the model and predict \(y\) when \(x = 5\).
Answer
\(\dfrac{48}{12} = \left(\dfrac{4}{2}\right)^b\), so \(2^b = 4\) and \(b = 2\). Then \(12 = a(2)^2\), so \(a = 3\). The model is \(y = 3x^2\), and \(y(5) = 75\).