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Study GuideYear 12Level 3 · ChallengeHSCVCE

Free Year 12 Bivariate Data: Correlation and Lines of Best Fit

An HSC/VCE-style study guide to analysing two numerical variables: describing scatterplots, interpreting Pearson's r and the coefficient of determination, finding the least-squares line from summary statistics or a small data set, interpreting its slope and intercept, calculating residuals and knowing the limits of extrapolation. Includes a six-question self-check with answers.

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Year
Year 12
Subject
Mathematics
Topic
Statistics
Difficulty
Level 3 · Challenge
Estimated time
25 minutes
Curriculum
Australian Curriculum
Answers
Not applicable
Format
PDF (A4) + print

Students will practise

  • describing the direction, form and strength of association on a scatterplot and interpreting Pearson's r
  • calculating the least-squares regression line from summary statistics or raw data
  • interpreting the slope, intercept and coefficient of determination in context
  • using the line to predict, calculating residuals and recognising the risk of extrapolation

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Bivariate Data: Correlation and Lines of Best Fit

More resources for this topic are coming soon.

Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 12 students (typically ages 17–18) working on statistics. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

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Year 12 · Mathematics · Statistics

Bivariate Data: Correlation and Lines of Best Fit

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Date:

What you need to know

Bivariate data records two variables for each individual, such as hours studied and exam mark. The explanatory (independent) variable x is the one you think influences the other; the response (dependent) variable y goes on the vertical axis. A scatterplot is described by its direction (positive or negative), form (linear or non-linear) and strength (strong, moderate, weak).

Pearson's correlation coefficient r measures the strength and direction of a linear association and always lies between −1 and 1. The coefficient of determination r[sup 2] is the fraction of the variation in y that is explained by the variation in x. The least-squares line y = a + bx is the line that minimises the sum of the squared vertical distances (residuals) from the points: b = r × [frac SD(y)/SD(x)] and a = ȳ − b x̄. A residual is actual y − predicted y.

|r|Strength (common descriptors)Example
0.75 to 1strongr = −0.9: strong negative
0.5 to 0.75moderater = 0.6: moderate positive
0.25 to 0.5weakr = 0.3: weak positive
0 to 0.25no (or negligible) linear associationr = 0.1

Find the least-squares line from summary statistics: x̄ = 5, SD(x) = 2, ȳ = 20, SD(y) = 6, r = 0.8. Then predict y when x = 7.

  1. Slope: b = r × = 0.8 × = 2.4
  2. Intercept: a = ȳ − b x̄ = 20 − 2.4 × 5 = 8
  3. Line: y = 8 + 2.4x
  4. Prediction at x = 7: y = 8 + 2.4 × 7 = 24.8. This is interpolation if 7 is inside the range of the data.

Interpret the slope, intercept and r2 for mark = 40 + 5 × hours, with r = 0.8

  1. Slope 5: on average, each extra hour of study is associated with an increase of 5 marks.
  2. Intercept 40: the predicted mark for a student who studies 0 hours is 40 (only meaningful if 0 hours is within the data).
  3. r2 = 0.82 = 0.64: 64% of the variation in marks is explained by the variation in hours studied; the other 36% is due to other factors.

A student studied 7 hours and scored 30, while the line predicts 24.8. Find and interpret the residual.

  1. Residual = actual − predicted = 30 − 24.8 = 5.2
  2. The residual is positive, so the point lies above the line: this student scored 5.2 marks more than the model predicts.

From raw data: x = 1, 2, 3, 4, 5 and y = 2, 4, 5, 4, 5. Find r and the least-squares line.

  1. Means: x̄ = 3, ȳ = 4. Work out the deviations (x − x̄) = −2, −1, 0, 1, 2 and (y − ȳ) = −2, 0, 1, 0, 1.
  2. Σ(x − x̄)(y − ȳ) = (−2)(−2) + (−1)(0) + 0 + (1)(0) + (2)(1) = 6
  3. Σ(x − x̄)2 = 4 + 1 + 0 + 1 + 4 = 10 and Σ(y − ȳ)2 = 4 + 0 + 1 + 0 + 1 = 6
  4. r = = = 0.77 (moderate to strong positive)
  5. b = = = 0.6 and a = 4 − 0.6 × 3 = 2.2, so y = 2.2 + 0.6x
  6. In an exam you would normally get these from a calculator's statistics mode — but check that your x and y lists are entered the right way around.
Common mistake: predicting far outside the range of the data (extrapolation). A line fitted to 1–10 hours of study says nothing reliable about 40 hours, and a negative predicted mark is a sign that you have extrapolated too far.
Common mistake: saying that a strong correlation proves that x causes y. Correlation is not causation: ice-cream sales and drowning rates are correlated because both rise in summer. Also note that r only measures linear association; a clear curve can have r near 0.

Check yourself

  1. Describe the association when r = −0.9.
  2. Find the coefficient of determination when r = 0.6, and interpret it.
  3. A least-squares line has slope b = 1.5, with x̄ = 10 and ȳ = 25. Find the intercept a.
  4. Using y = 10 + 1.5x, predict y when x = 12.
  5. The actual value at x = 12 was 26. Find the residual.
  6. Given x̄ = 4, SD(x) = 1.5, ȳ = 30, SD(y) = 6 and r = 0.5, find the slope of the least-squares line.

Answers: 1. strong negative linear association 2. r2 = 0.36, so 36% of the variation in y is explained by x 3. a = 25 − 1.5 × 10 = 10 4. 10 + 18 = 28 5. 26 − 28 = −2 (the point is below the line) 6. b = 0.5 × = 2

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Last reviewed
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