Two-variable data: models and scatterplots
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Separate observed values, fitted models, and predictions.
The central idea: observations and models are not the same object
A scatterplot represents paired observations (x, y). Read both axes, their units, and their scales. A dot at (5, 18) means an observed input of 5 was paired with an observed output of 18; it does not mean a fitted model necessarily predicts 18 at that input.
Describe direction, form, and strength
An association is positive when larger inputs tend to accompany larger outputs, and negative when they tend to accompany smaller outputs. Its form may be approximately linear or curved. Strength concerns how closely the points follow a pattern, not whether a line is steep. A strong nonlinear relationship can fail to look like a strong linear one.
Association alone does not establish causation. Another variable, selection effects, or reverse direction can help explain a pattern. Section 7 supplies the study-design reasoning needed for a causal conclusion.
Interpret a line in the language of the variables
For
If x is years since 2020, zero refers to 2020, not to the beginning of the object’s entire life. If x
Read a drawn model line, not two arbitrary observations
To find the equation of a displayed line of best fit, select two clear points on that line. They need not be data points. Compute the slope and then the intercept. If using regression from a table, the best-fit calculation uses all supplied observations; connecting two arbitrary observations is not generally the same procedure.
Three quantities to keep separate
The input is x. The model prediction is . The observed response is y. To predict, substitute x into the model. To evaluate the model at an observation, compare y with rather than replacing one with the other.
Select a model by its pattern of change
The official domain includes linear, quadratic, and exponential models for two-variable data.[2] Exact differences and ratios are especially useful for tables with equally spaced inputs.
Family | Model form | Pattern at equal input steps |
|---|---|---|
Linear | y | Constant first differences. |
Quadratic | y | Constant nonzero second differences. |
Exponential | y | Constant output ratios when values are nonzero. |
For real measured data, look for approximate rather than perfectly exact patterns. A finite table can be fit by other complicated formulas too; use the stated model family or the choices and the context rather than asserting an infinite rule from a few points alone.
Exponential time intervals matter
If a quantity multiplies by b every h input units, write y
Regression is fitting, not interpolation through every point
A least-squares line minimizes the sum of squared vertical residuals. Desmos can fit a line from a table using a tilde relation such as y_1 ~ m x_1 + b.[7] You should still identify the correct columns, units, and model family. The fitted prediction at an observed input can differ from the observed output. Round only after using sufficient parameter precision.
Residuals diagnose a mismatch
A residual is y−: observed minus predicted. A positive residual means the observation lies above the model, so the model underpredicts there. A negative residual means the model overpredicts. Residuals near zero indicate close local fit; a systematic curved residual pattern suggests that a line is missing structure.
Residual language and calculator regression are supporting tools in this guide, not a demand to memorize a new statistical formula sheet. The central skill is connecting data, equations, and predictions.
Decide how far the model can be trusted
Interpolation and extrapolation
Interpolation predicts within the observed input range. Extrapolation predicts outside it. An equation can return a numerical result far beyond the data without supplying evidence that the real relationship continues that far. Even interpolation can be uncertain when data are noisy or the model is inappropriate.
The proper response is not “extrapolation is always wrong.” It is that the proposed prediction requires an additional assumption that the modeled pattern continues. The farther the input lies outside the observed range, the more important that assumption can become.
Do not let a graph’s appearance replace its scale
Check axis starts and tick spacing. A bar chart that begins near the observed values can exaggerate a modest change. Calculate the change from the numerical values, not from the apparent ratio of bar heights above a truncated baseline. An axis need not always start at zero, but the visual must be interpreted with its actual scale.
Similarly, a visually steep line can have a small numerical slope if axes use different scales. Slope is a quotient of coordinate differences, with units, not an angle measured from a printed picture.
Compare models at the inputs the question permits
When comparing a linear and an exponential model, compute or graph both under the stated domain. If the question asks for the first whole-number year, an intersection near a fractional year is not itself the answer. Check the neighboring integer inputs and any initial equality. When a model represents a count, the equation may predict a noninteger; interpret or round only as the question directs.
A model‐reading routine
Read the axes. Identify whether a value is observed or predicted. Name the model family. Interpret the parameters with units. Substitute the requested input. Finally, ask whether the input and conclusion lie within the model’s intended scope.
What to practice next
Examples 4.01–4.06 develop scatterplot and parameter interpretation; 4.07–4.12 connect residuals and model families; and 4.13–4.15 address extrapolation, misleading displays, and model comparisons.
15 worked examples
Try the question before reading the solution. The examples progress from Foundation to Challenge.
4.01. Translate a plotted point into its context
The scatterplot shows observed seedling heights at different ages. What does the point (4, 18) represent?
A. A seedling was 4 millimeters tall at age 18 days.
B. A seedling was 18 millimeters tall at age 4 days.
C. Every seedling grows 18 millimeters in 4 days.
D. The predicted height at age 18 days is 4 millimeters.
Show worked solutionHide worked solution for example 4.01
Recognize the structure. Read the horizontal axis first, then the vertical axis. An observed point is not a universal growth rule.
Work it through. The horizontal coordinate 4 is an age in days. The vertical coordinate 18 is a height in millimeters. The point therefore records a seedling with height 18 mm at age 4 days. No rate or future prediction is established by this one point.
Answer: B.
Check. The axis labels supply both the variable meanings and their units. Reversing the coordinates reverses the interpretation.
Avoid the trap. A point describes an input-output pair. It does not mean that every seedling has that height, nor does it give the amount of height gained since birth.
4.02. Describe association without claiming causation
The scatterplot relates a machine’s age to its resale value for six machines. Which description best fits the data?
A. A roughly linear positive association.
B. A roughly linear negative association.
C. No apparent association.
D. An association proving that age alone determines resale value.
Show worked solutionHide worked solution for example 4.02
Recognize the structure. Describe direction and shape. Keep causal or deterministic claims separate.
Work it through. As age increases from left to right, resale value generally decreases. The points follow a roughly straight downward pattern, so the association is negative and approximately linear. The graph does not rule out other influences such as condition or model.
Answer: B.
Check. Larger horizontal values correspond to smaller vertical values, matching the sign of a negative slope.
Avoid the trap. A strong-looking association does not prove that one variable alone determines the other. Association is a description of paired data, not a complete causal explanation.
4.03. Use points on the model line
A scatterplot and a fitted line are shown. The labeled points (2, 14) and (8, 32) lie exactly on the fitted line. What value does the line predict when x = 6?
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Recognize the structure. Use the supplied points on the line, not nearby observed points that may lie off it.
Work it through. The slope is (32 − 14)/(8 − 2)
Answer: 26.
Check. Six is four units to the right of 2. The model therefore rises 4(3)
Avoid the trap. A fitted line is not required to pass through every observed point. Picking two arbitrary dots would generally produce a different line.
4.04. Evaluate a model and keep its units
For a particular data set, the model = 52 − 1.8x predicts temperature T, in degrees Celsius, at altitude x, in kilometers. What temperature does the model predict at an altitude of 10 kilometers?
Show worked solutionHide worked solution for example 4.04
Recognize the structure. The input is altitude; substitute the requested value for x. The hat denotes a prediction.
Work it through. At x
The output is in degrees Celsius because that is the model’s response unit.
Answer: 34 degrees Celsius.
Check. An altitude increase of 10 km corresponds to a modeled decrease of 18 degrees from the intercept value 52.
Avoid the trap. The number 1.8 is not the predicted temperature. Also, a model prediction is not a guarantee that an individual measurement will equal it.
4.05. Interpret a slope as a rate with two units
The model = 18 + 2.4t predicts a tank’s volume V, in liters, after t hours. What does 2.4 represent?
A. The predicted volume at time zero.
B. The predicted increase in volume, in liters, for each additional hour.
C. The number of hours needed to reach 18 liters.
D. A 2.4% increase in volume each hour.
Show worked solutionHide worked solution for example 4.05
Recognize the structure. The coefficient of the input is the change in predicted output per unit increase in input.
Work it through. Increasing t by 1 increases 18 + 2.4t by 2.4. Since output is measured in liters and input in hours, the rate is 2.4 liters per hour. It is an additive rate, not a percentage.
Answer: B.
Check. Predictions at times 0 and 1 are 18 and 20.4 liters; their difference is 2.4 liters.
Avoid the trap. A slope needs both units. “2.4 liters” alone is a change amount without the time interval that makes it a rate.
4.06. Interpret the intercept at the correct time origin
The equation = 3.2d + 18 models plant height H, in centimeters, where d is the number of days since an experiment began. What does 18 represent?
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Recognize the structure. An intercept is the predicted output at input zero. The context defines what zero means.
Work it through. When d
Answer: The predicted plant height, 18 cm, at the start of the experiment.
Check. The term 3.2d is zero at the stated origin, leaving only the intercept.
Avoid the trap. Do not automatically interpret an intercept as a physical “starting amount” before reading how the variable is defined. A shifted year or day variable shifts the meaning of zero.
4.07. Distinguish observed, predicted, and residual values
A fitted model is = 48 + 3x. One observation has x = 6 and y = 63. The residual is defined as observed value minus predicted value. What is its residual?
Show worked solutionHide worked solution for example 4.07
Recognize the structure. Find the prediction first; then subtract it from the observed output.
Work it through. The model predicts 48 + 3(6)
residual = y − = 63 − 66 = −3.
The negative residual means the observation lies 3 units below the model, so the model overpredicts by 3.
Answer: −3.
Check. Observed output equals predicted output plus residual: 66 + (−3)
Avoid the trap. Using 66−63 reverses the residual definition. A negative residual does not mean the observed output itself is negative.
4.08. Use residual structure to detect a poor linear fit
A linear model has the following residuals, using observed minus predicted values:
x
0
1
2
3
4
Residual
4
−1
−3
−1
4
Which observation most strongly suggests that a curved model may fit better?
A. Some residuals are negative.
B. The residuals follow a curved pattern: positive at both ends and negative in the middle.
C. The input values are integers.
D. The residuals are not all equal.
Show worked solutionHide worked solution for example 4.08
Recognize the structure. Look for a systematic pattern, not merely for nonzero residuals.
Work it through. The model underpredicts at both ends and overpredicts in the middle. This organized U-shaped residual pattern suggests curvature remains unexplained by the line. A curved model may capture that structure. Residuals of mixed signs or different sizes, by themselves, are normal for imperfect data.
Answer: B.
Check. Adding the residuals to the model’s predictions would raise the ends and lower the middle, producing a curved adjustment.
Avoid the trap. A few residuals do not prove one unique model is correct. The conclusion is that a curve is worth considering, not that the data establish a universal quadratic law.
4.09. Fit a line using all the data
The data are shown below. Using the least-squares linear model fitted to all five points, predict y when x = 3.
x
0
1
2
3
4
y
2
5
5
8
10
Show worked solutionHide worked solution for example 4.09
Recognize the structure. A regression model uses all observations. Its prediction need not equal the observed value at the same input.
Work it through. Enter the pairs in a calculator table and fit y1 ∼ mx1 + b. The fitted coefficients are m
The recorded value at x
Answer: 7.9.
Check. The mean point is (2, 6), and 1.9(2) + 2.2
Avoid the trap. A line through the first and last points is not automatically the least-squares line. Also, copying the observed 8 does not answer a model-prediction question.
Another route. For verification, the centered products sum to 19 and the squared input deviations sum to 10, giving slope 19/10
4.10. Compare constant differences with constant ratios
Two sequences are recorded at equally spaced times:
Time
0
1
2
3
A
80
90
100
110
B
80
100
125
156.25
Which description fits these values?
A. Both are linear.
B. A is linear; B grows exponentially by 25% per time unit.
C. A grows exponentially; B is linear.
D. Both grow exponentially by 10% per time unit.
Show worked solutionHide worked solution for example 4.10
Recognize the structure. Use differences for a linear pattern and ratios for an exponential pattern, after checking equal input spacing.
Work it through. A increases by 10 each time, matching A(t)
Answer: B.
Check. The exponential increases are 20, 25, and 31.25, not a constant difference. They become larger as the base amount grows.
Avoid the trap. An increasing sequence is not automatically exponential. Also, unequal time gaps would require adjusting the differences or factors to comparable intervals.
4.11. Respect the growth interval in a table
A quantity follows an exponential model. Its values are given in the table. What does the model predict at t = 8?
t
0
2
4
6
Quantity
120
180
270
405
Show worked solutionHide worked solution for example 4.11
Recognize the structure. The factor between adjacent entries applies to two time units, not one.
Work it through. Each two-unit interval multiplies the quantity by 1.5. From t
Q(t)
Answer: 607.5.
Check. At t
Avoid the trap. 120(1.5)t doubles the number of growth periods. The exponent must count the intervals associated with the stated factor.
4.12. Recognize a quadratic pattern without overfitting
A quadratic model fits all four entries in the table exactly. Which model is it?
x
0
1
2
3
y
3
6
13
24
A. y = 3x + 3 B. y = 3(2)x
C. y = 2x2 + x + 3 D. y = x2 + 2x + 3
Show worked solutionHide worked solution for example 4.12
Recognize the structure. The input steps are equal. Constant second differences support the stated quadratic structure.
Work it through. First differences are 3, 7, 11; second differences are 4, 4. For y
Answer: C: y
Check. At x
Avoid the trap. The task states the model is quadratic. Without a model assumption, finitely many points do not determine how an arbitrary relationship behaves everywhere.
4.13. Separate a computable prediction from a reliable one
A linear model was fitted using tree ages from 4 to 12 years. A researcher uses it to predict the height of a 40-year-old tree. Which statement is best supported?
A. The prediction must be correct because a line can be evaluated at 40.
B. The prediction is interpolation and is therefore exact.
C. The prediction is extrapolation and may be unreliable because the relationship may change outside the observed age range.
D. No prediction can be calculated at 40.
Show worked solutionHide worked solution for example 4.13
Recognize the structure. Compare the requested input with the range of observed inputs.
Work it through. Forty is outside the observed interval from 4 to 12, so the calculation is extrapolation. The model can be evaluated there, but the data do not establish that a young-tree growth pattern continues to age 40. A computable output is not the same as a well-supported prediction.
Answer: C.
Check. A prediction at age 8 would be interpolation, although even that would remain an estimate rather than an exact individual height.
Avoid the trap. Do not confuse mathematical domain with evidentiary support. A formula may be defined for an input even when the data offer little justification for using it there.
4.14. Read values rather than exaggerated visual heights
The graph shows output in two years. A caption claims, “Output doubled because the second bar is twice as tall above the displayed baseline.” What was the actual percent increase in output?
Show worked solutionHide worked solution for example 4.14
Recognize the structure. The displayed baseline is 90, not zero. Percent change uses the actual values, 100 and 110.
Work it through. The true increase is 110 − 100
percent increase = × 100% = 10%.
The visible heights above 90 are 10 and 20, but those are not the output quantities themselves.
Answer: 10%.
Check. Doubling the original output would produce 200, far above the displayed value of 110.
Avoid the trap. A truncated axis can magnify an apparent change. Read the tick labels and units before comparing bar lengths or estimating a percentage visually.
4.15. Compare linear and exponential models over integer times
Two models are A(t) = 100 + 30t and B(t) = 100(1.2)t, where t is a nonnegative integer. What is the least value of t for which B(t) > A(t)?
Show worked solutionHide worked solution for example 4.15
Recognize the structure. The question asks for the first integer time, so compare a table of model values, including earlier times.
Work it through. A calculator table gives
t | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
A(t) | 100 | 130 | 160 | 190 | 220 | 250 | 280 |
B(t) | 100 | 120 | 144 | 172.8 | 207.36 | 248.832 | 298.5984 |
The models are equal at 0, and B is below A for each integer from 1 through 5. At 6, 298.5984 > 280.
Answer: 6.
Check. At the immediately preceding integer, 248.832 < 250, so 5 is not sufficient. The table checks every allowed earlier input.
Avoid the trap. Exponential growth is not necessarily larger at the first positive input. Also, a noninteger intersection from a graph is not itself the requested integer time.
Section 4: exit questions
Try the five section-exit questions before checking the explanations. Use scratch paper for your reasoning and enter the requested numerical value for a question without choices.