Mathematics Advanced • Year 12 • Module 5 • Lesson 9
Bivariate Data Analysis
Build procedural fluency in describing scatter plots, calculating Pearson's r from raw data, and interpreting r² as the proportion of variation explained.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 State the range of Pearson's r:
______ ≤ r ≤ ______ and describe what each extreme value means: r = +1 → ______________________, r = −1 → ______________________.
Q1.2 Name the three features used to describe a scatter plot.
________________, ________________, ________________
Q1.3 If r = 0.7, state r² and complete the sentence: "____% of the variation in y can be explained by the linear relationship with x."
2. Worked example, Pearson's r for 5 (x, y) pairs
Data: (2, 55), (4, 62), (5, 70), (6, 75), (8, 88), study hours x vs test scores y.
Problem. Calculate Pearson's correlation coefficient r using the computational formula.
Step 1, Build the sums (set up a tidy table).
x: 2 + 4 + 5 + 6 + 8 = 25
y: 55 + 62 + 70 + 75 + 88 = 350
x²: 4 + 16 + 25 + 36 + 64 = 145
y²: 3025 + 3844 + 4900 + 5625 + 7744 = 25,138
xy: 110 + 248 + 350 + 450 + 704 = 1,862
n = 5
Step 2, Apply the computational formula.
r = [ n·Σxy − (Σx)(Σy) ] / √{ [ n·Σx² − (Σx)² ] [ n·Σy² − (Σy)² ] }
Step 3, Plug in.
Numerator = 5(1862) − (25)(350) = 9310 − 8750 = 560
Denominator = √{ [5(145) − 25²][5(25138) − 350²] }
= √{ [725 − 625][125,690 − 122,500] }
= √{ 100 × 3190 } = √319,000 ≈ 564.8
Step 4, Compute r and r².
r = 560 / 564.8 ≈ 0.992
r² ≈ 0.984
Conclusion. r ≈ 0.992 indicates a very strong positive linear relationship between study hours and test scores. About 98% of the variation in test scores is explained by the linear relationship with study hours.
3. Faded example, fill in the missing steps
Calculate r for the data (1, 3), (2, 5), (3, 4), (4, 8), (5, 9). Fill in each blank. 4 marks
Step 1, Sums.
Σx = 1+2+3+4+5 = ______ · Σy = 3+5+4+8+9 = ______
Σx² = 1+4+9+16+25 = ______ · Σy² = 9+25+16+64+81 = ______
Σxy = 1(3)+2(5)+3(4)+4(8)+5(9) = ______ · n = ______
Step 2, Numerator. n·Σxy − (Σx)(Σy) = ____·____ − ____·____ = ____________
Step 3, Denominator pieces.
n·Σx² − (Σx)² = ____·____ − ____² = ____________
n·Σy² − (Σy)² = ____·____ − ____² = ____________
Step 4, r and r². r = ______ / √( ____ × ____ ) ≈ ____________ · r² ≈ ____________
Conclusion. r ≈ ____ describes a ____________ correlation; r² ≈ ____ means ____% of the variation in y is explained.
4. Graduated practice, calculate or interpret r
Show your working. Quote r and r² to 2 decimal places.
Foundation, single-step calculations (4 questions)
| Q | Question | Answer |
|---|---|---|
| 4.1 1 | If r = 0.6, find r² and complete: "______% of variation explained". | |
| 4.2 1 | If r = −0.9, find r² and state the strength descriptor. | |
| 4.3 1 | State the value of r for: (1, 2), (2, 4), (3, 6), (4, 8). No formula needed, explain in one phrase. | |
| 4.4 1 | From the lesson's strength table, classify |r| = 0.42 (e.g. "weak", "moderate"). |
Standard, typical HSC difficulty (6 questions)
Compute r using the computational formula. Show the sums.
4.5 Calculate r for (1, 8), (2, 5), (3, 4), (4, 3), (5, 2). Describe the relationship and find r². 3 marks
4.6 Calculate r for (1, 1), (2, 4), (3, 9), (4, 16). Explain in one sentence why r is < 1 despite the clear pattern. 3 marks
4.7 A study reports r = 0.3 between hours of TV watched and exam scores. State r² and answer: is this a practically significant relationship? Justify in one sentence. 2 marks
4.8 For a data set with r = 0.95, sketch (in the space below) the kind of scatter plot you would expect, and state what percentage of variation in y is explained by x. 2 marks
4.9 The scatter plot of two variables looks like a perfect parabola (a "U" shape). What value of Pearson's r is consistent with this, and why does this not mean there is no relationship? 2 marks
4.10 A small data set has Σx = 30, Σy = 60, Σxy = 372, Σx² = 220, Σy² = 736, n = 5. Compute r. 2 marks
Extension, combine concepts (2 questions)
4.11 A data set has r = 0.4. An additional data point is added that lies exactly on the original trend line. State (without recomputing) whether r will increase, decrease, or stay roughly the same, and justify in one sentence. 3 marks
4.12 Two studies of the same phenomenon report r = 0.9 (Study 1) and r = 0.3 (Study 2). State the proportion of variation explained in each, then explain in one sentence what it tells us, geometrically, about the spread of points around the trend line in each study. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, Range of r
−1 ≤ r ≤ +1. r = +1: perfect positive linear relationship. r = −1: perfect negative linear relationship.
Q1.2, Three scatter-plot features
Direction (positive / negative / none), form (linear / curved / clustered), strength (strong / moderate / weak).
Q1.3, r = 0.7
r² = 0.49 → 49% of the variation in y is explained by the linear relationship with x.
Q3, Faded example: (1, 3), (2, 5), (3, 4), (4, 8), (5, 9)
Σx = 15, Σy = 29, Σx² = 55, Σy² = 195, Σxy = 101, n = 5.
Numerator = 5(101) − 15(29) = 505 − 435 = 70.
n·Σx² − (Σx)² = 275 − 225 = 50. n·Σy² − (Σy)² = 975 − 841 = 134.
r = 70 / √(50 × 134) = 70 / √6700 ≈ 0.855. r² ≈ 0.731.
Conclusion: strong positive correlation; ≈ 73% of variation in y is explained by x.
Q4.1, r = 0.6
r² = 0.36 → 36% of variation explained.
Q4.2, r = −0.9
r² = 0.81 (the sign drops out). |r| = 0.9 falls in the 0.90 – 1.00 band → very strong (negative) correlation.
Q4.3, (1, 2), (2, 4), (3, 6), (4, 8)
r = +1 the four points lie exactly on the line y = 2x (perfect positive linear relationship).
Q4.4, Classify |r| = 0.42
Falls in the 0.30 – 0.50 band → moderate correlation.
Q4.5, (1, 8), (2, 5), (3, 4), (4, 3), (5, 2)
Σx = 15, Σy = 22, Σx² = 55, Σy² = 118, Σxy = 51, n = 5.
Numerator = 5(51) − 15(22) = 255 − 330 = −75.
n·Σx² − (Σx)² = 275 − 225 = 50. n·Σy² − (Σy)² = 590 − 484 = 106.
r = −75 / √(50 × 106) = −75 / √5300 ≈ −1.03… rounding suggests we recheck. Σxy = 1(8)+2(5)+3(4)+4(3)+5(2) = 8 + 10 + 12 + 12 + 10 = 52. Numerator = 5(52) − 330 = 260 − 330 = −70. r = −70 / √(50 × 106) = −70 / 72.80 ≈ −0.96. r² ≈ 0.92 → very strong negative correlation; ≈ 92% of variation in y explained.
Q4.6, (1, 1), (2, 4), (3, 9), (4, 16)
The points lie on y = x² (a perfect parabola). Σx = 10, Σy = 30, Σx² = 30, Σy² = 354, Σxy = 1(1)+2(4)+3(9)+4(16) = 1+8+27+64 = 100, n = 4.
Numerator = 4(100) − 10(30) = 400 − 300 = 100.
n·Σx² − (Σx)² = 120 − 100 = 20. n·Σy² − (Σy)² = 1416 − 900 = 516.
r = 100 / √(20 × 516) = 100 / √10320 ≈ 0.984. r² ≈ 0.968. r is < 1 because the true relationship is curved (y = x²), so a straight line cannot fit perfectly, there are non-zero vertical gaps between the points and any line. r captures linear association, not the full curved fit.
Q4.7, r = 0.3 for TV vs exam scores
r² = 0.09 → only 9% of variation in exam scores is explained by TV hours; not practically significant in any meaningful sense: 91% of variation is driven by other factors (study time, sleep, etc.).
Q4.8, Scatter plot for r = 0.95
Sketch: tightly clustered points hugging an upward-sloping straight line, with very little vertical scatter. r² = 0.9025 → about 90% of variation in y is explained by x.
Q4.9, Perfect parabola
r ≈ 0 (or very small), the positive and negative correlations on the two arms of the parabola cancel out. This does not mean there is no relationship: there is a strong non-linear relationship that Pearson's r is simply blind to.
Q4.10, From sums Σx = 30, Σy = 60, Σxy = 372, Σx² = 220, Σy² = 736, n = 5
Numerator = 5(372) − 30(60) = 1860 − 1800 = 60. n·Σx² − (Σx)² = 1100 − 900 = 200. n·Σy² − (Σy)² = 3680 − 3600 = 80. r = 60 / √(200 × 80) = 60 / √16000 ≈ 0.474 (≈ moderate).
Q4.11, Adding a point exactly on the trend line
r will increase (or stay essentially the same, depending on the position). Adding a point that already fits the linear pattern perfectly reduces the relative weight of any off-line points, tightening the linear fit and pushing r closer to 1.
Q4.12, r = 0.9 vs r = 0.3
r² = 0.81 → 81% of variation in y explained (Study 1). r² = 0.09 → 9% explained (Study 2). Geometrically: Study 1's scatter plot has points tightly hugging the trend line; Study 2's scatter plot has points loosely scattered around the trend line, with most of the up-and-down variation driven by something other than the linear relationship with x.