Fit a straight line to a linear scatter plot and judge how well it fits
Several sessions: instruction with the class data set, modeled examples, a small data collection, independent practice and a written interpretation.
Ohio learning standards this draft targets
math-2017:8.SP.2
Read the standard text
math-2017:8.SP.2 Understand that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line. (GAISE Model, steps 3 and 4)
Provisional targets. Clause-by-clause educational review is still pending.
📘Section 1 · LearnLearn the idea
Ten hypothetical bean plants with daily sunlight from 2 to 10 hours and mature heights from 15 to 26 cm. The fitted line h = 1.5s + 12 rises 1.5 cm per hour from an intercept of 12 cm; every dot lies within 1 cm of it, with five dots above, three below and two on the line.
Learning goal: Understand why straight lines are widely used to model relationships between two quantitative variables, informally fit a straight line to a scatter plot that suggests a linear association, and informally assess the fit by judging how close the data points lie to the line. These are the analyze and interpret steps of the GAISE model applied to a linear pattern.
Why a straight line. A straight line is the simplest description of "one quantity goes up steadily as the other goes up." It has only two features, a slope and an intercept, so it is easy to write as an equation, easy to draw and easy to use for a prediction. Many real relationships are close enough to straight over the range that was measured for a line to be useful, even when the true relationship is more complicated. That is why straight lines are used so widely to model relationships between two quantitative variables. The condition matters: a line is a reasonable model only when the scatter plot suggests a linear association. Fitting a line to the cocoa cooling data from the previous lesson would be a mistake, because that plot curves.
The data set. A hypothetical school garden experiment grew ten bean plants of the same variety, giving them between 2 and 10 hours of direct sunlight per day (two plants both got 4 hours), and measured each plant's mature height in centimeters. The numbers are invented for teaching.
Sunlight 2 hours, height 15 cm. Sunlight 3 hours, height 17 cm. Sunlight 4 hours, height 17 cm. Sunlight 4 hours, height 19 cm. Sunlight 5 hours, height 20 cm. Sunlight 6 hours, height 22 cm. Sunlight 7 hours, height 22 cm. Sunlight 8 hours, height 24 cm. Sunlight 9 hours, height 26 cm. Sunlight 10 hours, height 26 cm.
The scatter plot with sunlight horizontal and height vertical shows a positive association: more sunlight goes with taller plants. The dots rise at a roughly steady rate with no curve and no outlier, so the association is linear and a straight-line model is appropriate.
Fitting a line informally. Informal fitting means drawing a line by judgement rather than by a formula. Lay a ruler across the plot and move it until the line follows the direction of the dots, with roughly as many dots above the line as below it, and with the dots on each side about as far from the line as those on the other. The line does not need to pass through any particular dot. In fact, a line forced through the first and last dots is often a poor choice, because those two dots may not represent the trend of the middle.
One line that satisfies these conditions passes through (2, 15) and (10, 27). Its slope is the rise over the run: (27 − 15) ÷ (10 − 2) = 12 ÷ 8 = 1.5 centimeters per hour. Moving back from (2, 15) by 2 hours lowers the height by 2 × 1.5 = 3, so the line crosses the vertical axis at 12. Its equation is h = 1.5s + 12, where s is daily hours of sunlight and h is predicted mature height in centimeters.
Assessing the fit by closeness. The model's quality is judged by how close the actual data points lie to the line. For each plant, compute the height the line predicts and compare it with the measured height. The difference, measured minus predicted, is the vertical gap between the dot and the line; a positive gap means the dot is above the line.
At 2 hours the line predicts 1.5 × 2 + 12 = 15, and the plant measured 15, gap 0. At 3 hours the line predicts 16.5, measured 17, gap +0.5. At 4 hours the line predicts 18; the two plants measured 17 and 19, gaps −1 and +1. At 5 hours: predicted 19.5, measured 20, gap +0.5. At 6 hours: predicted 21, measured 22, gap +1. At 7 hours: predicted 22.5, measured 22, gap −0.5. At 8 hours: predicted 24, measured 24, gap 0. At 9 hours: predicted 25.5, measured 26, gap +0.5. At 10 hours: predicted 27, measured 26, gap −1.
Every gap is within 1 cm on plants that range from 15 to 26 cm tall. Five dots are above the line, three are below and two are on it. The gaps do not drift: the line is not consistently high at one end and low at the other. On both counts, closeness and balance, this line fits well.
A line that fits badly. Suppose someone proposes h = 2s + 8 instead. It predicts 12, 14, 16, 16, 18, 20, 22, 24, 26 and 28 for the ten plants in the order listed. The gaps are +3, +3, +1, +3, +2, +2, 0, 0, 0 and −2. Six dots are above the line, three on it and only one below. More telling, the gaps are large and positive for every plant with 6 hours or less and then shrink to zero or negative. The line is too steep: it runs under the left side of the plot and over the right. Judged by closeness, this line fits worse, even though it touches three dots and the first line touches only two. Passing through dots is not the test; being close to all of them is.
A good check. The mean sunlight is 58 ÷ 10 = 5.8 hours and the mean height is 208 ÷ 10 = 20.8 cm. The first line predicts 1.5 × 5.8 + 12 = 20.7 at the mean sunlight, within 0.1 cm of the mean height. The formal fitting method taught in high school, the least-squares line, always passes through the point of means, so an informal line should pass near it; a miss that is large compared with the gaps between the dots and the line is a reason to investigate, and a small miss is not. This is a check that can rule a line out, not one that can rule it in: a line through the point of means with the wrong slope still fits badly, so use it alongside the gap table, never instead of it. The second line predicts 2 × 5.8 + 8 = 19.6 there, 1.2 cm low, another sign of a poor fit.
Two things a fitted line does not do. It does not make the association exact: the dots still scatter around it, and the scatter is part of the truth. And it does not prove that sunlight causes growth. The line models the pattern of association in these ten plants; explaining the pattern needs an experiment designed for that purpose, not a line on a plot.
✏️Section 2 · ModelWork through the examples
Worked slope from two points on the line: A fitted line passes through (4, 18) and (8, 24). Slope = (24 − 18) ÷ (8 − 4) = 6 ÷ 4 = 1.5 cm per hour. Going back 4 hours from (4, 18) lowers the height by 4 × 1.5 = 6, so the intercept is 12, and the equation is again h = 1.5s + 12. Different pairs of points on the same line give the same equation; the slope belongs to the line, not to the points chosen.
Worked alternative fit: A student draws a line through the first and last dots, (2, 15) and (10, 26). Slope = 11 ÷ 8 = 1.375, and the intercept is 15 − 2 × 1.375 = 12.25, giving h = 1.375s + 12.25. Its predictions for the ten plants are 15, 16.375, 17.75, 17.75, 19.125, 20.5, 21.875, 23.25, 24.625 and 26, so the gaps are 0, +0.625, −0.75, +1.25, +0.875, +1.5, +0.125, +0.75, +1.375 and 0. Every gap is within 1.5 cm, so the fit is acceptable, but seven dots are above the line and only one is below: the line runs slightly under the data through the middle. The line h = 1.5s + 12 balances the dots better, which is why fitting by eye across all the dots usually beats joining the end points.
Worked closeness comparison: Add the sizes of the gaps, ignoring sign. For h = 1.5s + 12 the total is 0 + 0.5 + 1 + 1 + 0.5 + 1 + 0.5 + 0 + 0.5 + 1 = 6.0 cm over ten plants, or 0.6 cm per plant. For h = 2s + 8 the total is 3 + 3 + 1 + 3 + 2 + 2 + 0 + 0 + 0 + 2 = 16 cm, or 1.6 cm per plant. For the end-point line the total is 7.25 cm, or about 0.73 cm per plant. Smaller average gap means closer fit, so the order from best to worst is 1.5s + 12, then the end-point line, then 2s + 8.
🧩Section 3 · DoPractice, demonstrate, and explain
Fit and record: Plot the ten bean plants on graph paper with sunlight from 0 to 12 hours across and height from 0 to 30 cm up. Draw your own fitted line with a ruler, choose two points on it that are easy to read, and write its equation. Make a table with columns for sunlight, measured height, predicted height and gap, then count the dots above and below your line and compute the average gap size. Compare with h = 1.5s + 12.
Independent performance: Use the bivariate data you collected in the previous lesson, or a new set of at least ten pairs whose scatter plot suggests a linear association. First state in writing why a straight line is a reasonable model for this plot; if the plot curves or shows no association, choose different data and say why. Draw a fitted line, derive its equation from two points on the line, and complete a gap table. Then write an assessment of the fit that reports the largest gap, the average gap size, the counts above and below, and whether the gaps drift from one end of the plot to the other. Success means the equation reproduces the line you drew, every gap is computed correctly, and the assessment says clearly whether the line fits well and why.
✅Section 4 · Check7-question lesson practice test
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Problem 1
Why are straight lines so widely used to model relationships between two quantitative variables, and under what condition is a line an appropriate model?
Show worked solution
A line is the simplest steady-change model, defined by just a slope and an intercept, so it is easy to write, draw and use for predictions, and many relationships are close to straight over the measured range. It is appropriate only when the scatter plot suggests a linear association; a curved plot such as the cocoa cooling data should not be fitted with a line.
Problem 2
Six hypothetical students recorded hours studied and test score: (1, 62), (2, 66), (3, 71), (4, 73), (5, 79) and (6, 82). Fit a straight line informally, write its equation from two points on your line, and give one piece of evidence that it fits well.
Show worked solution
Answers vary; any line that follows the direction of the dots with small, balanced gaps is acceptable. One good line passes through (1, 62) and (6, 82): slope (82 − 62) ÷ (6 − 1) = 20 ÷ 5 = 4, and going back 1 hour from (1, 62) gives an intercept of 58, so y = 4x + 58. Its predictions are 62, 66, 70, 74, 78 and 82, so the gaps are 0, 0, +1, −1, +1 and 0: every dot is within 1 point of the line, with two above, one below and three on it, and the gaps do not drift from one end to the other.
Problem 3
Someone proposes the line h = 1.4s + 12.5 for the bean plants. Find its prediction for the plant that got 9 hours of sunlight, the gap to that plant's measured height of 26 cm, and say whether the dot is above or below the line.
Show worked solution
Predicted 1.4 × 9 + 12.5 = 12.6 + 12.5 = 25.1 cm. Gap = 26 − 25.1 = +0.9 cm, so the dot is 0.9 cm above the line.
Problem 4
Two lines are proposed for the same plot. Line P touches three dots, but its other gaps run up to 3 cm and its average gap size is 1.5 cm. Line Q touches no dot, but every dot is within 1 cm of it and its average gap size is 0.6 cm. Which fits better and why?
Show worked solution
Line Q. Fit is judged by how close all the data points are to the line, not by how many the line touches: Q's largest gap (1 cm) and its average gap (0.6 cm) are both smaller than P's (3 cm and 1.5 cm). Small gaps for every dot beat a line that hits a few dots and misses others by 3 cm.
Problem 5
For a line fitted to another data set, the gaps (measured minus predicted) in order of the horizontal value are −2, −1.5, −1, 0, +0.5, +1, +1.5 and +2.5. What do the signs and sizes of these gaps tell you about the line?
Show worked solution
At the left the gaps are negative, so the dots are below the line there; at the right they are positive, so the dots are above it. The gaps drift steadily from negative to positive, so the line runs above the data on the left and below it on the right: it is not steep enough. A well-fitted line has small gaps with no drift from one end to the other, so a steeper line would fit better.
Problem 6
A data set has mean horizontal value 4 and mean vertical value 30, and its dots lie within about 1 unit of a good line. Line A is y = 3x + 19 and line B is y = 5x + 6. Use the point of means to check each line, and say what the check can and cannot show.
Show worked solution
At x = 4, line A predicts 3 × 4 + 19 = 31, which misses the mean height 30 by 1, about the size of the scatter, so the check gives no reason to doubt A. Line B predicts 5 × 4 + 6 = 26, which misses by 4, large compared with the scatter, so B should be investigated. The least-squares line always passes through the point of means, so a large miss is a warning sign; but passing near it does not prove a good fit, because a line through that point can still have the wrong slope. The gap table decides.
Problem 7
After fitting the line, a student writes: "The line shows that sunlight makes bean plants grow 1.5 cm taller per hour." Rewrite the sentence so that it claims only what the fitted line supports.
Show worked solution
For example: "In these ten plants, each additional hour of daily sunlight is associated with about 1.5 cm more mature height." Any sentence that describes an association ("associated with", "tends to go with") and claims no cause is correct. The line models a pattern of association; it does not show that sunlight caused the growth, and the scatter around the line remains.