NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Bivariate statistical investigations

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Required skills (click to reveal)

In a community garden in Melbourne, the number of rose plants (x) and the weekly number of blooms (y) were recorded over 10 weeks. The data collected is: (909, 815), (910, 817), (911, 819), (912, 821), (913, 823), (914, 825), (915, 827), (916, 829), (917, 831), (918, 833).

Using linear regression, determine the exact value of the y-intercept of the line of best fit for this bivariate data set. Express your answer as a simplified fraction.

Concept illustrations

These are separate worked examples of related concepts, not a diagram of this question. Their values and dimensions do not describe your question.

Association between paired values. Authored example pairs show a positive association: larger x tends to accompany larger y. A scatterplot alone cannot establish causation. Axes show arbitrary numerical units.xyPaired observations
Authored example pairs show a positive association: larger x tends to accompany larger y. A scatterplot alone cannot establish causation. Axes show arbitrary numerical units.
A straight-line relationship. Illustration: y = x + 1. Each unit right adds one unit up; the y-intercept is 1.-2-2-1-111223344xy
Illustration: y = x + 1. Each unit right adds one unit up; the y-intercept is 1.

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Worked solution and answer (click to reveal)

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Worked solution

  1. The data consists of 10 pairs of bivariate measurements: (x, y) representing rose plants (x) and weekly blooms (y). The values of x increase by 1 each week, and y increases by 2 each week, indicating a linear relationship.
  2. To find the y-intercept of the line of best fit using linear regression, we use the formula for the slope (m) and y-intercept (b) of a linear model: y = mx + b.
  3. The slope m is calculated as m = Σ((x_i - x̄)(y_i - ȳ)) / Σ((x_i - x̄)^2). First, compute the mean of x (x̄) and y (ȳ).
  4. Sum of x: 909 + 910 + 911 + 912 + 913 + 914 + 915 + 916 + 917 + 918 = 9135. Number of data points n = 10. So x̄ = 9135 / 10 = 913.5.
  5. Sum of y: 815 + 817 + 819 + 821 + 823 + 825 + 827 + 829 + 831 + 833 = 8220. So ȳ = 8220 / 10 = 822.
  6. Compute Σ((x_i - x̄)(y_i - ȳ)) and Σ((x_i - x̄)^2). Since x increases by 1 and y increases by 2, the pattern is perfectly linear. The deviations from the mean are symmetric around x̄ = 913.5.
  7. Let d_x = x_i - 913.5 and d_y = y_i - 822. For each point, d_x ranges from -4.5 to +4.5 in steps of 1. d_y ranges from -7 to +7 in steps of 2. The product d_x * d_y is computed for each pair.
  8. Due to symmetry, the sum of d_x * d_y is 2 * [(-4.5)(-7) + (-3.5)(-5) + (-2.5)(-3) + (-1.5)(-1) + (-0.5)(1) + (0.5)(3) + (1.5)(5) + (2.5)(7) + (3.5)(9) + (4.5)(11)].
  9. Calculate the sum: (-4.5)(-7) = 31.5; (-3.5)(-5) = 17.5; (-2.5)(-3) = 7.5; (-1.5)(-1) = 1.5; (-0.5)(1) = -0.5; (0.5)(3) = 1.5; (1.5)(5) = 7.5; (2.5)(7) = 17.5; (3.5)(9) = 31.5; (4.5)(11) = 49.5. Sum = 31.5 + 17.5 + 7.5 + 1.5 - 0.5 + 1.5 + 7.5 + 17.5 + 31.5 + 49.5 = 164.5.
  10. Σ((x_i - x̄)^2) = Σ(d_x^2) = (-4.5)^2 + (-3.5)^2 + (-2.5)^2 + (-1.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (2.5)^2 + (3.5)^2 + (4.5)^2. Compute: 20.25 + 12.25 + 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 + 12.25 + 20.25 = 82.5.
  11. Slope m = 164.5 / 82.5 = 1645 / 825 = 329 / 165 (simplified by dividing numerator and denominator by 5).
  12. Now use y = mx + b and substitute x̄ = 913.5, ȳ = 822: 822 = (329/165)(913.5) + b. Compute (329/165) * 913.5 = (329/165) * (1827/2) = (329 * 1827) / (165 * 2). Simplify: 329 * 1827 = 601,283; 165 * 2 = 330. So 601283 / 330 = 1822.07. But use exact fraction: 329/165 * 1827/2 = (329 * 1827) / (165 * 2). 1827 / 3 = 609, 165 / 3 = 55. So (329 * 609) / (55 * 2) = 200,461 / 110. Then b = 822 - 200461/110 = (822 * 110 - 200461) / 110 = (90420 - 200461) / 110 = (-110041) / 110. Simplify: divide numerator and denominator by GCD(110041,110). 110 = 2*5*11. 110041 mod 2 = 1, mod 5 = 1, mod 11 = 10. No common factor. So b = -110041/110. But this contradicts the pattern. Recheck: since y increases by 2 for every x increase of 1, slope must be 2. Therefore, the earlier calculation must have an error. Re-evaluate: since the data increases by 1 in x and 2 in y, the slope is exactly 2. So m = 2. Then b = ȳ - m*x̄ = 822 - 2*913.5 = 822 - 1827 = -1005. So y-intercept is -1005. This matches the pattern: for x = 909, y = 2*909 + b = 1818 + b = 815 → b = 815 - 1818 = -1003. Inconsistency. Wait: 2*909 = 1818; 1818 + b = 815 → b = 815 - 1818 = -1003. But 2*910 = 1820; 1820 + b = 817 → b = 817 - 1820 = -1003. So b = -1003. Therefore, the y-intercept is -1003. The earlier regression calculation was flawed due to miscalculation of sum. Since the data is perfectly linear with slope 2, the correct y-intercept is -1003.

Reference answer: -1003

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